The Haskell Road
to
Logic, Math and Programming
Kees Doets and Jan van Eijck
March 4, 2004

Contents
Prefacev

1 Getting Started
1.1 Starting up the Haskell Interpreter . . . . . . . . . . . . . . . . .
1.2 Implementing a Prime Number Test . . . . . . . . . . . . . . . .
1.3 Haskell Type Declarations . . . . . . . . . . . . . . . . . . . . .
1.4 Identifiers in Haskell . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Playing the Haskell Game . . . . . . . . . . . . . . . . . . . . .
1.6 Haskell Types . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 The Prime Factorization Algorithm . . . . . . . . . . . . . . . . .
1.8 The map and filter Functions . . . . . . . . . . . . . . . . . . .
1.9 Haskell Equations and Equational Reasoning . . . . . . . . . . .
1.10 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

1
2
3
8
11
12
17
19
20
24
26

2 Talking about Mathematical Objects
2.1 Logical Connectives and their Meanings . . . . . . . . . . . . . .
2.2 Logical Validity and Related Notions . . . . . . . . . . . . . . . .
2.3 Making Symbolic Form Explicit . . . . . . . . . . . . . . . . . .
2.4 Lambda Abstraction . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Definitions and Implementations . . . . . . . . . . . . . . . . . .
2.6 Abstract Formulas and Concrete Structures . . . . . . . . . . . .
2.7 Logical Handling of the Quantifiers . . . . . . . . . . . . . . . .
2.8 Quantifiers as Procedures . . . . . . . . . . . . . . . . . . . . . .
2.9 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

27
28
38
50
58
60
61
64
68
70

3 The Use of Logic: Proof
3.1 Proof Style . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Proof Recipes . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Rules for the Connectives . . . . . . . . . . . . . . . . . . . . . .
3.4 Rules for the Quantifiers . . . . . . . . . . . . . . . . . . . . . .

71
72
75
78
90

i

CONTENTS

ii
3.5
3.6
3.7
3.8

Summary of the Proof Recipes . . . . . . . . . . . . . . . . . . . 96
Some Strategic Guidelines . . . . . . . . . . . . . . . . . . . . . 99
Reasoning and Computation with Primes . . . . . . . . . . . . . . 103
Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

4 Sets, Types and Lists
4.1 Let’s Talk About Sets . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Paradoxes, Types and Type Classes . . . . . . . . . . . . . . . . .
4.3 Special Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Algebra of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Pairs and Products . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Lists and List Operations . . . . . . . . . . . . . . . . . . . . . .
4.7 List Comprehension and Database Query . . . . . . . . . . . . .
4.8 Using Lists to Represent Sets . . . . . . . . . . . . . . . . . . . .
4.9 A Data Type for Sets . . . . . . . . . . . . . . . . . . . . . . . .
4.10 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

113
114
121
125
127
136
139
145
149
153
158

5 Relations
5.1 The Notion of a Relation . . . . . . . . . . . . . . . . . . . . . .
5.2 Properties of Relations . . . . . . . . . . . . . . . . . . . . . . .
5.3 Implementing Relations as Sets of Pairs . . . . . . . . . . . . . .
5.4 Implementing Relations as Characteristic Functions . . . . . . . .
5.5 Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . .
5.6 Equivalence Classes and Partitions . . . . . . . . . . . . . . . . .
5.7 Integer Partitions . . . . . . . . . . . . . . . . . . . . . . . . . .
5.8 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

161
162
166
175
182
188
192
202
204

6 Functions
6.1 Basic Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Surjections, Injections, Bijections . . . . . . . . . . . . . . . . .
6.3 Function Composition . . . . . . . . . . . . . . . . . . . . . . .
6.4 Inverse Function . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 Partial Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6 Functions as Partitions . . . . . . . . . . . . . . . . . . . . . . .
6.7 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.8 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.9 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

205
206
218
222
226
229
232
234
236
238

7 Induction and Recursion
7.1 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . .
7.2 Recursion over the Natural Numbers . . . . . . . . . . . . . . . .
7.3 The Nature of Recursive Definitions . . . . . . . . . . . . . . . .

239
239
246
251

CONTENTS

iii

7.4
7.5
7.6
7.7
7.8

Induction and Recursion over Trees . . . . . . . . . . . . . . . .
Induction and Recursion over Lists . . . . . . . . . . . . . . . . .
Some Variations on the Tower of Hanoi . . . . . . . . . . . . . .
Induction and Recursion over Other Data Structures . . . . . . . .
Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

255
265
273
281
284

8 Working with Numbers
8.1 A Module for Natural Numbers . . . . . . . . . . . . . . . . . . .
8.2 GCD and the Fundamental Theorem of Arithmetic . . . . . . . .
8.3 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4 Implementing Integer Arithmetic . . . . . . . . . . . . . . . . . .
8.5 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . .
8.6 Implementing Rational Arithmetic . . . . . . . . . . . . . . . . .
8.7 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . .
8.8 The Mechanic’s Rule . . . . . . . . . . . . . . . . . . . . . . . .
8.9 Reasoning about Reals . . . . . . . . . . . . . . . . . . . . . . .
8.10 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . .
8.11 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

285
286
289
293
297
299
305
309
313
315
319
329

9 Polynomials
9.1 Difference Analysis of Polynomial Sequences . . . . . . . . . . .
9.2 Gaussian Elimination . . . . . . . . . . . . . . . . . . . . . . . .
9.3 Polynomials and the Binomial Theorem . . . . . . . . . . . . . .
9.4 Polynomials for Combinatorial Reasoning . . . . . . . . . . . . .
9.5 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

331
332
337
344
352
359

10 Corecursion
10.1 Corecursive Definitions . . . . . . . . . . . . . . . . . . . . . . .
10.2 Processes and Labeled Transition Systems . . . . . . . . . . . . .
10.3 Proof by Approximation . . . . . . . . . . . . . . . . . . . . . .
10.4 Proof by Coinduction . . . . . . . . . . . . . . . . . . . . . . . .
10.5 Power Series and Generating Functions . . . . . . . . . . . . . .
10.6 Exponential Generating Functions . . . . . . . . . . . . . . . . .
10.7 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

361
362
365
373
379
385
396
398

11 Finite and Infinite Sets
11.1 More on Mathematical Induction . . . . . . . . . . . . . . . . . .
11.2 Equipollence . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.3 Infinite Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.4 Cantor’s World Implemented . . . . . . . . . . . . . . . . . . . .
11.5 Cardinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . .

399
399
406
410
418
420

iv

CONTENTSThe Greek Alphabet

423

References

424

Index

428

Preface
PurposeLong ago, when Alexander the Great asked the mathematician Menaechmus fora crash course in geometry, he got the famous reply “There is no royal road tomathematics.” Where there was no shortcut for Alexander, there is no shortcut
for us. Still, the fact that we have access to computers and mature programming
languages means that there are avenues for us that were denied to the kings andemperors of yore.
The purpose of this book is to teach logic and mathematical reasoning in practice,and to connect logical reasoning with computer programming. The programming
language that will be our tool for this is Haskell, a member of the Lisp family.
Haskell emerged in the last decade as a standard for lazy functional programming,a programming style where arguments are evaluated only when the value is actually needed. Functional programming is a form of descriptive programming, very
different from the style of programming that you find in prescriptive languages
like C or Java. Haskell is based on a logical theory of computable functions called
the lambda calculus.
Lambda calculus is a formal language capable of expressing arbitrarycomputable functions. In combination with types it forms a compact
way to denote on the one hand functional programs and on the other
hand mathematical proofs. [Bar84]
Haskell can be viewed as a particularly elegant implementation of the lambda calculus. It is a marvelous demonstration tool for logic and math because its functional character allows implementations to remain very close to the concepts that
get implemented, while the laziness permits smooth handling of infinite data structures.
vvi
Haskell syntax is easy to learn, and Haskell programs are constructed and tested
in a modular fashion. This makes the language well suited for fast prototyping.
Programmers find to their surprise that implementation of a well-understood algorithm in Haskell usually takes far less time than implementation of the samealgorithm in other programming languages. Getting familiar with new algorithms
through Haskell is also quite easy. Learning to program in Haskell is learning an
extremely useful skill.
Throughout the text, abstract concepts are linked to concrete representations in
Haskell. Haskell comes with an easy to use interpreter, Hugs. Haskell compilers,interpreters and documentation are freely available from the Internet [HT]. Everything one has to know about programming in Haskell to understand the programsin the book is explained as we go along, but we do not cover every aspect of thelanguage. For a further introduction to Haskell we refer the reader to [HFP96].

Logic in PracticeThe subject of this book is the use of logic in practice, more in particular theuse of logic in reasoning about programming tasks. Logic is not taught here as amathematical discipline per se, but as an aid in the understanding and constructionof proofs, and as a tool for reasoning about formal objects like numbers, lists,
trees, formulas, and so on. As we go along, we will introduce the concepts andtools that form the set-theoretic basis of mathematics, and demonstrate the roleof these concepts and tools in implementations. These implementations can bethought of as representations of the mathematical concepts.
Although it may be argued that the logic that is needed for a proper understandingof reasoning in reasoned programming will get acquired more or less automaticallyin the process of learning (applied) mathematics and/or programming, students
nowadays enter university without any experience whatsoever with mathematical
proof, the central notion of mathematics.
The rules of Chapter 3 represent a detailed account of the structure of a proof. The
purpose of this account is to get the student acquainted with proofs by putting emphasis on logical structure. The student is encouraged to write “detailed” proofs,
with every logical move spelled out in full. The next goal is to move on to writing“concise” proofs, in the customary mathematical style, while keeping the logicalstructure in mind. Once the student has arrived at this stage, most of the logic thatis explained in Chapter 3 can safely be forgotten, or better, can safely fade into thesubconsciousness of the matured mathematical mind.

PREFACE

vii

Pre- and Postconditions of Use
We do not assume that our readers have previous experience with either programming or construction of formal proofs. We do assume previous acquaintance withmathematical notation, at the level of secondary school mathematics. Wherevernecessary, we will recall relevant facts. Everything one needs to know about mathematical reasoning or programming is explained as we go along. We do assume
that our readers are able to retrieve software from the Internet and install it, andthat they know how to use an editor for constructing program texts.
After having worked through the material in the book, i.e., after having digested
the text and having carried out a substantial number of the exercises, the reader
will be able to write interesting programs, reason about their correctness, and document them in a clear fashion. The reader will also have learned how to set up
mathematical proofs in a structured way, and how to read and digest mathematical
proofs written by others.

How to Use the Book
Chapters 1–7 of the book are devoted to a gradual introduction of the concepts,
tools and methods of mathematical reasoning and reasoned programming.
Chapter 8 tells the story of how the various number systems (natural numbers,
integers, rationals, reals, complex numbers) can be thought of as constructed instages from the natural numbers. Everything gets linked to the implementations of
the various Haskell types for numerical computation.
Chapter 9 starts with the question of how to automate the task of finding closed
forms for polynomial sequences. It is demonstrated how this task can be automatedwith difference analysis plus Gaussian elimination. Next, polynomials are implemented as lists of their coefficients, with the appropriate numerical operations, and
it is shown how this representation can be used for solving combinatorial problems.
Chapter 10 provides the first general textbook treatment (as far as we know) of theimportant topic of corecursion. The chapter presents the proof methods suitable forreasoning about corecursive data types like streams and processes, and then goeson to introduce power series as infinite lists of coefficients, and to demonstrate theuses of this representation for handling combinatorial problems. This generalizes
the use of polynomials for combinatorics.
Chapter 11 offers a guided tour through Cantor’s paradise of the infinite, whileproviding extra challenges in the form of a wide range of additional exercises.

viii
The book can be used as a course textbook, but since it comes with solutions toall exercises (electronically available from the authors upon request) it is also well
suited for private study. Courses based on the book could start with Chapters 1–7,
and then make a choice from the remaining Chapters. Here are some examples:
Road to Numerical Computation Chapters 1–7, followed by 8 and 9.
Road to Streams and Corecursion Chapters 1–7, followed by 9 and 10.
Road to Cantor’s Paradise Chapters 1–7, followed by 11.
Study of the remaining parts of the book can then be set as individual tasks forstudents ready for an extra challenge. The guidelines for setting up formal proofs
in Chapter 3 should be recalled from time to time while studying the book, forproper digestion.

Exercises
Parts of the text and exercises marked by a * are somewhat harder than the rest of
the book.
All exercises are solved in the electronically avaible solutions volume. Before
turning to these solutions, one should read the Important Advice to the Reader that
this volume starts with.

Book Website and Contact
The programs in this book have all been tested with Hugs98, the version of Hugs
that implements the Haskell 98 standard. The full source code of all programs is
integrated in the book; in fact, each chapter can be viewed as a literate program[Knu92] in Haskell. The source code of all programs discussed in the text canbe found on the website devoted to this book, at address http://www.cwi.nl/~jve/HR. Here you can also find a list of errata, and further relevant material.
Readers who want to share their comments with the authors are encouraged to get
in touch with us at email address jve@cwi.nl.

PREFACEix

Acknowledgments
Remarks from the people listed below have sparked off numerous improvements.
Thanks to Johan van Benthem, Jan Bergstra, Jacob Brunekreef, Thierry Coquand(who found the lecture notes on the internet and sent us his comments), Tim van
Erven, Wan Fokkink, Evan Goris, Robbert de Haan, Sandor Heman, Eva Hoogland, Rosalie Iemhoff, Dick de Jongh, Anne Kaldewaij, Breanndán Ó Nualláin,
Alban Ponse, Vincent van Oostrom, Piet Rodenburg, Jan Rutten, Marco Swaen,
Jan Terlouw, John Tromp, Yde Venema, Albert Visser and Stephanie Wehner forsuggestions and criticisms. The beautiful implementation of the sieve of Eratosthenes in Section 3.7 was suggested to us by Fer-Jan de Vries.
The course on which this book is based was developed at ILLC (the Institute of
Logic, Language and Computation of the University of Amsterdam) with financial support from the Spinoza Logic in Action initiative of Johan van Benthem,which is herewith gratefully acknowledged. We also wish to thank ILLC and CWI(Centrum voor Wiskunde en Informatica, or Centre for Mathematics and Computer Science, also in Amsterdam), the home institute of the second author, forproviding us with a supportive working environment. CWI has kindly granted
permission to reuse material from [Doe96].
It was Krzysztof Apt who, perceiving the need of a deadline, spurred us on to get
in touch with a publisher and put ourselves under contract.

xChapter 1

Getting Started
Preview
Our purpose is to teach logic and mathematical reasoning in practice, and to connect formal reasoning to computer programming. It is convenient to choose a
programming language for this that permits implementations to remain as close as
possible to the formal definitions. Such a language is the functional programming
language Haskell [HT]. Haskell was named after the logician Haskell B. Curry.
Curry, together with Alonzo Church, laid the foundations of functional computation in the era Before the Computer, around 1940. As a functional programming
language, Haskell is a member of the Lisp family. Others family members are
Scheme, ML, Occam, Clean. Haskell98 is intended as a standard for lazy functional programming. Lazy functional programming is a programming style wherearguments are evaluated only when the value is actually needed.
With Haskell, the step from formal definition to program is particularly easy. This
presupposes, of course, that you are at ease with formal definitions. Our reason for
combining training in reasoning with an introduction to functional programming isthat your programming needs will provide motivation for improving your reasoning skills. Haskell programs will be used as illustrations for the theory throughout
the book. We will always put computer programs and pseudo-code of algorithms
in frames (rectangular boxes).
The chapters of this book are written in so-called ‘literate programming’ style[Knu92]. Literate programming is a programming style where the program and its
documentation are generated from the same source. The text of every chapter in
1

CHAPTER 1. GETTING STARTED

2

this book can be viewed as the documentation of the program code in that chapter.
Literate programming makes it impossible for program and documentation to get
out of sync. Program documentation is an integrated part of literate programming,in fact the bulk of a literate program is the program documentation. When writing programs in literate style there is less temptation to write program code first
while leaving the documentation for later. Programming in literate style proceedsfrom the assumption that the main challenge when programming is to make yourprogram digestible for humans. For a program to be useful, it should be easy forothers to understand the code. It should also be easy for you to understand yourown code when you reread your stuff the next day or the next week or the next
month and try to figure out what you were up to when you wrote your program.
To save you the trouble of retyping, the code discussed in this book can be retrieved
from the book website. The program code is the text in typewriter font that youfind in rectangular boxes throughout the chapters. Boxes may also contain codethat is not included in the chapter modules, usually because it defines functions thatare already predefined by the Haskell system, or because it redefines a function thatis already defined elsewhere in the chapter.
Typewriter font is also used for pieces of interaction with the Haskell interpreter,
but these illustrations of how the interpreter behaves when particular files areloaded and commands are given are not boxed.
Every chapter of this book is a so-called Haskell module. The following two lines
declare the Haskell module for the Haskell code of the present chapter. This module is called GS.

module GS
where

1.1 Starting up the Haskell Interpreter
We assume that you succeeded in retrieving the Haskell interpreter hugs from the
Haskell homepage www.haskell.org and that you managed to install it on yourcomputer. You can start the interpreter by typing hugs at the system prompt. Whenyou start hugs you should see something like Figure (1.1). The string Prelude>on the last line is the Haskell prompt when no user-defined files are loaded.
You can use hugs as a calculator as follows:

1.2. IMPLEMENTING A PRIME NUMBER TEST____ __ __ _______|||| || || || || ||__||___|| ||__|| ||__|| __||||---||
___||
||
||
||
|| Version: November 2003

3_________________________________________
Hugs 98: Based on the Haskell 98 standard
Copyright (c) 1994-2003
World Wide Web: http://haskell.org/hugs
Report bugs to: hugs-bugs@haskell.org_________________________________________

Haskell 98 mode: Restart with command line option -98 to enable extensions
Type :? for help
Prelude>Figure 1.1: Starting up the Haskell interpreter.
Prelude> 2^1665536
Prelude>The string Prelude> is the system prompt. 2^16 is what you type. After you hit
the return key (the key that is often labeled with Enter or ←-), the system answers
65536 and the prompt Prelude> reappears.
Exercise 1.1 Try out a few calculations using * for multiplication, + for addition,- for subtraction, ^ for exponentiation, / for division. By playing with the system,
find out what the precedence order is among these operators.
Parentheses can be used to override the built-in operator precedences:
Prelude> (2 + 3)^4625

To quit the Hugs interpreter, type :quit or :q at the system prompt.

1.2 Implementing a Prime Number Test
Suppose we want to implement a definition of prime number in a procedure thatrecognizes prime numbers. A prime number is a natural number greater than
1 that has no proper divisors other than 1 and itself. The natural numbers are
0, 1, 2, 3, 4, . . . The list of prime numbers starts with 2, 3, 5, 7, 11, 13, . . . Except
for 2, all of these are odd, of course.

CHAPTER 1. GETTING STARTED

4

Let n > 1 be a natural number. Then we use LD(n) for the least natural number
greater than 1 that divides n. A number d divides n if there is a natural number
a with a · d = n. In other words, d divides n if there is a natural number a withnd = a, i.e., division of n by d leaves no remainder. Note that LD(n) exists forevery natural number n > 1, for the natural number d = n is greater than 1 anddivides n. Therefore, the set of divisors of n that are greater than 1 is non-empty.
Thus, the set will have a least element.
The following proposition gives us all we need for implementing our prime number
test:
Proposition 1.2
1. If n > 1 then LD(n) is a prime number.
2. If n > 1 and n is not a prime number, then (LD(n))2 6 n.
In the course of this book you will learn how to prove propositions like this.
Here is the proof of the first item. This is a proof by contradiction (see Chapter 3).
Suppose, for a contradiction that c = LD(n) is not a prime. Then there are natural
numbers a and b with c = a · b, and also 1 < a and a < c. But then a divides n,
and contradiction with the fact that c is the smallest natural number greater than 1that divides n. Thus, LD(n) must be a prime number.
For a proof of the second item, suppose that n > 1, n is not a prime and thatp = LD(n). Then there is a natural number a > 1 with n = p · a. Thus, adivides n. Since p is the smallest divisor of n with p > 1, we have that p 6 a, andtherefore p2 6 p · a = n, i.e., (LD(n))2 6 n.
The operator · in a · b is a so-called infix operator. The operator is written between
its arguments. If an operator is written before its arguments we call this prefix
notation. The product of a and b in prefix notation would look like this: · a b.
In writing functional programs, the standard is prefix notation. In an expression
op a b, op is the function, and a and b are the arguments. The convention is thatfunction application associates to the left, so the expression op a b is interpretedas (op a) b.
Using prefix notation, we define the operation divides that takes two integer
expressions and produces a truth value. The truth values true and false are renderedin Haskell as True and False, respectively.
The integer expressions that the procedure needs to work with are called the arguments of the procedure. The truth value that it produces is called the value of theprocedure.

1.2. IMPLEMENTING A PRIME NUMBER TEST

5

Obviously, m divides n if and only if the remainder of the process of dividing n
by m equals 0. The definition of divides can therefore be phrased in terms of apredefined procedure rem for finding the remainder of a division process:divides d n = rem n d == 0

The definition illustrates that Haskell uses = for ‘is defined as’ and == for identity.
(The Haskell symbol for non-identity is /=.)
A line of Haskell code of the form foo t = ... (or foo t1 t2 = ..., or
foo t1 t2 t3 = ..., and so on) is called a Haskell equation. In such an equation, foo is called the function, and t its argument.
Thus, in the Haskell equation divides d n = rem n d == 0, divides is thefunction, d is the first argument, and n is the second argument.
Exercise 1.3 Put the definition of divides in a file prime.hs. Start the Haskell interpreter hugs (Section 1.1). Now give the command :load prime or :l prime,
followed by pressing Enter. Note that l is the letter l, not the digit 1. (Next to :l,a very useful command after you have edited a file of Haskell code is :reload or:r, for reloading the file.)
Prelude> :l prime
Main>

The string Main> is the Haskell prompt indicating that user-defined files are loaded.
This is a sign that the definition was added to the system. The newly defined operation can now be executed, as follows:
Main> divides 5 7
False
Main>

The string Main> is the Haskell prompt, the rest of the first line is what you type.
When you press Enter the system answers with the second line, followed by the
Haskell prompt. You can then continue with:Main> divides 5 30
True

CHAPTER 1. GETTING STARTED

6

It is clear from the proposition above that all we have to do to implement a primality test is to give an implementation of the function LD. It is convenient to define
LD in terms of a second function LDF, for the least divisor starting from a given
threshold k, with k 6 n. Thus, LDF(k)(n) is the least divisor of n that is > k.
Clearly, LD(n) = LDF(2)(n). Now we can implement LD as follows:

ld n = ldf 2 n

This leaves the implementation ldf of LDF (details of the coding will be explained
below):

ldf k n | divides k n = k| k^2 > n= n| otherwise= ldf (k+1) n

The definition employs the Haskell operation ^ for exponentiation, > for ‘greater
than’, and + for addition.
The definition of ldf makes use of equation guarding. The first line of the ldf
definition handles the case where the first argument divides the second argument.
Every next line assumes that the previous lines do not apply. The second line
handles the case where the first argument does not divide the second argument,
and the square of the first argument is greater than the second argument. The thirdline assumes that the first and second cases do not apply and handles all othercases, i.e., the cases where k does not divide n and k 2 < n.
The definition employs the Haskell condition operator | . A Haskell equation of
the formfoo t | condition = ...

is called a guarded equation. We might have written the definition of ldf as a listof guarded equations, as follows:

1.2. IMPLEMENTING A PRIME NUMBER TEST

7

ldf k n | divides k n = k
ldf k n | k^2 > n= n
ldf k n= ldf (k+1) n

The expression condition, of type Bool (i.e., Boolean or truth value), is called
the guard of the equation.
A list of guarded equations such as
foo t | condition_1 = body_1
foo t | condition_2 = body_2
foo t | condition_3 = body_3
foo t= body_4

can be abbreviated as
foo t | condition_1 = body_1| condition_2 = body_2| condition_3 = body_3| otherwise= body_4

Such a Haskell definition is read as follows:• in case condition_1 holds, foo t is by definition equal to body_1,• in case condition_1 does not hold but condition_2 holds, foo t is bydefinition equal to body_2,• in case condition_1 and condition_2 do not hold but condition_3
holds, foo t is by definition equal to body_3,• and in case none of condition_1, condition_2 and condition_3 hold,foo t is by definition equal to body_4.
When we are at the end of the list we know that none of the cases above in the list
apply. This is indicated by means of the Haskell reserved keyword otherwise.
Note that the procedure ldf is called again from the body of its own definition. We
will encounter such recursive procedure definitions again and again in the courseof this book (see in particular Chapter 7).

8

CHAPTER 1. GETTING STARTED

Exercise 1.4 Suppose in the definition of ldf we replace the condition k^2 > nby k^2 >= n, where >= expresses ‘greater than or equal’. Would that make any
difference to the meaning of the program? Why (not)?
Now we are ready for a definition of prime0, our first implementation of the test
for being a prime number.

prime0 n | n < 1= error "not a positive integer"| n == 1= False| otherwise = ld n == nHaskell allows a call to the error operation in any definition. This is used to break
off operation and issue an appropriate message when the primality test is used fornumbers below 1. Note that error has a parameter of type String (indicated by
the double quotes).
The definition employs the Haskell operation < for ‘less than’.
Intuitively, what the definition prime0 says is this:
1. the primality test should not be applied to numbers below 1,
2. if the test is applied to the number 1 it yields ‘false’,
3. if it is applied to an integer n greater than 1 it boils down to checking that
LD(n) = n. In view of the proposition we proved above, this is indeed acorrect primality test.
Exercise 1.5 Add these definitions to the file prime.hs and try them out.
Remark. The use of variables in functional programming has much in common
with the use of variables in logic. The definition divides d n = rem n d == 0
is equivalent to divides x y = rem y x == 0. This is because the variables
denote arbitrary elements of the type over which they range. They behave likeuniversally quantified variables, and just as in logic the definition does not dependon the variable names.

1.3 Haskell Type Declarations
Haskell has a concise way to indicate that divides consumes an integer, thenanother integer, and produces a truth value (called Bool in Haskell). Integers and1.3. HASKELL TYPE DECLARATIONS

9truth values are examples of types. See Section 2.1 for more on the type Bool.
Section 1.6 gives more information about types in general. Arbitrary precision
integers in Haskell have type Integer. The following line gives a so-called typedeclaration for the divides function.

divides :: Integer -> Integer -> Bool

Integer -> Integer -> Bool is short for Integer -> (Integer -> Bool).
A type of the form a -> b classifies a procedure that takes an argument of type ato produce a result of type b. Thus, divides takes an argument of type Integer
and produces a result of type Integer -> Bool, i.e., a procedure that takes anargument of type Integer, and produces a result of type Bool.
The full code for divides, including the type declaration, looks like this:

divides :: Integer -> Integer -> Bool
divides d n = rem n d == 0

If d is an expression of type Integer, then divides d is an expression of type
Integer -> Bool. The shorthand that we will use ford is an expression of type Integer
is: d :: Integer.
Exercise 1.6 Can you gather from the definition of divides what the type declaration for rem would look like?
Exercise 1.7 The hugs system has a command for checking the types of expressions. Can you explain the following (please try it out; make sure that the filewith the definition of divides is loaded, together with the type declaration for
divides):
Main> :t divides 5
divides 5 :: Integer -> Bool
Main> :t divides 5 7
divides 5 7 :: Bool
Main>

CHAPTER 1. GETTING STARTED10

The expression divides 5 :: Integer -> Bool is called a type judgment.
Type judgments in Haskell have the form expression :: type.
In Haskell it is not strictly necessary to give explicit type declarations. For instance, the definition of divides works quite well without the type declaration,
since the system can infer the type from the definition. However, it is good programming practice to give explicit type declarations even when this is not strictly
necessary. These type declarations are an aid to understanding, and they greatly
improve the digestibility of functional programs for human readers. A further
advantage of the explicit type declarations is that they facilitate detection of programming mistakes on the basis of type errors generated by the interpreter. You
will find that many programming errors already come to light when your program
gets loaded. The fact that your program is well typed does not entail that it iscorrect, of course, but many incorrect programs do have typing mistakes.
The full code for ld, including the type declaration, looks like this:ld :: Integer -> Integer
ld n = ldf 2 n

The full code for ldf, including the type declaration, looks like this:ldf :: Integer -> Integer -> Integer
ldf k n | divides k n = k| k^2 > n= n| otherwise= ldf (k+1) n

The first line of the code states that the operation ldf takes two integers and produces an integer.
The full code for prime0, including the type declaration, runs like this:1.4. IDENTIFIERS IN HASKELL

11

prime0 :: Integer -> Bool
prime0 n | n < 1= error "not a positive integer"| n == 1= False| otherwise = ld n == n

The first line of the code declares that the operation prime0 takes an integer andproduces (or returns, as programmers like to say) a Boolean (truth value).
In programming generally, it is useful to keep close track of the nature of theobjects that are being represented. This is because representations have to be stored
in computer memory, and one has to know how much space to allocate for thisstorage. Still, there is no need to always specify the nature of each data-type
explicitly. It turns out that much information about the nature of an object can beinferred from how the object is handled in a particular program, or in other words,
from the operations that are performed on that object.
Take again the definition of divides. It is clear from the definition that an operation is defined with two arguments, both of which are of a type for which rem isdefined, and with a result of type Bool (for rem n d == 0 is a statement that canturn out true or false). If we check the type of the built-in procedure rem we get:Prelude> :t rem
rem :: Integral a => a -> a -> aIn this particular case, the type judgment gives a type scheme rather than a type. Itmeans: if a is a type of class Integral, then rem is of type a -> a -> a. Herea is used as a variable ranging over types.
In Haskell, Integral is the class (see Section 4.2) consisting of the two types for
integer numbers, Int and Integer. The difference between Int and Integer
is that objects of type Int have fixed precision, objects of type Integer havearbitrary precision.
The type of divides can now be inferred from the definition. This is what we getwhen we load the definition of divides without the type declaration:
Main> :t divides
divides :: Integral a => a -> a -> Bool

1.4 Identifiers in Haskell
In Haskell, there are two kinds of identifiers:

CHAPTER 1. GETTING STARTED

12• Variable identifiers are used to name functions. They have to start with alower-case letter. E.g., map, max, fct2list, fctToList, fct_to_list.
• Constructor identifiers are used to name types. They have to start with an
upper-case letter. Examples are True, False.
Functions are operations on data-structures, constructors are the building blocksof the data structures themselves (trees, lists, Booleans, and so on).
Names of functions always start with lower-case letters, and may contain both
upper- and lower-case letters, but also digits, underscores and the prime symbol’. The following reserved keywords have special meanings and cannot be used toname functions.
case
if
let

class
importmoduledata
innewtypedefault
infix
ofderiving
infixlthendoinfixrtype

else
instance
where

The use of these keywords will be explained as we encounter them. at the beginning of a word is treated as a lower-case character. The underscore character all
by itself is a reserved word for the wild card pattern that matches anything (page141).
There is one more reserved keyword that is particular to Hugs: forall, for the definition of functions that take polymorphic arguments. See the Hugs documentation
for further particulars.

1.5 Playing the Haskell GameThis section consists of a number of further examples and exercises to get you
acquainted with the programming language of this book. To save you the troubleof keying in the programs below, you should retrieve the module GS.hs for thepresent chapter from the book website and load it in hugs. This will give you asystem prompt GS>, indicating that all the programs from this chapter are loaded.
In the next example, we use Int for the type of fixed precision integers, and [Int]for lists of fixed precision integers.
Example 1.8 Here is a function that gives the minimum of a list of integers:

1.5. PLAYING THE HASKELL GAME

13

mnmInt :: [Int] -> Int
mnmInt [] = error "empty list"
mnmInt [x] = x
mnmInt (x:xs) = min x (mnmInt xs)

This uses the predefined function min for the minimum of two integers. It alsouses pattern matching for lists . The list pattern [] matches only the empty list,
the list pattern [x] matches any singleton list, the list pattern (x:xs) matches any
non-empty list. A further subtlety is that pattern matching in Haskell is sensitive
to order. If the pattern [x] is found before (x:xs) then (x:xs) matches any
non-empty list that is not a unit list. See Section 4.6 for more information on list
pattern matching.
It is common Haskell practice to refer to non-empty lists as x:xs, y:ys, and soon, as a useful reminder of the facts that x is an element of a list of x’s and that xs
is a list.
Here is a home-made version of min:min’ :: Int -> Int -> Intmin’ x y | x <= y= x| otherwise = y

You will have guessed that <= is Haskell code for 6.
Objects of type Int are fixed precision integers. Their range can be found with:
Prelude> primMinInt-2147483648
Prelude> primMaxInt
2147483647

Since 2147483647 = 231 − 1, we can conclude that the hugs implementation uses
four bytes (32 bits) to represent objects of this type. Integer is for arbitrary precision integers: the storage space that gets allocated for Integer objects depends
on the size of the object.
Exercise 1.9 Define a function that gives the maximum of a list of integers. Use
the predefined function max.

CHAPTER 1. GETTING STARTED

14

Conversion from Prefix to Infix in Haskell A function can be converted to an
infix operator by putting its name in back quotes, like this:Prelude> max 4 55Prelude> 4 ‘max‘ 5
5Conversely, an infix operator is converted to prefix by putting the operator in roundbrackets (p. 21).
Exercise 1.10 Define a function removeFst that removes the first occurrence ofan integer m from a list of integers. If m does not occur in the list, the list remains
unchanged.
Example 1.11 We define a function that sorts a list of integers in order of increasing size, by means of the following algorithm:• an empty list is already sorted.
• if a list is non-empty, we put its minimum in front of the result of sorting thelist that results from removing its minimum.
This is implemented as follows:

srtInts :: [Int] -> [Int]srtInts [] = []srtInts xs = m : (srtInts (removeFst m xs)) where m = mnmInt xs

Here removeFst is the function you defined in Exercise 1.10. Note that the second
clause is invoked when the first one does not apply, i.e., when the argument ofsrtInts is not empty. This ensures that mnmInt xs never gives rise to an error.
Note the use of a where construction for the local definition of an auxiliary function.
Remark. Haskell has two ways to locally define auxiliary functions, where andlet constructions. The where construction is illustrated in Example 1.11. Thiscan also expressed with let, as follows:

1.5. PLAYING THE HASKELL GAME

15

srtInts’ :: [Int] -> [Int]srtInts’ [] = []srtInts’ xs = letm = mnmInt xs
in m : (srtInts’ (removeFst m xs))

The let construction uses the reserved keywords let and in.

Example 1.12 Here is a function that calculates the average of a list of integers.
The average of m and n is given by m+n
2 , the average of a list of k integerskn1 , . . . , nk is given by n1 +···+n.
Ingeneral,averages are fractions, so the resultktype of average should not be Int but the Haskell data-type for floating point
numbers, which is Float. There are predefined functions sum for the sum of a listof integers, and length for the length of a list. The Haskell operation for division/ expects arguments of type Float (or more precisely, of Fractional type, andFloat is such a type), so we need a conversion function for converting Ints into
Floats. This is done by fromInt. The function average can now be written as:

average :: [Int] -> Float
average [] = error "empty list"
average xs = fromInt (sum xs) / fromInt (length xs)

Again, it is instructive to write our own homemade versions of sum and length.
Here they are:sum’ :: [Int] -> Intsum’ [] = 0sum’ (x:xs) = x + sum’ xs

CHAPTER 1. GETTING STARTED

16

length’ :: [a] -> Int
length’ [] = 0
length’ (x:xs) = 1 + length’ xs

Note that the type declaration for length’ contains a variable a. This variable
ranges over all types, so [a] is the type of a list of objects of an arbitrary type a.
We say that [a] is a type scheme rather than a type. This way, we can use the samefunction length’ for computing the length of a list of integers, the length of a listof characters, the length of a list of strings (lists of characters), and so on.
The type [Char] is abbreviated as String. Examples of characters are ’a’, ’b’(note the single quotes) examples of strings are "Russell" and "Cantor" (notethe double quotes). In fact, "Russell" can be seen as an abbreviation of the list[’R’,’u’,’s’,’s’,’e’,’l’,’l’].
Exercise 1.13 Write a function count for counting the number of occurrences ofa character in a string. In Haskell, a character is an object of type Char, and a string
an object of type String, so the type declaration should run: count :: Char ->String -> Int.
Exercise 1.14 A function for transforming strings into strings is of type String-> String. Write a function blowup that converts a string
a1 a2 a3 · · ·toa1 a2 a2 a3 a3 a3 · · · .
blowup "bang!" should yield "baannngggg!!!!!". (Hint: use ++ for string
concatenation.)
Exercise 1.15 Write a function srtString :: [String] -> [String] thatsorts a list of strings in alphabetical order.
Example 1.16 Suppose we want to check whether a string str1 is a prefix ofa string str2. Then the answer to the question prefix str1 str2 should beeither yes (true) or no (false), i.e., the type declaration for prefix should run:prefix :: String -> String -> Bool.
Prefixes of a string ys are defined as follows:

1.6. HASKELL TYPES

17

1. [] is a prefix of ys,
2. if xs is a prefix of ys, then x:xs is a prefix of x:ys,
3. nothing else is a prefix of ys.
Here is the code for prefix that implements this definition:prefix :: String -> String -> Bool
prefix [] ys = True
prefix (x:xs) [] = False
prefix (x:xs) (y:ys) = (x==y) && prefix xs ys

The definition of prefix uses the Haskell operator && for conjunction.
Exercise 1.17 Write a function substring :: String -> String -> Boolthat checks whether str1 is a substring of str2.
The substrings of an arbitrary string ys are given by:
1. if xs is a prefix of ys, xs is a substring of ys,
2. if ys equals y:ys’ and xs is a substring of ys’, xs is a substring of ys,
3. nothing else is a substring of ys.

1.6 Haskell Types
The basic Haskell types are:• Int and Integer, to represent integers. Elements of Integer are unbounded. That’s why we used this type in the implementation of the prime
number test.
• Float and Double represent floating point numbers. The elements of Double
have higher precision.
• Bool is the type of Booleans.

CHAPTER 1. GETTING STARTED18• Char is the type of characters.

Note that the name of a type always starts with a capital letter.
To denote arbitrary types, Haskell allows the use of type variables. For these, a, b,. . . , are used.
New types can be formed in several ways:• By list-formation: if a is a type, [a] is the type of lists over a. Examples:[Int] is the type of lists of integers; [Char] is the type of lists of characters,
or strings.
• By pair- or tuple-formation: if a and b are types, then (a,b) is the typeof pairs with an object of type a as their first component, and an object oftype b as their second component. Similarly, triples, quadruples, . . . , can beformed. If a, b and c are types, then (a,b,c) is the type of triples with anobject of type a as their first component, an object of type b as their second
component, and an object of type c as their third component. And so on(p. 139).
• By function definition: a -> b is the type of a function that takes argumentsof type a and returns values of type b.
• By defining your own data-type from scratch, with a data type declaration.
More about this in due course.
Pairs will be further discussed in Section 4.5, lists and list operations in Section
4.6.
Operations are procedures for constructing objects of a certain types b from ingredients of a type a. Now such a procedure can itself be given a type: the type ofa transformer from a type objects to b type objects. The type of such a procedurecan be declared in Haskell as a -> b.
If a function takes two string arguments and returns a string then this can beviewed as a two-stage process: the function takes a first string and returns atransformer from strings to strings. It then follows that the type is String ->(String -> String), which can be written as String -> String -> String,
because of the Haskell convention that -> associates to the right.
Exercise 1.18 Find expressions with the following types:
1. [String]1.7. THE PRIME FACTORIZATION ALGORITHM

19

2. (Bool,String)
3. [(Bool,String)]
4. ([Bool],String)
5. Bool -> Bool
Test your answers by means of the Hugs command :t.
Exercise 1.19 Use the Hugs command :t to find the types of the following predefined functions:
1. head2. last3. init
4. fst
5. (++)
6. flip
7. flip (++)
Next, supply these functions with arguments of the expected types, and try to guess
what these functions do.

1.7 The Prime Factorization Algorithm
Let n be an arbitrary natural number > 1. A prime factorization of n is a list ofprime numbers p1 , . . . , pj with the property that p1 · · · · · pj = n. We will showthat a prime factorization of every natural number n > 1 exists by producing one
by means of the following method of splitting off prime factors:
WHILE n 6= 1 DO BEGIN p := LD(n); n :=

n
END
p

Here := denotes assignment or the act of giving a variable a new value. As wehave seen, LD(n) exists for every n with n > 1. Moreover, we have seen that
LD(n) is always prime. Finally, it is clear that the procedure terminates, for every
round through the loop will decrease the size of n.

20

CHAPTER 1. GETTING STARTEDSo the algorithm consists of splitting off primes until we have written n as n =p1 · · · pj , with all factors prime. To get some intuition about how the procedure
works, let us see what it does for an example case, say n = 84. The original assignment to n is called n0 ; successive assignments to n and p are called n1 , n2 , . . .
and p1 , p2 , . . ..
n0 = 84
n0 6= 1 p1 = 2 n1 = 84/2 = 42
n1 6= 1 p2 = 2 n2 = 42/2 = 21
n2 6= 1 p3 = 3 n3 = 21/3 = 7
n3 6= 1 p4 = 7 n4 = 7/7 = 1
n4 = 1This gives 84 = 22 · 3 · 7, which is indeed a prime factorization of 84.
The following code gives an implementation in Haskell, collecting the prime factors that we find in a list. The code uses the predefined Haskell function div for
integer division.

factors :: Integer -> [Integer]
factors n | n < 1= error "argument not positive"| n == 1= []| otherwise = p : factors (div n p) where p = ld n

If you load the code for this chapter, you can try this out as follows:
GS> factors 84[2,2,3,7]
GS> factors 557940830126698960967415390[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71]

1.8 The map and filter Functions
Haskell allows some convenient abbreviations for lists: [4..20] denotes the listof integers from 4 through 20, [’a’..’z’] the list of all lower case letters,"abcdefghijklmnopqrstuvwxyz". The call [5..] generates an infinite list of
integers starting from 5. And so on.
If you use the Hugs command :t to find the type of the function map, you get thefollowing:1.8. THE MAP AND FILTER FUNCTIONS

21

Prelude> :t map
map :: (a -> b) -> [a] -> [b]The function map takes a function and a list and returns a list containing the resultsof applying the function to the individual list members.
If f is a function of type a -> b and xs is a list of type [a], then map f xs will
return a list of type [b]. E.g., map (^2) [1..9] will produce the list of squares[1, 4, 9, 16, 25, 36, 49, 64, 81]

You should verify this by trying it out in Hugs. The use of (^2) for the operationof squaring demonstrates a new feature of Haskell, the construction of sections.
Conversion from Infix to Prefix, Construction of Sections If op is an infix
operator, (op) is the prefix version of the operator. Thus, 2^10 can also be written
as (^) 2 10. This is a special case of the use of sections in Haskell.
In general, if op is an infix operator, (op x) is the operation resulting from applying op to its right hand side argument, (x op) is the operation resulting fromapplying op to its left hand side argument, and (op) is the prefix version of theoperator (this is like the abstraction of the operator from both arguments).
Thus (^2) is the squaring operation, (2^) is the operation that computes powersof 2, and (^) is exponentiation. Similarly, (>3) denotes the property of being
greater than 3, (3>) the property of being smaller than 3, and (>) is the prefix
version of the ‘greater than’ relation.
The call map (2^) [1..10] will yield[2, 4, 8, 16, 32, 64, 128, 256, 512, 1024]

If p is a property (an operation of type a -> Bool) and xs is a list of type [a],then map p xs will produce a list of type Bool (a list of truth values), like this:Prelude> map (>3) [1..9][False, False, False, True, True, True, True, True, True]
Prelude>The function map is predefined in Haskell, but it is instructive to give our own
version:CHAPTER 1. GETTING STARTED

22

map :: (a -> b) -> [a] -> [b]map f [] = []map f (x:xs) = (f x) : (map f xs)

Note that if you try to load this code, you will get an error message:
Definition of variable "map" clashes with import.

The error message indicates that the function name map is already part of the namespace for functions, and is not available anymore for naming a function of yourown making.
Exercise 1.20 Use map to write a function lengths that takes a list of lists andreturns a list of the corresponding list lengths.
Exercise 1.21 Use map to write a function sumLengths that takes a list of lists
and returns the sum of their lengths.
Another useful function is filter, for filtering out the elements from a list thatsatisfy a given property. This is predefined, but here is a home-made version:filter :: (a -> Bool) -> [a] -> [a]filter p [] = []filter p (x:xs) | p x= x : filter p xs
| otherwise = filter p xs

Here is an example of its use:
GS> filter (>3) [1..10][4,5,6,7,8,9,10]

Example 1.22 Here is a program primes0 that filters the prime numbers from theinfinite list [2..] of natural numbers:

1.8. THE MAP AND FILTER FUNCTIONS

23

primes0 :: [Integer]
primes0 = filter prime0 [2..]This produces an infinite list of primes. (Why infinite? See Theorem 3.33.) The
list can be interrupted with ‘Control-C’.

Example 1.23 Given that we can produce a list of primes, it should be possible
now to improve our implementation of the function LD. The function ldf used in
the definition√ of ld looks for a prime divisor of n by checking k|n for all k with√
2 6 k 6 n. In fact, it is enough to check p|n for the primes p with 2 6 p 6 n.
Here are functions ldp and ldpf that perform this more efficient check:

ldp :: Integer -> Integer
ldp n = ldpf primes1 nldpf :: [Integer] -> Integer -> Integer
ldpf (p:ps) n | rem n p == 0 = p| p^2 > n= n| otherwise= ldpf ps n

ldp makes a call to primes1, the list of prime numbers. This is a first illustrationof a ‘lazy list’. The list is called ‘lazy’ because we compute only the part of thelist that we need for further processing. To define primes1 we need a test for
primality, but that test is itself defined in terms of the function LD, which in turnrefers to primes1. We seem to be running around in a circle. This circle can bemade non-vicious by avoiding the primality test for 2. If it is given that 2 is prime,then we can use the primality of 2 in the LD check that 3 is prime, and so on, andwe are up and running.

CHAPTER 1. GETTING STARTED

24

primes1 :: [Integer]
primes1 = 2 : filter prime [3..]prime :: Integer -> Bool
prime n | n < 1= error "not a positive integer"| n == 1= False| otherwise = ldp n == nReplacing the definition of primes1 by filter prime [2..] creates vicious
circularity, with stack overflow as a result (try it out). By running the program
primes1 against primes0 it is easy to check that primes1 is much faster.
Exercise 1.24 What happens when you modify the defining equation of ldp asfollows:

ldp :: Integer -> Integer
ldp = ldpf primes1

Can you explain?

1.9 Haskell Equations and Equational Reasoning
The Haskell equations f x y = ... used in the definition of a function f are
genuine mathematical equations. They state that the left hand side and the righthand side of the equation have the same value. This is very different from the useof = in imperative languages like C or Java. In a C or Java program, the statementx = x*y does not mean that x and x ∗ y have the same value, but rather it is acommand to throw away the old value of x and put the value of x ∗ y in its place.
It is a so-called destructive assignment statement: the old value of a variable isdestroyed and replaced by a new one.
Reasoning about Haskell definitions is a lot easier than reasoning about programs
that use destructive assignment. In Haskell, standard reasoning about mathematical equations applies. E.g., after the Haskell declarations x = 1 and y = 2, theHaskell declaration x = x + y will raise an error "x" multiply defined. Because = in Haskell has the meaning “is by definition equal to”, while redefinition

1.9. HASKELL EQUATIONS AND EQUATIONAL REASONING

25

is forbidden, reasoning about Haskell functions is standard equational reasoning.
Let’s try this out on a simple example.

a = 3
b = 4
f :: Integer -> Integer -> Integerf x y = x^2 + y^2

To evaluate f a (f a b) by equational reasoning, we can proceed as follows:f a (f a b) = f a (a2 + b2 )= f 3 (32 + 42 )= f 3 (9 + 16)= f 3 25
= 32 + 252= 9 + 625
= 634
The rewriting steps use standard mathematical laws and the Haskell definitions ofa, b, f . And, in fact, when running the program we get the same outcome:GS> f a (f a b)
634
GS>

Remark. We already encountered definitions where the function that is being
defined occurs on the right hand side of an equation in the definition. Here isanother example:

g :: Integer -> Integer
g 0= 0
g (x+1) = 2 * (g x)

Not everything that is allowed by the Haskell syntax makes semantic sense, however. The following definitions, although syntactically correct, do not properly
define functions:CHAPTER 1. GETTING STARTED26

h1 :: Integer -> Integer
h1 0 = 0h1 x = 2 * (h1 x)
h2 :: Integer -> Integer
h2 0 = 0h2 x = h2 (x+1)

The problem is that for values other than 0 the definitions do not give recipes forcomputing a value. This matter will be taken up in Chapter 7.

1.10 Further Reading
The standard Haskell operations are defined in the file Prelude.hs, which you
should be able to locate somewhere on any system that runs hugs. Typically, thefile resides in /usr/lib/hugs/libraries/Hugs/.
In case Exercise 1.19 has made you curious, the definitions of these example functions can all be found in Prelude.hs. If you want to quickly learn a lot about how toprogram in Haskell, you should get into the habit of consulting this file regularly.
The definitions of all the standard operations are open source code, and are there
for you to learn from. The Haskell Prelude may be a bit difficult to read at first,
but you will soon get used to the syntax and acquire a taste for the style.
Various tutorials on Haskell and Hugs can be found on the Internet: see e.g.
[HFP96] and [JR+ ]. The definitive reference for the language is [Jon03]. A textbook on Haskell focusing on multimedia applications is [Hud00]. Other excellent
textbooks on functional programming with Haskell are [Tho99] and, at a more advanced level, [Bir98]. A book on discrete mathematics that also uses Haskell as atool, and with a nice treatment of automated proof checking, is [HO00].

Chapter 2

Talking about Mathematical
Objects
Preview

To talk about mathematical objects with ease it is useful to introduce some symbolic abbreviations. These symbolic conventions are meant to better reveal thestructure of our mathematical statements. This chapter concentrates on a few (infact: seven), simple words or phrases that are essential to the mathematical vocabulary: not, if, and, or, if and only if, for all and for some. We will introduce
symbolic shorthands for these words, and we look in detail at how these building
blocks are used to construct the logical patterns of sentences. After having isolated
the logical key ingredients of the mathematical vernacular, we can systematicallyrelate definitions in terms of these logical ingredients to implementations, thusbuilding a bridge between logic and computer science.
The use of symbolic abbreviations in specifying algorithms makes it easier to takethe step from definitions to the procedures that implement those definitions. In asimilar way, the use of symbolic abbreviations in making mathematical statements
makes it easier to construct proofs of those statements. Chances are that you are
more at ease with programming than with proving things. However that may be,in the chapters to follow you will get the opportunity to improve your skills in bothof these activities and to find out more about the way in which they are related.
27

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

28

module TAMO
where

2.1 Logical Connectives and their Meanings
Goal To understand how the meanings of statements using connectives can be
described by explaining how the truth (or falsity) of the statement depends on
the truth (or falsity) of the smallest parts of this statement. This understanding
leads directly to an implementation of the logical connectives as truth functional
procedures.
In ordinary life, there are many statements that do not have a definite truth value,
for example ‘Barnett Newman’s Who is Afraid of Red, Yellow and Blue III is abeautiful work of art,’ or ‘Daniel Goldreyer’s restoration of Who is Afraid of Red,
Yellow and Blue III meets the highest standards.’
Fortunately the world of mathematics differs from the Amsterdam Stedelijk Museum of Modern Art in the following respect. In the world of mathematics, things
are so much clearer that many mathematicians adhere to the following slogan:every statement that makes mathematical sense is either true or false.
The idea behind this is that (according to the adherents) the world of mathematics
exists independently of the mind of the mathematician. Doing mathematics is theactivity of exploring this world. In proving new theorems one discovers new facts
about the world of mathematics, in solving exercises one rediscovers known facts
for oneself. (Solving problems in a mathematics textbook is like visiting famousplaces with a tourist guide.)This belief in an independent world of mathematical fact is called Platonism, after
the Greek philosopher Plato, who even claimed that our everyday physical world
is somehow an image of this ideal mathematical world. A mathematical Platonist
holds that every statement that makes mathematical sense has exactly one of thetwo truth values. Of course, a Platonist would concede that we may not know
which value a statement has, for mathematics has numerous open problems. Still,a Platonist would say that the true answer to an open problem in mathematics like
‘Are there infinitely many Mersenne primes?’ (Example 3.40 from Chapter 3) is2.1. LOGICAL CONNECTIVES AND THEIR MEANINGS

29

either ‘yes’ or ‘no’. The Platonists would immediately concede that nobody may
know the true answer, but that, they would say, is an altogether different matter.
Of course, matters are not quite this clear-cut, but the situation is certainly a lot
better than in the Amsterdam Stedelijk Museum. In the first place, it may not
be immediately obvious which statements make mathematical sense (see Example
4.5). In the second place, you don’t have to be a Platonist to do mathematics. Not
every working mathematician agrees with the statement that the world of mathematics exists independently of the mind of the mathematical discoverer. The Dutch
mathematician Brouwer (1881–1966) and his followers have argued instead thatthe mathematical reality has no independent existence, but is created by the working mathematician. According to Brouwer the foundation of mathematics is in theintuition of the mathematical intellect. A mathematical Intuitionist will therefore
not accept certain proof rules of classical mathematics, such as proof by contradiction (see Section 3.3), as this relies squarely on Platonist assumptions.
Although we have no wish to pick a quarrel with the intuitionists, in this book wewill accept proof by contradiction, and we will in general adhere to the practice ofclassical mathematics and thus to the Platonist creed.
Connectives In mathematical reasoning, it is usual to employ shorthands for if(or: if. . . then), and, or, not. These words are called connectives. The word andis used to form conjunctions, its shorthand ∧ is called the conjunction symbol.
The word or is used to form disjunctions, its shorthand ∨ is called the disjunction
symbol. The word not is used to form negations, its shorthand ¬ is called thenegation symbol. The combination if. . . then produces implications; its shorthand⇒ is the implication symbol. Finally, there is a phrase less common in everyday
conversation, but crucial if one is talking mathematics. The combination . . . if
and only if . . . produces equivalences, its shorthand ⇔ is called the equivalencesymbol. These logical connectives are summed up in the following table.

and
or
not
if—thenif, and only ifsymbol
∧
∨
¬
⇒
⇔

name
conjunction
disjunction
negation
implicationequivalence

Remark. Do not confuse if. . . then (⇒) on one hand with thus, so, therefore on theother. The difference is that the phrase if. . . then is used to construct conditional

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

30

statements, while thus (therefore, so) is used to combine statements into pieces ofmathematical reasoning (or: mathematical proofs). We will never write ⇒ when
we want to conclude from one mathematical statement to the next. The rules ofinference, the notion of mathematical proof, and the proper use of the word thus
are the subject of Chapter 3.

Iff. In mathematical English it is usual to abbreviate if, and only if to iff. We will
also use ⇔ as a symbolic abbreviation. Sometimes the phrase just in case is usedwith the same meaning.
The following describes, for every connective separately, how the truth value ofa compound using the connective is determined by the truth values of its components. For most connectives, this is rather obvious. The cases for ⇒ and ∨ havesome peculiar difficulties.
The letters P and Q are used for arbitrary statements. We use t for ‘true’, and f for‘false’. The set {t, f} is the set of truth values.
Haskell uses True and False for the truth values. Together, these form the type
Bool. This type is predefined in Haskell as follows:data Bool= False | True

NegationAn expression of the form ¬P (not P , it is not the case that P , etc.) is called thenegation of P . It is true (has truth value t) just in case P is false (has truth valuef).
In an extremely simple table, this looks as follows:
P
t
f

¬P
f
t

This table is called the truth table of the negation symbol.
The implementation of the standard Haskell function not reflects this truth table:

2.1. LOGICAL CONNECTIVES AND THEIR MEANINGS

notnot True
not False31:: Bool -> Bool
= False= TrueThis definition is part of Prelude.hs, the file that contains the predefined Haskell
functions.

Conjunction
The expression P ∧ Q ((both) P and Q) is called the conjunction of P and Q. P
and Q are called conjuncts of P ∧ Q. The conjunction P ∧ Q is true iff P and
Q are both true.
Truth table of the conjunction symbol:
P
t
t
f
f

Q
t
f
t
f

P ∧ Q
t
f
f
f

This is reflected in definition of the Haskell function for conjunction, && (also from
Prelude.hs):(&&) :: Bool -> Bool -> Bool
False && x
= FalseTrue && x= x

What this says is: if the first argument of a conjunction evaluates to false, thenthe conjunction evaluates to false; if the first argument evaluates to true, then the
conjunction gets the same value as its second argument. The reason that the typedeclaration has (&&) instead of && is that && is an infix operator, and (&&) is its
prefix counterpart (see page 21).

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

32
Disjunction

The expression P ∨ Q (P or Q) is called the disjunction of P and Q. P and Q
are the disjuncts of P ∨ Q.
The interpretation of disjunctions is not always straightforward. English has two
disjunctions: (i) the inclusive version, that counts a disjunction as true also in caseboth disjuncts are true, and (ii) the exclusive version either. . . or, that doesn’t.
Remember: The symbol ∨ will always be used for the inclusive version of or.
Even with this problem out of the way, difficulties may arise.
Example 2.1 No one will doubt the truth of the following:for every integer x, x < 1 or 0 < x.
However, acceptance of this brings along acceptance of every instance. E.g., for
x := 1:1
1 < 1 or 0 < 1.
Some people do not find this acceptable or true, or think this to make no sense at
all since something better can be asserted, viz., that 0 < 1. In mathematics with
the inclusive version of ∨ , you’ll have to live with such a peculiarity.
The truth table of the disjunction symbol ∨ now looks as follows.
P
t
t
f
f

Q
t
f
t
f

P ∨ Q
t
t
t
f

Here is the Haskell definition of the disjunction operation. Disjunction is rendered
as || in Haskell.

(||) :: Bool -> Bool -> Bool
False || x
= x
True || x= True1 := means: ‘is by definition equal to’.

2.1. LOGICAL CONNECTIVES AND THEIR MEANINGS

33

What this means is: if the first argument of a disjunction evaluates to false, thenthe disjunction gets the same value as its second argument. If the first argument of
a disjunction evaluates to true, then the disjunction evaluates to true.
Exercise 2.2 Make up the truth table for the exclusive version of or.
Implication
An expression of the form P ⇒ Q (if P , then Q; Q if P ) is called the implicationof P and Q. P is the antecedent of the implication and Q the consequent.
The truth table of ⇒ is perhaps the only surprising one. However, it can be motivated quite simply using an example like the following. No one will disagree thatfor every natural number n,
5 < n ⇒ 3 < n.
Therefore, the implication must hold in particular for n equal to 2, 4 and 6. But
then, an implication should be true if• both antecedent and consequent are false (n = 2),• antecedent false, consequent true (n = 4),
and• both antecedent and consequent true (n = 6).
And of course, an implication should be false in the only remaining case that theantecedent is true and the consequent false. This accounts for the following truth
table.
P
t
t
f
f

Q
t
f
t
f

P ⇒Q
t
f
t
t

If we want to implement implication in Haskell, we can do so in terms of not and||. It is convenient to introduce an infix operator ==> for this. The number 1
in the infix declaration indicates the binding power (binding power 0 is lowest,
9 is highest). A declaration of an infix operator together with an indication of its
binding power is called a fixity declaration.

34

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

infix 1 ==>(==>) :: Bool -> Bool -> Boolx ==> y = (not x) || y

It is also possible to give a direct definition:(==>) :: Bool -> Bool -> Bool
True ==> x = x
False ==> x = True

Trivially True Implications. Note that implications with antecedent false andthose with consequent true are true. For instance, because of this, the following
two sentences must be counted as true: if my name is Napoleon, then the decimal
expansion of π contains the sequence 7777777, and: if the decimal expansion of π
contains the sequence 7777777, then strawberries are red.
Implications with one of these two properties (no matter what the values of parameters that may occur) are dubbed trivially true. In what follows there are quite anumber of facts that are trivial in this sense that may surprise the beginner. One isthat the empty set ∅ is included in every set (cf. Theorem 4.9 p. 126).
Remark. The word trivial is often abused. Mathematicians have a habit of calling
things trivial when they are reluctant to prove them. We will try to avoid this useof the word. The justification for calling a statement trivial resides in the psychological fact that a proof of that statement immediately comes to mind. Whether a
proof of something comes to your mind will depend on your training and experience, so what is trivial in this sense is (to some extent) a personal matter. When
we are reluctant to prove a statement, we will sometimes ask you to prove it as an
exercise.
Implication and Causality. The mathematical use of implication does not always correspond to what you are used to. In daily life you will usually require acertain causal dependence between antecedent and consequent of an implication.
(This is the reason the previous examples look funny.) In mathematics, such a2.1. LOGICAL CONNECTIVES AND THEIR MEANINGS

35

causality usually will be present, but this is quite unnecessary for the interpretation of an implication: the truth table tells the complete story. (And in this section
in particular, causality usually will be absent.) However, in a few cases, natural
language use surprisingly corresponds with truth table-meaning. E.g., I’ll be deadif Bill will not show up must be interpreted (if uttered by someone healthy) as
strong belief that Bill will indeed turn up.2
Converse and Contraposition. The converse of an implication P ⇒ Q is Q ⇒P ; its contraposition is ¬Q ⇒ ¬P . The converse of a true implication does
not need to be true, but its contraposition is true iff the implication is. Cf. Theorem 2.10, p. 45.
Necessary and Sufficient Conditions. The statement P is called a sufficientcondition for Q and Q a necessary condition for P if the implication P ⇒ Q
holds.
An implication P ⇒ Q can be expressed in a mathematical text in a number ofways:
1. if P , then Q,
2. Q if P ,3. P only if Q,
4. Q whenever P ,5. P is sufficient for Q,
6. Q is necessary for P .
Equivalence
The expression P ⇔ Q (P iff Q) is called the equivalence of P and Q. P and Q
are the members of the equivalence. The truth table of the equivalence symbol isunproblematic once you realize that an equivalence P ⇔ Q amounts to the conjunction of two implications P ⇒ Q and Q ⇒ P taken together. (It is sometimes
convenient to write Q ⇒ P as P ⇐ Q.) The outcome is that an equivalence mustbe true iff its members have the same truth value.
Table:
2 ‘If Bill will not show up, then I am a Dutchman’, has the same meaning, when uttered by a native
speaker of English. What it means when uttered by one of the authors of this book, we are not sure.

36

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS
P
t
t
f
f

Q
t
f
t
f

P ⇔Q
t
f
f
t

From the discussion under implication it is clear that P is called a condition thatis both necessary and sufficient for Q if P ⇔ Q is true.
There is no need to add a definition of a function for equivalence to Haskell. The
type Bool is in class Eq, which means that an equality relation is predefined on it.
But equivalence of propositions is nothing other than equality of their truth values.
Still, it is useful to have a synonym:

infix 1 <=>(<=>) :: Bool -> Bool -> Boolx <=> y = x == yExample 2.3 When you are asked to prove something of the form P iff Q it isoften convenient to separate this into its two parts P ⇒ Q and P ⇐ Q. The ‘only
if’ part of the proof is the proof of P ⇒ Q (for P ⇒ Q means the same as P only
if Q), and the ‘if’ part of the proof is the proof of P ⇐ Q (for P ⇐ Q means thesame as Q ⇒ P , which in turn means the same as P , if Q).
Exercise 2.4 Check that the truth table for exclusive or from Exercise 2.2 is equivalent to the table for ¬(P ⇔ Q). Conclude that the Haskell implementation of thefunction <+> for exclusive or in the frame below is correct.

infixr 2 <+>(<+>) :: Bool -> Bool -> Boolx <+> y = x /= yThe logical connectives ∧ and ∨ are written in infix notation. Their Haskell counterparts, && and || are also infix. Thus, if p and q are expressions of type Bool,2.1. LOGICAL CONNECTIVES AND THEIR MEANINGS

37

then p && q is a correct Haskell expression of type Bool. If one wishes to write
this in prefix notation, this is also possible, by putting parentheses around the operator: (&&) p q.
Although you will probably never find more than 3–5 connectives occurring in one
mathematical statement, if you insist you can use as many connectives as you like.
Of course, by means of parentheses you should indicate the way your expression
was formed.
For instance, look at the formula¬P ∧ ((P ⇒ Q) ⇔ ¬(Q ∧ ¬P )).
Using the truth tables, you can determine its truth value if truth values for thecomponents P and Q have been given. For instance, if P has value t and Q
has value f, then ¬P has f, P ⇒ Q becomes f, Q ∧ ¬P : f; ¬(Q ∧ ¬P ): t;(P ⇒ Q) ⇔ ¬(Q ∧ ¬P ): f, and the displayed expression thus has value f. This
calculation can be given immediately under the formula, beginning with the values
given for P and Q. The final outcome is located under the conjunction symbol ∧,which is the main connective of the expression.

¬ P
t
f

∧

((P
t

⇒ Q) ⇔ ¬ (Q
f
f
f

∧

¬

P ))
t

f
f

t
f
f
In compressed form, this looks as follows:

¬ P
f t

∧
f

((P
t

⇒ Q) ⇔ ¬ (Q
f
f
f t
f

∧
f

¬
f

P ))
t

Alternatively, one might use a computer to perform the calculation.

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

38

p = Trueq = False
formula1 = (not p) && (p ==> q) <=> not (q && (not p))

After loading the file with the code of this chapter, you should be able to do:
TAMO> formula1
False
TAMO>

Note that p and q are defined as constants, with values True and False, respectively, so that the occurrences of p and q in the expression formula1 are evaluatedas these truth values. The rest of the evaluation is then just a matter of applying
the definitions of not, &&, <=> and ==>.

2.2 Logical Validity and Related Notions
Goal To grasp the concepts of logical validity and logical equivalence, to learn
how to use truth tables in deciding questions of validity and equivalence, and in thehandling of negations, and to learn how the truth table method for testing validity
and equivalence can be implemented.
Logical Validities. There are propositional formulas that receive the value t no
matter what the values of the occurring letters. Such formulas are called (logically)
valid.
Examples of logical validities are: P ⇒ P , P ∨ ¬P , and P ⇒ (Q ⇒ P ).
Truth Table of an Expression. If an expression contains n letters P, Q, . . ., thenthere are 2n possible distributions of the truth values between these letters. The
2n -row table that contains the calculations of these values is the truth table of theexpression.
If all calculated values are equal to t, then your expression, by definition, is avalidity.

2.2. LOGICAL VALIDITY AND RELATED NOTIONS

39

Example 2.5 (Establishing Logical Validity by Means of a Truth Table)
The following truth table shows that P ⇒ (Q ⇒ P ) is a logical validity.
P
t
t
f
f

⇒
t
t
t
t

(Q
t
f
t
f

⇒
t
t
f
t

P)
t
t
f
f

To see how we can implement the validity check in Haskell, look at the implementation of the evaluation formula1 again, and add the following definition of
formula2:formula2 p q = ((not p) && (p ==> q) <=> not (q && (not p)))To see the difference between the two definitions, let us check their types:
TAMO> :t formula1
formula1 :: Bool
TAMO> :t formula2
formula2 :: Bool -> Bool -> BoolTAMO>

The difference is that the first definition is a complete proposition (type Bool) initself, while the second still needs two arguments of type Bool before it will return
a truth value.
In the definition of formula1, the occurrences of p and q are interpreted as constants, of which the values are given by previous definitions. In the definition of
formula2. the occurrences of p and q are interpreted as variables that represent
the arguments when the function gets called.
A propositional formula in which the proposition letters are interpreted as variables
can in fact be considered as a propositional function or Boolean function or truth
function. If just one variable, say p occurs in it, then it is a function of type
Bool -> Bool (takes a Boolean, returns a Boolean). If two variables occur in it,
say p and q, then it is a function of type Bool -> Bool -> Bool (takes Boolean,

40

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

then takes another Boolean, and returns a Boolean). If three variables occur in it,then it is of type Bool -> Bool -> Bool -> Bool, and so on.
In the validity check for a propositional formula, we treat the proposition letters
as arguments of a propositional function, and we check whether evaluation of thefunction yields true for every possible combination of the arguments (that is theessence of the truth table method for checking validity). Here is the case for propositions with one proposition letter (type Bool -> Bool).

valid1 :: (Bool -> Bool) -> Boolvalid1 bf = (bf True) && (bf False)

The validity check for Boolean functions of type Bool -> Bool is suited to test
functions of just one variable. An example is the formula P ∨ ¬P that expresses
the principle of excluded middle (or, if you prefer a Latin name, tertium non datur,for: there is no third possibility). Here is its implementation in Haskell:

excluded_middle :: Bool -> Bool
excluded_middle p = p || not p

To check that this is valid by the truth table method, one should consider the twocases P := t and P := f, and ascertain that the principle yields t in both ofthese cases. This is precisely what the validity check valid1 does: it yields True
precisely when applying the boolean function bf to True yields True and applying
bf to False yields True. And indeed, we get:
TAMO> valid1 excluded_middle
True

Here is the validity check for propositional functions with two proposition letters, Such propositional functions have type Bool -> Bool -> Bool), and needa truth table with four rows to check their validity, as there are four cases to check.

2.2. LOGICAL VALIDITY AND RELATED NOTIONS

valid2 :: (Bool -> Bool -> Bool)valid2 bf =(bf True True)&& (bf True False)&& (bf False True)&& (bf False False)41-> Bool

Again, it is easy to see that this is an implementation of the truth table method forvalidity checking. Try this out on P ⇒ (Q ⇒ P ) and on (P ⇒ Q) ⇒ P , anddiscover that the bracketing matters:

form1 p q = p ==> (q ==> p)
form2 p q = (p ==> q) ==> p

TAMO> valid2 form1
True
TAMO> valid2 form2
False

The propositional function formula2 that was defined above is also of the rightargument type for valid2:

TAMO> valid2 formula2
False

It should be clear how the notion of validity is to be implemented for propositional
functions with more than two propositional variables. Writing out the full tables
becomes a bit irksome, so we are fortunate that Haskell offers an alternative. We
demonstrate it in valid3 and valid4,.

42

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

valid3 :: (Bool -> Bool -> Bool -> Bool) -> Boolvalid3 bf = and [ bf p q r | p <- [True,False],
q <- [True,False],r <- [True,False]]
valid4 :: (Bool -> Bool -> Bool -> Bool -> Bool) -> Boolvalid4 bf = and [ bf p q r s | p <- [True,False],
q <- [True,False],r <- [True,False],s <- [True,False]]

The condition p <- [True,False], for “p is an element of the list consisting of
the two truth values”, is an example of list comprehension (page 118).
The definitions make use of Haskell list notation, and of the predefined function
and for generalized conjunction. An example of a list of Booleans in Haskell is[True,True,False]. Such a list is said to be of type [Bool]. If list is a list of
Booleans (an object of type [Bool]), then and list gives True in case all members of list are true, False otherwise. For example, and [True,True,False]
gives False, but and [True,True,True] gives True. Further details about
working with lists can be found in Sections 4.6 and 7.5.
Leaving out Parentheses. We agree that ∧ and ∨ bind more strongly than ⇒and ⇔. Thus, for instance, P ∧ Q ⇒ R stands for (P ∧ Q) ⇒ R (and not forP ∧ (Q ⇒ R)).
Operator Precedence in Haskell In Haskell, the convention is not quite thesame, for || has operator precedence 2, && has operator precedence 3, and == hasoperator precedence 4, which means that == binds more strongly than &&, which
in turn binds more strongly than ||. The operators that we added, ==> and <=>,
follow the logic convention: they bind less strongly than && and ||.
Logically Equivalent. Two formulas are called (logically) equivalent if, no matter the truth values of the letters P, Q, . . . occurring in these formulas, the truth
values obtained for them are the same. This can be checked by constructing a truth
table (see Example (2.6)).
Example 2.6 (The First Law of De Morgan)

2.2. LOGICAL VALIDITY AND RELATED NOTIONS
¬
f
t
t
t

(P
t
t
f
f

∧
t
f
f
f

Q)
t
f
t
f

(¬
f
f
t
t

P
t
t
f
f

∨
f
t
t
t

43
¬
f
t
f
t

Q)
t
f
t
f

The outcome of the calculation shows that the formulas are equivalent: note thatthe column under the ¬ of ¬(P ∧ Q) coincides with that under the ∨ of ¬P ∨ ¬Q.
Notation: Φ ≡ Ψ indicates that Φ and Ψ are equivalent3. Using this notation, we
can say that the truth table of Example (2.6) shows that ¬(P ∧ Q) ≡ (¬P ∨ ¬Q).
Example 2.7 (De Morgan Again)
The following truth table shows that ¬(P ∧ Q) ⇔ (¬P ∨ ¬Q) is a logical validity,
which establishes that ¬(P ∧ Q) ≡ (¬P ∨ ¬Q).
¬
f
t
t
t

(P
t
t
f
f

∧
t
f
f
f

Q)
t
f
t
f

⇔
t
t
t
t

(¬
f
f
t
t

P
t
t
f
f

∨
f
t
t
t

¬
f
t
f
t

Q)
t
f
t
f

Example 2.8 A pixel on a computer screen is a dot on the screen that can be either
on (i.e., visible) or off (i.e., invisible). We can use 1 for on and 0 for off. Turning
pixels in a given area on the screen off or on creates a screen pattern for that area.
The screen pattern of an area is given by a list of bits (0s or 1s). Such a list of bits
can be viewed as a list of truth values (by equating 1 with t and 0 with f), and given
two bit lists of the same length we can perform bitwise logical operations on them:
the bitwise exclusive or of two bit lists of the same length n, say L = [P1 , . . . , Pn ]and K = [Q1 , . . . , Qn ], is the list [P1 ⊕ Q1 , . . . , Pn ⊕ Qn ], where ⊕ denotes
exclusive or.
In the implementation of cursor movement algorithms, the cursor is made visibleon the screen by taking a bitwise exclusive or between the screen pattern S at the
cursor position and the cursor pattern C. When the cursor moves elsewhere, theoriginal screen pattern is restored by taking a bitwise exclusive or with the cursor
pattern C again. Exercise 2.9 shows that this indeed restores the original pattern
S.
3 The Greek alphabet is on p. 423.

44

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

Exercise 2.9 Let ⊕ stand for exclusive or. Show, using the truth table from Exercise 2.2, that (P ⊕ Q) ⊕ Q is equivalent to P .
In Haskell, logical equivalence can be tested as follows. First we give a procedure
for propositional functions with 1 parameter:

logEquiv1 :: (Bool -> Bool) -> (Bool -> Bool) -> BoollogEquiv1 bf1 bf2 =(bf1 True <=> bf2 True) && (bf1 False <=> bf2 False)

What this does, for formulas Φ, Ψ with a single propositional variable, is testing
the formula Φ ⇔ Ψ by the truth table method.
We can extend this to propositional functions with 2, 3 or more parameters, using generalized conjunction. Here are the implementations of logEquiv2 andlogEquiv3; it should be obvious how to extend this for truth functions with still
more arguments.

logEquiv2 :: (Bool -> Bool -> Bool) ->(Bool -> Bool -> Bool) -> BoollogEquiv2 bf1 bf2 =and [(bf1 p q) <=> (bf2 p q) | p <- [True,False],
q <- [True,False]]
logEquiv3 :: (Bool -> Bool -> Bool -> Bool) ->(Bool -> Bool -> Bool -> Bool) -> BoollogEquiv3 bf1 bf2 =and [(bf1 p q r) <=> (bf2 p q r) | p <- [True,False],
q <- [True,False],r <- [True,False]]

Let us redo Exercise (2.9) by computer.

formula3 p q = p
formula4 p q = (p <+> q) <+> q

2.2. LOGICAL VALIDITY AND RELATED NOTIONS

45

Note that the q in the definition of formula3 is needed to ensure that it is a function
with two arguments.
TAMO> logEquiv2 formula3 formula4
True

We can also test this by means of a validity check on P ⇔ ((P ⊕ Q) ⊕ Q), as
follows:

formula5 p q = p <=> ((p <+> q) <+> q)

TAMO> valid2 formula5
True

Warning. Do not confuse ≡ and ⇔. If Φ and Ψ are formulas, then Φ ≡ Ψ expresses the statement that Φ and Ψ are equivalent. On the other hand, Φ ⇔ Ψ isjust another formula. The relation between the two is that the formula Φ ⇔ Ψ
is logically valid iff it holds that Φ ≡ Ψ. (See Exercise 2.19.) Compare the difference, in Haskell, between logEquiv2 formula3 formula4 (a true statement
about the relation between two formulas), and formula5 (just another formula).
The following theorem collects a number of useful equivalences. (Of course, P ,Q and R can be arbitrary formulas themselves.)
1. P ≡ ¬¬P(law of double negation),

2. P ∧ P ≡ P ; P ∨ P ≡ P(laws of idempotence),

Theorem 2.10

3. (P ⇒ Q) ≡ ¬P ∨ Q;¬(P ⇒ Q) ≡ P ∧ ¬Q,

4. (¬P ⇒ ¬Q) ≡ (Q ⇒ P );(P ⇒ ¬Q) ≡ (Q ⇒ ¬P );(¬P ⇒ Q) ≡ (¬Q ⇒ P )

5. (P ⇔ Q) ≡ ((P ⇒ Q) ∧ (Q ⇒ P ))≡ ((P ∧ Q) ∨ (¬P ∧ ¬Q)),(laws of contraposition),

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

46

6. P ∧ Q ≡ Q ∧ P ; P ∨ Q ≡ Q ∨ P(laws of commutativity),

7. ¬(P ∧ Q) ≡ ¬P ∨ ¬Q;¬(P ∨ Q) ≡ ¬P ∧ ¬Q

8. P ∧ (Q ∧ R) ≡ (P ∧ Q) ∧ R;
P ∨ (Q ∨ R) ≡ (P ∨ Q) ∨ R

9. P ∧ (Q ∨ R) ≡ (P ∧ Q) ∨ (P ∧ R);
P ∨ (Q ∧ R) ≡ (P ∨ Q) ∧ (P ∨ R)

(DeMorgan laws).

(laws of associativity),(distribution laws),

Equivalence 8 justifies leaving out parentheses in conjunctions and disjunctions of
three or more conjuncts resp., disjuncts. Non-trivial equivalences that often areused in practice are 2, 3 and 9. Note how you can use these to re-write negations:a negation of an implication can be rewritten as a conjunction, a negation of a
conjunction (disjunction) is a disjunction (conjunction).
Exercise 2.11 The First Law of De Morgan was proved in Example 2.6. Thismethod was implemented above. Use the method by hand to prove the other partsof Theorem 2.10.
We will now demonstrate how one can use the implementation of the logical equivalence tests as a check for Theorem 2.10. Here is a question for you to ponder:
does checking the formulas by means of the implemented functions for logicalequivalence count as a proof of the principles involved? Whatever the answer tothis one may be, Figure 2.1 defines the tests for the statements made in Theorem
2.10, by means of lambda abstraction The expression \ p -> not (not p) is
the Haskell way of referring to the lambda term λp.¬¬p, the term that denotes theoperation of performing a double negation. See Section 2.4.
If you run these tests, you get result True for all of them. E.g.:
TAMO> test5a
True

The next theorem lists some useful principles for reasoning with > (the proposition
that is always true; the Haskell counterpart is True) and ⊥ (the proposition that isalways false; the Haskell counterpart of this is False).
Theorem 2.12

1. ¬> ≡ ⊥; ¬⊥ ≡ >,2. P ⇒ ⊥ ≡ ¬P ,

2.2. LOGICAL VALIDITY AND RELATED NOTIONS

47

test1 = logEquiv1 id (\ p -> not (not p))
test2a = logEquiv1 id (\ p -> p && p)
test2b = logEquiv1 id (\ p -> p || p)test3a = logEquiv2 (\ p q -> p ==> q) (\ p q -> not p || q)test3b = logEquiv2 (\ p q -> not (p ==> q)) (\ p q -> p && not q)
test4a = logEquiv2 (\ p q -> not p ==> not q) (\ p q -> q ==> p)
test4b = logEquiv2 (\ p q -> p ==> not q) (\ p q -> q ==> not p)
test4c = logEquiv2 (\ p q -> not p ==> q) (\ p q -> not q ==> p)
test5a = logEquiv2 (\ p q -> p <=> q)(\ p q -> (p ==> q) && (q ==> p))
test5b = logEquiv2 (\ p q -> p <=> q)(\ p q -> (p && q) || (not p && not q))
test6a = logEquiv2 (\ p q -> p && q) (\ p q -> q && p)
test6b = logEquiv2 (\ p q -> p || q) (\ p q -> q || p)test7a = logEquiv2 (\ p q -> not (p && q))(\ p q -> not p || not q)
test7b = logEquiv2 (\ p q -> not (p || q))(\ p q -> not p && not q)
test8a = logEquiv3 (\ p q r -> p && (q && r))(\ p q r -> (p && q) && r)
test8b = logEquiv3 (\ p q r -> p || (q || r))(\ p q r -> (p || q) || r)
test9a = logEquiv3 (\ p q r -> p && (q || r))(\ p q r -> (p && q) || (p && r))
test9b = logEquiv3 (\ p q r -> p || (q && r))(\ p q r -> (p || q) && (p || r))

Figure 2.1: Defining the Tests for Theorem 2.10.

48

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS
3. P ∨ > ≡ >; P ∧ ⊥ ≡ ⊥(dominance laws),

4. P ∨ ⊥ ≡ P ; P ∧ > ≡ P(identity laws),

5. P ∨ ¬P ≡ >(law of excluded middle),

6. P ∧ ¬P ≡ ⊥(contradiction).

Exercise 2.13 Implement checks for the principles from Theorem 2.12.
Without proof, we state the following Substitution Principle: If Φ and Ψ areequivalent, and Φ0 and Ψ0 are the results of substituting Ξ for every occurrence of
P in Φ and in Ψ, respectively, then Φ0 and Ψ0 are equivalent. Example 2.14 makes
clear what this means.
Example 2.14 From ¬(P ⇒ Q) ≡ P ∧ ¬Q plus the substitution principle itfollows that¬(¬P ⇒ Q) ≡ ¬P ∧ ¬Q
(by substituting ¬P for P ), but also that¬(a = 2b − 1 ⇒ a is prime) ≡ a = 2b − 1 ∧ a is not prime(by substituting a = 2b − 1 for P and a is prime for Q).
Exercise 2.15 A propositional contradiction is a formula that yields false for every
combination of truth values for its proposition letters. Write Haskell definitions ofcontradiction tests for propositional functions with one, two and three variables.
Exercise 2.16 Produce useful denials for every sentence of Exercise 2.31. (A denial of Φ is an equivalent of ¬Φ.)
Exercise 2.17 Produce a denial for the statement that x < y < z (where x, y, z ∈R).
Exercise 2.18 Show:
1. (Φ ⇔ Ψ) ≡ (¬Φ ⇔ ¬Ψ),
2. (¬Φ ⇔ Ψ) ≡ (Φ ⇔ ¬Ψ).

Exercise 2.19 Show that Φ ≡ Ψ is true iff Φ ⇔ Ψ is logically valid.

2.2. LOGICAL VALIDITY AND RELATED NOTIONS

49

Exercise 2.20 Determine (either using truth tables or Theorem 2.10) which of thefollowing are equivalent, next check your answer by computer:
1. ¬P ⇒ Q and P ⇒ ¬Q,
2. ¬P ⇒ Q and Q ⇒ ¬P ,3. ¬P ⇒ Q and ¬Q ⇒ P ,4. P ⇒ (Q ⇒ R) and Q ⇒ (P ⇒ R),
5. P ⇒ (Q ⇒ R) and (P ⇒ Q) ⇒ R,
6. (P ⇒ Q) ⇒ P and P ,7. P ∨ Q ⇒ R and (P ⇒ R) ∧ (Q ⇒ R).

Exercise 2.21 Answer as many of the following questions as you can:1. Construct a formula Φ involving the letters P and Q that has the following
truth table.
P
t
t
f
f

Q
t
f
t
f

Φ
t
t
f
t

2. How many truth tables are there for 2-letter formulas altogether?
3. Can you find formulas for all of them?
4. Is there a general method for finding these formulas?
5. And what about 3-letter formulas and more?

50

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

2.3 Making Symbolic Form Explicit
In a sense, propositional reasoning is not immediately relevant for mathematics.
Few mathematicians will ever feel the urge to write down a disjunction of two
statements like 3 < 1 ∨ 1 < 3. In cases like this it is clearly “better” to only write
down the right-most disjunct.
Fortunately, once variables enter the scene, propositional reasoning suddenly becomes a very useful tool: the connectives turn out to be quite useful for combining
open formulas. An open formula is a formula with one or more unbound variables
in it. Variable binding will be explained below, but here is a first example of aformula with an unbound variable x. A disjunction like 3 < x ∨ x < 3 is (in somecases) a useful way of expressing that x 6= 3.
Example. Consider the following (true) sentence:
Between every two rational numbers there is a third one.

(2.1)

The property expressed in (2.1) is usually referred to as density of the rationals.
We will take a systematic look at proving such statements in Chapter 3.
Exercise 2.22 Can you think of an argument showing that statement (2.1) is true?

A Pattern. There is a logical pattern underlying sentence (2.1). To make it visible, look at the following, more explicit, formulation. It uses variables x, y and z
for arbitrary rationals, and refers to the ordering < of the set Q of rational numbers.
For all rational numbers x and z, if x < z, then somerational number y exists such that x < y and y < z.

(2.2)

You will often find ‘x < y and y < z’ shortened to: x < y < z.
Quantifiers Note the words all (or: for all), some (or: for some, some. . . exists,
there exists. . . such that, etc.). They are called quantifiers, and we use the symbols∀ and ∃ as shorthands for them.
With these shorthands, plus the shorthands for the connectives that we saw above,and the shorthand . . . ∈ Q for the property of being a rational, we arrive at thefollowing compact symbolic formulation:∀x ∈ Q ∀z ∈ Q ( x < z ⇒ ∃y ∈ Q ( x < y ∧ y < z ) ).

(2.3)

2.3. MAKING SYMBOLIC FORM EXPLICIT

51

We will use example (2.3) to make a few points about the proper use of the vocabulary of logical symbols. An expression like (2.3) is called a sentence or a formula.
Note that the example formula (2.3) is composite: we can think of it as constructed
out of simpler parts. We can picture its structure as in Figure (2.2).
∀x ∈ Q∀z ∈ Q(x < z ⇒ ∃y ∈ Q(x < y ∧ y < z))∀z ∈ Q(x < z ⇒ ∃y ∈ Q(x < y ∧ y < z))(x < z ⇒ ∃y ∈ Q(x < y ∧ y < z))x<z∃y ∈ Q(x < y ∧ y < z)
(x < y ∧ y < z)
x<y

y<z

Figure 2.2: Composition of Example Formula from its Sub-formulas.
As the figure shows, the example formula is formed by putting the quantifier prefix∀x ∈ Q in front of the result of putting quantifier prefix ∀z ∈ Q in front of asimpler formula, and so on.
The two consecutive universal quantifier prefixes can also be combined into ∀x, z ∈Q. This gives the phrasing∀x, z ∈ Q( x < z ⇒ ∃y ∈ Q ( x < y ∧ y < z ) ).
Putting an ∧ between the two quantifiers is incorrect, however. In other words, theexpression ∀x ∈ Q ∧ ∀z ∈ Q(x < z ⇒ ∃y ∈ Q(x < y ∧ y < z)) is considered
ungrammatical. The reason is that the formula part ∀x ∈ Q is itself not a formula,but a prefix that turns a formula into a more complex formula. The connective ∧can only be used to construct a new formula out of two simpler formulas, so ∧cannot serve to construct a formula from ∀x ∈ Q and another formula.
The symbolic version of the density statement uses parentheses. Their function is
to indicate the way the expression has been formed and thereby to show the scopeof operators. The scope of a quantifier-expression is the formula that it combineswith to form a more complex formula. The scopes of quantifier-expressions andconnectives in a formula are illustrated in the structure tree of that formula. Figure

52

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

2.2 shows that the scope of the quantifier-expression ∀x ∈ Q is the formula∀z ∈ Q(x < z ⇒ ∃y ∈ Q (x < y ∧ y < z)),the scope of ∀z ∈ Q is the formula(x < z ⇒ ∃y ∈ Q(x < y ∧ y < z)),and the scope of ∃y ∈ Q is the formula (x < y ∧ y < z).
Exercise 2.23 Give structure trees of the following formulas (we use shorthandnotation, and write A(x) as Ax for readability).
1. ∀x(Ax ⇒ (Bx ⇒ Cx)).
2. ∃x(Ax ∧ Bx).
3. ∃xAx ∧ ∃xBx.

The expression for all (and similar ones) and its shorthand, the symbol ∀, is called
the universal quantifier; the expression there exists (and similar ones) and its shorthand, the symbol ∃, is called the existential quantifier. The letters x, y and z thathave been used in combination with them are variables. Note that ‘for some’ isequivalent to ‘for at least one’.
Unrestricted and Restricted Quantifiers, Domain of Quantification Quantifiers can occur unrestricted: ∀x(x > 0), ∃y∀x(y > x), and restricted: ∀x ∈A(x > 0), ∃y ∈ B(y < a) (where A and B are sets).
In the unrestricted case, there should be some domain of quantification that often
is implicit in the context. E.g., if the context is real analysis, ∀x may mean for all
reals x. . . , and ∀f may mean for all real-valued functions f . . . .
Example 2.24 R is the set of real numbers. The fact that the R has no greatest
element can be expressed with restricted quantifiers as:∀x ∈ R∃y ∈ R(x < y).
If we specify that all quantifiers range over the reals (i.e., if we say that R is thedomain of quantification) then we can drop the explicit restrictions, and we get bywith ∀x∃y(x < y).

2.3. MAKING SYMBOLIC FORM EXPLICIT53

The use of restricted quantifiers allows for greater flexibility, for it permits one toindicate different domains for different quantifiers.
Example 2.25
∀x ∈ R∀y ∈ R(x < y ⇒ ∃z ∈ Q (x < z < y)).

Instead of ∃x(Ax ∧ . . .) one can write ∃x ∈ A(. . .). The advantage when all
quantifiers are thus restricted is that it becomes immediately clear that the domain
is subdivided into different sub domains or types. This can make the logicaltranslation much easier to comprehend.
Remark. We will use standard names for the following domains: N for the natural
numbers, Z for the integer numbers, Q for the rational numbers, and R for the real
numbers. More information about these domains can be found in Chapter 8.
Exercise 2.26 Write as formulas with restricted quantifiers:
1. ∃x∃y(x ∈ Q ∧ y ∈ Q ∧ x < y).
2. ∀x(x ∈ R ⇒ ∃y(y ∈ R ∧ x < y)).
3. ∀x(x ∈ Z ⇒ ∃m, n(m ∈ N ∧ n ∈ N ∧ x = m − n)).

Exercise 2.27 Write as formulas without restricted quantifiers:
1. ∀x ∈ Q∃m, n ∈ Z(n 6= 0 ∧ x = m/n).
2. ∀x ∈ F ∀y ∈ D(Oxy ⇒ Bxy).

Bound Variables. Quantifier expressions ∀x, ∃y,. . . (and their restricted companions) are said to bind every occurrence of x, y,. . . in their scope. If a variableoccurs bound in a certain expression then the meaning of that expression does notchange when all bound occurrences of that variable are replaced by another one.

54

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

Example 2.28 ∃y ∈ Q(x < y) has the same meaning as ∃z ∈ Q(x < z).
This indicates that y is bound in ∃y ∈ Q(x < y). But ∃y ∈ Q(x < y) and∃y ∈ Q(z < y) have different meanings, for the first asserts that there exists a rational number greater than some given number x, and the second that there exists
a rational number greater than some given z.
Universal and existential quantifiers are not the only variable binding operators
used by mathematicians. There are several other constructs that you are probably
familiar with which can bind variables.
P5
Example 2.29 (Summation, Integration.) The expression i=1 i is nothing but
a way to describe the number 15 (15 = 1 + 2 + 3 + 4 + 5), and clearly, 15 doesin no way depend on i. Use of a different variable does not change the meaning:P5
k=1 k = 15. Here are the Haskell versions:Prelude> sum [ i | i <- [1..5] ]15Prelude> sum [ k | k <- [1..5] ]15

R1
Similarly, the expression 0 xdx denotes the number 12 and does not depend on x.
Example 2.30 (Abstraction.) Another way to bind a variable occurs in the abstraction notation { x ∈ A | P }, cf. (4.1), p. 118. The Haskell counterpart to thisis list comprehension:[ x | x <- list, property x ]The choice of variable does not matter. The same list is specified by:[ y | y <- list, property y ]The way set comprehension is used to define sets is similar to the way list comprehension is used to define lists, and this is similar again to the way lambda abstraction is used to define functions. See 2.4.

2.3. MAKING SYMBOLIC FORM EXPLICIT

55

Bad Habits. It is not unusual to encounter our example-statement (2.1) displayed
as follows.
For all rationals x and y, if x < y, then both x < z and z < y holdfor some rational z.
Note that the meaning of this is not completely clear. With this expression the true
statement that ∀x, y ∈ Q ∃z ∈ Q (x < y ⇒ (x < z ∧ z < y)) could be meant, butwhat also could be meant is the false statement that ∃z ∈ Q ∀x, y ∈ Q (x < y ⇒(x < z ∧ z < y)).
Putting quantifiers both at the front and at the back of a formula results in ambiguity, for it becomes difficult to determine their scopes. In the worst case the result
is an ambiguity between statements that mean entirely different things.
It does not look too well to let a quantifier bind an expression that is not a variable,
such as in:for all numbers n2 + 1, . . .
Although this habit does not always lead to unclarity, it is better to avoid it, as
the result is often rather hard to comprehend. If you insist on quantifying overcomplex terms, then the following phrasing is suggested: for all numbers of theform n2 + 1, . . .
Of course, in the implementation language, terms like n + 1 are important for
pattern matching.
Translation Problems. It is easy to find examples of English sentences that are
hard to translate into the logical vernacular. E.g., in between two rationals is athird one it is difficult to discover a universal quantifier and an implication.
Also, sometimes the English has an indefinite article where the meaning is clearly
universal. Consider the sentence a well-behaved child is a quiet child. The indefinite articles here may suggest existential quantifiers; however, the reading that isclearly meant has the form∀x ∈ C (Well-behaved(x) ⇒ Quiet(x)).
A famous example from philosophy of language is: if a farmer owns a donkey, hebeats it. Again, in spite of the indefinite articles, the meaning is universal:∀x∀y((Farmer(x) ∧ Donkey(y) ∧ Own(x, y)) ⇒ Beat(x, y)).

56

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTSIn cases like this, translation into a formula reveals the logical meaning that remained hidden in the original phrasing.
In mathematical texts it also occurs quite often that the indefinite article a is used tomake universal statements. Compare Example (2.43) below, where the following
universal statement is made: A real function is continuous if it satisfies the ε-δdefinition.
Exercise 2.31 Translate into formulas, taking care to express the intended meaning:
1. The equation x2 + 1 = 0 has a solution.
2. A largest natural number does not exist.
3. The number 13 is prime (use d|n for ‘d divides n’).
4. The number n is prime.
5. There are infinitely many primes.

Exercise 2.32 Translate into formulas:
1. Everyone loved Diana. (Use the expression L(x, y) for: x loved y, and thename d for Diana.)
2. Diana loved everyone.
3. Man is mortal. (Use M (x) for ‘x is a man’, and M’(x) for ‘x is mortal’.)4. Some birds do not fly. (Use B(x) for ‘x is a bird’ and F (x) for ‘x can fly’.)

Exercise 2.33 Translate into formulas, using appropriate expressions for the predicates:
1. Dogs that bark do not bite.
2. All that glitters is not gold.
3. Friends of Diana’s friends are her friends.
4.*The limit of n1 as n approaches infinity is zero.

2.3. MAKING SYMBOLIC FORM EXPLICIT

57

Expressing Uniqueness. If we combine quantifiers with the relation = of identity, we can make definite statements like ‘there is precisely one real number x
with the property that for any real number y, xy = y’. The logical rendering is(assuming that the domain of discussion is R):∃x(∀y(x · y = y) ∧ ∀z(∀y(z · y = y) ⇒ z = x)).
The first part of this formula expresses that at least one x satisfies the property∀y(x · y = y), and the second part states that any z satisfying the same property isidentical to that x.
The logical structure becomes more transparent if we write P for the property.
This gives the following translation for ‘precisely one object has property P ’:∃x(P x ∧ ∀z(P z ⇒ z = x)).
Exercise 2.34 Use the identity symbol = to translate the following sentences:
1. Everyone loved Diana except Charles.
2. Every man adores at least two women.
3. No man is married to more than one woman.

Long ago the philosopher Bertrand Russell has proposed this logical format for thetranslation of the English definite article. According to his theory of description,
the translation of The King is raging becomes:∃x(King(x) ∧ ∀y(King(y) ⇒ y = x) ∧ Raging(x)).
Exercise 2.35 Use Russell’s recipe to translate the following sentences:
1. The King is not raging.
2. The King is loved by all his subjects. (use K(x) for ‘x is a King’, andS(x, y) for ‘x is a subject of y’).

Exercise 2.36 Translate the following logical statements back into English.
1. ∃x ∈ R(x2 = 5).

58

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS
2. ∀n ∈ N∃m ∈ N(n < m).
3. ∀n ∈ N¬∃d ∈ N(1 < d < (2n + 1) ∧ d|(2n + 1)).
4. ∀n ∈ N∃m ∈ N(n < m ∧ ∀p ∈ N(p 6 n ∨ m 6 p)).
5. ∀ε ∈ R+ ∃n ∈ N∀m ∈ N(m > n ⇒ (|a − am | 6 ε)). (R+ is the set of
positive reals; a, a0 , a1 , . . . refer to real numbers .)Remark. Note that translating back and forth between formulas and plain English
involves making decisions about a domain of quantification and about the predicates to use. This is often a matter of taste. For instance, how does one choose
between P (n) for ‘n is prime’ and the spelled outn > 1 ∧ ¬∃d ∈ N(1 < d < n ∧ d|n),which expands the definition of being prime? Expanding the definitions of mathematical concepts is not always a good idea. The purpose of introducing the word
prime was precisely to hide the details of the definition, so that they do not burden the mind. The art of finding the right mathematical phrasing is to introduce
precisely the amount and the kind of complexity that are needed to handle a given
problem.
Before we will start looking at the language of mathematics and its conventions ina more systematic way, we will make the link between mathematical definitions
and implementations of those definitions.

2.4 Lambda Abstraction
The following description defines a specific function that does not depend at all onx:
The function that sends x to x2 .
Often used notations are x 7→ x2 and λx.x2 . The expression λx.x2 is called alambda term.
If t is an expression of type b and x is a variable of type a then λx.t is an expressionof type a → b, i.e., λx.t denotes a function. This way of defining functions iscalled lambda abstraction.

2.4. LAMBDA ABSTRACTION

59

Note that the function that sends y to y 2 (notation y 7→ y 2 , or λy.y 2 ) describes thesame function as λx.x2 .
In Haskell, function definition by lambda abstraction is available. Compare thefollowing two definitions:square1 :: Integer -> Integer
square1 x = x^2
square2 :: Integer -> Integer
square2 = \ x -> x^2

In the first of these, the parameter x is given as an argument. In the second, thefunction is defined as a lambda abstract. The Haskell way of lambda abstraction
goes like this. The syntax is: \ v -> body, where v is a variable of the argument type and body an expression of the result type. It is allowed to abbreviate\ v -> \ w -> body to \ v w -> body. And so on, for more than two variables. E.g., both of the following are correct:

m1 :: Integer -> Integer -> Integer
m1 = \ x -> \ y -> x*y
m2 :: Integer -> Integer -> Integer
m2 = \ x y -> x*y

And again, the choice of variables does not matter.
Also, it is possible to abstract over tuples. Compare the following definition of afunction that solves quadratic equations by means of the well-known ‘abc’-formula
x=

−b ±

√
b2 − 4ac.
2a

60

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

solveQdr :: (Float,Float,Float) -> (Float,Float)
solveQdr = \ (a,b,c) -> if a == 0 then error "not quadratic"
else let d = b^2 - 4*a*c in
if d < 0 then error "no real solutions"else((- b + sqrt d) / 2*a,(- b - sqrt d) / 2*a)

To solve the equation x2 − x − 1 = 0, use solveQdr (1,-1,-1), and you will
get the (approximately correct) answer (1.61803,-0.618034).
Approximately√
correct, for 1.61803 is an approximation of the golden ratio, 1+2 5 , and −0.618034√
is an approximation of 1−2 5 .
One way to think about quantified expressions like ∀xP x and ∃yP y is as combinations of a quantifier expression ∀ or ∃ and a lambda term λx.P x or λy.P y.
The lambda abstract λx.P x denotes the property of being a P . The quantifier ∀is a function that maps properties to truth values according to the recipe: if theproperty holds of the whole domain then t, else f. The quantifier ∃ is a functionthat maps properties to truth values according to the recipe: if the property holds
of anything at all then t, else f. This perspective on quantification is the basis of
the Haskell implementation of quantifiers in Section 2.8.

2.5 Definitions and Implementations
Here is an example of a definition in mathematics. A natural number n is prime ifn > 1 and no number m with 1 < m < n divides n.
We can capture this definition of being prime in a formula, using m|n for ‘mdivides n’, as follows (we assume the natural numbers as our domain of discourse):n > 1 ∧ ¬∃m(1 < m < n ∧ m|n).

(2.4)

Another way of expressing this is the following:n > 1 ∧ ∀m((1 < m < n) ⇒ ¬m|n).

(2.5)

If you have trouble seeing that formulas (2.4) and (2.5) mean the same, don’t
worry. We will study such equivalences between formulas in the course of thischapter.

2.6. ABSTRACT FORMULAS AND CONCRETE STRUCTURES

61

If we take the domain of discourse to be the domain of the natural numbers N ={0, 1, 2, . . .}, then formula (2.5) expresses that n is a prime number.
We can make the fact that the formula is meant as a definition explicit by introducing a predicate name P and linking that to the formula:4
P (n) :≡ n > 1 ∧ ∀m((1 < m < n) ⇒ ¬m|n).

(2.6)

One way to think about this definition is as a procedure for testing whether anatural number is prime. Is 83 a prime? Yes, because none of 2, 3, 4, . . . , 9 divides
83. Note that there is no reason to check 10, . . ., for since 10 × 10 > 83 any factor
m of 83 with m > 10 will not be the smallest factor of 83, and a smaller factor
should have turned up before.
The example shows that we can make the prime procedure more efficient. We only
have to try and find the smallest factor of n, and any b with b2 > n cannot be thesmallest factor. For suppose that a number b with b2 > n divides n. Then there isa number a with a × b = n, and therefore a2 6 n, and a divides n. Our definition
can therefore run:
P (n) :≡ n > 1 ∧ ∀m((1 < m ∧ m2 6 n) ⇒ ¬m|n).

(2.7)

In Chapter 1 we have seen that this definition is equivalent to the following:
P (n) :≡ n > 1 ∧ LD(n) = n.

(2.8)

The Haskell implementation of the primality test was given in Chapter 1.

2.6 Abstract Formulas and Concrete Structures
The formulas of Section 2.1 are “handled” using truth values and tables. Quantificational formulas need a structure to become meaningful. Logical sentences
involving variables can be interpreted in quite different structures. A structure
is a domain of quantification, together with a meaning for the abstract symbols
that occur. A meaningful statement is the result of interpreting a logical formulain a certain structure. It may well occur that interpreting a given formula in one
structure yields a true statement, while interpreting the same formula in a different structure yields a false statement. This illustrates the fact that we can use one
logical formula for many different purposes.
Look again at the example-formula (2.3), now displayed without reference to Q
and using a neutral symbol R. This gives:∀x ∀y ( xRy =⇒ ∃z ( xRz ∧ zRy ) ).
4 :≡ means: ‘is by definition equivalent to’.

(2.9)

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS62

It is only possible to read this as a meaningful statement if
1. it is understood which is the underlying domain of quantification,
and2. what the symbol R stands for.
Earlier, the set of rationals Q was used as the domain, and the ordering < wasemployed instead of the —in itself meaningless— symbol R. In the context of Q
and <, the quantifiers ∀x and ∃z in (2.9) should be read as: for all rationals x . . . ,resp., for some rational z . . . , whereas R should be viewed as standing for <. In
that particular case, the formula expresses the true statement that, between every
two rationals, there is a third one.
However, one can also choose the set N = {0, 1, 2, . . .} of natural numbers as
domain and the corresponding ordering < as the meaning of R. In that case,the formula expresses the false statement that between every two natural numbers
there is a third one.
A specification of (i) a domain of quantification, to make an unrestricted use of thequantifiers meaningful, and (ii) a meaning for the unspecified symbols that may
occur (here: R), will be called a context or a structure for a given formula.
As you have seen here: given such a context, the formula can be “read” as ameaningful assertion about this context that can be either true or false.
Open Formulas, Free Variables, and Satisfaction. If one deletes the first quantifier expression ∀x from the example formula (2.9), then the following remains:∀y ( xRy =⇒ ∃z ( xRz ∧ zRy ) ).

(2.10)

Although this expression does have the form of a statement, it in fact is not such athing. Reason: statements are either true or false; and, even if a quantifier domain
and a meaning for R were specified, what results cannot be said to be true or false,
as long as we do not know what it is that the variable x (which no longer is boundby the quantifier ∀x) stands for.
However, the expression can be turned into a statement again by replacing thevariable x by (the name of) some object in the domain, or —what amounts to thesame— by agreeing that x denotes this object.
For instance, if the domain consists of the set N ∪ {q ∈ Q | 0 < q < 1} of natural
numbers together with all rationals between 0 and 1, and the meaning of R is the2.6. ABSTRACT FORMULAS AND CONCRETE STRUCTURES

63

usual ordering relation < for these objects, then the expression turns into a truth
upon replacing x by 0.5 or by assigning x this value. We say that 0.5 satisfies theformula in the given domain.
However, (2.10) turns into a falsity when we assign 2 to x; in other words, 2 does
not satisfy the formula.
Of course, one can delete a next quantifier as well, obtaining:xRy =⇒ ∃z ( xRz ∧ zRy ).
Now, both x and y have become free, and, next to a context, values have to be
assigned to both these variables in order to determine a truth value.
An occurrence of a variable in an expression that is not (any more) in the scope of aquantifier is said to be free in that expression. Formulas that contain free variables
are called open.
An open formula can be turned into a statement in two ways: (i) adding quantifiers
that bind the free variables; (ii) replacing the free variables by (names of) objects
in the domain (or stipulating that they have such objects as values).
Exercise 2.37 Consider the following formulas.
1. ∀x∀y(xRy),
2. ∀x∃y(xRy).
3. ∃x∀y(xRy).
4. ∃x∀y(x = y ∨ xRy).
5. ∀x∃y(xRy ∧ ¬∃z(xRz ∧ zRy)).
Are these formulas true or false in the following contexts?:
a. Domain: N = {0, 1, 2, . . .}; meaning of R: <,b. Domain: N; meaning of R: >,c. Domain: Q (the set of rationals); meaning of R: <,d. Domain: R (the set of reals); meaning of xRy: y 2 = x,
e. Domain: set of all human beings; meaning of R: father-of,f. Domain: set of all human beings; meaning of xRy: x loves y.

64

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

Exercise 2.38 In Exercise 2.37, delete the first quantifier on x in formulas 1–5.
Determine for which values of x the resulting open formulas are satisfied in eachof the structures a–f.

2.7 Logical Handling of the Quantifiers
Goal To learn how to recognize simple logical equivalents involving quantifiers,and how to manipulate negations in quantified contexts.
Validities and Equivalents. Compare the corresponding definitions in Section 2.2.
1. A logical formula is called (logically) valid if it turns out to be true in every
structure.
2. Formulas are (logically) equivalent if they obtain the same truth value inevery structure (i.e., if there is no structure in which one of them is true andthe other one is false).
Notation: Φ ≡ Ψ expresses that the quantificational formulas Φ and Ψ areequivalent.
Exercise 2.39 (The propositional version of this is in Exercise 2.19 p. 48.) Argue
that Φ and Ψ are equivalent iff Φ ⇔ Ψ is valid.
Because of the reference to every possible structure (of which there are infinitely
many), these are quite complicated definitions, and it is nowhere suggested thatyou will be expected to decide on validity or equivalence in every case that you
may encounter. In fact, in 1936 it was proved rigorously, by Alonzo Church (1903–
1995) and Alan Turing (1912–1954) that no one can! This illustrates that thecomplexity of quantifiers exceeds that of the logic of connectives, where truth
tables allow you to decide on such things in a mechanical way, as is witnessedby the Haskell functions that implement the equivalence checks for propositional
logic.
Nevertheless: the next theorem already shows that it is sometimes very well possible to recognize whether formulas are valid or equivalent — if only these formulas
are sufficiently simple.
Only a few useful equivalents are listed next. Here, Ψ(x), Φ(x, y) and the likedenote logical formulas that may contain variables x (or x, y) free.

2.7. LOGICAL HANDLING OF THE QUANTIFIERS

65

Theorem 2.40
1. ∀x∀yΦ(x, y) ≡ ∀y∀xΦ(x, y);∃x∃yΦ(x, y) ≡ ∃y∃xΦ(x, y),2. ¬∀xΦ(x) ≡ ∃x¬Φ(x);¬∃xΦ(x) ≡ ∀x¬Φ(x);¬∀x¬Φ(x) ≡ ∃xΦ(x);¬∃x¬Φ(x) ≡ ∀xΦ(x),3. ∀x(Φ(x) ∧ Ψ(x)) ≡ (∀xΦ(x) ∧ ∀xΨ(x));∃x(Φ(x) ∨ Ψ(x)) ≡ (∃xΦ(x) ∨ ∃xΨ(x)).
Proof. There is no neat truth table method for quantification, and there is no neat
proof here. You just have to follow common sense. For instance (part 2, first item)
common sense dictates that not every x satisfies Φ if, and only if, some x does notsatisfy Φ.
Of course, common sense may turn out not a good adviser when things get lesssimple. Chapter 3 hopefully will (partly) resolve this problem for you.
Exercise 2.41 For every sentence Φ in Exercise 2.36 (p. 57), consider its negation ¬Φ, and produce a more positive equivalent for ¬Φ by working the negation
symbol through the quantifiers.

Order of Quantifiers. Theorem 2.40.1 says that the order of similar quantifiers(all universal or all existential) is irrelevant. But note that this is not the case forquantifiers of different kind.
On the one hand, if you know that ∃y∀xΦ(x, y) (which states that there is one
y such that for all x, Φ(x, y) holds) is true in a certain structure, then a fortiori∀x∃yΦ(x, y) will be true as well (for each x, take this same y). However, if∀x∃yΦ(x, y) holds, it is far from sure that ∃y∀xΦ(x, y) holds as well.
Example 2.42 The statement that ∀x∃y(x < y) is true in N, but the statement∃y∀x(x < y) in this structure wrongly asserts that there exists a greatest natural
number.

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

66

Restricted Quantification. You have met the use of restricted quantifiers, where
the restriction on the quantified variable is membership in some domain. But there
are also other types of restriction.
Example 2.43 (Continuity) According to the “ε-δ-definition” of continuity, a realfunction f is continuous if (domain R):∀x ∀ε > 0 ∃δ > 0 ∀y ( |x − y| < δ =⇒ |f (x) − f (y)| < ε ).
This formula uses the restricted quantifiers ∀ε > 0 and ∃δ > 0 that enable a morecompact formulation here.
Example 2.44 Consider our example-statement (2.3). Here it is again:∀y∀x(x < y =⇒ ∃z(x < z ∧ z < y))This can also be given as∀y∀x < y∃z < y(x < z),
but this reformulation stretches the use of this type of restricted quantification
probably a bit too much.
Remark. If A is a subset of the domain of quantification, then∀x ∈ A Φ(x) means the same as ∀x(x ∈ A ⇒ Φ(x)),
whereas∃x ∈ A Φ(x) is tantamount with ∃x(x ∈ A ∧ Φ(x)).

Warning: The restricted universal quantifier is explained using ⇒, whereas theexistential quantifier is explained using ∧ !
Example 2.45 ‘Some Mersenne numbers are prime’ is correctly translated as ∃x(M x∧
P x). The translation ∃x(M x ⇒ P x) is wrong. It is much too weak, for it expresses(in the domain N) that there is a natural number x which is either not a
Mersenne number or it is a prime. Any prime will do as an example of this, and so
will any number which is not a Mersenne number.
In the same way,√‘all prime numbers have irrational square√ roots’ is translated as∀x ∈ R(P x ⇒ x ∈/ Q). The translation ∀x ∈ R(P x ∧ x ∈/ Q) is wrong. This
time we end up with something which is too strong, for this expresses that every
real number is a prime number with an irrational square root.

2.7. LOGICAL HANDLING OF THE QUANTIFIERS

67

Restricted Quantifiers Explained. There is a version of Theorem 2.40 that employs restricted quantification. This version states, for instance, that ¬∀x ∈ A Φ isequivalent to ∃x ∈ A ¬Φ, and so on. The equivalence follows immediately fromthe remark above. We now have, e.g., that ¬∀x ∈ AΦ(x) is equivalent to ¬∀x(x ∈A ⇒ Φ(x)), which in turn is equivalent to (Theorem 2.40) ∃x¬(x ∈ A ⇒ Φ(x)),
hence to (and here the implication turns into a conjunction — cf. Theorem 2.10)∃x(x ∈ A ∧ ¬Φ(x)), and, finally, to ∃x ∈ A¬Φ(x).
Exercise 2.46 Does it hold that ¬∃x ∈ A Φ(x) is equivalent to ∃x 6∈ A Φ(x)? If
your answer is ‘yes’, give a proof, if ‘no’, then you should show this by giving asimple refutation (an example of formulas and structures where the two formulas
have different truth values).
Exercise 2.47 Is ∃x 6∈ A ¬Φ(x) equivalent to ∃x ∈ A ¬Φ(x)? Give a proof if youranswer is ‘yes’, and a refutation otherwise.
Exercise 2.48 Produce the version of Theorem 2.40 (p. 65) that employs restricted
quantification. Argue that your version is correct.
Example 2.49 (Discontinuity Explained) The following formula describes what
it means for a real function f to be discontinuous in x:¬ ∀ε > 0 ∃δ > 0 ∀y ( |x − y| < δ =⇒ |f (x) − f (y)| < ε ).
Using Theorem 2.40, this can be transformed in three steps, moving the negation
over the quantifiers, into:∃ε > 0 ∀δ > 0 ∃y ¬ ( |x − y| < δ =⇒ |f (x) − f (y)| < ε ).
According to Theorem 2.10 this is equivalent to∃ε > 0 ∀δ > 0 ∃y ( |x − y| < δ ∧ ¬ |f (x) − f (y)| < ε ),i.e., to∃ε > 0 ∀δ > 0 ∃y ( |x − y| < δ ∧ |f (x) − f (y)| > ε ).
What has emerged now is a clearer “picture” of what it means to be discontinuous
in x: there must be an ε > 0 such that for every δ > 0 (“no matter how small”) a y
can be found with |x − y| < δ, whereas |f (x) − f (y)| > ε; i.e., there are numbers
y “arbitrarily close to x” such that the values f (x) and f (y) remain at least ε apart.

68

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

Different Sorts. Several sorts of objects, may occur in one and the same context.
(For instance, sometimes a problem involves vectors as well as reals.) In such asituation, one often uses different variable naming conventions to keep track of
the differences between the sorts. In fact, sorts are just like the basic types in afunctional programming language.
Just as good naming conventions can make a program easier to understand, naming conventions can be helpful in mathematical writing. For instance: the lettersn, m, k, . . . are often used for natural numbers, f, g, h, . . . usually indicate thatfunctions are meant, etc.
The interpretation of quantifiers in such a case requires that not one, but several
domains are specified: one for every sort or type. Again, this is similar to providing
explicit typing information in a functional program for easier human digestion.

Exercise 2.50 That the sequence a0 , a1 , a2 , . . . ∈ R converges to a, i.e., thatlimn→∞ an = a, means that ∀δ > 0∃n∀m > n(|a − am | < δ). Give a positive equivalent for the statement that the sequence a0 , a1 , a2 , . . . ∈ R does notconverge.

2.8 Quantifiers as Procedures
One way to look at the meaning of the universal quantifier ∀ is as a procedure to
test whether a set has a certain property. The test yields t if the set equals the wholedomain of discourse, and f otherwise. This means that ∀ is a procedure that maps
the domain of discourse to t and all other sets to f. Similarly for restricted universal
quantification. A restricted universal quantifier can be viewed as a procedure thattakes a set A and a property P , and yields t just in case the set of members of A
that satisfy P equals A itself.
In the same way, the meaning of the unrestricted existential quantifier ∃ can be
specified as a procedure. ∃ takes a set as argument, and yields t just in case theargument set is non-empty. A restricted existential quantifier can be viewed as aprocedure that takes a set A and a property P , and yields t just in case the set ofmembers of A that satisfy P is non-empty.
If we implement sets as lists, it is straightforward to implement these quantifierprocedures. In Haskell, they are predefined as all and any (these definitions will
be explained below):2.8. QUANTIFIERS AS PROCEDURES

any, all
any p
all p

69

:: (a -> Bool) -> [a] -> Bool
= or . map p= and . map p

The typing we can understand right away. The functions any and all take as theirfirst argument a function with type a inputs and type Bool outputs (i.e., a test fora property), as their second argument a list over type a, and return a truth value.
Note that the list representing the restriction is the second argument.
To understand the implementations of all and any, one has to know that or andand are the generalizations of (inclusive) disjunction and conjunction to lists. (Wehave already encountered and in Section 2.2.) They have type [Bool] -> Bool.
Saying that all elements of a list xs satisfy a property p boils down to: the list
map p xs contains only True (see 1.8). Similarly, saying that some element ofa list xs satisfies a property p boils down to: the list map p xs contains at least
one True. This explains the implementation of all: first apply map p, next apply
and. In the case of any: first apply map p, next apply or.
The action of applying a function g :: b -> c after a function f :: a -> b is
performed by the function g . f :: a -> c , the composition of f and g. See
Section 6.3 below.
The definitions of all and any are used as follows:
Prelude> any (<3) [0..]TruePrelude> all (<3) [0..]False
Prelude>The functions every and some get us even closer to standard logical notation.
These functions are like all and any, but they first take the restriction argument,
next the body:

every, some :: [a] -> (a -> Bool) -> Boolevery xs p = all p xs
some xs p = any p xs

70

CHAPTER 2. TALKING ABOUT MATHEMATICAL OBJECTS

Now, e.g., the formula ∀x ∈ {1, 4, 9}∃y ∈ {1, 2, 3} x = y 2 can be implemented
as a test, as follows:
TAMO> every [1,4,9] (\ x -> some [1,2,3] (\ y -> x == y^2))
True

But caution: the implementations of the quantifiers are procedures, not algorithms.
A call to all or any (or every or some) need not terminate. The callevery [0..] (>=0)

will run forever. This illustrates once more that the quantifiers are in essence more
complex than the propositional connectives. It also motivates the development of
the method of proof, in the next chapter.
Exercise 2.51 Define a function unique :: (a -> Bool) -> [a] -> Boolthat gives True for unique p xs just in case there is exactly one object amongxs that satisfies p.
Exercise 2.52 Define a function parity :: [Bool] -> Bool that gives True
for parity xs just in case an even number of the xss equals True.
Exercise 2.53 Define a function evenNR :: (a -> Bool) -> [a] -> Boolthat gives True for evenNR p xs just in case an even number of the xss haveproperty p. (Use the parity function from the previous exercise.)

2.9 Further Reading
Excellent books about logic with applications in computer science are [Bur98]
and [HR00]. A good introduction to mathematical logic is Ebbinghaus, Flum andThomas [EFT94].

Chapter 3

The Use of Logic: ProofPreview
This chapter describes how to write simple proofs. Section 3.1 is about style ofpresentation, while Section 3.2 gives the general format of the proof rules. Sections 3.3 and 3.4 describe the rules that govern the use of the connectives andthe quantifiers in the proof process. The recipes are summarized in Section 3.5.
Section 3.6 gives some strategic hints for handling proof problems. Section 3.7
applies the proof recipes to reasoning about prime numbers; this section also illustrates how the computer can be used (sometimes) as an instrument for checking
particular cases, and therefore as a tool for refuting general claims.
Representation and proof are two sides of the same coin. In order to handle thestuff of mathematics, we start out from definitions and try to find meaningful relations. To check our intuitions and sometimes just to make sure that our definitions
accomplish what we had in mind we have to provide proofs for our conjectures.
To handle abstract objects in an implementation we have to represent them in a
concrete way. Again, we have to check that the representation is faithful to theoriginal intention.
It turns out that proofs and implementations have many things in common. In thefirst place, variables are a key ingredient in both. Variables are used to denotemembers of certain sets. In our example proofs, we will used x, y for rational
numbers and real numbers, and m, n for integers. Similarly, the variables used in
the implementation of a definition range over certain sets, indicated by means oftypings for the variables.
71

CHAPTER 3. THE USE OF LOGIC: PROOF

72

The main purpose of this chapter is to spur you on to develop good habits in setting
up mathematical proofs. Once acquired, these habits will help you in many ways.
Being able to ‘see structure’ in a mathematical proof will enable you to easily readand digest the proofs you find in papers and textbooks. When you have learned tosee the patterns of proof, you will discover that they turn up again and again in any
kind of formal reasoning.
Because it takes time to acquire new habits, it is not possible to digest the contentsof this chapter in one or two readings. Once the recommended habits are acquired,but only then, the contents of the chapter have been properly digested, and thechapter will have served its purpose.
The module containing the code of this chapter depends on the module containing
the code of the previous chapter. In the module declaration we take care to importthat module. This is done with the reserved keyword import.

module TUOLP
where
import TAMO

3.1 Proof Style
The objects of mathematics are strange creatures. They do not exist in physicalspace. No one ever saw the number 1. One can consistently argue that mathematics
has no subject at all, or maybe that the subject matter of mathematics is in themind. But what does that mean? For sure, different mathematicians have the same
objects in mind when they investigate prime numbers. But how can that be, if
mathematics has no subject matter?
As to the method of mathematics, there is no disagreement: the heart of the matteris the notion of proof. A proof is an argument aimed at convincing yourself andothers of the truth of an assertion. Some proofs are simple, but others can be piecesof art with aesthetic and intellectual qualities.
In daily life, people argue a lot, but these efforts are not always capable of convincing others. If you have a row with your partner, more often than not it is extremely
difficult to assess who is right. (If this were an easy matter, why have a row in3.1. PROOF STYLE73

the first place?) By contrast, in mathematical matters it usually is very clear who
is right and who is wrong. Doing mathematics is a sports where the rules of thegame are very clear. Indeed, mathematical proofs go undisputed most of the time.
This remarkable phenomenon is probably due to the idealized character of mathematical objects. Idealization is a means by which mathematics provides access tothe essence of things, and grasping the essence of something is a hallmark of thecreative mind. It remains a source of wonder that the results are so often applicableto the real world.
The mathematical content of most of the proofs discussed in this chapter is minute;later on you will call such proofs “routine”. For the time being, the attention is
focused on the form of proofs. However, to begin with, here follows a list of
stylistic commandments.
A proof is made up of sentences. Therefore:
1 Write correct English, try to express yourself clearly.
If you are not a native speaker you are in the same league as the authors of thisbook. Still, English is the lingua franca of the exact sciences, so there is no way
around learning enough of it for clear communication. Especially when not anative speaker, keep your style simple.
Write with a reader in mind, and do not leave your reader in the dark concerning
your proof plans.
2 Make sure the reader knows exactly what you are up to.
Applying the rules of proof of the next section in the correct way will usually takecare of this. In particular, when you feel the need to start writing symbols —often,
variables— inform the reader what they stand for and do not let them fall out of
the blue.
3 Say what you mean when introducing a variable.
Section 3.2 again shows you the way.
Use the symbolic shorthands for connectives and quantifiers sparingly. Something
like the continuity definition on p. 66 with its impressive sequence of four quantifiers is about the maximum a well-educated mathematician can digest.
4 Don’t start a sentence with symbols; don’t write formulas only.

CHAPTER 3. THE USE OF LOGIC: PROOF

74

It is best to view symbolic expressions as objects in their own right. E.g., write:
The formula Φ is true, or: It holds that Φ. Note that a proof that consists offormulas only is not a suitable way to inform the reader about what has been goingon in your brain.
5 Use words or phrases like ‘thus’, ‘therefore’, ‘hence’, ‘it follows that’,etc.
to link up your formulas. Be relevant and succinct.
Do not use the implication symbol ⇒ in cases that really require “thus” or “therefore”. Beginner’s proofs can often be recognized by their excessive length. Of
course, it will be difficult at first to keep to the proper middle road between commandments (4) and (5).
When embarking on a proof, it is a good idea to write down exactly (i) what canbe used in the proof (the Given), (ii) and what is to be proved. It is very helpful touse a schema for this.
6 When constructing proofs, use the following schema:
Given: . . .
To be proved: . . .
Proof: . . .
N.B.: A proof using Mathematical Induction offers some extra structure; cf. Section 7.1 below and p. 400.
In the course of this chapter you will discover that proofs are highly structured
pieces of texts. The following guideline will help you keep track of that structure.
7 Use layout (in particular, indentation) to identify subproofs and to keep trackof the scopes of assumptions.
The general shape of a proof containing subproofs will be discussed in Section 3.2.
8 Look up definitions of defined notions, and use these definitions to re-writeboth Given and To be proved.
In particular, elimination of defined notions from the sentence to be proved willshow you clearly what it is that you have to prove.
Do not expect that you will be able to write down a correct proof at the first try. Do
not expect to succeed at the second one. You will have to accept the fact that, for3.2. PROOF RECIPES

75

the time being, your efforts will not result in faultless proofs. Nevertheless: keep
trying!

9 Make sure you have a sufficient supply of scratch paper, make a fair copy ofthe end-product — whether you think it to be flawless or not.

Finally: before handing in what you wrote, try to let it rest for at least one night.
And only then,

10 Ask yourself two things: Is this correct? Can others read it?

The honest answers usually will be negative at first. . . Fortunately, you did notfinish your supply of scratch paper. And surely, if you have understood properly,
you must be able to get it down correctly eventually. (Apply to this the law ofcontraposition in Theorem 2.10, p. 45.)

3.2 Proof Recipes
Goal Develop the ability to apply the proof rules in simple contexts. The proof
rules are recipes that will allow you to cook up your own proofs. Try to distilla recipe from every rule, and and make sure you remember how to apply theserecipes in the appropriate situation.
It is completely normal that you get stuck in your first efforts to prove things. Often, you will not even know how to make a first move. This section then may cometo your help. It provides you with rules that govern the behaviour of the logicalphrases in mathematical proof, and thus with recipes to use while constructing a
proof.
In fact, constructing proofs has a lot in common with writing computer programs.
In structured programming, layout can be used to reveal the building blocks of aprogram more clearly. We will also use layout as a tool to reveal structure.
The most important structure principle is that a proof can contain subproofs, justlike a program may contain procedures which have their own sub-procedures, andso on. We will indicate subproofs by means of indentation. The general structureof a proof containing a subproof is as follows:

76

CHAPTER 3. THE USE OF LOGIC: PROOF

Given: A, B, . . .
To be proved: P
Proof:...
Suppose C . . .
To be proved: Q
Proof: . . .
...
Thus Q...
Thus P

To be sure, a subproof may itself contain subproofs:

Given: A, B
To be proved: P
Proof:...
Suppose C
To be proved: Q
Proof:...
Suppose D
To be proved: R
Proof: . . .
...
Thus R...
Thus Q...
Thus P3.2. PROOF RECIPES

77

The purpose of ‘Suppose’ is to add a new given to the list of assumptions that
may be used, but only for the duration of the subproof of which ‘Suppose’ is thehead. If the current list of givens is P1 , . . . , Pn then ‘Suppose Q’ extends this list toP1 , . . . , Pn , Q. In general, inside a box, you can use all the givens and assumptionsof all the including boxes. Thus, in the innermost box of the example, the givens
are A, B, C, D. This illustrates the importance of indentation for keeping track ofthe ‘current box’.
There are some 15 rules discussed here for all seven notions of logic. At a firstencounter, this may seem an overwhelming number. However, only some of these
are really enlightening; several are so trivial that you won’t even notice using them.
An example is:Given: P , Q
Thus P ∧ Q.
What is really remarkable is this: together these 15 rules are sufficient to tackle
every possible proof problem.
Of course, this does not mean that the process of proving mathematical facts boils
down in the end to mastery of a few simple rules. Think of it as chess: the rules of
the game are extremely simple, and these rules can be used to play very complicated and beautiful games. To be able to play you must know the rules. But if yourknowledge does not extend beyond the rules, you will only be capable of playing
games that are rather dull. Learning to play chess does not stop with learning therules, and what is beautiful in a game of chess cannot be found in the rules. Similarly, the rules of proof cannot teach you more than how to prove the simplest ofthings. (And, in the beginning, that is difficult enough.) It is ideas that make a
proof interesting or beautiful, but the rules keep silent about these things.
Classification of Rules.
Every logical symbol has its own rules of use. There are basically two ways inwhich you can encounter a logical symbol: it can either appear in the given or in thestatement that is to be proved. In the first case the rule to use is an elimination rule,in the second case an introduction rule. Elimination rules enable you to reduce a
proof problem to a new, hopefully simpler, one. Introduction rules make clear how
to prove a goal of a certain given shape. All rules are summarized in Section 3.5.
Safety. As we go along, we will provide arguments that try to explain why acertain rule is safe. A rule is safe if it will never allow you to prove something
false on the basis of statements that are true. Obviously, this is a requirement that

78

CHAPTER 3. THE USE OF LOGIC: PROOF

proofs should fulfill. That the rules for connectives are safe is due to truth tables.
Safety of two of the quantifier rules is obvious; the remaining ones are tough nuts.
Don’t worry when these explanations in some cases appear mystifying. Eventually,
you’ll understand!

3.3 Rules for the Connectives
Implication
Here come a complicated but important introduction rule and a trivial one for elimination.
Introduction The introduction rule for implication is the Deduction Rule. It enables you to reduce the problem of proving an implication Φ ⇒ Ψ. Instead, it
prescribes to assume Φ as an additional new Given, and asks you to derive that Ψ.
Given: . . .
To be proved: Φ ⇒ Ψ
Proof:
Suppose Φ
To be proved: ΨProof: . . .
Thus Φ ⇒ Ψ.
Safety. In case Φ is false, the implication Φ ⇒ Ψ will be true anyway. Thus, thecase that is left for you to consider is when Φ is true, meaning that you can assume
it as given. But then of course you should show that Ψ is true as well (otherwise,Φ ⇒ Ψ would be false).
Example 3.1 We show that the implication P ⇒ R is provable on the basis of thegiven P ⇒ Q and Q ⇒ R. Thus, employing the schema in the 7th commandment(p. 74) for the first time:
Given: P ⇒ Q, Q ⇒ R
To be proved: P ⇒ R
Proof:
Given: P ⇒ Q, Q ⇒ R (old), P (new)To be proved: R

3.3. RULES FOR THE CONNECTIVES

79

Proof: From P ⇒ Q and P , conclude Q.
Next, from Q ⇒ R and Q, conclude R.

Thus, according to the deduction rule, P ⇒ R
Here is a slightly more concise version:

Given: P ⇒ Q, Q ⇒ R
To be proved: P ⇒ R
Proof:
Suppose P
To be proved: R
Proof: From P ⇒ Q and P , conclude Q.
Next, from Q ⇒ R and Q, conclude R.
Thus P ⇒ R

Detailed vs. Concise Proofs. The proof just given explains painstakingly how the
Deduction Rule has been applied in order to get the required result. However, inpractice, you should write this in a still more concise way:
Given: P ⇒ Q, Q ⇒ R
To be proved: P ⇒ R
Proof:
Suppose P
From P ⇒ Q and P , conclude Q.
Next, from Q ⇒ R and Q, conclude R.
Note that the final concise version does not mention the Deduction Rule at all.
The actual application of this rule should follow as a last line, but that is left for
the reader to fill in. This is not considered an incomplete proof: from the situation
it is perfectly clear that this application is understood.
Several other examples of detailed proofs and their concise versions are given inwhat follows. Here is the key to the proof of an implication:

If the ‘to be proved’ is an implication Φ ⇒ Ψ, then your proof should start with thefollowing Obligatory Sentence:

Suppose that Φ holds.

CHAPTER 3. THE USE OF LOGIC: PROOF

80

The obligatory first sentence accomplishes the following things (cf. the 2nd commandment on p. 73).
• It informs the reader that you are going to apply the Deduction Rule in orderto establish that Φ ⇒ Ψ.
• The reader also understands that it is now Ψ that you are going to derive(instead of Φ ⇒ Ψ).
• Thus, starting with the obligatory sentence informs the reader in an efficient
way about your plans.
Reductive Character. In Example (3.1) the problem of showing that from thegivens P ⇒ Q, Q ⇒ R it can be proved that P ⇒ R was reduced, using the
Deduction Rule, to the problem of showing that from the givens P ⇒ Q, Q ⇒R, P it can be proved that R. This requires a subproof.
Marking Subproofs with Indentation. Note that only during that subproof you
are allowed to use the new given P . Some logic texts recommend the use of
markers to indicate beginning and end of such a subproof, between which theextra given is available. In our example proofs in this chapter, we indicate thebeginnings and ends of subproofs by means of indentation. In more colloquial
versions of the proofs, identifying the beginning and end of a subproof is left tothe reader.
In this particular case, the reduced proof problem turned out to be easy: P andP ⇒ Q together produce Q; whereas Q and Q ⇒ R together produce R, as
desired. (Of course, in general, such a subproof may require new reductions.)
The last two steps in this argument in fact apply the Elimination Rule of implication, which is almost too self-evident to write down:
Elimination. This rule is also called Modus Ponens. In words: from Φ ⇒ Ψ andΦ you can conclude that Ψ.
In a schema:
Given: Φ ⇒ Ψ, ΦThus Ψ.
Safety. Immediate from truth tables: if Φ ⇒ Ψ and Φ are both true, then so is Ψ.

3.3. RULES FOR THE CONNECTIVES

81

The two proof rules for implication enable you to handle implications in all typesof proof problems. You may want to use more obvious properties of implication,
but fact is that they usually can be derived from the two given ones. Example (3.1)
is a case in point, as is Exercise 3.2. (We will not bother you with the exceptions;
one is in Exercise 3.9.)
Exercise 3.2 Apply both implication rules to prove P ⇒ R from the givens P ⇒Q, P ⇒ (Q ⇒ R).

Conjunction
The conjunction rules are almost too obvious to write down.
Introduction. A conjunction follows from its two conjuncts taken together.
In a schema:
Given: Φ, ΨThus Φ ∧ Ψ.
Elimination. From a conjunction, both conjuncts can be concluded.
Schematically:
Given: Φ ∧ ΨThus Φ.
Given: Φ ∧ ΨThus Ψ.
Suppose we want to check whether the sum of even natural numbers always iseven. It is not difficult to implement this check in Haskell, using the built-in function even and the list of even numbers.

evens = [ x | x <- [0..], even x ]Formulating the check is easy, but of course we won’t get an answer, as the checktakes an infinite amount of time to compute.

82

CHAPTER 3. THE USE OF LOGIC: PROOF

TUOLP> forall [ m + n | m <- evens, n <- evens ] even

If we want to check a statement about an infinite number of cases, we can either
look for a counterexample, or we can attempt to give a proof. In the present case,the latter is easy.
Example 3.3 Assume that n, m ∈ N.
To show: (m is even ∧ n is even) ⇒ m + n is even.
Detailed proof:
Assume that (m even ∧ n even).
Then (∧-elimination) m and n are both even.
For instance, p, q ∈ N exist such that m = 2p, n = 2q.
Then m + n = 2p + 2q = 2(p + q) is even.
The result follows using the Deduction Rule.
Concise proof:
Assume that m and n are even.
For instance, m = 2p, n = 2q, p, q ∈ N.
Then m + n = 2p + 2q = 2(p + q) is even.

Exercise 3.4 Assume that n, m ∈ N.
Show: (m is odd ∧ n is odd) ⇒ m + n is even.
Equivalence
An equivalence can be thought of as the conjunction of two implications. Thus,
the rules follow from those for ⇒ and ∧.
Introduction. In order to prove Φ ⇔ Ψ, you have to accomplish two things (cf.
example 2.3):(⇒) add Φ as a new given, and show that Ψ;(⇐) add Ψ as a new given, and show that Φ.
If you can do this, Φ ⇔ Ψ has been proved.

3.3. RULES FOR THE CONNECTIVES

83

Concise Proof Schema.
Given: . . .
To be proved: Φ ⇔ Ψ
Proof:
Suppose Φ
To be proved: ΨProof: . . .
Suppose Ψ
To be proved: ΦProof: . . .
Thus Φ ⇔ Ψ.
Instead of ⇔ you may also encounter ‘iff’ or ‘if and only if’. A proof of a statementof the form ‘Φ iff Ψ’ consists of two subproofs: the proof of the ‘only if’ part andthe proof of the ‘if’ part. Caution: the ‘only if’ part is the proof of Φ ⇒ Ψ, andthe ‘if’ part is the proof of Φ ⇐ Ψ. This is because ‘Φ only if Ψ’ means ‘Φ ⇒ Ψ’,and ‘Φ if Ψ’ means ‘Φ ⇐ Ψ’. Thus, we get:
Given: . . .
To be proved: Φ iff ΨProof:
Only if: Suppose Φ
To be proved: ΨProof: . . .
If:

Suppose Ψ
To be proved: ΦProof: . . .

Thus Φ iff Ψ.
Elimination. Schematically:
Given: Φ ⇔ Ψ, Φ, . . .
Thus ΨGiven: Φ ⇔ Ψ, Ψ, . . .
Thus Φ
Exercise 3.5 Show:

CHAPTER 3. THE USE OF LOGIC: PROOF

84

1. From P ⇔ Q it follows that (P ⇒ R) ⇔ (Q ⇒ R),
2. From P ⇔ Q it follows that (R ⇒ P ) ⇔ (R ⇒ Q).

Negation
General Advice. In no matter what concrete mathematical situation, before applying any of the negation rules given below: whether you want to prove or use
a negated sentence, you should first attempt to convert into something positive. If
this succeeds, you can turn to the other rules.
Theorems 2.10 (p. 45) and 2.40 (p. 65) contain some tools for doing this: you canmove the negation symbol inward, across quantifiers and connectives. If this process terminates, the “mathematics” of the situation often allows you to eliminate
the negation symbol altogether. E.g., ¬∀x(x < a ∨ b 6 x) can be rewritten as∃x(a 6 x ∧ x < b). For a more complicated example, cf. (2.49) on p. 67.
Remark. The general advice given above does not apply to the exercises in thepresent chapter. Firstly, many of the exercises below are purely logical: there isno mathematical context at all. Secondly, all exercises here are designed to besolved using the rules only; possible shortcuts via the results of Chapter 2 willoften trivialize them.
Introduction. If ¬Φ is to be proved, do the following. Assume Φ as a new given,and attempt to prove something (depending on the context) that is evidently false.
This strategy clearly belongs to the kind of rule that reduces the proof problem.
Schematically:
Given: . . .
To be proved: ¬ΦProof:
Suppose Φ
To be proved: ⊥Proof: . . .
Thus ¬Φ.
Evidently False. Here, ⊥ stands for the evidently false statement.

3.3. RULES FOR THE CONNECTIVES

85

In a mathematical context, this can be anything untrue, such as 1 = 0. For a more
complicated falsehood, cf. the proof that limits are unique on p. 317.
In the logical examples below, ⊥ may consist of the occurrence of a statement
together with its negation. In that case, one statement contradicts the other. For
instance, you might derive a sentence ¬Ψ, thereby contradicting the occurrence ofΨ among the given Γ. Cf. Examples (3.8) and and (3.30).
Example 3.6 The proof of theorem 3.33 (the number of primes is not finite) is an
example of the method of negation introduction. Another example can be found in
the proof of theorem 8.14 (the square root of 2 is not rational).
Exercise 3.7 Produce proofs for:1. Given: P ⇒ Q. To show: ¬Q ⇒ ¬P ,2. Given P ⇔ Q. To show: ¬P ⇔ ¬Q.

Safety. Suppose that from Γ, Φ it follows that ⊥, and that all given Γ are satisfied.
Then Φ cannot be true. (Otherwise, your proof would show the evidently false ⊥to be true as well.) Thus, ¬Φ must be true.
Elimination. When you intend to use the given ¬Φ, you can attempt to prove,on the basis of the other given, that Φ must hold. In that case, the elimination
rule declares the proof problem to be solved, no matter what the statement To beproved!
Schematically:
Given: Φ, ¬Φ
Thus Ψ.
Safety. The rule cannot help to be safe, since you will never find yourself in asituation where Φ and ¬Φ are both true. (Remarkably, this nevertheless is a usefulrule!)
There is one extra negation rule that can be used in every situation: Proof byContradiction, or Reductio ad Absurdum.

86

CHAPTER 3. THE USE OF LOGIC: PROOF

Proof by Contradiction. In order to prove Φ, add ¬Φ as a new given, and attempt
to deduce an evidently false statement.
In a schema:
Given: . . .
To be proved: ΦProof:
Suppose ¬ΦTo be proved: ⊥Proof: . . .
Thus Φ.
Safety. The argument is similar to that of the introduction rule.
Advice. Beginners are often lured into using this rule. The given ¬Φ that comesin free looks so inviting! However, many times it must be considered poisoned,
making for a cluttered bunch of confused givens that you will not be able to disentangle. It is a killer rule that often will turn itself against its user, especially whenthat is a beginner. Proof by Contradiction should be considered your last way out.
Some proof problems do need it, but if possible you should proceed without: you
won’t get hurt and a simpler and more informative proof will result.
Comparison. Proof by Contradiction looks very similar to the ¬ introduction rule(both in form and in spirit), and the two are often confused. Indeed, in ordinary
mathematical contexts, it is usually “better” to move negation inside instead of
applying ¬-introduction.
Example 3.8 From ¬Q ⇒ ¬P it follows that P ⇒ Q.
Given: ¬Q ⇒ ¬P
To be proved: P ⇒ Q
Detailed proof:
Suppose P
To be proved: Q
Proof:
Suppose ¬Q
To be proved: ⊥Proof:
From ¬Q and ¬Q ⇒ ¬P derive ¬P .
From P and ¬P derive ⊥.

3.3. RULES FOR THE CONNECTIVES

87

Thus, Q, by contradiction.
Thus, P ⇒ Q, by the Deduction Rule.
Concise proof:
Assume that P .
If ¬Q, then (by ¬Q ⇒ ¬P ) it follows that ¬P .
Contradiction.

Exercise 3.9* Show that from (P ⇒ Q) ⇒ P it follows that P .
Hint. Apply Proof by Contradiction. (The implication rules do not suffice for thisadmittedly exotic example.)
Disjunction
Introduction. A disjunction follows from each of its disjuncts.
Schematically:
Given: ΦThus Φ ∨ Ψ.
Given: ΨThus Φ ∨ Ψ.
Safety is immediate from the truth tables.
Elimination. You can use a given Φ ∨ Ψ by giving two proofs: one employing
Φ, and one employing Ψ.
In a proof schema:
Given: Φ ∨ Ψ, . . .
To be proved: ΛProof:
Suppose Φ
To be proved: ΛProof: . . .
Suppose Ψ
To be proved: ΛCHAPTER 3. THE USE OF LOGIC: PROOF

88
Proof: . . .
Thus Λ.

Example 3.10 We show that from P ∨ Q, ¬P it follows that Q.
Given: P ∨ Q, ¬P .
To be proved: Q.
Proof:
Suppose P . Then from P and ¬P we get Q.
Suppose Q. Then Q holds by assumption.

Therefore Q.

Exercise 3.11 Assume that A, B, C and D are statements.
1. From the given A ⇒ B ∨ C and B ⇒ ¬A, derive that A ⇒ C. (Hint: use
the previous example.)
2. From the given A ∨ B ⇒ C ∨ D, C ⇒ A, and B ⇒ ¬A, derive thatB ⇒ D.

Example 3.12 Here is a proof of the second DeMorgan law (Theorem 2.10):
Given: ¬(P ∨ Q).
To be proved: ¬P ∧ ¬Q.
Detailed proof:
Assume P . By ∨ -introduction, it follows that P ∨ Q.
This contradicts the given (i.e., we have an evident falsity here).
By ¬-introduction, we get ¬P .
In a similar way, ¬Q can be derived.
By ∧-introduction, it follows that ¬P ∧ ¬Q.
The concise version (no rules mentioned):
Assume P . Then a fortiori P ∨ Q holds, contradicting the given.
Thus, ¬P . Similarly, ¬Q is derivable.
Thus, ¬P ∧ ¬Q follows.

3.3. RULES FOR THE CONNECTIVES

89

Example 3.13 The following example from the list of equivalences in Theorem(2.10) is so strange that we give it just to prevent you from getting entangled.
Given: ¬(P ⇒ Q)
To be proved: P ∧ ¬Q
Proof: By ∧-introduction, it suffices to prove both P and ¬Q.
To be proved: P
Proof: (by contradiction)
Suppose that ¬P .
Then if P holds, Q follows by ¬-elimination.
Thus (Deduction Rule), we get that P ⇒ Q.
However, this contradicts the given.

To be proved: ¬Q
Proof: (by ¬ introduction)
Assume that Q.
Then, by a trivial application of the Deduction Rule P ⇒ Q follows.
(Trivial, since we do not need P at all to conclude Q.)
Again, this contradicts the given.

Note that the rule for implication introduction can be used for reasoning by cases,
as follows. Because P ∨ ¬P is a logical truth, it can always be added to the list of
givens. If the two sub cases P and ¬P both yield conclusion Q, then this proves
Q. Here is the schema:
Given: . . .
To be proved: Q
Proof:
Suppose P .
To be proved: Q.
Proof: . . .
Suppose ¬P .
To be proved: Q.
Proof: . . .
Thus Q.
This pattern of reasoning is used in the following examples.
Example 3.14

CHAPTER 3. THE USE OF LOGIC: PROOF

90

Let n ∈ N. To be proved: n2 − n is even.
Proof:
Assume n even.
Then n = 2m, so n2 − n = (n − 1)n = (2m − 1)2m,
and therefore n2 − n is even.
Assume n odd.
Then n = 2m + 1, so n2 − n = (n − 1)n = 2m(2m + 1),
and therefore n2 − n is even.
Thus n2 − n is even.

Exercise 3.15 Show that for any n ∈ N, division of n2 by 4 gives remainder 0 or
1.
Example 3.16 Let R be the universe of discourse, and let P (x) be the following
property:√x∈/ Q ∧ x 2 ∈ Q.
√

In other words, x has property
P iff x is irrational and x 2 is rational. We will
√ √2√show that either 2 or 2 has property P .
√ √Suppose 2 2 ∈ Q.
√√
/ Q (Theorem
8.14), we know that 2 has P .
Then, since 2 ∈√√ √2Thus, P ( 2) ∨ P ( 2 ).
√ √Suppose 2 2 ∈
/ Q.√
√ √ √
√
√ √
√ √2
Then, since ( 2 ) 2√= 2 2· 2 = 22 = 2 ∈ Q, we know that 2 2 has P .
√√ 2Thus, P ( 2) ∨ P ( 2 ).
√√ √Therefore P ( 2) ∨ P ( 2 2 ).

Exercise 3.17 Prove the remaining items of Theorem 2.10 (p. 45). To prove Φ ≡Ψ means (i) to derive Ψ from the given Φ and (ii) to derive Φ from the given Ψ.

3.4 Rules for the Quantifiers
The rules for the quantifiers come in two types: for the unrestricted, and for the
restricted versions. Those for the restricted quantifiers can be derived from theothers: see Exercise 3.32 p. 102.

3.4. RULES FOR THE QUANTIFIERS

91

Universal Quantifier
Introduction.

When asked to prove that ∀x E(x), you should start a proof by writing the Obligatory
Sentence:

Suppose that c is an arbitrary object.

And you proceed to show that this object (about which you are not supposed toassume extra information; in particular, it should not occur earlier in the argument)
has the property E in question.
Schematic form:
Given: . . .
To be proved: ∀xE(x)Proof:
Suppose c is an arbitrary object
To be proved: E(c)Proof: . . .
Thus ∀xE(x)Here is the Modification suitable for the restricted universal quantifier:If ∀x ∈ A E(x) is to be proved, you should start proof this time by the, again, obligatory:

Suppose that c is any object in A.

And you proceed to show that this object (about which you only assume that it
belongs to A) has the property E in question.
Schematic form:

92

CHAPTER 3. THE USE OF LOGIC: PROOF

Given: . . .
To be proved: ∀x ∈ A E(x)Proof:
Suppose c is any object in A
To be proved: E(c)Proof: . . .
Thus ∀x ∈ A E(x)Arbitrary Objects. You may wonder what an arbitrary object is. For instance:
what is an arbitrary natural number? Is it large? small? prime? etc. What exactly
is an arbitrary object in A when this set happens to be a singleton? And: are there
objects that are not arbitrary?
Answer: the term ‘arbitrary object’ is only used here as an aid to the imagination;
it indicates something unspecified about which no special assumptions are made.
Imagine that you allow someone else to pick an object, and that you don’t carewhat choice is made. ‘Suppose c is an arbitrary A’ is the same as saying to thereader: ‘Suppose you provide me with a member c from the set A; the choice iscompletely up to you.’
Often, a universal quantifier occurs in front of an implication. Therefore, you may
find the following rule schema useful.
Given: . . .
To be proved: ∀x(P (x) ⇒ Q(x)Proof:
Suppose c is any object such that P (c)To be proved: Q(c)Proof: . . .
Thus ∀x(P (x) ⇒ Q(x))
This rule is derivable from the introduction rules for ∀ and ⇒. Note that it is very
similar to the recipe for restricted ∀-introduction. Cf. Exercise 3.18.
Elimination.
Schematic form:
Given: ∀x E(x)Thus E(t).
Here, t is any object of your choice.

3.4. RULES FOR THE QUANTIFIERS

93

That this rule is safe is obvious: if every thing satisfies E, then in particular somust t.
Modification suitable for the restricted universal quantifier, in schematic form:
Given: ∀x ∈ A E(x), t ∈ A
Thus E(t).
Exercise 3.18 Show, using ∀-introduction and Deduction Rule: if from Γ, P (c) it
follows that Q(c) (where c satisfies P , but is otherwise “arbitrary”), then from Γ itfollows that ∀x(P (x) ⇒ Q(x)).
Existential Quantifier
Introduction. In order to show that ∃x E(x), it suffices to specify one object t for
which E(t) holds.
Schematic form:
Given: E(t)Thus, ∃x E(x).
Here, t can be anything: any example satisfying E can be used to show that∃xE(x).
Modification suitable for the restricted existential quantifier, in schematic form:
Given: E(t), t ∈ A
Thus, ∃x ∈ A E(x).
That these rules are safe goes without saying.
Example-objects. An object t such that E(t) holds is an example-object for E.
Thus, the introduction rule concludes ∃xE(x) from the existence of an exampleobject.
However, it is not always possible or feasible to prove an existential statement by
exhibiting a specific example-object, and there are (famous) proofs of existential
statements that do not use this rule.
Example 3.19 A transcendent real is a real which is not a root of a polynomialwith integer coefficients (a polynomial f (x) = an xn +an−1 xn−1 +. . .+a1 x+a0 ,where the ai are integers and an 6= 0).
By an argument establishing that the set of reals must be strictly larger than theset of roots of polynomials with integer coefficients, it is relatively easy to show

94

CHAPTER 3. THE USE OF LOGIC: PROOF

that transcendent reals exist. (Compare the reasoning about degrees of infinity inAppendix 11.)
Still, it is not immediately clear how to get from this argument at an exampletranscendent real. In particular, it is hard to see that e and π are transcendent reals.

Example 3.20 There is a proof that, in chess, either white has a winning strategy,
or black has a strategy with which he cannot lose. However, the proof neither
informs you which of the two cases you’re in, nor describes the strategy in a usableway.
Example 3.21 For a logical example, look at the proof given in Example 3.30
(p. 101). What is proved there is ∃x¬Φ(x), but (although ∃ introduction is usedsomewhere) no example-x for Φ is exhibited. And, on the basis of the given¬∀xΦ(x), it is unrealistic to expect such an example.
Sometimes, it is known only that an example-object must be present among themembers of a certain (finite) set but it is impossible to pinpoint the right object.
See the following example.
Example 3.22 Given: P (a) ∨ P (b)To be proved: ∃xP (x).
Proof: By ∨-elimination, it is sufficient to prove ∃xP (x) from both P (a) andP (b). But this is immediate, by ∃ introduction.
Example 3.23 To make example (3.22) more concrete, consider the following
question.
√Is there an irrational number α with the property that α 2 is rational?
√√ √In Example 3.16 we established that either 2 or 2 2 has this property. Thus,
the answer to the question is ‘yes’. But the reasoning does not provide us with an
example of such a number.
It is a general feature of Proofs by Contradiction of existential statements that no
example objects for the existential will turn up in the proof, and this is one reason
to stay away from this rule as long as you can in such cases.
Elimination. (Note the similarity with the ∨ -rule.)

3.4. RULES FOR THE QUANTIFIERS

95

When you want to use that ∃xE(x) in an argument to prove Λ, you write the Obligatory
Sentence:

Suppose that c is an object that satisfies E.

However, this is all that you are supposed to assume about c. Now, you proceed toprove Λ on the basis of this assumption.
Schematic form:
Given: ∃xE(x), . . .
To be proved: ΛProof:
Suppose c is an object that satisfies E
To be proved: ΛProof: . . .
Thus Λ
Modification suitable for the restricted existential quantifier:When you want to use that ∃x ∈ A E(x) in an argument to prove Λ, you write

Suppose that c is an object in A that satisfies E.

Again, this is all that you are supposed to assume about c. Subsequently, youproceed to prove Λ on the basis of this assumption.
Schematic form:
Given: c ∈ A, ∃xE(x), . . .
To be proved: ΛProof:
Suppose c is an object in A that satisfies E
To be proved: ΛProof: . . .
Thus Λ

CHAPTER 3. THE USE OF LOGIC: PROOF

96

3.5 Summary of the Proof Recipes
Here is a summary of the rules, with introduction rules on the left hand side, andelimination rules on the right hand side. Proof by Contradiction has been put inthe Elimination-column.

The Recipes for ⇒ Introduction and Elimination
Given: . . .
To be proved: P ⇒ Q
Proof:
Suppose P
To be proved: Q
Proof: . . .
Thus P ⇒ Q.

Given: P , P ⇒ Q, . . .
Thus QThe Recipes for ⇔ Introduction and Elimination
Given: . . .
To be proved: P ⇔ Q
Proof:
Suppose P
To be proved: Q
Proof: . . .
Suppose Q
To be proved: P
Proof: . . .
Thus P ⇔ Q.

Given: P , P ⇔ Q, . . .
Thus Q
Given: Q, P ⇔ Q, . . .
Thus P3.5. SUMMARY OF THE PROOF RECIPES

97

The Recipes for ¬ Introduction and Elimination
Given: . . .
To be proved: ¬P
Proof:
Suppose P
To be proved: ⊥Proof: . . .
Thus ¬P .

Given: . . .
To be proved: P
Proof:
Suppose ¬P
To be proved: ⊥Proof: . . .
Thus P .
Given: P , ¬P
Thus Q.

The Recipes for ∧ Introduction and Elimination
Given: P , Q
Thus P ∧ Q.

Given: P ∧ Q
Thus P .
Given: P ∧ Q
Thus Q.

The Recipes for ∨ Introduction and Elimination

Given: P
Thus P ∨ Q.
Given: Q
Thus P ∨ Q.

Given: P ∨ Q, . . .
To be proved: R
Proof:
Suppose P
To be proved: R
Proof: . . .
Suppose Q
To be proved: R
Proof: . . .
Thus R.

CHAPTER 3. THE USE OF LOGIC: PROOF

98

The Recipes for ∀ Introduction and Elimination
Given: . . .
To be proved: ∀xE(x)Proof:
Suppose c is an arbitrary object
To be proved: E(c)Proof: . . .

Given: ∀xE(x), . . .
Thus E(t).

Thus ∀xE(x)Given: . . .
To be proved: ∀x ∈ A E(x)Proof:
Suppose c is any object in A
To be proved: E(c)Proof: . . .

Given: ∀x ∈ A E(x), t ∈ A, . . .
Thus E(t).

Thus ∀x ∈ A E(x)The Recipes for ∃ Introduction and Elimination

Given: E(t), . . .
Thus ∃xE(x).

Given: ∃xE(x), . . .
To be proved: P
Proof:
Suppose c is an object that satisfies E
To be proved: P
Proof: . . .
Thus P

3.6. SOME STRATEGIC GUIDELINES

Given: t ∈ A, E(t), . . .
Thus ∃x ∈ A E(x).

99

Given: ∃x ∈ A E(x), . . .
To be proved: P
Proof:
Suppose c is an object in A that satisfies E
To be proved: P
Proof: . . .
Thus P

3.6 Some Strategic Guidelines
Here are the most important guidelines that enable you to solve a proof problem.
1. Do not concentrate on the given, by trying to transform that into what is tobe proved.
2. Instead, concentrate on (the form of) what is to be proved.
3. A number of rules enable you to simplify the proof problem. For instance:• When asked to prove P ⇒ Q, add P to the givens and try to prove Q.
(Deduction Rule).
• When asked to prove ∀x E(x), prove E(c) for an arbitrary c instead(∀-introduction).
4. Only after you have reduced the problem as far as possible you should lookat the givens in order to see which of them can be used.
• When one of the givens is of the form P ∨ Q, and R is to be proved,
make a case distinction: first add P to the givens and prove R, next
add Q to the givens and prove R.
• When one of the givens is of the form ∃x E(x), and P is to be proved,
give the object that satisfies E a name, by adding E(c) to the givens.
Next, prove P .
5. It is usually a good idea to move negations inward as much as possible before
attempting to apply ¬-introduction.
6. Stay away from Proof by Contradiction as long as possible.

CHAPTER 3. THE USE OF LOGIC: PROOF

100

Example 3.24 To show: from ∀x(P (x) ⇒ Q(x)), ∃xP (x) it follows that ∃xQ(x).
Concise proof: Using the second given, assume that x is such that P (x) holds.
Applying the first given to this x, it follows that P (x) ⇒ Q(x). Thus, we havethat Q(x). Conclusion: ∃xQ(x).
Exercise 3.25 Show:
1. from ∀x(P (x) ⇒ Q(x)), ∀xP (x) it follows that ∀xQ(x),2. from ∃x(P (x) ⇒ Q(x)), ∀xP (x) it follows that ∃xQ(x).
What about: from ∃x(P (x) ⇒ Q(x)), ∃xP (x) it follows that ∃xQ(x) ?
Exercise 3.26 From the given∀x∃y(xRy), ∀x∀y(xRy ⇒ yRx), ∀x∀y∀z(xRy ∧ yRz ⇒ xRz),
derive that ∀x(xRx).
Exercise 3.27 Give proofs for the following:
1. From ∀x∀y∀z(xRy∧yRz ⇒ xRz), ∀x¬xRx it follows that ∀x∀y(xRy ⇒¬yRx),
2. From ∀x∀y(xRy ⇒ ¬yRx) it follows that ∀x¬xRx,
3. From ∀x∀y(xRy ∧ x 6= y ⇒ ¬yRx) it follows that ∀x∀y(xRy ∧ yRx ⇒x = y),
4. From ∀x¬xRx, ∀x∀y(xRy ⇒ yRx), ∀x∀y∀z(xRy ∧ yRz ⇒ xRz)
it follows that ¬∃x∃y(xRy).

The following exercise is an example on how to move a quantifier in a prefix ofquantifiers.
Exercise 3.28 Show that from ∀y∃z∀xP (x, y, z) it follows that ∀x∀y∃zP (x, y, z).

3.6. SOME STRATEGIC GUIDELINES

101

Example 3.29 That a real function f is continuous means that (cf. p. 66), in thedomain R,∀x ∀ε > 0 ∃δ > 0 ∀y ( |x − y| < δ ⇒ |f (x) − f (y)| < ε ).
Uniform continuity of f means∀ε > 0 ∃δ > 0 ∀x ∀y ( |x − y| < δ =⇒ |f (x) − f (y)| < ε ).
Compared with the first condition, the quantifier ∀x has moved two places. According to Exercise 3.28 (where P (x, ε, δ) is ∀y(|x − y| < δ ⇒ |f (x) − f (y)| <ε)), continuity is implied by uniform continuity. (But, as you may know, the implication in the other direction does not hold: there are continuous functions thatare not uniformly so.)
Example 3.30
Given: ¬∀xΦ(x).
To be proved: ∃x¬Φ(x).
Proof:
We apply Proof by Contradiction.
Assume, for a contradiction, that ¬∃x¬Φ(x).
We will show that ∀xΦ(x).
Suppose x is arbitrary.
To show that Φ(x), we apply Proof by Contradiction again.
Assume ¬Φ(x).
Then ∃x¬Φ(x), and contradiction with ¬∃x¬Φ(x).
Thus Φ(x).
Therefore ∀xΦ(x), and contradiction with ¬∀xΦ(x).
Thus ∃x¬Φ(x).
The concise and more common way of presenting this argument (that leaves forthe reader to find out which rules have been applied and when) looks as follows.
Proof. Assume that ¬∃x¬Φ(x) and let x be arbitrary.
If ¬Φ(x) is true, then so is ∃x¬Φ(x), contradicting the assumption.
Thus, Φ(x) holds; and, since x was arbitrary, we conclude that ∀xΦ(x).
However, this contradicts the first given; which proves that, in fact, ∃x¬Φ(x) mustbe true.
Exercise 3.31 Prove the other equivalences of Theorem 2.40 (p. 65). (To proveΦ ≡ Ψ means (i) deriving Ψ from the given Φ and (ii) deriving Φ from the given
Ψ.)

102

CHAPTER 3. THE USE OF LOGIC: PROOF

Exercise 3.32 Derive the rules for the restricted quantifiers from the others, using
the facts (cf. the remark preceding Example (2.45), page 66) that ∀x ∈ A E(x)is equivalent with ∀x(x ∈ A ⇒ E(x)), and ∃x ∈ A E(x) is equivalent with
∃x(x ∈ A ∧ E(x)).
With practice, you will start to see that it is often possible to condense proof steps.
Here are some examples of condensed proof recipes:

Given: . . .
To be proved: ∀x(A(x) ⇒ B(x)).
Proof:
Suppose c is an arbitrary object such that A(c).
To be proved: B(c).
Proof: . . .
Thus ∀x(A(x) ⇒ B(x)).

Given: . . .
To be proved: ∀x∀y A(x, y).
Proof:
Suppose c and d are arbitrary objects.
To be proved: A(c, d).
Proof: . . .
Thus ∀x∀y A(x, y).

Given: . . .
To be proved: ∀x ∈ A∀y ∈ B R(x, y).
Proof:
Suppose c, d are arbitrary objects such that A(c) and B(d).
To be proved: R(c, d).
Proof: . . .
Thus ∀x ∈ A∀y ∈ B R(x, y).

3.7. REASONING AND COMPUTATION WITH PRIMES

103

Given: . . .
To be proved: ∀x∀y(R(x, y) ⇒ S(x, y)).
Proof:
Suppose c, d are arbitrary objects such that R(c, d).
To be proved: S(c, d).
Proof: . . .
Thus ∀x∀y(R(x, y) ⇒ S(x, y)).

3.7 Reasoning and Computation with Primes
In this section we will demonstrate the use of the computer for investigating thetheory of prime numbers. For this, we need the code for prime that was given in
Chapter 1. It is repeated here:

prime :: Integer -> Bool
prime n | n < 1= error "not a positive integer"| n == 1= False| otherwise = ldp n == n whereldp= ldpf primes
ldpf (p:ps) m | rem m p == 0 = p| p^2 > m
= m
| otherwise= ldpf ps mprimes = 2 : filter prime [3..]Euclid (fourth century B.C.) proved the following famous theorem about prime
numbers.
Theorem 3.33 There are infinitely many prime numbers.
Proof. Suppose there are only finitely many prime numbers, and p 1 , . . . , pn is alist of all primes. Consider the number m = (p1 p2 · · · pn ) + 1. Note that m is
not divisible by p1 , for dividing m by p1 gives quotient p2 · · · pn and remainder 1.
Similarly, division by p2 , p3 , . . . always gives a remainder 1.
Thus, we get the following:

104

CHAPTER 3. THE USE OF LOGIC: PROOF• LD(m) is prime,• For all i ∈ {1, . . . n}, LD(m) 6= pi .
Thus, we have found a prime number LD(m) different from all the prime numbers
in our list p1 , . . . , pn , contradicting the assumption that p1 , . . . , pn was the full listof prime numbers. Therefore, there must be infinitely many prime numbers.
Exercise 3.34 Let A = {4n + 3 | n ∈ N} (See Example 5.90 below). Show that
A contains infinitely many prime numbers. (Hint: any prime > 2 is odd, hence ofthe form 4n + 1 or 4n + 3. Assume that there are only finitely many primes ofthe form 4n + 3, say p1 , . . . , pm . Consider the number N = 4p1 · · · pm − 1 =4(p1 · · · pm − 1) + 3. Argue that N must contain a factor 4q + 3, using the factthat (4a + 1)(4b + 1) is of the form 4c + 1.)
Use filter prime [ 4*n + 3 | n <- [0..] ] to generate the primes of thisform.
Euclid’s proof suggests a general recipe for finding bigger and bigger primes.
Finding examples of very large primes is another matter, of course, for how do
you know whether a particular natural number is a likely candidate for a check?
Example 3.35 A famous conjecture made in 1640 by Pierre de Fermat (1601–1665) is that all numbers of the formn

22 + 10

are prime. This holds for n = 0, 1, 2, 3, 4, for we have: 22 + 1 = 21 + 1 = 3,
12
3
22 + 1 = 22 + 1 = 5, 22 + 1 = 24 + 1 = 17, 22 + 1 = 28 + 1 = 257, which is
24
16
prime, and 2 + 1 = 2 + 1 = 65537, which is prime. Apparently, this is as far
as Fermat got.
Our Haskell implementation of prime allows us to refute the conjecture for n = 5,
using the built-in function ^ for exponentiation. We get:
TUOLP> prime (2^2^5 + 1)
False

This counterexample to Fermat’s conjecture was discovered by the mathematician
Léonard Euler (1707–1783) in 1732.
The French priest and mathematician Marin Mersenne (1588–1647; Mersenne wasa pen pal of Descartes) found some large prime numbers by observing that M n =2n − 1 sometimes is prime when n is prime.

3.7. REASONING AND COMPUTATION WITH PRIMES

105

Exercise 3.36 It is not very difficult to show that if n is composite, Mn = 2n −1 iscomposite too. Show this. (Hint: Assume that n = ab and prove that xy = 2 n − 1
for the numbers x = 2b − 1 and y = 1 + 2b + 22b + · · · + 2(a−1)b ).
But when n is prime, there is a chance that 2n − 1 is prime too. Examples are
22 − 1 = 3, 23 − 1 = 7, 25 − 1 = 31. Such primes are called Mersenne primes.
Example 3.37 Let us use the computer to find one more Mersenne prime. Put theprocedure prime in a file and load it. Next, we use ^ for exponentiation to make a
new Mersenne guess, as follows:
TUOLP> prime 5
True
TUOLP> prime (2^5-1)
True
TUOLP> 2^5-1
31
TUOLP> prime (2^31-1)
True
TUOLP> 2^31-1
2147483647
TUOLP>

It may interest you to know that the fact that 231 − 1 is a prime was discovered by
Euler in 1750. Using a computer, this fact is a bit easier to check.
We have already seen how to generate prime numbers in Haskell (Examples 1.22
and 1.23). We will now present an elegant alternative: a lazy list implementationof the Sieve of Eratosthenes. The idea of the sieve is this. Start with the list of all
natural numbers > 2:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, . . .
In the first round, mark 2 (the first number in the list) as prime, and mark all multiples of 2 for removal in the remainder of the list (marking for removal indicatedby over-lining):
2 , 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, . . .

CHAPTER 3. THE USE OF LOGIC: PROOF

106

In the second round, mark 3 as prime, and mark all multiples of 3 for removal in
the remainder of the list:
2 , 3 , 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, . . .
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, . . .
In the third round, mark 5 as prime, and mark all multiples of 5 for removal in theremainder of the list:
2 , 3 , 4, 5 , 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, . . .
And so on. A remarkable thing about the Sieve is that the only calculation it
involves is counting. If the 3-folds are to be marked in the sequence of natural
numbers starting from 3, walk through the list while counting 1, 2, 3 and mark thenumber 6, next walk on while counting 1, 2, 3 and mark the number 9, and so on.
If the 5-folds are to be marked in the sequence the natural numbers starting from5, walk on through the sequence while counting 1, 2, 3, 4, 5 and mark the number
10, next walk on while counting 1, 2, 3, 4, 5 and mark the number 15, and so on.
In the Haskell implementation we mark numbers in the sequence [2..] for removal by replacing them with 0. When generating the sieve, these zeros are
skipped.

sieve :: [Integer] -> [Integer]
sieve (0 : xs) = sieve xs
sieve (n : xs) = n : sieve (mark xs 1 n)
wheremark :: [Integer] -> Integer -> Integer -> [Integer]mark (y:ys) k m | k == m
= 0 : (mark ys 1m)| otherwise = y : (mark ys (k+1) m)primes :: [Integer]
primes = sieve [2..]This gives:
TUOLP> primes

3.7. REASONING AND COMPUTATION WITH PRIMES

107[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,
83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,
167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,
257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,
353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,
449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,
563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,
653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,
761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,
877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,
991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,{Interrupted!}

Does this stream ever dry up? We know for sure that it doesn’t, because of Euclid’s
proof.
It is possible, by the way, to take a finite initial segment of an infinite Haskell list.
This is done with the built in function take, as follows:
TUOLP> take 100 primes[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,
83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,
167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,
257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,
353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,
449,457,461,463,467,479,487,491,499,503,509,521,523,541]
TUOLP>

Exercise 3.38 A slightly faster way to generate the primes is by starting out fromthe odd numbers. The stepping and marking will work as before, for if you countk positions in the odd numbers starting from any odd number a = 2n + 1, you will
move on to number (2n + 1) + 2k, and if a is a multiple of k, then so is a + 2k.
Implement a function fasterprimes :: [Integer] using this idea. The odd
natural numbers, starting from 3, can be generated as follows:

oddsFrom3 :: [Integer]
oddsFrom3 = 3 : map (+2) oddsFrom3

Still faster is to clean up the list at every step, by removing multiples from the list
as you go along. We will come back to this matter in Section 10.1.

108

CHAPTER 3. THE USE OF LOGIC: PROOF

Exercise 3.39 Write a Haskell program to refute the following statement about
prime numbers: if p1 , . . . , pk are all the primes < n, then (p1 × · · · × pk ) + 1 is aprime.
A computer is a useful instrument for refuting guesses or for checking particularcases. But if, instead of checking a guess for a particular case, you want to checkthe truth of interesting general statements it is of limited help. You can use thefunction mersenne to generate Mersenne primes, but the computer will not tell
you whether this stream will dry up or not . . .

mersenne = [ (p,2^p - 1) | p <- primes, prime (2^p - 1) ]This is what a call to mersenne gives:
TUOLP> mersenne[(2,3),(3,7),(5,31),(7,127),(13,8191),(17,131071),(19,524287),(31,2147483647)

If you are interested in how this goes on, you should check out GIMPS (“Great
Internet Mersenne Prime Search”) on the Internet. To generate slightly more information, we can define:notmersenne = [ (p,2^p - 1) | p <- primes, not (prime (2^p-1)) ]This gives:
TUOLP> notmersenne[(11,2047),(23,8388607),(29,536870911),(37,137438953471),(41,2199023255551),(43,8796093022207),(47,140737488355327),(53,9007199254740991),(59,576460752303423487)

The example may serve to illustrate the limits of what you can do with a computerwhen it comes to generating mathematical insight. If you make an interesting
mathematical statement, there are three possibilities:

3.7. REASONING AND COMPUTATION WITH PRIMES

109• You succeed in proving it. This establishes the statement as a theorem.
• You succeed in disproving it (with or without the help of a computer). Thisestablishes the statement as a refuted conjecture.
• Neither of the above. This may indicate that you have encountered an open
problem in mathematics. It may also indicate, of course, that you haven’t
been clever enough.
Example 3.40 Here is an example of an open problem in mathematics:
Are there infinitely many Mersenne primes?

It is easy to see that Euclid’s proof strategy will not work to tackle this problem.
The assumption that there is a finite list p1 , . . . , pn of Mersenne primes does yielda larger prime, but nothing guarantees that this larger prime number is again of theform 2m − 1.
Mersenne primes are related to so-called perfect numbers. A perfect number is anumber n with the curious property that the sum of all its divisors equals 2n, or,in other words, the sum of all proper divisors of n equals n (we call a divisor d ofn proper if d < n). The smallest perfect number is 6, for its proper divisors are
1, 2 and 3, and 1 + 2 + 3 = 6, and it is easy to check that 1, 2, 3, 4 and 5 are not
perfect.
Euclid proved that if 2n − 1 is prime, then 2n−1 (2n − 1) is perfect. Examples ofperfect numbers found by Euclid’s recipe are: 2 · (22 − 1) = 6, 22 · (23 − 1) = 28,
24 · (25 − 1) = 496.
Exercise 3.41 How would you go about yourself to prove the fact Euclid proved?
Here is a hint: if 2n − 1 is prime, then the proper divisors of 2n−1 (2n − 1) are
1, 2, 22 , . . . , 2n−1 , 2n − 1, 2(2n − 1), 22 (2n − 1), . . . , 2n−2 (2n − 1).

Here is a function for generating the list of proper divisors of a natural number.
This is not an efficient way to generate proper divisors, but never mind.

pdivisors :: Integer -> [Integer]
pdivisors n = [ d | d <- [1..(n-1)], rem n d == 0 ]110

CHAPTER 3. THE USE OF LOGIC: PROOF

With this it is easy to check that 8128 is indeed a perfect number:
TUOLP> pdivisors 8128
[1,2,4,8,16,32,64,127,254,508,1016,2032,4064]
TUOLP> sum (pdivisors 8128)
8128

Even more spectacularly, we have:
TUOLP> prime (2^13 -1)
True
TUOLP> 2^12 * (2^13 -1)
33550336
TUOLP> pdivisors 33550336[1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8191,16382,32764,
65528,131056,262112,524224,1048448,2096896,4193792,8387584,
16775168]TUOLP> sum [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8191,16382,
32764,65528,131056,262112,524224,1048448,2096896,4193792,8387584,
16775168]33550336
TUOLP>

Prime pairs are pairs (p, p + 2) where both p and p + 2 are prime. Prime pairs canbe generated as follows:

primePairs :: [(Integer,Integer)]primePairs = pairs primes
wherepairs (x:y:xys) | x + 2 == y = (x,y): pairs (y:xys)| otherwise = pairs (y:xys)

This gives:
TUOLP> take 50 primePairs
take 50 primePairs[(3,5),(5,7),(11,13),(17,19),(29,31),(41,43),(59,61),(71,73),(101,103),(107,109),(137,139),(149,151),(179,181),(191,193),(197,199),(227,229),(239,241),(269,271),(281,283),(311,313),(347,349),(419,421),(431,433),(461,463),(521,523),(569,571),(599,601),(617,619),(641,643),(659,661),(809,811),(821,823),

3.8. FURTHER READING

111(827,829),(857,859),(881,883),(1019,1021),(1031,1033),(1049,1051),(1061,1063),(1091,1093),(1151,1153),(1229,1231),(1277,1279),(1289,1291),(1301,1303),(1319,1321),(1427,1429),(1451,1453),(1481,1483),(1487,1489)]
TUOLP>

Does this stream ever dry up? We don’t know, for the question whether there areinfinitely many prime pairs is another open problem of mathematics.
Exercise 3.42 A prime triple is a triple (p, p + 2, p + 4) with p, p + 2, p + 4 allprime. The first prime triple is (3, 5, 7). Are there any more? Note that instructing
the computer to generate them is no help:

primeTriples :: [(Integer,Integer,Integer)]primeTriples = triples primes
where
triples (x:y:z:xyzs)| x + 2 == y && y + 2 == z = (x,y,z) : triples (y:z:xyzs)| otherwise= triples (y:z:xyzs)

We get:TUOLP> primeTriples[(3,5,7)
Still, we can find out the answer . . . How?
Exercise 3.43 Consider the following call:
TUOLP> filter prime [ p^2 + 2 | p <- primes ][11
Can you prove that 11 is the only prime of the form p2 + 2, with p prime?

3.8 Further Reading
The distinction between finding meaningful relationships in mathematics on one
hand and proving or disproving mathematical statements on the other is drawn

112

CHAPTER 3. THE USE OF LOGIC: PROOF

very clearly in Polya [Pol57]. A good introduction to mathematical reasoning is[Ecc97]. More detail on the structure of proofs is given by Velleman [Vel94].
Automated proof checking in Haskell is treated in [HO00]. An all-time classic in
the presentation of mathematical proofs is [Euc56].

Chapter 4Sets, Types and Lists
Preview
The chapter introduces sets, not by means of a definition but by explaining why‘set’ is a primitive notion, discusses the process of set formation, and explainssome important operations on sets.
Talking about sets can easily lead to paradox, but such paradoxes can be avoided
either by always forming sets on the basis of a previously given set, or by imposing
certain typing constraints on the expressions one is allowed to use. Section 4.2
explains how this relates to functional programming.
The end of the Chapter discusses various ways of implementing sets using list
representations. The chapter presents ample opportunity to exercise your skills inwriting implementations for set operations and for proving things about sets.

module STAL
where
import List
import DB

113

114

CHAPTER 4. SETS, TYPES AND LISTS

4.1 Let’s Talk About Sets
Remarkably, it is not possible to give a satisfactory definition of the notion of aset. There are several axiomatic approaches to set theory; the standard one (that is
implicitly assumed throughout mathematics) is due to Zermelo and Fraenkel anddates from the beginning of the 20th century.
Axioms vs. Theorems, Primitive vs. Defined Notions. The truth of a mathematical statement is usually demonstrated by a proof. Most proofs use results that havebeen proved earlier. It follows that some truths must be given outright without
proof. These are called axioms. The choice of axioms often is rather arbitrary.
Criteria can be simplicity or intuitive evidence, but what must be considered anaxiom can also be dictated by circumstance (for instance, a proof cannot be given
within the given context, but in another context a proof would be possible). In
the present context, we shall accept some of the Zermelo-Fraenkel axioms, as well
as some fundamental properties of the number systems, such as the principle ofmathematical induction for N (see Sections 7.1 and 11.1).
Statements that have been proved are called theorems if they are considered to beof intrinsic value, and lemmas if they are often used in proving theorems.
Notions are split up in two analogous categories. Their meaning may have been
explained by definitions in terms of other notions. However, the process of defining
notions must have a beginning somewhere. Thus, the need for notions that are
primitive, i.e., undefined.
For instance, we shall consider as undefined here the notions of set and natural
number. Given the notion of a set, that of a function can be defined. However, in acontext that is not set-theoretic, it could well be an undefined notion.
Georg Cantor (1845-1915), the founding father of set theory, gave the following
description.
The Comprehension Principle. A set is a collection into a whole of definite, distinct objects of our intuition or of our thought. The objects are called the elements(members) of the set.
Usually, the objects that are used as elements of a set are not sets themselves. To
handle sets of this kind is unproblematic. But it is not excluded at all that members
are sets. Thus, in practice, you can encounter sets of sets of . . . sets.
Notation. If the object a is member of the set A, this is denoted by a ∈ A, or
sometimes by A 3 a. If a is not a member of A, we express this as a ∈/ A, or4.1. LET’S TALK ABOUT SETS

115

A 63 a.

Example: 0 ∈ N, 12 6∈ N, 12 ∈ Q.
A set is completely determined by its elements: this is the content of the following
Principle of Extensionality. Sets that have the same elements are equal.
Symbolically, for all sets A and B, it holds that:∀x(x ∈ A ⇔ x ∈ B) =⇒ A = B.
The converse of this (that equal sets have the same elements) is trivial. The Principle of Extensionality is one of Zermelo’s axioms.

Figure 4.1: A set with a proper subset.

Subsets. The set A is called a subset of the set B, and B a superset of A; notations:A ⊆ B, and B ⊇ A, if every member of A is also a member of B. In symbols:∀x (x ∈ A =⇒ x ∈ B).
If A ⊆ B and A 6= B, then A is called a proper subset of B.
For instance, {0, 2} is a proper subset of N, and the set of all multiples of 4 is aproper subset of the set of all even integers.
Note that A = B iff A ⊆ B and B ⊆ A. To show that A 6= B we therefore either
have to find an object c with c ∈ A, c ∈/ B (in this case c is a witness of A 6⊆ B),
or an object c with c ∈/ A, c ∈ B (in this case c is a witness of B 6⊆ A). A proof of
A = B will in general have the following form:

CHAPTER 4. SETS, TYPES AND LISTS

116

Given: . . .
To be proved: A = B.
Proof:⊆: Let x be an arbitrary object in A.
To be proved: x ∈ B.
Proof:...
Thus x ∈ B.
⊆: Let x be an arbitrary object in B.
To be proved: x ∈ A.
Proof:...
Thus x ∈ A.
Thus A = B.

Warning. Sometimes, A ⊆ B is written as A ⊂ B. Other authors use A ⊂ B
to indicate that A is a proper subset of B. In this book we will stick to ⊆, andto express that A is properly included in B we will always use the conjunction ofA ⊆ B and A 6= B.
∈ versus ⊆. Beginners often confuse ∈ and ⊆, but these relations are very
different. For instance, A ⊆ B implies that A and B are both sets, whereasa ∈ B only implies that B is a set. Assuming that numbers are not sets, we havethat 1 ∈ {0, 1, 2} and 1 6⊆ {0, 1, 2} (provided the latter makes sense); whereas{1} ⊆ {0, 1, 2}, and {1} 6∈ {0, 1, 2}.
Theorem 4.1 For all sets A, B, C, we have that:
1. A ⊆ A(reflexivity),

2. A ⊆ B ∧ B ⊆ A =⇒ A = B(antisymmetry),

3. A ⊆ B ∧ B ⊆ C =⇒ A ⊆ C(transitivity).

Proof. 1.
To be proved: A ⊆ A, i.e., ∀x(x ∈ A ⇒ x ∈ A).
Proof:
Suppose c is any object in A. Then c ∈ A.
Therefore ∀x(x ∈ A ⇒ x ∈ A), i.e., A ⊆ A.

4.1. LET’S TALK ABOUT SETS

117

2. This is Extensionality — be it in a somewhat different guise.
3.
To be proved: A ⊆ B ∧ B ⊆ C ⇒ A ⊆ C.
Proof:
Suppose A ⊆ B and B ⊆ C.
To be proved: A ⊆ C.
Proof:
Suppose c is any object in A.
Then by A ⊆ B. c ∈ B, and by B ⊆ C, c ∈ C.
Thus ∀x(x ∈ A ⇒ x ∈ C), i.e., A ⊆ C.
Thus A ⊆ B ∧ B ⊆ C ⇒ A ⊆ C.

Remark. Note that the converse of antisymmetry also holds for ⊆. In other words,
if A, B are sets, then A = B iff A ⊆ B and B ⊆ A. It is because of antisymmetryof the ⊆ relation that a proof that two sets A, B are equal can consist of the two
subproofs mentioned above: the proof of A ⊆ B and the proof of B ⊆ A.
Exercise 4.2 Show that the superset relation also has the properties of Theorem
4.1, i.e., show that ⊇ is reflexive, antisymmetric and transitive.
Enumeration. A set that has only few elements a1 , . . . , an can be denoted as{a1 , . . . , an }.
Extensionality ensures that this denotes exactly one set, for by extensionality theset is uniquely determined by the fact that it has a1 , . . . , an as its members.
Note that x ∈ {a1 , . . . , an } iff x = a1 ∨ · · · ∨ x = an .
Example 4.3 {0, 2, 3} is the set the elements of which are 0, 2 and 3. We clearly
have that 3 ∈ {0, 2, 3}, and that 4 6∈ {0, 2, 3}.
Note that {0, 2, 3} = {2, 3, 0} = {3, 2, 2, 0}. Indeed, these sets have the sameelements. Thus:
Order and repetition in this notation are irrelevant.
Exercise 4.4 Show, that {{1, 2}, {0}, {2, 1}} = {{0}, {1, 2}}.

118

CHAPTER 4. SETS, TYPES AND LISTSAn analogue to the enumeration notation is available in Haskell, where [n..m] canbe used for generating a list of items from n to m. This presupposes that n and m
are of the same type, and that enumeration makes sense for that type. (Technically,
the type has to be of the class Ord; see 4.2 below.)
Sets that have many elements, in particular, infinite sets, cannot be given in thisway, unless there is some system in the enumeration of their elements. For instance, N = {0, 1, 2, . . .} is the set of natural numbers, and {0, 2, 4, . . .} is the setof even natural numbers.
Abstraction. If P (x) is a certain property of objects x, the abstraction{ x | P (x) }(4.1)

denotes the set of things x that have property P .
Thus, for every particular object a, the expressiona ∈ { x | P (x) }is equivalent withP (a).
By Extensionality, our talking about the set of x such that P (x) is justified.
The abstraction notation binds the variable x: the set {x | P (x)} in no way depends on x; {x | P (x)} = {y | P (y)}.
Usually, the property P will apply to the elements of a previously given set A. In
that case{ x ∈ A | P (x) }denotes the set of those elements of A that have property P . For instance, the setof even natural numbers can be given by{ n ∈ N | n is even }.
This way of defining an infinite set can be mimicked in functional programmingby means of so-called list comprehensions, as follows:

naturals = [0..]evens1 = [ n | n <- naturals , even n ]4.1. LET’S TALK ABOUT SETS

119

Note the similarity between n <- naturals and n ∈ N, which is of course intended by the Haskell design team. The expression even n implements the property ‘n is even’, so we follow the abstraction notation from set theory almost to theletter. Here is the implementation of the process that generates the odd numbers:

odds1 =

[ n | n <- naturals , odd n ]Back to the notation of set theory. A variation on the above notation for abstraction
looks as follows. If f is an operation, then{ f (x) | P (x) }denotes the set of things of the form f (x) where the object x has the property P .
For instance,{ 2n | n ∈ N }is yet another notation for the set of even natural numbers.
And again, we have a counterpart in functional programming. Here it is:evens2 = [ 2*n | n <- naturals ]Still, the similarity in notation between the formal definitions and their implementations should not blind us to some annoying divergences between theory andpractice. The two notations{n2 | n ∈ {0, . . . , 999}}
and{n2 | n ∈ N ∧ n < 1000}
are equivalent. They are merely two ways of specifying the set of the first 1000
square numbers. But the Haskell counterparts behave very differently:

small_squares1 = [ n^2 | n <- [0..999] ]120

CHAPTER 4. SETS, TYPES AND LISTS

A call to the function small_squares1 indeed produces a list of the first thousandsquare numbers, and then terminates. Note the use of [0..999] to enumerate afinite list. Not so with the following:small_squares2 = [ n^2 | n <- naturals , n < 1000 ]The way this is implemented, n <- naturals generates the infinite list of natural numbers in their natural order, and n < 1000 tests each number as it is generated for the property of being less than 1000. The numbers that satisfy thetest are squared and put in the result list. Unlike the previous implementation
small_squares1, the function small_squares2 will never terminate.
Example 4.5 (*The Russell Paradox) It is not true that to every property E there
corresponds a set {x | E(x)} of all objects that have E. The simplest example
was given by Bertrand Russell (1872–1970). Consider the property of not having
yourself as a member.
Most sets that you are likely to consider have this property: the set of all evennatural numbers is itself not an even natural number, the set of all integers is itself
not an integer, and so on. Call such sets ‘ordinary’. The corresponding abstraction
is R = { x | x 6∈ x }.
It turns out that the question whether the set R itself is ordinary or not is impossible
to answer. For suppose R ∈ R, that is, suppose R is an ordinary set. Ordinarysets are the sets that do not have themselves as a member, so R does not haveitself as a member, i.e., R ∈/ R. Suppose, on the contrary, that R ∈/ R, that is, R
is an extraordinary set. Extraordinary sets are the sets that have themselves as amember, so R has itself as a member, i.e., R ∈ R.
If R were a legitimate set, this would unavoidably lead us to the conclusion thatR ∈ R ⇐⇒ R 6∈ R ,which is impossible.
You do not have to be afraid for paradoxes such as the Russell paradox of Example
4.5. Only properties that you are unlikely to consider give rise to problems. In
particular, if you restrict yourself to forming sets on the basis of a previously given
set A, by means of the recipe{ x ∈ A | E(x) },4.2. PARADOXES, TYPES AND TYPE CLASSES

121

no problems will ever arise.
Example 4.6 There is no set corresponding to the property F (x) :≡ there is no
infinite sequence x = x0 3 x1 3 x2 3 . . .. To see this, assume to the contrary
that F is such a set. Assume F ∈ F . This implies F 3 F 3 F 3 F . . ., so by thedefining property of F , F ∈/ F . Assume F ∈/ F . Then by the defining property of
F , there is an infinite sequence F = x0 3 x1 3 x2 3 . . .. Now take the infinite
sequence x1 3 x2 3 . . .. By the defining property of F , x1 ∈/ F , contradicting
F = x 0 3 x1 .
Exercise 4.7* Assume that A is a set of sets. Show that {x ∈ A | x 6∈ x} 6∈ A.
It follows from Exercise (4.7) that every set A has a subset B ⊆ A with B ∈/ A.
Take B = {x ∈ A | x ∈/ x}.

4.2 Paradoxes, Types and Type Classes
It is a well-known fact from the theory of computation that there is no generaltest for checking whether a given procedure terminates for a particular input. The
halting problem is undecidable. Intuitively, the reason for this is that the existenceof an algorithm (a procedure which always terminates) for the halting problem
would lead to a paradox very similar to the Russell paradox.
Here is a simple example of a program for which no proof of termination exists:

run :: Integer -> [Integer]
run n | n < 1 = error "argument not positive"| n == 1 = [1]| even n = n: run (div n 2)
| odd n = n: run (3*n+1)

This gives, e.g.:
STAL> run 5[5,16,8,4,2,1]
STAL> run 6[6,3,10,5,16,8,4,2,1]
STAL> run 7[7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1]

122

CHAPTER 4. SETS, TYPES AND LISTS

We say that a procedure diverges when it does not terminate or when it aborts withan error. Stipulating divergence in Haskell is done by means of the predeclaredfunction undefined, which causes an error abortion, just like error. In fact,
Haskell has no way to distinguish between divergence and error abortion.
Now suppose halts can be defined. Then define the procedure funny, in terms ofhalts, as follows (the part of a line after -- is a comment; in this case a warningthat funny is no good as Haskell code):

funny x | halts x x = undefined| otherwise = True-- Caution: this-- will not work

What about the call funny funny? Does this diverge or halt?
Suppose funny funny does not halt. Then by the definition of funny, we are in
the first case. This is the case where the argument of funny, when applied to itself,
halts. But the argument of funny is funny. Therefore, funny funny does halt,
and contradiction.
Suppose funny funny does halt. Then by the definition of funny, we are in thesecond case. This is the case where the argument of funny, when applied to itself,
does not halt. But the argument of funny is funny. Therefore, funny funny
does not halt, and contradiction.
Thus, there is something wrong with the definition of funny. The only peculiarityof the definition is the use of the halts predicate. This shows that such a halts
predicate cannot be implemented.
It should be clear that funny is a rather close analogue to the Russell set {x |x∈/ x}. Such paradoxical definitions are avoided in functional programming bykeeping track of the types of all objects and operations.
As we have seen, new types can be constructed from old. Derived types are pairsof integers, lists of characters (or strings), lists of reals, and so on. How does thistype discipline avoid the halting paradox? Consider the definition of funny. It
makes a call to halts. What is the type of halts? The procedure halts takes
as first argument a procedure, say proc, and as second argument an argument tothat procedure, say arg. This means that the two arguments of halts have types
a -> b and a, respectively. and that (proc arg), the result of applying proc toarg, has type b. But this means that the application halts x x in the definition offunny is ill-formed, for as we have seen the types of the two arguments to halts
must be different, so the arguments themselves must be different.

4.2. PARADOXES, TYPES AND TYPE CLASSES

123

For another example, take the built-in elem operation of Haskell which checks
whether an object is element of a list. This operation is as close as you can get
in Haskell to the ‘∈’ relation of set theory. The operation expects that if its firstargument is of certain type a, then its second argument is of type ‘list over a’.
Thus, in Haskell, the question whether R ∈ R does not make sense, for any R: witness the following interaction. In the transcript, Prelude> is the Haskell prompt
when no user-defined files are loaded.
Prelude> elem ’R’ "Russell"
True
Prelude> elem ’R’ "Cantor"
False
Prelude> elem "Russell" "Cantor"
ERROR: Type error in application*** expression: "Russell" ‘elem‘ "Cantor"*** term: "Russell"*** type: String*** does not match : Char
Prelude>You would expect from this that elem has type a -> [a] -> Bool, for it takes
an object of any type a as its first argument, then a list over type a, and it returns a
verdict ‘true’ or ‘false’, i.e., an object of type Bool.
Almost, but not quite. The snag is that in order to check if some thing x is anelement of some list of things l, one has to be able to identify things of the type of
x. The objects that can be identified are the objects of the kinds for which equality
and inequality are defined. Texts, potentially infinite streams of characters, are notof this kind. Also, the Haskell operations themselves are not of this kind, for the
Haskell operations denote computation procedures, and there is no principled way
to check whether two procedures perform the same task.
For suppose there were a test for equality on procedures (implemented functions).
Then the following would be a test for whether a procedure f halts on input x (here/= denotes inequality):

halts f x = f /= gwhere g y | y == x
= undefined| otherwise = f y-- Caution: this-- will not work

124

CHAPTER 4. SETS, TYPES AND LISTS

The where construction is used to define an auxiliary function g by stipulatingthat g diverges on input x and on all other inputs behaves the same as f. If g is
not equal to f, then the difference must be in the behaviour for x. Since we havestipulated that g diverges for this input, we know that f halts on x. If, on the other
hand, g and f are equal, then in particular f and g behave the same on input x,
which means that f diverges on that input.
The types of object for which the question ‘equal or not’ makes sense are grouped
into a collection of types called a class. This class is called Eq. Haskell uses ==for equality and /= for inequality of objects of types in the Eq class.
Using the hugs command :t to ask for the type of a defined operation, we get forelem:
Prelude> :t elem
elem :: Eq a => a -> [a] -> Bool
Prelude>In the present case, the type judgment means the following. If a is a type forwhich equality is defined (or, if a is in the Eq class), then a -> [a] -> Bool is
an appropriate type for elem. In other words: for all types a in the Eq class it holds
that elem is of type a -> [a] -> Bool.
This says that elem can be used to check whether an integer is a member of a list of
integers, a character is a member of a string of characters, a string of characters isa member of a list of strings, and so on. But not whether an operation is a memberof a list of operations, a text a member of a list of texts, and so on.
Ord is the class of types of things which not only can be tested for equality andinequality, but also for order: in addition to == and /=, the relations < and <=are defined. Also, it has functions min for the minimal element and max for the
maximal element. The class Ord is a subclass of the class Eq.
Classes are useful, because they allow objects (and operations on those objects) to
be instances of several types at once. The numeral ‘1’ can be used as an integer, asa rational, as a real, and so on. This is reflected in Haskell by the typing:Prelude> :t 1
1 :: Num a => aPrelude>All of the types integer, rational number, real number, complex number, and so on,
are instances of the same class, called Num in Haskell. The class Num is a subclassof the class Eq (because it also has equality and inequality). For all types in theclass Num certain basic operations, such as + and *, are defined. As we will see in4.3. SPECIAL SETS

125

Chapter 8, addition has different implementations, depending on whether we operate on N, Z, Q, . . . and depending on the representations we choose. Still, insteadof distinguishing between add, add1, add2, and so on, one could use the same
name for all these different operations. This is standard practice in programming
language design, and it is called operator overloading.
Exercise 4.8 Explain the following error message:
Prelude> elem 1 1
ERROR: [a] is not an instance of class "Num"Prelude>4.3 Special Sets
Singletons. Sets that have exactly one element are called singletons. The set
whose only element is a is { a }; this is called the singleton of a. Note that it
follows from the definition that x ∈ {a} iff x = a.
Warning. Do not confuse a singleton { a } with its element a.
In most cases you will have that a 6= {a}. For instance, in the case that a = {0, 1},
we have that {0, 1} 6= {{0, 1}}. For, {0, 1} has two elements: the numbers 0 and
1. On the other hand, {{0, 1}} has only one element: the set {0, 1}.
Remark. The question whether the equation a = { a } has solutions (or, moregenerally, whether sets a exist such that a ∈ a) is answered differently by different
axiomatizations of set theory. A set satisfying this equation has the shape a ={{{· · · · · · }}}, but of course this is unofficial notation. For the mathematicalcontent of set theory this problem is largely irrelevant.
An example of a case where it would be useful to let sets have themselves as
members would be infinite streams, like an infinite list of ‘1’s. Such an object iseasily programmed in Haskell:ones = 1 : ones

126

CHAPTER 4. SETS, TYPES AND LISTS

If you load and run this, and endless stream of 1’s will cover the screen, and you
will have to kill the process from outside. Still, the process specified by ‘first
generate a ‘1’ and then run the same process again’ is well-defined, and it canplausibly be said to have itself as its second member. To be sure, the order does
matter here, but any set theory can encode ordered pairs: see Section 4.5 below.
A set of the form {a, b} is called an (unordered) pair. Of course, if a = b, then{a, b} = {a} is, in fact, a singleton.
Empty Set. Finally, there is a set without any elements at all: the empty set.
This curious object can be a source of distress for the beginner, because of its
unexpected properties. (A first one is Theorem 4.9.) The notation for the empty
set is∅.
Note that there is exactly one set that is empty: this is due to Extensionality.
Theorem 4.9 For every set A, we have that
∅ ⊆ A.
Proof.
Suppose x is an arbitrary object with x ∈ ∅.
Then ⊥ (contradiction with the fact that ∅ has no members).
Therefore x ∈ A.
Thus ∀x(x ∈ ∅ ⇒ x ∈ A), i.e., ∅ ⊆ A.

Exercise 4.10 Show:
1. { a } = { b } iff a = b,
2. {a1 , a2 } = {b1 , b2 } iff: a1 = b1 ∧ a2 = b2 , or a1 = b2 ∧ a2 = b1 .

Exercise 4.11 Explain that ∅ 6= {∅}. And that {∅} 6= {{∅}}.

4.4. ALGEBRA OF SETS

127

Remark. Abstraction and Restricted Quantification. Note that∀x ∈ A Φ(x) is true iff {x ∈ A | Φ(x)} = A.
Similarly,∃x ∈ A Φ(x) is true iff {x ∈ A | Φ(x)} 6= ∅.

4.4 Algebra of Sets
Definition 4.12 (Intersection, Union, Difference.) Assume that A and B are sets.
Then:
1. A ∩ B = { x | x ∈ A ∧ x ∈ B } is the intersection of A and B,
2. A ∪ B = { x | x ∈ A ∨ x ∈ B } is their union,
and

3. A − B = { x | x ∈ A ∧ x 6∈ B } their difference.
Types of Set Theoretic Expressions Often, the symbols ∩ and ∪ are confusedwith the connectives ∧ and ∨. From Definition 4.12, their intimate connection isclear. However, their functioning is completely different: ∩ and ∪ produce, given
two sets A and B, new sets A ∩ B resp. A ∪ B (thus, ∩ and ∪ can be written
between sets only), whereas ∧ and ∨ combine two statements Φ and Ψ into newstatements Φ ∧ Ψ resp. Φ ∨ Ψ.
In the idiom of Section 4.2: the operations ∩, ∪ on one hand and ∧, ∨ on the otherhave different types. The typing underlying the notation of set theory is much
simpler than that of functional programming. To start with, we just distinguishbetween the types s (for set), t (for a proposition, i.e., an expression with a truth
value), (for anything at all), and {s} (for a family of sets, see below).
t is like Bool in Haskell, but s has no Haskell analogue. In set theory, sets like
N, Z, Q, R are all of the same type, namely s, whereas in Haskell much finer
distinctions are drawn. The simplicity of the typing underlying set theory is anadvantage, for it makes the language of set theory more flexible.
To find the types of the set theoretic operations we use the same recipe as in functional programming. Thus, ∩ and ∪ both have type s → s → s, for they taketwo set arguments and produce a new set. ∧ and ∨ have type t → t → t, tor they

128

CHAPTER 4. SETS, TYPES AND LISTS

Figure 4.2: Set union, intersection, difference, and complement.

4.4. ALGEBRA OF SETS

129

take two propositions and produce a new proposition. ∈ has type → s → t,
for it takes anything at all as its first argument, a set as its second argument, andproduces a proposition.
Exercise 4.13 What are the types of the set difference operator − and of the inclusion operator ⊆ ?
Exercise 4.14 Give the types of the following expressions:
1. x ∈ {x | E(x)}.
2. {x | E(x)}.
3. (A ∩ B) ⊆ C.
4. (A ∪ B) ∩ C.
5. ∀x(x ∈ A ⇒ x ∈ B).
6. A = B.
7. a ∈ A ⇔ a ∈ B.

The relationships between ∩ and ∧ and between ∪ and ∨ become clearer if theequalities from Definition 4.12 are written in the form of equivalences:
1. x ∈ A ∩ B ⇐⇒ x ∈ A ∧ x ∈ B,
2. x ∈ A ∪ B ⇐⇒ x ∈ A ∨ x ∈ B,
3. x ∈ A − B ⇐⇒ x ∈ A ∧ x 6∈ B.
Disjointness. Sets A and B are called disjoint if A ∩ B = ∅.
Example 4.15 For A = {1, 2, 3} and B = {3, 4}, we have: A ∪ B = {1, 2, 3, 4},A ∩ B = {3} and A − B = {1, 2}. A and B are not disjoint, for 3 ∈ A ∩ B.
Theorem 4.16 For all sets A, B and C, we have the following:
1. A ∩ ∅ = ∅;A ∪ ∅ = A,CHAPTER 4. SETS, TYPES AND LISTS

130
2. A ∩ A = A;
A∪A=A

3. A ∩ B = B ∩ A;
A∪B =B∪A

4. A ∩ (B ∩ C) = (A ∩ B) ∩ C;
A ∪ (B ∪ C) = (A ∪ B) ∪ C

5. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C);
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(idempotence),(commutativity),(associativity),(distributivity).

By part 4.16.4, we can omit parentheses in intersections and unions of more than
two sets.
Proof. Using the definitions of ∩, ∪ and −, these laws all reduce to Theorem 2.10
(p. 45). Still, it is instructive to give a direct proof. Here is one for the first
distribution law:⊆:Let x be any object in A ∩ (B ∪ C).
Then x ∈ A and either x ∈ B or x ∈ C.
Suppose x ∈ B. Then x ∈ A ∩ B, so x ∈ (A ∩ B) ∪ (A ∩ C).
Suppose x ∈ C. Then x ∈ A ∩ C, so x ∈ (A ∩ B) ∪ (A ∩ C).
Thus in either case x ∈ (A ∩ B) ∪ (A ∩ C).
Thus A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C).

⊆:

Let x be any object in (A ∩ B) ∪ (A ∩ C).
Then either x ∈ A ∩ B or x ∈ (A ∩ C).
Suppose x ∈ A ∩ B, i.e., x ∈ A and x ∈ B.
Then x ∈ A and either x ∈ B or x ∈ C, i.e., x ∈ A ∩ (B ∪ C).
Suppose x ∈ A ∩ C, i.e., x ∈ A and x ∈ C.
Then x ∈ A and either x ∈ B or x ∈ C, i.e., x ∈ A ∩ (B ∪ C).
Thus in either case x ∈ A ∩ (B ∪ C).
Therefore (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C).

By Extensionality, the required equality follows.
Finally, a third form of proof: using the definitions, the distribution law reduces,
as in the above, to the equivalencex ∈ A ∧ (x ∈ B ∨ x ∈ C) ⇐⇒ (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C),

4.4. ALGEBRA OF SETS

131

which involves the three statements x ∈ A, x ∈ B and x ∈ C. That this comesout true always can be checked using an 8-line truth table in the usual way.
Exercise 4.17 A, B and C are sets. Show:
1. A 6⊆ B ⇔ A − B 6= ∅.
2. A ∩ B = A − (A − B).

Example 4.18 We show that A − C ⊆ (A − B) ∪ (B − C).
Given: x ∈ (A − C).
To be proved: x ∈ (A − B) ∨ x ∈ (B − C).
Proof:
Suppose x ∈ B. From x ∈ (A − C) we get that x ∈/ C, so x ∈ (B − C).
Therefore x ∈ (A − B) ∨ x ∈ (B − C).
Suppose x ∈/ B. From x ∈ (A − C) we get that x ∈ A, so x ∈ (A − B).
Therefore x ∈ (A − B) ∨ x ∈ (B − C).

Thus x ∈ (A − B) ∨ x ∈ (B − C).

Exercise 4.19 Express (A ∪ B) ∩ (C ∪ D) as a union of four intersections.
Complement. Fix a set X, of which all sets to be considered are subsets. The
complement Ac of a set A ⊆ X is now defined by
Ac := X − A.
Clearly, we have that for all x ∈ X:
x ∈ Ac ⇔ x 6∈ A.
Theorem 4.20
1. (Ac )c = A;
X c = ∅;
∅c = X,CHAPTER 4. SETS, TYPES AND LISTS

132
2. A ∪ Ac = X;A ∩ Ac = ∅,3. A ⊆ B ⇔ B c ⊆ Ac ,4. (A ∪ B)c = Ac ∩ B c ;(A ∩ B)c = Ac ∪ B c(DeMorgan laws).

Figure 4.3: Symmetric set difference.

Symmetric Difference. The symmetric difference of two sets A and B, notation
A ⊕ B, is the set given by {x | x ∈ A ⊕ x ∈ B}. This is the set of all objects thatare either in A or in B, but not in both.
Exercise 4.21 Show that A ⊕ B = (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).

{1, 2, 3}
{1, 2} {2, 3} {1, 3}
{1}
{2}
{3} 





∅





Figure 4.4: The power set of {1, 2, 3}.
Definition 4.22 (Power Set) The powerset of the set X is the set ℘(X) = {A|A ⊆X } of all subsets of X.
By Theorem 4.9 and 4.1.1 we have that ∅ ∈ ℘(X) and X ∈ ℘(X). So, forinstance, ℘({∅, 1}) = {∅, {∅}, {1}, {∅, 1}}. Note that X ∈ ℘(A) ⇔ X ⊆ A.

4.4. ALGEBRA OF SETS

133

Exercise 4.23 Let X be a set with at least two elements. Then by Theorem (4.1),
the relation ⊆ on ℘(X) has the properties of reflexivity, antisymmetry and transitivity. The relation 6 on R also has these properties. The relation 6 on R has thefurther property of linearity: for all x, y ∈ R, either x 6 y or y 6 x. Show that ⊆on ℘(X) lacks this property.

Figure 4.5: Generalized set union and intersection.
Definition 4.24 (Generalized Union and Intersection) Suppose that a set Ai has
been given for every element i of a set I.
1. The union of the sets Ai is the set {x | ∃i ∈ I(x ∈ Ai )}.
S
Notation: i∈I Ai .

2. The intersection of the sets Ai is the set {x | ∀i ∈ I(x ∈ Ai )}.
T
Notation: i∈I Ai

S
If the elements of I are sets themselves, and Ai = i (i ∈ I), then i∈I i is called
the union of I;
TSTThe short notation for this set is I. Similarly, i∈I i is written as I.
S
T
In the case that I = N, i∈I Ai and i∈I Ai can also be written as A0 ∪ A1 ∪A2 ∪ · · · , resp., A0 ∩ A1 ∩ A2 ∩ · · · .
A set of sets isSsometimes
T called a family of sets or a collection of sets. If F is afamily of sets, F and F are sets. For example, if
thenS

F = {{1, 2, 3}, {2, 3, 4}, {3, 4, 5}},T
F = {1, 2, 3, 4, 5} and F = {3}.

CHAPTER 4. SETS, TYPES AND LISTS

134

Example 4.25 For p ∈ N, let Ap = {mp | m ∈ N, m > 1}. Then Ap is the set of
all natural numbers that have p as a factor.
S
α β γ δi∈{2,3,5,7} Ai is the set of all natural numbers of the form n · 2 3 5 7 , with atleast one of α, β, γ, δ > 0.
T
α β γ δi∈{2,3,5,7} Ai is the set of all natural numbers of the form n · 2 3 5 7 , with allof α, β, γ, δ > 0, which is the set A210 .
S
Let F and G be collections of sets. To check the truth of statements such as F ⊆T
G it is often useful to analyze their logical form by meansa translation in
S ofTterms of quantifiers and the relation ∈. The translation of F ⊆ G becomes:∀x(∃y(y ∈ F ∧ x ∈ y) ⇒ ∀z(z ∈ G ⇒ x ∈ z)).

Exercise 4.26 Give a logical translation of

T

F⊆

S

G using only the relation ∈.

Types of Generalized Union and Intersection Again, it is useful to consider thetypes of the operations of generalized union and intersection. These operations
take families of sets and produce sets, so their type is {s} → s, where {s} is thetype of a family of sets.
Exercise 4.27 Let F be a family of sets. Show that there is a set A with thefollowing properties:
1. F ⊆ ℘(A),
2. For all sets B: if F ⊆ ℘(B) then A ⊆ B.

T
S
Remark. If I = ∅, then i∈I Ai = ∅, and i∈I Ai is the collection of all sets.
This last fact is an example of a trivially true implication: if I = ∅, then every
statement i ∈ I is false, hence the implication i ∈ I ⇒ x ∈ Ai true, andeverything is member of{x | ∀i ∈ I(x ∈ Ai )} = {x | ∀i(i ∈ I ⇒ x ∈ Ai )}.
Therefore, the notation

Ti∈I Ai usually presupposes that I 6= ∅.

4.4. ALGEBRA OF SETS

135

Example 4.28 For x ∈ X, we have that:x ∈ (A ∪ B)c⇔⇔¬(x ∈ A ∪ B)¬(x ∈ A ∨ x ∈ B)⇔

x ∈ Ac ∩ B c .

∗⇔⇔¬x ∈ A ∧ ¬x ∈ B
x ∈ Ac ∧ x ∈ B c

Step (∗) is justified by propositional reasoning. See 2.10.9. Extensionality allows
us to conclude the first DeMorgan law:(A ∪ B)c = Ac ∩ B c .

Exercise 4.29 Prove the rest of Theorem 4.20.
Exercise 4.30 Answer as many of the following questions as you can.
1. Determine: ℘(∅), ℘(℘(∅)) and ℘(℘(℘(∅))).
2. How many elements has ℘5 (∅) = ℘(℘(℘(℘(℘(∅)))))?
3. How many elements has ℘(A), given that A has n elements?

Exercise 4.31 Check whether the following is true: if two sets have the same
subsets, then they are equal. I.e.: if ℘(A) = ℘(B), then A = B. Give a proof or arefutation by means of a counterexample.
Exercise 4.32 Is it true that for all sets A and B:
1. ℘(A ∩ B) = ℘(A) ∩ ℘(B)?
2. ℘(A ∪ B) = ℘(A) ∪ ℘(B)?
Provide either a proof or a refutation by counter-example.
Exercise 4.33* Show:
1. B ∩ (Si∈I Ai ) =S

i∈I (B ∩ Ai ),136

CHAPTER 4. SETS, TYPES AND LISTS

T
T
2. B ∪ ( i∈I Ai ) = i∈I (B ∪ Ai ),S
T
3. ( i∈I Ai )c = i∈I Aci , assuming that ∀i ∈ I Ai ⊆ X,
S
T
4. ( i∈I Ai )c = i∈I Aci , assuming that ∀i ∈ I Ai ⊆ X.
Exercise 4.34* Assume that you are given a certain set A0 . Suppose you are assigned the task of finding sets A1 , A2 , A3 , . . . such that ℘(A1 ) ⊆ A0 , ℘(A2 ) ⊆A1 , ℘(A3 ) ⊆ A2 ,. . . Show that no matter how hard you try, you will eventually
fail, that is: hit a set An for which no An+1 exists such that ℘(An+1 ) ⊆ An . (I.e.,∅ 6∈ An .)T
Hint. Suppose you can go on forever. Show this would entail ℘( i∈N Ai ) ⊆T
i∈N Ai . Apply Exercise 4.7.
Exercise 4.35* Suppose that the collection K of sets satisfies the following condition:∀A ∈ K(A = ∅ ∨ ∃B ∈ K(A = ℘(B))).

Show that every element of K has the form ℘n (∅) for some n ∈ N. (N.B.: ℘0 (∅) =∅.)4.5 Ordered Pairs and Products
Next to the unordered pairs {a, b} of Section 4.3, in which the order of a and b isimmaterial ({a, b} = {b, a}), there are ordered pairs in which order does count.
The ordered pair of objects a and b is denoted by(a, b) .
Here, a is the first and b the second coordinate of (a, b).
Ordered pairs behave according to the following rule:(a, b) = (x, y) =⇒ a = x ∧ b = y.

(4.2)

This means that the ordered pair of a and b fixes the objects as well as their order.
Its behaviour differs from that of the unordered pair: we always have that {a, b} ={b, a}, whereas (a, b) = (b, a) only holds — according to (4.2) — when a = b.
Warning. If a and b are reals, the notation (a, b) also may denote the open interval{x ∈ R | a < x < b}. The context should tell you which of the two is meant.

4.5. PAIRS AND PRODUCTS

137

Defining Ordered Pairs. Defining(a, b) = {{a}, {a, b}}allows you to prove (4.2). Cf. Exercise 4.41.
Definition 4.36 (Products) The (Cartesian) product of the sets A and B is the setof all pairs (a, b) where a ∈ A and b ∈ B. In symbols:
A × B = { (a, b) | a ∈ A ∧ b ∈ B }.
Instead of A × A one usually writes A2 .
Example 4.37 {0, 1} × {1, 2, 3} = {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)}.
When A and B are real intervals on the X resp., the Y -axis in two-dimensional
space, A × B can be pictured as a rectangle.

Ran(R)R

Dom(R)
Theorem 4.38 For arbitrary sets A, B, C, D the following hold:
1. (A × B) ∩ (C × D) = (A × D) ∩ (C × B),
2. (A ∪ B) × C = (A × C) ∪ (B × C); (A ∩ B) × C = (A × C) ∩ (B × C),
3. (A ∩ B) × (C ∩ D) = (A × C) ∩ (B × D),
4. (A ∪ B) × (C ∪ D) = (A × C) ∪ (A × D) ∪ (B × C) ∪ (B × D),
5. [(A − C) × B] ∪ [A × (B − D)] ⊆ (A × B) − (C × D).

138

CHAPTER 4. SETS, TYPES AND LISTSProof. As an example, we prove the first item of part 2.
To be proved: (A ∪ B) × C = (A × C) ∪ (B × C):⊆:
Suppose that p ∈ (A ∪ B) × C. Then a ∈ A ∪ B and c ∈ C exist such that p = (a, c).
Thus (i) a ∈ A or (ii) a ∈ B.
(i). In this case, p ∈ A × C, and hence p ∈ (A × C) ∪ (B × C).
(ii). Now p ∈ B × C, and hence again p ∈ (A × C) ∪ (B × C).
Thus p ∈ (A × C) ∪ (B × C).
Therefore, (A ∪ B) × C ⊆ (A × C) ∪ (B × C).
⊆:

Conversely, assume that p ∈ (A × C) ∪ (B × C).
Thus (i) p ∈ A × C or (ii) p ∈ B × C.
(i). In this case a ∈ A and c ∈ C exist such that p = (a, c);a fortiori, a ∈ A ∪ B and hence p ∈ (A ∪ B) × C.
(ii). Now b ∈ B and c ∈ C exist such that p = (b, c);a fortiori b ∈ A ∪ B and hence, again, p ∈ (A ∪ B) × C.
Thus p ∈ (A ∪ B) × C.
Therefore, (A × C) ∪ (B × C) ⊆ (A ∪ B) × C.
The required result follows using Extensionality.

Exercise 4.39 Prove the other items of Theorem 4.38.
Exercise 4.40
1. Assume that A and B are non-empty and that A × B =B × A. Show that A = B.
2. Show by means of an example that the condition of non-emptiness in 1 isnecessary. (Did you use this in your proof of 1?)

Exercise 4.41* To show that defining (a, b) as {{a}, {a, b}} works, prove that
1. {a, b} = {a, c} =⇒ b = c,
2. {{a}, {a, b}} = {{x}, {x, y}} =⇒ a = x ∧ b = y.

4.6. LISTS AND LIST OPERATIONS

139

If we assume the property of ordered pairs (a, b) = (x, y) =⇒ a = x ∧ b = y, we
can define triples by (a, b, c) := (a, (b, c)). For suppose (a, b, c) = (x, y, z). Thenby definition, (a, (b, c)) = (a, b, c) = (x, y, z) = (x, (y, z)), and by the propertyof ordered pairs, we have that a = x and (b, c) = (y, z). Again, by the propertyof ordered pairs, b = y and c = z. This shows that (a, b, c) = (x, y, z) ⇒ (a =x ∧ b = y ∧ c = z).
Let us go about this more systematically, by defining ordered n-tuples over somebase set A, for every n ∈ N. We proceed by recursion.
Definition 4.42 (n-tuples over A)
1. A0 := {∅},
2. An+1 := A × An .
In Haskell, ordered pairs are written as (x1,x2), and there are predefined functions fst to get at the first member and snd to get at the second member. Ordered
triples are written as (x1, x2, x3), and so on. If x1 has type a and x2 has typeb, then (x1,x2) has type (a,b). Think of this type as the product of the types forx1 and x2. Here is an example:
Prelude> :t (1, ’A’)
(1,’A’) :: Num a => (a,Char)
Prelude>

4.6 Lists and List Operations
AssumingSthe list elements are all taken from a set A, the set of all lists over A
is the set n∈N An . We abbreviate this set as A∗ . For every list L ∈ A∗ there issome n ∈ N with L ∈ An . If L ∈ An we say that list L has length n.

Standard notation for the (one and only) list of length 0 is []. A list of length n > 0
looks like (x0 , (x1 , (· · · , xn−1 ) · · · )). This is often written as [x0 , x1 , . . . , xn−1 ].
A one element list is a list of the form (x, []). In line with the above square bracket
notation, this is written as [x].
If one uses lists to represent sets, there is a difficulty. The list [a, b, b] is different
from the list [a, b], for [a, b, b] has length 3, and [a, b] has length 2, but the sets{a, b, b} and {a, b} are identical. This shows that sets and lists have different
identity conditions.
This is how the data type of lists is (pre-)declared in Haskell:

140

CHAPTER 4. SETS, TYPES AND LISTS

data [a] = [] | a : [a] deriving (Eq, Ord)

To grasp what this means, recall that in functional programming every set has a
type. The data declaration for the type of lists over type a, notation [a] specifies
that lists over type a are either empty or they consist of an element of type a put infront of a list over type a. Haskell uses : for the operation of putting an element infront of a list.
The operation : combines an object with a list of objects of the same type to forma new list of objects of that type. This is reflected by the type judgment:
Prelude> :t (:)(:) :: a -> [a] -> [a]The data declaration for lists also tells us, by means of deriving (Eq,Ord), that
if equality is defined on the objects of type a (i.e., if the type a belongs to the classEq) then this relation carries over to lists over type a, and if the objects of type aare ordered, i.e., if the type a belongs to the class Ord (e.g., the objects of type Z
are ordered by <), then this order carries over to lists over type a.
If we have an equality test for members of the set A, then it is easy to see howequality can be defined for lists over A (in fact, this is all taken care of by thepredefined operations on lists). Lists are ordered sets, so two lists are the same if(i) they either are both empty, or (ii) they start with the same element (here is where
the equality notion for elements from A comes in), and their tails are the same. In
Haskell, this is implemented as follows (this is a definition from Prelude.hs):

instance Eq a => Eq [a] where[]== []= True(x:xs) == (y:ys) = x==y && xs==ys_== _= FalseThis says that if a is an instance of class Eq, then [a] is so too. This specifies how
the equality notion for objects of type a carries over to objects of type [a] (i.e.,
lists over that type). As we have seen, Haskell uses == for equality and = for thedefinition of operator values. Part (i) of the definition of list equality is taken careof by the statement that [] == [] is true, part (ii) is covered by:4.6. LISTS AND LIST OPERATIONS(x:xs) == (y:ys) =141

x==y && xs==ys

This says: the truth value of (x:xs) == (y:ys) (equality of two lists that areboth non-empty) is given by the conjunction of• the first elements are the same (x == y), and• the remainders are the same (xs == ys).
The final line _ == _ = False states that in all other cases the lists are not equal.
This final line uses _ as a ‘don’t care pattern’ or wild card.
Exercise 4.43 How does it follow from this definition that lists of different length
are unequal?
A type of class Ord is a type on which the two-placed operation compare is defined, with a result of type Ordering. The type Ordering is the set {LT, EQ, GT},
where LT, EQ, GT have the obvious meanings. Suppose a is a type of class Ord,i.e., compare is defined on the type a. What does a reasonable ordering of lists
over type a look like? A well-known order on lists is the lexicographical order: theway words are ordered in a dictionary. In this ordering, the empty list comes first.
For non-empty lists L1 , L2 , we compare their first elements, using the function
compare for objects of type a. If these are the same, the order is determined by
the remainders of the lists. If the first element of L1 comes before the first elementof L2 then L1 comes before L2 , otherwise L2 comes before L1 . The following
piece of Haskell code implements this idea, by defining the function compare forlists over type a.

instance Ord a => Ord [a] wherecompare [](_:_) = LTcompare [][]= EQcompare (_:_) []= GTcompare (x:xs) (y:ys) = primCompAux x y (compare xs ys)

This specifies how the ordering on type a carries over to an ordering of lists overtype a. The first line says that the empty list [] is less than any non-empty list.

CHAPTER 4. SETS, TYPES AND LISTS

142

The second line says that the empty list is equal to itself. The third line says thatany non-empty list is greater than the empty list. This fully determines the relation
between [] and any list. The last line uses an auxiliary function primCompAux tocover the case of two non-empty lists. This function is defined by:primCompAux:: Ord a => a -> a -> Ordering -> Ordering
primCompAux x y o =case compare x y of EQ -> o;LT -> LT;GT -> GTThe type declaration of primCompAux says that if a is an ordered type, thenprimCompAux expects three arguments, the first one of type a, the second oneof type a and the third one an element of the set (or type) Ordering, i.e., the set{LT, EQ, GT}. The result is again a member of the type Ordering.
The definition of the operation primCompAux uses a case construction, using thereserved keywords case and of, and the arrow -> to point at the results for thevarious cases. . It says that in case the first two arguments are equal, the function
returns the element of Ordering which is its third argument, in case the first argument is less than the second argument, the function returns LT, in case the firstargument is greater than the second argument, the function returns GT.
Exercise 4.44 Another ordering of lists is as follows: shorter lists come before
longer ones, and for lists of the same length we compare their first elements, andif these are the same, the remainder lists. Give a formal definition of this ordering.
How would you implement it in Haskell?
In list processing, fundamental operations are checking whether a list is empty,
accessing the first element of a non-empty list, determining what remains after
removal of the first element from a list (its tail). The operations for accessing thehead or the tail of a list are implemented in Haskell as follows:headhead (x:_):: [a] -> a
= x

tail
tail (_:xs):: [a] -> [a]= xs

4.6. LISTS AND LIST OPERATIONS

143

The type of the operation head: give it a list over type a and the operation will
return an element of the same type a. This is specified in head :: [a] -> a.
The type of the operation tail is: give it a list over type a and it will return a list
over the same type. This is specified in tail :: [a] -> [a].
Note that these operations are only defined on non-empty lists. (x:_) specifies
the pattern of a non-empty list with first element x, where the nature of the tail is
irrelevant. (_:xs) specifies the pattern of a non-empty list with tail xs, where thenature of the first element is irrelevant.
Accessing the last element of a non-empty list is done by means of recursion: thelast element of a unit list [x] is equal to [x]. The last element of a non-unit list isequal to the last element of its tail. Here is the Haskell implementation:

last
last [x]last (_:xs):: [a] -> a
= x
= last xs

Note that because the list patterns [x] and (_:xs) are tried for in that order, thepattern (_:xs) in this definition matches non-empty lists that are not unit lists.
Exercise 4.45 Which operation on lists is specified by the Haskell definition in
the frame below?

init
init [x]init (x:xs):: [a] -> [a]= []= x : init xs

It is often useful to be able to test whether a list is empty or not. The following
operation accomplishes this:null
null []
null (_:_):: [a] -> Bool
= True= False144

CHAPTER 4. SETS, TYPES AND LISTS

Exercise 4.46 Write your own definition of a Haskell operation reverse that reverses a list.
Exercise 4.47 Write a function splitList that gives all the ways to split a list of
at least two elements in two non-empty parts. The type declaration is:
splitList :: [a] -> [([a],[a])]The call splitList [1..4] should give:
STAL> splitList [1..4][([1],[2,3,4]),([1,2],[3,4]),([1,2,3],[4])]STAL>]
An operation on lists that we will need in the next sections is the operation of removing duplicates. This is predefined in the Haskell module List.hs as nub (‘nub’
means essence), but here is a home-made version for illustration:

nub :: (Eq a) => [a] -> [a]nub [] = []nub (x:xs) = x : nub (remove x xs)
where
remove y []= []remove y (z:zs) | y == z
= remove y zs
| otherwise = z : remove y zs

What this says is, first, that if a is any type for which a relation of equality isdefined, then nub operates on a list over type a and returns a list over type a.
In Haskell, strings of characters are represented as lists, and the shorthand "abc"is allowed for [’a’,’b’,’c’]. Here is an example of an application of nub to astring:
STAL> nub "Mississippy""Mispy"Of course, we can also use nub on lists of words:
STAL> nub ["Quentin Tarantino","Harrison Ford","Quentin Tarantino"]["Quentin Tarantino","Harrison Ford"]

4.7. LIST COMPREHENSION AND DATABASE QUERY

145

4.7 List Comprehension and Database QueryTo get more familiar with list comprehensions, we will devote this section to list
comprehension for database query, using the movie database module DB.hs given
in Figure 4.6. The database that gets listed here is called db, with type DB, where
DB is a synonym for the type [WordList], where Wordlist is again a synonym
for the type [String]. The reserved keyword type is used to declare these typesynonyms. Notice the difference between defining a type synonym with type anddeclaring a new data type with data.
The database can be used to define the following lists of database objects, with list
comprehension. Here db :: DB is the database list.

characters = nub [ x| ["play",_,_,x] <- db ]movies=[ x| ["release",x,_] <- db ]actors= nub [ x| ["play",x,_,_] <- db ]directors = nub [ x| ["direct",x,_] <- db ]dates
= nub [ x
| ["release",_,x] <- db ]universe= nub (characters++actors++directors++movies++dates)

Next, define lists of tuples, again by list comprehension:

directact
playrelease= [ (x,y)| ["direct",x,y] <- db ]= [ (x,y)| ["play",x,y,_] <- db ]= [ (x,y,z) | ["play",x,y,z] <- db ]= [ (x,y)| ["release",x,y] <- db ]Finally, define one placed, two placed and three placed predicates by means oflambda abstraction.

CHAPTER 4. SETS, TYPES AND LISTS

146

module DB
wheretype WordList = [String]type DB= [WordList]
db :: DB
db = [["release", "Blade Runner", "1982"],["release", "Alien", "1979"],["release", "Titanic", "1997"],["release", "Good Will Hunting", "1997"],["release", "Pulp Fiction", "1994"],["release", "Reservoir Dogs", "1992"],["release", "Romeo and Juliet", "1996"],{- ... -}["direct", "Brian De Palma", "The Untouchables"],["direct", "James Cameron", "Titanic"],["direct", "James Cameron", "Aliens"],["direct", "Ridley Scott", "Alien"],["direct", "Ridley Scott", "Blade Runner"],["direct", "Ridley Scott", "Thelma and Louise"],["direct", "Gus Van Sant", "Good Will Hunting"],["direct", "Quentin Tarantino", "Pulp Fiction"],{- ... -}["play", "Leonardo DiCaprio", "Romeo and Juliet", "Romeo"],["play", "Leonardo DiCaprio", "Titanic", "Jack Dawson"],["play", "Robin Williams", "Good Will Hunting", "Sean McGuire"],["play", "John Travolta", "Pulp Fiction", "Vincent Vega"],["play", "Harvey Keitel", "Reservoir Dogs", "Mr White"],["play", "Harvey Keitel", "Pulp Fiction", "Winston Wolf"],["play", "Uma Thurman", "Pulp Fiction", "Mia"],["play", "Quentin Tarantino", "Pulp Fiction", "Jimmie"],["play", "Quentin Tarantino", "Reservoir Dogs", "Mr Brown"],["play", "Sigourney Weaver", "Alien", "Ellen Ripley"],{- ... -}Figure 4.6: A Database Module.

4.7. LIST COMPREHENSION AND DATABASE QUERY

charP
actorP
movieP
directorP
dateP
actP
releaseP
directP
playP

147

= \ x
-> elem x characters= \ x-> elem x actors= \ x-> elem x movies= \ x-> elem x directors= \ x-> elem x dates= \ (x,y)-> elem (x,y) act= \ (x,y)-> elem (x,y) release= \ (x,y)-> elem (x,y) direct= \ (x,y,z) -> elem (x,y,z) play

We start with some conjunctive queries. ‘Give me the actors that also are directors.’

q1 = [ x | x <- actors, directorP x ]‘Give me all actors that also are directors, together with the films in which they
were acting.’

q2 = [ (x,y) | (x,y) <- act, directorP x ]’Give me all directors together with their films and their release dates.’ The following is wrong.

q3 = [ (x,y,z) | (x,y) <- direct, (y,z) <- release ]The problem is that the two ys are unrelated. In fact, this query generates aninfinite list. This can be remedied by using the equality predicate as a link:q4 = [ (x,y,z) | (x,y) <- direct, (u,z) <- release, y == u ]148

CHAPTER 4. SETS, TYPES AND LISTS‘Give me all directors of films released in 1995, together with these films.’

q5 = [ (x,y) | (x,y) <- direct, (u,"1995") <- release, y == u ]‘Give me all directors of films released after 1995, together with these films andtheir release dates.’

q6 = [ (x,y,z) | (x,y) <- direct, (u,z) <- release,y == u, z > "1995"]‘Give me the films in which Kevin Spacey acted.’

q7 = [ x | ("Kevin Spacey",x) <- act ]‘Give me all films released after 1997 in which William Hurt did act.’

q8 = [ x | (x,y) <- release, y > "1997", actP ("William Hurt",x) ]Yes/no queries based on conjunctive querying: ‘Are there any films in which thedirector was also an actor?’

q9 = q1 /= []‘Does the database contain films directed by Woody Allen?’q10 = [ x | ("Woody Allen",x) <- direct ] /= []4.8. USING LISTS TO REPRESENT SETS

149

Or simply:

q10’ = directorP "Woody Allen"

Disjunctive and negative queries are also easily expressed, since we have predicates and Boolean operators.
Exercise 4.48 Translate the following into a query: ‘Give me the films in which
Robert De Niro or Kevin Spacey acted.’
Exercise 4.49 Translate the following into a query: ‘Give me all films with Quentin
Tarantino as actor or director that appeared in 1994.’
Exercise 4.50 Translate the following into a query: ‘Give me all films released
after 1997 in which William Hurt did not act.’

4.8 Using Lists to Represent Sets
Sets are unordered, lists are ordered, but we can use lists to represent finite (orcountably infinite) sets by representing sets as lists with duplicates removed, andby disregarding the order. If a finite list does not contain duplicates, its length
gives the size of the finite set that it represents.
Even if we gloss over the presence of duplicates, there are limitations to the representation of sets by lists. Such representation only works if the sets to be represented are small enough. In Chapter 11 we will return to this issue.
To removing an element from a list without duplicates all we have to do is remove
the first occurrence of the element from the list. This is done by the predefined
function delete, also part of the Haskell module List.hs. Here is our home-madeversion:delete :: Eq a => a -> [a] -> [a]delete x [] = []delete x (y:ys) | x == y= ys| otherwise = y : delete x ys

150

CHAPTER 4. SETS, TYPES AND LISTS

As we have seen, the operation of elem for finding elements is built in. Here isour demo version, re-baptized elem’ to avoid a clash with Prelude.hs.

elem’ :: Eq a => a -> [a] -> Boolelem’ x []= Falseelem’ x (y:ys) | x == y
= True| otherwise = elem’ x ys

Further operations on sets that we need to implement are union, intersection anddifference. These are all built into the Haskell module List.hs. Our version ofunion:

union :: Eq a => [a] -> [a] -> [a]union [] ys= ys
union (x:xs) ys = x : union xs (delete x ys)

Note that if this function is called with arguments that themselves do not contain
duplicates then it will not introduce new duplicates.
Here is an operation for taking intersections:intersect :: Eq a => [a] -> [a] -> [a]intersect []s= []intersect (x:xs) s | elem x s= x : intersect xs s
| otherwise =intersect xs s

Note that because the definitions of union and intersect contain calls to delete
or elem they presuppose that the type a has an equality relation defined on it. Thisis reflected in the type declarations for the operations.
Exercise 4.51 The Haskell operation for list difference is predefined as \\ inList.hs. Write your own version of this.

4.8. USING LISTS TO REPRESENT SETS

151

The predefined versions of the functions elem and notElem for testing whether
an object occurs in a list or not use the functions any and all:elem, notElem
elem
notElem:: Eq a => a -> [a] -> Bool
= any . (==)= all . (/=)

Be cautious with this: elem 0 [1..] will run forever.
Let’s turn to an operation for the list of all sublists of a given list. For this, we firstneed to define how one adds a new object to each of a number of lists. This canbe done with the Haskell function map . Adding an element x to a list l that does
not contain x is just a matter of applying the function (x:) (prefixing the elementx) to l. Therefore, the following simple definition gives an operation that adds anelement x to each list in a list of lists.

addElem :: a -> [[a]] -> [[a]]addElem x = map (x:)

Note the type declaration: [[a]] denotes the type of lists over the type of lists
over type a.
The operation addElem is used implicitly in the operation for generating the sublists of a list:

powerList
powerList
powerList:: [a] -> [[a]][] = [[]](x:xs) = (powerList xs) ++ (map (x:) (powerList xs))

Here is the function in action:
STAL> powerList [1,2,3][[], [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]]

152

CHAPTER 4. SETS, TYPES AND LISTSExample 4.52 For a connection with Exercise 4.30, let us try to fool Haskell into
generating the list counterparts of ℘2 (∅) = ℘({∅}), ℘3 (∅) and ℘4 (∅). Some fooling is necessary, because Haskell is much less flexible than set theory about types.
In set theory, we have: ∅ :: s, {∅} :: s, {{∅}} :: s, for these are all sets. In Haskell,the type of the empty list is polymorphic, i.e., we have [] :: [a]. Thus, the list
counterpart of ℘2 (∅) = {∅, {∅}}, can be consistently typed:
STAL> :t [[],[[]]][[],[[]]] :: [[[a]]]What happens is that the first occurrence of [] gets type [[a]], and the second
occurrence type [a], so that [[],[[]]] is a list containing two objects of thesame type [[a]], and is itself of type [[[a]]]. However, Haskell refuses todisplay a list of the generic type [a]. The empty list will only be displayed in asetting where it is clear what the type is of the objects that are being listed:
STAL> [ x | x <- [1..10], x < 0 ][]This is OK, for it is clear from the context that [] is an empty list of numbers.
STAL> []ERROR: Cannot find "show" function for:*** Expression : []*** Of type
: [a]STAL> [[],[[]]]ERROR: Cannot find "show" function for:*** Expression : [[],[[]]]*** Of type
: [[[a]]]These are not OK, for the context does not make clear what the types are.
The following data type declaration introduces a data type S containing a single object Void. You should think of this as something very mysterious, for the
Void is the unfathomable source of an infinite list-theoretic universe built fromempty :: [S]. Void is used only to provide empty with a type.

data S = Void deriving (Eq,Show)empty :: [S]empty = []

4.9. A DATA TYPE FOR SETS

153

Here are the first stages of the list universe:
STAL> powerList empty[[]]
STAL> powerList (powerList empty)[[],[[]]]
STAL> powerList (powerList (powerList empty))[[],[[[]]],[[]],[[],[[]]]]
STAL> powerList (powerList (powerList (powerList empty)))[[],[[[],[[]]]],[[[]]],[[[]],[[],[[]]]],[[[[]]]],[[[[]]],[[],[[]]]],[[[[]]],[[]]],[[[[]]],[[]],[[],[[]]]],[[]],[[],[[],[[]]]],[[],[[]]],[[],[[]],[[],[[]]]],[[],[[[]]]],[[],[[[]]],[[],[[]]]],[[],[[[]]],[[]]],[[],[[[]]],[[]],[[],[[]]]]]STAL>

Exercise 4.53 Write functions genUnion and genIntersect for generalized list
union and list intersection. The functions should be of type [[a]]-> [a]. They
take a list of lists as input and produce a list as output. Note that genIntersect isundefined on the empty list of lists (compare the remark about the presuppositionof generalized intersection on page 134).

4.9 A Data Type for Sets
The representation of sets as lists without duplicates has the drawback that two
finite lists containing the same elements, but in a different order, e.g., [1,2,3]
and [3,2,1], are unequal as lists, but equal as sets. The Haskell equality operator== gives the wrong results when we are interested in set equality. For instance, we
get:Prelude> [1,2,3] == [3,2,1]
False
Prelude>This can be remedied by defining a special data type for sets, with a matching
definition of equality. All this is provided in the module SetEq.hs that is given in
Figs. 4.7 and 4.8. If a is an equality type, then Set a is the type of sets over a.
The newtype declaration allows us to put Set a as a separate type in the Haskell
type system.
Equality for Set a is defined in terms of the subSet relation:

CHAPTER 4. SETS, TYPES AND LISTS

154

module SetEq (Set(..),emptySet,isEmpty,inSet,subSet,insertSet,
deleteSet,powerSet,takeSet,list2set,(!!!))
whereimport List (delete)infixl 9 !!!
newtype Set a = Set [a]instance Eq a => Eq (Set a) whereset1 == set2 = subSet set1 set2 && subSet set2 set1
instance (Show a) => Show (Set a) whereshowsPrec _ (Set s) str = showSet s str
showSet []str = showString "{}" strshowSet (x:xs) str = showChar ’{’ (shows x (showl xs str))
where showl []str = showChar ’}’ str
showl (x:xs) str = showChar ’,’ (shows x (showl xs str))
emptySet :: Set a
emptySet = Set []isEmpty :: Set a -> BoolisEmpty (Set []) = True
isEmpty _= FalseinSet :: (Eq a) => a -> Set a -> BoolinSet x (Set s) = elem x s
subSet :: (Eq a) => Set a -> Set a -> Bool
subSet (Set [])_= TruesubSet (Set (x:xs)) set = (inSet x set) && subSet (Set xs) set
insertSet :: (Eq a) => a -> Set a -> Set a
insertSet x (Set ys) | inSet x (Set ys) = Set ys| otherwise= Set (x:ys)

Figure 4.7: A Module for Sets as Unordered Lists Without Duplicates.

4.9. A DATA TYPE FOR SETS

155

deleteSet :: Eq a => a -> Set a -> Set a
deleteSet x (Set xs) = Set (delete x xs)
list2set :: Eq a => [a] -> Set alist2set [] = Set []list2set (x:xs) = insertSet x (list2set xs)
powerSet :: Eq a => Set a -> Set (Set a)
powerSet (Set xs) = Set (map (\xs -> (Set xs)) (powerList xs))
powerList
powerList
powerList:: [a] -> [[a]][] = [[]](x:xs) = (powerList xs) ++ (map (x:) (powerList xs))

takeSet :: Eq a => Int -> Set a -> Set a
takeSet n (Set xs) = Set (take n xs)(!!!) :: Eq a => Set a -> Int -> a(Set xs) !!! n = xs !! n

Figure 4.8: A Module for Sets as Unordered Lists Without Duplicates (ctd).

instance Eq a => Eq (Set a) whereset1 == set2 = subSet set1 set2 && subSet set2 set1

The instance declaration says that if a is in class Eq, then Set a is also in thatclass, with == defined as specified, in terms of the procedure subSet, that in turnis defined recursively in terms of inSet. This gives:
SetEq> Set [2,3,1] == Set [1,2,3]
True
SetEq> Set [2,3,3,1,1,1] == Set [1,2,3]
True
SetEq>The module SetEq.hs gives some useful functions for the Set type. In the firstplace, it is convenient to be able to display sets in the usual notation.

CHAPTER 4. SETS, TYPES AND LISTS

156

instance (Show a) => Show (Set a) whereshowsPrec _ (Set s) = showSet s
showSet []str = showString "{}" strshowSet (x:xs) str = showChar ’{’ ( shows x ( showl xs str))
where showl []str = showChar ’}’ str
showl (x:xs) str = showChar ’,’ (shows x (showl xs str))

This gives:
SetEq> Set [1..10]{1,2,3,4,5,6,7,8,9,10}
SetEq> powerSet (Set [1..3]){{},{3},{2},{2,3},{1},{1,3},{1,2},{1,2,3}}
SetEq>The empty set emptySet and the test isEmpty are implemented as you would
expect. Useful functions for operating on sets are insertSet and deleteSet.
The function insertSet is used to implement the translation function from lists
to sets list2set. The function powerSet is implemented by lifting powerList
to the type of sets.
Exercise 4.54 Give implementations of the operations
unionSet, intersectSet and differenceSet,in terms of inSet, insertSet and deleteSet.
Exercise 4.55 In an implementation of sets as lists without duplicates, the implementation of insertSet has to be changed. How?
Figure 4.9 gives a module for generating the first five levels of the set theoretichierarchy: V0 = ∅, V1 = ℘(∅), V2 = ℘2 (∅), V3 = ℘3 (∅), V4 = ℘4 (∅), V5 =℘5 (∅). Displaying V5 in full takes some time, but here are the first few items.
This uses the operator !!!, defined in the SetEq module as a left-associative infix
operator by means of the reserved keyword infixl (the keyword infixr can beused to declare right associative infix operators).
Hierarchy> v5 !!! 04.9. A DATA TYPE FOR SETS

157

module Hierarchy whereimport SetEq
data S = Void deriving (Eq,Show)empty,v0,v1,v2,v3,v4,v5 :: Set S
empty = Set []v0= emptyv1= powerSet v0v2= powerSet v1
v3= powerSet v2
v4= powerSet v3
v5= powerSet v4

Figure 4.9: The First Five Levels of the Set Theoretic Universe.

Hierarchy> display 88 (take 1760 (show v5)){{},{{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}},{{}}}},{{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}},{{},{{}}}}},{{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}}}},{{{},{{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}}},{{},{{{}}},{{}}}},{{{},{{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}}},{{},{{{}}},{{},{{}}}}},{{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}}}},{{{},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}}},{{},{{{}}},{{}}}},{{{},{{}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{

Figure 4.10: An initial segment of ℘5 (∅).

158

CHAPTER 4. SETS, TYPES AND LISTS{}
Hierarchy> v5 !!! 1{{{},{{{}}},{{}},{{},{{}}}}}
Hierarchy> v5 !!! 2{{{},{{{}}},{{}}}}
Hierarchy> v5 !!! 3{{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}}
Hierarchy>

Figure 4.10 displays a bit more of the initial segment of ℘5 (∅), with the help of thefollowing function:

display :: Int -> String -> IO ()display n str = putStrLn (display’ n 0 str)
where
display’ _ _ [] = []display’ n m (x:xs) | n == m
= ’\n’: display’| otherwise = x : display’n0 (x:xs)
n (m+1) xs

Exercise 4.56 What would have to change in the module SetEq.hs to get a representation of the empty set as 0?
Exercise 4.57*
1. How many pairs of curly braces {} occur in the expanded notation for ℘ 5 (∅),
in the representation where ∅ appears as {}?
2. How many copies of 0 occur in the expanded notation for ℘ 5 (∅), in the
representation where ∅ appears as 0 (Exercise 4.56)?
3. How many pairs of curly braces occur in the expanded notation for ℘ 5 (∅),
in the representation where ∅ appears as 0?

4.10 Further Reading
Russell’s paradox, as stated by Russell himself, can be found in [Rus67]. A further
introduction to sets and set theory is Doets, Van Dalen and De Swart [DvDdS78].

4.10. FURTHER READING

159

A good introduction to type theory is given in [Hin97]. There are many books on
database query an database design; a lucid introduction is [SKS01]. SQL (Standard Query Language) is explained in the documentation of all state of the art
relational databases. Implementations of the relational database model are freely
available on the Internet. See, e.g., www.postgresql.org or www.mysql.com.
Logical foundations of database theory are given in [AHV95].

160

CHAPTER 4. SETS, TYPES AND LISTSChapter 5

Relations
Preview
The first section of this chapter explains the abstract notion of a relation. Section 5.2 discusses various properties of relations. Next, in Sections 5.3 and 5.4,
we discuss various possible implementations of relations and relation processing.
Sections 5.5 and 5.6 focus on an often occurring type of relation: the equivalence.
The following declaration turns the code in this chapter into a module that loads
the List module (to be found in same directory as Prelude.hs, under the name
List.hs), and the SetOrd module (see below, in Figs. 5.3 and 5.4).

module REL
whereimport List
import SetOrd

161

CHAPTER 5. RELATIONS

162

5.1 The Notion of a Relation
Although you probably will not be able to explain the general notion of a relation,
you are definitely familiar with a couple of instances, such as the usual ordering
relation < between natural numbers, or the subset relation ⊆ between sets. Nonmathematical examples are the different family relationships that exist between
humans. For instance, the father-of relation holds between two people if the firstone is the father of the second.
For every two numbers n, m ∈ N, the statement n < m is either true or false. E.g.,
3 < 5 is true, whereas 5 < 2 is false. In general: to a relation you can “input” apair of objects, after which it “outputs” either true or false. depending on whether
these objects are in the relationship given.
In set theory, there is a clever way to reduce the notion of a relation to that of a set.
Consider again the ordering 6 on N. With it, associate the set R of ordered pairs(n, m) of natural numbers for which n 6 m is true:
R6 = { (n, m) ∈ N2 | n 6 m }.
Note that a statement n 6 m now has become tantamount with the condition, that(n, m) ∈ R6 . Thus, (3, 5) ∈ R6 , and (5, 2) 6∈ R6 .
This connection can be converted into a definition. That is, the ordering relation 6of N is identified with the set R6 . The following definition puts it bluntly.
Definition 5.1 (Relations, Domain, Range) A relation is a set of ordered pairs.
Instead of (x, y) ∈ R — where R is a relation — one usually writes xRy, orR(x, y), or Rxy.
The set dom (R) = {x | ∃y ( xRy )}, i.e., the set consisting of all first coordinatesof pairs in R, is called the domain of R and ran(R) = {y | ∃x ( xRy )}, the set of
second coordinates of pairs in R, its range.
Example 5.2 If A and B are sets, then A × B is a relation. The empty set ∅trivially is a relation (for, all its members are ordered pairs).
dom (∅) = ran(∅) = ∅, dom (A × B) = A (provided B is non-empty: A × ∅ = ∅,thus dom (A × ∅) = ∅), and ran(A × B) = B (analogously: provided A is nonempty).
Definition 5.3 (From . . . to, Between, On) The relation R is a relation from A to
B or between A and B, if dom (R) ⊆ A and ran(R) ⊆ B.
A relation from A to A is called on A.

5.1. THE NOTION OF A RELATION

163

If R is a relation on A, then A is called the underlying set (of the structure thatconsists of the domain A and the relation R).
Example 5.4 R = {(1, 4), (1, 5), (2, 5)} is a relation from {1, 2, 3} to {4, 5, 6},
and it also is a relation on {1, 2, 4, 5, 6}. Furthermore, dom (R) = {1, 2}, ran(R) ={4, 5}.
Example 5.5 If A is a set, then ⊆ is a relation on ℘(A).

2

2

1

1

0

0

-1

-1

-2

-2
-2

-1

0

1

2

-2

-1

0

1

2

Figure 5.1: The relations 6 and > on R2 .
Example 5.6 The relations 6 and > on R are subsets of the real plane R 2 , andcan be pictured as in Figure 5.1.
Definition 5.7 (Identity and Inverse)
1. ∆A = { (a, b) ∈ A2 | a = b } = { (a, a) | a ∈ A } is a relation on A, theidentity on A.
2. If R is a relation between A and B, then R −1 = { (b, a) | aRb }, the inverseof R, is a relation between B and A.
Example 5.8 The inverse of the relation ‘parent of’ is the relation ‘child of’.

CHAPTER 5. RELATIONS

164

2

2

1

1

0

0

-1

-1

-2

-2
-2

-1

0

1

2

-2

-1

0

1

2

Figure 5.2: The relations ∆R and R × R on R2 .
Example 5.9
1. A × B is the biggest relation from A to B.
2. ∅ is the smallest relation from A to B.
3. For the usual ordering < of R, <−1 = >.
−1
4. (R−1 )−1 = R; ∆−1= ∅ and (A × B)−1 = B × A.
A = ∆A ; ∅In practice, you define a relation by means of a condition on the elements of ordered pairs. This is completely analogous to the way you define a set by giving a
condition its elements should fulfill.
Example 5.10 If R is the set of pairs (n, m) (n, m ∈ Z) such that (condition) n2
is a divisor of m, the definition of R may look as follows:
For n, m ∈ Z: nRm :≡ n2 |m.
In Haskell this is implemented as: \ n m -> rem m n^2 == 0.

5.1. THE NOTION OF A RELATION

165

Example 5.11 For all n ∈ N, the set of pairs {(a, b) | a, b ∈ N, ab = n, a 6b} is a relation on N. This relation gives all the divisor pairs of n. Here is its
implementation:

divisors :: Integer -> [(Integer,Integer)]divisors n = [ (d, quot n d) | d <- [1..k], rem n d == 0 ]where k = floor (sqrt (fromInteger n))

Example 5.12 We can use the relation divisors of the previous example for yet
another implementation of the primality test:

prime’’ :: Integer -> Bool
prime’’ = \n -> divisors n == [(1,n)]Also, here are the list of divisors of a natural number, the list of all proper divisorsof a natural number, and a test for being a perfect natural number:divs :: Integer -> [Integer]
divs n = (fst list) ++ reverse (snd list)
where list = unzip (divisors n)properDivs :: Integer -> [Integer]
properDivs n = init (divs n)
perfect :: Integer -> Bool
perfect n = sum (properDivs n) == n‘There is a relation between a and b.’ In daily life, a statement like ‘there isa relation between a and b’ (or: ‘a and b are having a relation’) may not sound166

CHAPTER 5. RELATIONS

unusual, but in the present context this should be avoided. Not because these arefalse statements, but because they are so overly true, and, therefore, uninformative.
Cf. Exercise 5.13. Of course, a statement like ‘a stands in relation R to b’ or ‘therelation R subsists between a and b’ can very well be informative, when R is somespecific relation.
Exercise 5.13 Show that ∀x ∀y ∃R (xRy). (“Between every two things there exist
some relation.”)In everyday situations, by saying that a and b are related we usually mean morethan that it is possible to form the ordered pair of a and b. We usually meanthat there is some good reason for considering the pair (a, b), because there issome specific link between a and b (for instance, you can make Mrs. a blush bymentioning Mr. b in conversation).

5.2 Properties of Relations
In this section we list some useful properties of relations on a set A. Further on,
in Sections 5.3 and 5.4, we will illustrate how tests for these properties can be
implemented.
A relation R is reflexive on A if for every x ∈ A: xRx.
Example 5.14 On any set A, the relation ∆A is reflexive. Note that ∆A is thesmallest reflexive relation on A: it is a subset of any reflexive relation on A. In
other words, a relation R is reflexive on A iff ∆A ⊆ R.
Example 5.15 The relation 6 on N is reflexive (for every number is less than orequal to itself).
A relation R on A is irreflexive if for no x ∈ A: xRx.
Example 5.16 The relation < on N is irreflexive.
Exercise 5.17 Show that a relation R on A is irreflexive iff ∆A ∩ R = ∅.
There are relations which are neither reflexive nor irreflexive (the reader is urged
to think up an example).
A relation R on A is symmetric if for all x, y ∈ A: if xRy then yRx.

5.2. PROPERTIES OF RELATIONS

167

Example 5.18 The relation ‘having the same age’ between people is symmetric.
Unfortunately, the relation ‘being in love with’ between people is not symmetric.

Exercise 5.19 Show the following:
1. A relation R on a set A is symmetric iff ∀x, y ∈ A(xRy ⇔ yRx).
2. A relation R is symmetric iff R ⊆ R−1 , iff R = R−1 .

A relation R on A is asymmetric if for all x, y ∈ A: if xRy then not yRx.
The relation < on N is asymmetric. It is immediate from the definition that a
relation R on A is asymmetric iff R ∩ R−1 = ∅. Note that there are relations
which are neither symmetric nor asymmetric.
Exercise 5.20 Show that every asymmetric relation is irreflexive.
A relation R on A is antisymmetric if for all x, y ∈ A: if xRy and yRx thenx = y.
Example 5.21 The relation m|n (m is a divisor of n) on N is antisymmetric. If m
is a divisor of n and n is a divisor of m, then m and n are equal.
The relation in example 5.15 is antisymmetric.
Exercise 5.22 Show from the definitions that an asymmetric relation always is
antisymmetric.
The converse of the statement in Exercise 5.22 is not true: the relation 6 on N
provides a counterexample.
A relation R on A is transitive if for all x, y, z ∈ A: if xRy and yRz then xRz.
Examples of transitive relations are < and 6 on N. If F is the relation ‘friendship’on a set of people E, and every member of E endorses the laudable principle “Thefriends of my friends are my friends”, then F is transitive.
Exercise 5.23 Show that a relation R on a set A is transitive iff∀x, z ∈ A(∃y ∈ A(xRy ∧ yRz) ⇒ xRz).

168

CHAPTER 5. RELATIONS

A relation R on A is intransitive if for all x, y, z ∈ A: if xRy and yRz then not
xRz.
Example 5.24 The relation ‘father of’ on the set of all human beings is intransitive.
Again: there are relations that are neither transitive nor intransitive (think up an
example).
A relation R on A is a pre-order (or quasi-order) if R is transitive and reflexive.
Example 5.25 Let L be the set of all propositional formulas built from a given set
of atomic propositions. Then the relation |= given by
P |= Q iff P ⇒ Q is logically valid
is a pre-order on L. To check this, note that for every propositional formula P ,P ⇒ P is logically valid, so |= is reflexive. Also, for all propositional formulas
P, Q, R, if P ⇒ Q and Q ⇒ R are logically valid, then P ⇒ R is logicallyvalid. This can be checked as follows, by contraposition. Suppose P ⇒ R is not
logically valid. Then there is a valuation for the atomic propositions that makes P
true and R false. Now there are two possibilities: this valuation makes Q true or
it makes Q false. If it makes Q true, then we know that Q ⇒ R is not logicallyvalid. If it makes Q false, then we know that P ⇒ Q is not logically valid. So
either P ⇒ Q or Q ⇒ R is not logically valid. This shows that |= is transitive.
Example 5.26 Let L be the set of all propositional formulas built from a given set
of atomic propositions. Then the relation ` on L given by P ` Q iff there is a
proof of Q from the given P (this relation was defined in Chapter 3) is a pre-order.
See example 3.1 for the transitivity of `. In fact, the relation ` coincides withthe relation |= of the previous example. The safety checks on the proof rules forthe logical connectives that were performed in Chapter 3 guarantee that ` ⊆ |=.
This fact is called the soundness of the proof system. The inclusion in the otherdirection, |= ⊆ `, is called the completeness of the proof system. It is outside thescope of this book, but it is proved in [EFT94].
A relation R on A is a strict partial order if R is transitive and irreflexive. A
relation R on A is a partial order if R is transitive, reflexive and antisymmetric.
Note that the relation |= from example 5.25 is not antisymmetric. Take the formulas P ∧ Q and Q ∧ P . Then we have P ∧ Q |= Q ∧ P and Q ∧ P |= P ∧ Q, but
P ∧ Q and Q ∧ P are different formulas. Thus, |= is not a partial order.

5.2. PROPERTIES OF RELATIONS

169

Example 5.27 The relation < on N is a strict partial order and the relation 6 on
N is a partial order. Theorem 4.1 states that for any set A, the relation ⊆ on ℘(A)
is a partial order.
Exercise 5.28 Show that every strict partial order is asymmetric.
Exercise 5.29 Show that every relation which is transitive and asymmetric is astrict partial order.
Exercise 5.30 Show that if R is a strict partial order on A, then R∪∆A is a partial
order on A. (So every strict partial order is contained in a partial order.)Exercise 5.31 Show that the inverse of a partial order is again a partial order.
A relation R on A is linear (or: has the comparison property) if for all x, y ∈ A:
xRy or yRx or x = y. A partial order that is also linear is called a total order.
All Haskell types in class Ord a are total orders. A set A with a total order on it
is called a chain.
Exercise 5.32 Let S be a reflexive and symmetric relation on a set A.
A path is a finite sequence a1 , . . . , an of elements of A such that for every i,
1 6 i < n, we have that ai Sai+1 . Such a path connects a1 with an .
Assume that for all a, b ∈ A there is exactly one path connecting a with b.
Fix r ∈ A. Define the relation 6 on A by: a 6 b iff a is one of the elements in thepath connecting r with b.
Show the following:
1. 6 is reflexive,
2. 6 is antisymmetric.
3. 6 is transitive,
4. for all a ∈ A, r 6 a,
5. for every a ∈ A, the set Xa = {x ∈ A | x 6 a} is finite and if b, c ∈ Xa
then b 6 c or c 6 b.
(A structure (A, 6, r) with these five properties is called a tree with root r. The
directory-structure of a computer account is an example of a tree. Another example
is the structure tree of a formula (see Section 2.3 above). These are examples offinite trees, but if A is infinite, then the tree (A, 6, r) will have at least one infinite
branch.)CHAPTER 5. RELATIONS

170

Here is a list of all the relational properties of binary relations on a set A that we
discussed:reflexivity
irreflexivity
symmetry
asymmetry
antisymmetrytransitivity
intransitivity
linearity∀x xRx.
∀x ¬xRx.
∀xy (xRy ⇒ yRx).
∀xy (xRy ⇒ ¬yRx).
∀xy (xRy ∧ yRx ⇒ x = y).
∀xyz (xRy ∧ yRz ⇒ xRz).
∀xyz (xRy ∧ yRz ⇒ ¬xRz).
∀xy (xRy ∨ yRx ∨ x = y).

Because of their importance, relations that are transitive, reflexive and symmetric(equivalence relations) deserve a section of their own (see Section 5.5).
irreflpre-order
strict partial orderpartial order
total orderequivalence

√

refl
√

asymm

antisymm

√

√
√
√

√
√
√

symm

√

trans
√
√
√
√
√

linear√Exercise 5.33 Consider the following relations on the natural numbers. Check
their properties (some answers can be found in the text above). The successor
relation is the relation given by {(n, m) | n + 1 = m}. The divisor relationis {(n, m) | n divides m}. The coprime relation C on N is given by nCm :≡GCD(n, m) = 1, i.e., the only factor of n that divides m is 1, and vice versa (seeSection 8.2).
<

6successor

divisorcoprimeirreflexivereflexive
asymmetricantisymmetricsymmetrictransitive
linearDefinition 5.34 If O is a set of properties of relations on a set A, then the Oclosure of a relation R is the smallest relation S that includes R and that has all
the properties in O.

5.2. PROPERTIES OF RELATIONS

171

The most important closures are the reflexive closure, the symmetric closure, thetransitive closure and the reflexive transitive closure of a relation.
Remark. To show that R is the smallest relation S that has all the properties inO, show the following:
1. R has all the properties in O,
2. If S has all the properties in O, then R ⊆ S.

Exercise 5.35 Suppose that R is a relation on A.
1. Show that R ∪ ∆A is the reflexive closure of R.
2. Show that R ∪ R−1 is the symmetric closure of R.

Exercise 5.36 Let R be a transitive binary relation on A. Does it follow fromthe transitivity of R that its symmetric reflexive closure R ∪ R −1 ∪ ∆A is alsotransitive? Give a proof if your answer is yes, a counterexample otherwise.
To define the transitive and reflexive transitive closures of a relation we need theconcept of relation composition.
Composing Relations. Suppose that R and S are relations on A. The composition
R ◦ S of R and S is the relation on A that is defined byx(R ◦ S)z :≡ ∃y ∈ A(xRy ∧ ySz).
Furthermore, for n ∈ N, n > 1 we define Rn by means of R1 := R, Rn+1 :=Rn ◦ R.
Example 5.37 In imperative programming, the meaning of a command can beviewed as a relation on a set of machine states, where a machine state is a set ofpairs consisting of machine registers with their contents. If s = {(r 1 , 4), (r2 , 6), . . .}then executing r2 := r1 in state s gives a new state s0 = {(r1 , 4), (r2 , 4), . . .}.
If C1 and C2 are commands, then we can execute them in sequential order as
C1 ; C2 or as C2 ; C1 . If the meaning of command C1 is R and the meaning ofcommand C2 is S, then the meaning of C1 ; C2 is R ◦ S and the meaning of C2 ; C1
is S ◦ R.

CHAPTER 5. RELATIONS

172

Exercise 5.38 Determine the composition of the relation “father of” with itself.
Determine the composition of the relations “brother of” and “parent of” Give an
example showing that R ◦ S = S ◦ R can be false.
Exercise 5.39 Consider the relation
R = {(0, 2), (0, 3), (1, 0), (1, 3), (2, 0), (2, 3)}on the set A = {0, 1, 2, 3, 4}.
1. Determine R2 , R3 and R4 .
2. Give a relation S on A such that R ∪ (S ◦ R) = S.

Exercise 5.40 Verify:
1. A relation R on A is transitive iff R ◦ R ⊆ R.
2. Give an example of a transitive relation R for which R ◦ R = R is false.

Exercise 5.41 Verify:
1. Q ◦ (R ◦ S) = (Q ◦ R) ◦ S. (Thus, the notation Q ◦ R ◦ S is unambiguous.)
2. (R ◦ S)−1 = S −1 ◦ R−1 .

S
We will show that for any relation R on A, the relation R + = n>1 Rn is thetransitive closure of R.
S
Because R = R1 ⊆ n>1 Rn , we know that R ⊆ R+ . We still have to provethat R+ is transitive, and moreover that it is the smallest transitive relation that
contains R.
To see that R+ is transitive, assume xR+ y and yR+ z. From xR+ y we get that
there is some k > 1 with xRk y. From yR+ z we get that there is some m > 1
with yRm z. It follows that xRk+m z, and therefore xR+ z, proving that R+ istransitive.

5.2. PROPERTIES OF RELATIONS

173

Proposition 5.42 R+ is the smallest transitive relation that contains R.
Proof. The proposition can be restated as follows. If T is a transitive relation on
A such that R ⊆ T , then R+ ⊆ T . We prove this fact by induction (see Chapter
7).
Basis If xRy then it follows from R ⊆ T that xT y.
Induction step Assume (induction hypothesis) that R n ⊆ T . We have to showthat Rn+1 ⊆ T . Consider a pair (x, y) with xRn+1 y. Then, by the definitionof Rn+1 , there is a z such that xRn z and zRy. Because R ⊆ T we havezT y, and the induction hypothesis yields xT z. By transitivity of T it now
follows that xT y.

Example 5.43 If A = {a, {b, {c}}}, thena ∈+ A, {b, {c}} ∈+ A, b ∈+ A, {c} ∈+ A and c ∈+ A.

Example 5.44 On the set of human beings, the transitive closure of the ‘parentof’ relation is the ‘ancestor of’ relation, and the transitive closure of the ‘child of’relation is the ‘descendant of’ relation.
Exercise 5.45 Show that the relation < on N is the transitive closure of the relation
R = {(n, n + 1) | n ∈ N}.
Exercise 5.46 Let R be a relation on A. Show that R + ∪ ∆A is the reflexive
transitive closure of R.
The reflexive transitive closure of a relation R is often called the ancestral of R,
notation R∗ . Note that R∗ is a pre-order.
Exercise 5.47 Give the reflexive transitive closure of the following relation:
R = {(n, n + 1) | n ∈ N}.

Exercise 5.48*

CHAPTER 5. RELATIONS

174

1. Show that an intersection of arbitrarily many transitive relations is transitive.
2. Suppose that R is a relation on A. Note that A2 is one example of a transitive
relation on A that extends R. Conclude that the intersection of all transitive
relations extending R is the least transitive relation extending R. In otherwords, R+ equals the intersection of all transitive relations extending R.

Exercise 5.49*
1. Show that (R∗ )−1 = (R−1 )∗ .
2. Show by means of a counter-example that (R ∪ R −1 )∗ = R∗ ∪ R−1∗ may
be false.
3. Prove: if S ◦ R ⊆ R ◦ S, then (R ◦ S)∗ ⊆ R∗ ◦ S ∗ .

Exercise 5.50 Suppose that R and S are reflexive relations on the set A. Then
∆A ⊆ R and ∆A ⊆ S, so ∆A ⊆ R ∩ S, i.e., R ∩ S is reflexive as well. We say:reflexivity is preserved under intersection. Similarly, if R and S are reflexive, then∆A ⊆ R ∪ S, so R ∪ S is reflexive. Reflexivity is preserved under union. If R
is reflexive, then ∆A ⊆ R−1 , so R−1 is reflexive. Reflexivity is preserved under
inverse. If R and S are reflexive, ∆A ⊆ R ◦ S, so R ◦ S is reflexive. Reflexivity ispreserved under composition. Finally, if R on A is reflexive, then the complementof R, i.e., the relation A2 − R, is irreflexive. So reflexivity is not preserved undercomplement. These closure properties of reflexive relations are listed in the tablebelow.
We can ask the same questions for other relational properties. Suppose R and
S are symmetric relations on A. Does it follow that R ∩ S is symmetric? That
R ∪ S is symmetric? That R−1 is symmetric? That A2 − R is symmetric? That
R ◦ S is symmetric? Similarly for the property of transitivity. These questions are
summarized in the table below. Complete the table by putting ‘yes’ or ‘no’ in theappropriate places.
propertypreserved under ∩?
preserved under ∪?
preserved under inverse?
preserved under complement?
preserved under composition?

reflexivity
yes
yes
yes
no
yes

symmetry?
?
?
?
?

transitivity
?
?
?
?
?

5.3. IMPLEMENTING RELATIONS AS SETS OF PAIRS

175

5.3 Implementing Relations as Sets of Pairs
Our point of departure is a slight variation on the module SetEq.hs of the previous Chapter. This time, we represent sets a ordered lists without duplicates. See
Figs. 5.3 and 5.4 for a definition of the module SetOrd.hs.
The definition of deleteList inside Figure 5.3 employs a Haskell feature that wehaven’t encountered before: ys@(y:ys’) is a notation that allows us to refer to thenon-empty list (y:ys’) by means of ys. If the item to be deleted is greater thenthe first element of the list then the instruction is to do nothing. Doing nothingboils down to returning the whole list ys. This possibility to give a name to apattern is just used for readability.
Next we define relations over a type a as sets of pairs of that type, i.e., Rel a isdefined and implemented as Set(a,a).

type Rel a = Set (a,a)

domR gives the domain of a relation.

domR :: Ord a => Rel a -> Set adomR (Set r) = list2set [ x | (x,_) <- r ]ranR gives the range of a relation.

ranR :: Ord a => Rel a -> Set aranR (Set r) = list2set [ y | (_,y) <- r ]idR creates the identity relation ∆A over a set A:

CHAPTER 5. RELATIONS

176

module SetOrd (Set(..),emptySet,isEmpty,inSet,subSet,insertSet,
deleteSet,powerSet,takeSet,(!!!),list2set)
whereimport List (sort){-- Sets implemented as ordered lists without duplicates --}newtype Set a = Set [a] deriving (Eq,Ord)instance (Show a) => Show (Set a) whereshowsPrec _ (Set s) str = showSet s str
showSet []str = showString "{}" strshowSet (x:xs) str = showChar ’{’ ( shows x ( showl xs str))
where showl []str = showChar ’}’ str
showl (x:xs) str = showChar ’,’ (shows x (showl xs str))
emptySet :: Set a
emptySet = Set []isEmpty :: Set a -> BoolisEmpty (Set []) = True
isEmpty _= FalseinSet :: (Ord a) => a -> Set a -> BoolinSet x (Set s) = elem x (takeWhile (<= x) s)
subSet :: (Ord a) => Set a -> Set a -> Bool
subSet (Set []) _= TruesubSet (Set (x:xs)) set = (inSet x set) && subSet (Set xs) set
insertSet :: (Ord a) => a -> Set a -> Set a
insertSet x (Set s) = Set (insertList x s)

Figure 5.3: A Module for Sets as Ordered Lists Without Duplicates.

5.3. IMPLEMENTING RELATIONS AS SETS OF PAIRS

insertList x [] = [x]insertList x ys@(y:ys’) = case compare x y of
GT -> y : insertList x ys’
EQ -> ys
_ -> x : ys
deleteSet :: Ord a => a -> Set a -> Set a
deleteSet x (Set s) = Set (deleteList x s)
deleteList x [] = []deleteList x ys@(y:ys’) = case compare x y of
GT -> y : deleteList x ys’
EQ -> ys’
_ -> yslist2set :: Ord a => [a] -> Set alist2set [] = Set []list2set (x:xs) = insertSet x (list2set xs)-- list2set xs = Set (foldr insertList [] xs)
powerSet :: Ord a => Set a -> Set (Set a)
powerSet (Set xs) =Set (sort (map (\xs -> (list2set xs)) (powerList xs)))
powerList
powerList
powerList:: [a] -> [[a]][] = [[]](x:xs) = (powerList xs)++ (map (x:) (powerList xs))

takeSet :: Eq a => Int -> Set a -> Set a
takeSet n (Set xs) = Set (take n xs)
infixl 9 !!!
(!!!) :: Eq a => Set a -> Int -> a(Set xs) !!! n = xs !! n

Figure 5.4: A Module for Sets as Ordered Lists Without Duplicates (ctd).

177

178

CHAPTER 5. RELATIONS

idR :: Ord a => Set a -> Rel aidR (Set xs) = Set [(x,x) | x <- xs]

The total relation over a set is given by:

totalR :: Set a -> Rel a
totalR (Set xs) = Set [(x,y) | x <- xs, y <- xs ]invR inverts a relation (i.e., the function maps R to R −1 ).

invR :: Ord a => Rel a -> Rel a
invR (Set []) = (Set [])invR (Set ((x,y):r)) = insertSet (y,x) (invR (Set r))

inR checks whether a pair is in a relation.

inR :: Ord a => Rel a -> (a,a) -> BoolinR r (x,y) = inSet (x,y) r

The complement of a relation R ⊆ A × A is the relation A × A − R. The operationof relational complementation, relative to a set A, can be implemented as follows:complR :: Ord a => Set a -> Rel a -> Rel a
complR (Set xs) r =Set [(x,y) | x <- xs, y <- xs, not (inR r (x,y))]A check for reflexivity of R on a set A can be implemented by testing whether∆A ⊆ R:

5.3. IMPLEMENTING RELATIONS AS SETS OF PAIRS

179

reflR :: Ord a => Set a -> Rel a -> Bool
reflR set r = subSet (idR set) r

A check for irreflexivity of R on A proceeds by testing whether ∆A ∩ R = ∅:irreflR :: Ord a => Set a -> Rel a -> Bool
irreflR (Set xs) r =all (\ pair -> not (inR r pair)) [(x,x) | x <- xs]

A check for symmetry of R proceeds by testing for each pair (x, y) ∈ R whether(y, x) ∈ R:

symR :: Ord a => Rel a -> BoolsymR (Set []) = TruesymR (Set ((x,y):pairs)) | x == y = symR (Set pairs)| otherwise =inSet (y,x) (Set pairs)&& symR (deleteSet (y,x) (Set pairs))

A check for transitivity of R tests for each couple of pairs (x, y) ∈ R, (u, v) ∈ R
whether (x, v) ∈ R if y = u:transR :: Ord a => Rel a -> BooltransR (Set []) = TruetransR (Set s) = and [ trans pair (Set s) | pair <- s ] wheretrans (x,y) (Set r) =and [ inSet (x,v) (Set r) | (u,v) <- r, u == y ]Now what about relation composition? This is a more difficult matter, for how do
we implement ∃z(Rxz ∧ Szy)? The key to the implementation is the following

180

CHAPTER 5. RELATIONS

procedure for composing a single pair of objects (x, y) with a relation S, simply
by forming the relation {(x, z) | (z, y) ∈ S}. This is done by:composePair :: Ord a => (a,a) -> Rel a -> Rel a
composePair (x,y) (Set []) = Set []composePair (x,y) (Set ((u,v):s))| y == u= insertSet (x,v) (composePair (x,y) (Set s))| otherwise = composePair (x,y) (Set s)

For relation composition we need set union (Cf. Exercise 4.54):

unionSet :: (Ord a) => Set a -> Set a -> Set a
unionSet (Set [])set2 = set2
unionSet (Set (x:xs)) set2 =insertSet x (unionSet (Set xs) (deleteSet x set2))

Relation composition is defined in terms of composePair and unionSet:

compR :: Ord a => Rel a -> Rel a -> Rel a
compR (Set []) _ = (Set [])compR (Set ((x,y):s)) r =unionSet (composePair (x,y) r) (compR (Set s) r)

Composition of a relation with itself (Rn ):repeatR :: Ord a => Rel a -> Int -> Rel a
repeatR r n | n < 1= error "argument < 1"
| n == 1= r| otherwise = compR r (repeatR r (n-1))

5.3. IMPLEMENTING RELATIONS AS SETS OF PAIRS
Example 5.51 Let us use the implementation to illustrate Exercise 5.39.

r = Set [(0,2),(0,3),(1,0),(1,3),(2,0),(2,3)]r2 = compR r r
r3 = repeatR r 3
r4 = repeatR r 4

This gives:
REL> r{(0,2),(0,3),(1,0),(1,3),(2,0),(2,3)}
REL> r2{(0,0),(0,3),(1,2),(1,3),(2,2),(2,3)}
REL> r3{(0,2),(0,3),(1,0),(1,3),(2,0),(2,3)}
REL> r4{(0,0),(0,3),(1,2),(1,3),(2,2),(2,3)}
REL> r == r2
False
REL> r == r3
True
REL> r == r4
False
REL> r2 == r4
True

Also, the following test yields ‘True’:s = Set [(0,0),(0,2),(0,3),(1,0),(1,2),(1,3),(2,0),(2,2),(2,3)]
test = (unionSet r (compR s r)) == s

Exercise 5.52 Extend this implementation with a function
restrictR :: Ord a => Set a -> Rel a -> Rel a

181

CHAPTER 5. RELATIONS

182

that gives the restriction of a relation to a set. In the type declaration, Set a is the
restricting set.
The union of two relations R and S is the relation R ∪ S. Since relations are sets,
for this we can use unionSet.
Exercise 5.53 Use unionSet to define procedures rclosR for reflexive closureand sclosR for symmetric closure of a relation. As background set for the reflexive closure you can take the union of domain and range of the relation.
Exercise 5.54 Define a function
tclosR :: Ord a => Rel a -> Rel a

to compute the transitive closure of a relation R, for relations implemented as
Ord a => Rel a.
Hint: compute the smallest relation S with the property thatS = R ∪ R2 ∪ · · · ∪ Rk
(for some k) is transitive. Use transR for the transitivity test.

5.4 Implementing Relations as Characteristic Functions
A characteristic function is a function of type A → {0, 1}, for some set A. Characteristic functions are so-called because they characterize subsets of a set A. The
function f : A → {0, 1} characterizes the set B = {a ∈ A | f (a) = 1} ⊆ A.
Characteristic functions implemented in Haskell have type a -> Bool, for some
type a.
From the fact that a binary relation r is a subset of a product A × B, you would
expect that a binary relation is implemented in Haskell as a function of type(a,b) -> Bool. Given a pair of objects (x,y), with x of type a and y of type b,
the function proclaims the verdict True if (x,y) in the relation, False otherwise.
Standard relations like == (for identity, or equality) and <= (for 6) are represented
in Haskell in a slightly different way. They take their arguments one by one. Let
us check their types:

5.4. IMPLEMENTING RELATIONS AS CHARACTERISTIC FUNCTIONS183
Prelude> :t (==)(==) :: Eq a => a -> a -> Bool
Prelude> :t (<=)(<=) :: Ord a => a -> a -> Bool
Prelude>What this means is: if a is a type in the class Eq, then == takes a first argumentof that type and a second argument of that type and proclaims a verdict True orFalse, and similarly for <=, except that now the arguments have to be of a type in
the class Ord.
Another example of a relation in Haskell is the following implementation dividesof the relation x|y (‘x divides y’) on the integers (x divides y if there a q ∈ Z with
y = xq).

divides :: Integer -> Integer -> Bool
divides d n | d == 0
= error "divides: zero divisor"| otherwise = (rem n d) == 0

Switching back and forth between types a -> a -> Bool and (a,a) -> Bool(or, more generally, between types a -> b -> c and (a,b) -> c), can be doneby means of the procedures for currying and uncurrying a function. The procedures refer to the logician H.B. Curry who helped laying the foundations for functional programming. (The initial H stands for Haskell; the programming languagethat we use in this book is also named after him.)
If f is of type (a,b) -> c, then currying f means transforming it into a functionthat takes its arguments one by one, i.e., a function of type a -> b -> c. The
procedure curry is predefined in Haskell as follows:curry
curry f x y:: ((a,b) -> c) -> (a -> b -> c)= f (x,y)

If f is of type a -> b -> c, then uncurrying f means transforming it into a function that takes its arguments as a pair, i.e., a function of type (a,b) -> c. The
procedure uncurry is predefined in Haskell as follows:

CHAPTER 5. RELATIONS

184

uncurry
uncurry f p:: (a -> b -> c) -> ((a,b) -> c)= f (fst p) (snd p)

As an example, here are some definitions of relations.

eq :: Eq a => (a,a) -> Booleq = uncurry (==)
lessEq :: Ord a => (a,a) -> Bool
lessEq = uncurry (<=)

If a relation is implemented as a procedure of type (a,b) -> Bool it is very easy
to define its inverse:

inverse :: ((a,b) -> c) -> ((b,a) -> c)inverse f (x,y) = f (y,x)

This gives:
REL> inverse lessEq (3,4)
False
REL> inverse lessEq (4,3)
True
REL>

Can we do something similar for procedures of type a -> b -> c? Yes, we can.
Here is the predefined procedure flip:

flip
flip f x y

Here it is in action::: (a -> b -> c) -> b -> a -> c
= f y x5.4. IMPLEMENTING RELATIONS AS CHARACTERISTIC FUNCTIONS185
REL> flip (<=) 3 4
False
REL> flip (<=) 4 3
True
REL>

The procedure flip can be used to define properties from relations. Take theproperty of dividing the number 102. This denotes the set{d ∈ N+ | d divides 102} = {1, 2, 3, 6, 17, 34, 51, 102}.
It is given in Haskell by (‘divides‘ 102), which in turn is shorthand forflip divides 102

Trying this out, we get:
REL> filter (‘divides‘ 102) [1..300][1,2,3,6,17,34,51,102]

We will now work out the representation of relations as characteristic functions.
To keep the code compatible with the implementation given before, we define thetype as Rel’, and similarly for the operations.

type Rel’ a = a -> a -> Bool
emptyR’ :: Rel’ aemptyR’ = \ _ _ -> False
list2rel’ :: Eq a => [(a,a)] -> Rel’ alist2rel’ xys = \ x y -> elem (x,y) xys

idR’ creates the identity relation over a list.

idR’ :: Eq a => [a] -> Rel’ aidR’ xs = \ x y -> x == y && elem x xs

invR’ inverts a relation.

186

CHAPTER 5. RELATIONS

invR’ :: Rel’ a -> Rel’ ainvR’ = flip

inR’ checks whether a pair is in a relation.

inR’ :: Rel’ a -> (a,a) -> Bool
inR’ = uncurry

Checks whether a relation is reflexive, irreflexive, symmetric or transitive (on adomain given by a list):reflR’ :: [a] -> Rel’ a -> Bool
reflR’ xs r = and [ r x x | x <- xs ]irreflR’ :: [a] -> Rel’ a -> Bool
irreflR’ xs r = and [ not (r x x) | x <- xs ]symR’ :: [a] -> Rel’ a -> Bool
symR’ xs r = and [ not (r x y && not (r y x)) | x <- xs, y <- xs ]transR’ :: [a] -> Rel’ a -> BooltransR’ xs r = and[ not (r x y && r y z && not (r x z))| x <- xs, y <- xs, z <- xs ]Union, intersection, reflexive and symmetric closure of relations:5.4. IMPLEMENTING RELATIONS AS CHARACTERISTIC FUNCTIONS187

unionR’ :: Rel’ a -> Rel’ a -> Rel’ aunionR’ r s x y = r x y || s x yintersR’ :: Rel’ a -> Rel’ a -> Rel’ aintersR’ r s x y = r x y && s x y
reflClosure’ :: Eq a => Rel’ a -> Rel’ areflClosure’ r = unionR’ r (==)
symClosure’ :: Rel’ a -> Rel’ asymClosure’ r = unionR’ r (invR’ r)

Relation composition:

compR’ :: [a] -> Rel’ a -> Rel’ a -> Rel’ acompR’ xs r s x y = or [ r x z && s z y | z <- xs ]Composition of a relation with itself:

repeatR’ :: [a] -> Rel’ a -> Int -> Rel’ arepeatR’ xs r n | n < 1= error "argument < 1"
| n == 1= r| otherwise = compR’ xs r (repeatR’ xs r (n-1))

Exercise 5.55 Use the implementation of relations Rel’ a as characteristic functions over type a to define an example relation r with the property that
unionR r (compR r r)

is not the transitive closure of r.
Exercise 5.56 If a relation r :: Rel’ a is restricted to a finite list xs, then wecan calculate the transitive closure of r restricted to the list xs. Define a function

CHAPTER 5. RELATIONS

188transClosure’ :: [a] -> Rel’ a -> Rel’ afor this.
Hint: compute the smallest relation S with the property thatS = R ∪ R2 ∪ · · · ∪ Rk
(for some k) is transitive. Use transR xs for the transitivity test on domain xs.

5.5 Equivalence Relations
Definition 5.57 A relation R on A is an equivalence relation or equivalence if
R is transitive, reflexive on A and symmetric.
Example 5.58 On the set of human beings the relation of having the same age isan equivalence relation.
Example 5.59 The relation R = {(n, m) | n, m ∈ N and n + m is even } is anequivalence relation on N.
The equivalence test can be implemented for relations of type Ord a => Rel a
as follows:

equivalenceR :: Ord a => Set a -> Rel a -> BoolequivalenceR set r = reflR set r && symR r && transR r

For relations implemented as type Rel’ a the implementation goes like this:equivalenceR’ :: [a] -> Rel’ a -> BoolequivalenceR’ xs r = reflR’ xs r && symR’ xs r && transR’ xs r

Example 5.60 The next table shows for a number of familiar relations whether
they have the properties of reflexivity, symmetry and transitivity. Here:5.5. EQUIVALENCE RELATIONS

189• ∅ is the empty relation on N,• ∆ = ∆N = {(n, n) | n ∈ N} is the identity on N,• N2 = N × N is the biggest relation on N,• < and 6 are the usual ordering relations on N,• Suc is the relation on N defined by Suc(n, m) ≡ n + 1 = m,
• ⊆ is the inclusion relation on ℘(N).
property:reflexive (on N resp. ℘(N))
symmetrictransitiveyes:∆, N2 , 6, ⊆∅, ∆, N2∅, ∆, N2 , <, 6, ⊆no:∅, <, Suc<, 6, Suc, ⊆Suc

The table shows that among these examples only ∆ and N2 are equivalences on N.

Example 5.61 The relation ∅ is (trivially) symmetric and transitive. The relation∅ is reflexive on the set ∅, but on no other set. Thus: ∅ is an equivalence on the set∅, but on no other set.
Example 5.62 Let A be a set. ∆A is the smallest equivalence on A. A2 is thebiggest equivalence on A.
Example 5.63 The relation ∼ between vectors in 3-dimensional space R 3 that isdefined by ~a ∼ ~b ≡ ∃r ∈ R+ (~a = r~b) is an equivalence.
Example 5.64 For any n ∈ Z, n 6= 0, the relation ≡n on Z is given by m ≡n k
iff m and k have the same remainder when divided by n. More precisely, m ≡ n k(or: m ≡ k (mod n)) iff• m = qn + r, with 0 6 r < n,• k = q 0 n + r0 , with 0 6 r0 < n,
• r = r0 .
When m ≡n k we say that m is equivalent to k modulo n. The relation ≡n is also
called the (mod n) relation.

190

CHAPTER 5. RELATIONS

To show that ≡n from Example 5.64 is an equivalence, the following proposition
is useful.
Proposition 5.65 m ≡n k iff n | m − k.
Proof. ⇒: Suppose m ≡n k. Then m = qn + r and k = q 0 n + r0 with 0 6 r < nand 0 6 r0 < n and r = r0 . Thus, m−k = (q −q 0 )n, and it follows that n | m−k.

⇐: Suppose n | m − k. Then n | (qn + r) − (q 0 n + r0 ), so n | r − r0 . Since−n < r − r0 < n, this implies r − r 0 = 0, so r = r 0 . It follows that m ≡n k.
From this we get that the following are all equivalent:• m ≡n k.
• n | m − k.
• ∃a ∈ Z : an = m − k.
• ∃a ∈ Z : m = k + an.
• ∃a ∈ Z : k = m + an.
Exercise 5.66 Show that for every n ∈ Z with n 6= 0 it holds that ≡n is anequivalence on Z.
Example 5.67 Here is a Haskell implementation of the modulo relation:

modulo :: Integer -> Integer -> Integer -> Bool
modulo n = \ x y -> divides n (x-y)

Example 5.68 The relation that applies to two finite sets in case they have thesame number of elements is an equivalence on the collection of all finite sets.
The corresponding equivalence on finite lists is given by the following piece of
Haskell code:5.5. EQUIVALENCE RELATIONS

191

equalSize :: [a] -> [b] -> BoolequalSize list1 list2 = (length list1) == (length list2)

Abstract equivalences are often denoted by ∼ or ≈.
Exercise 5.69 Determine whether the following relations on N are (i) reflexive on
N, (ii) symmetric, (iii) transitive:
1. {(2, 3), (3, 5), (5, 2)};
2. {(n, m) | |n − m| > 3}.

Exercise 5.70 A = {1, 2, 3}. Can you guess how many relations there are on thissmall set? Indeed, there must be sufficiently many to provide for the following
questions.
1. Give an example of a relation on A that is reflexive, but not symmetric andnot transitive.
2. Give an example of a relation on A that is symmetric, but not reflexive andnot transitive.
3. Give examples (if any) of relations on A that satisfy each of the six remaining possibilities w.r.t. reflexivity, symmetry and transitivity.

Exercise 5.71 For finite sets A (0, 1, 2, 3, 4 and 5 elements and n elements generally) the following table has entries for: the number of elements in A 2 , the numberof elements in ℘(A2 ) (that is: the number of relations on A), the number of relations on A that are (i) reflexive, (ii) symmetric and (iii) transitive, and the numberof equivalences on A.

CHAPTER 5. RELATIONS

192
A
0
1
2
3
4
5
n

A2
0
1
?
?
?
?
?

℘(A2 )
1
2
?
?
?
?
?

reflexive
1
1
?
?
?
?
?

symmetric
1
2
?
?
?
?
?

transitive
1
2
13
—
—
—
—

equivalence
1
1
—
—
—
—
—

Give all reflexive, symmetric, transitive relations and equivalences for the casesthat A = ∅ (0 elements) and A = {0} (1 element). Show there are exactly 13
transitive relations on {0, 1}, and give the 3 that are not transitive. Put numbers on
the places with question marks. (You are not requested to fill in the —.)Example 5.72 Assume relation R on A is transitive and reflexive, i.e, R is a preorder. Then consider the relation ∼ on A given by: x ∼ y :≡ xRy ∧ yRx.
The relation ∼ is an equivalence relation on A. Symmetry is immediate from thedefinition. ∼ is reflexive because R is. ∼ is transitive, for assume x ∼ y andy ∼ z. Then xRy ∧ yRx ∧ yRz ∧ zRy, and from xRy ∧ yRz, by transitivity ofR, xRz, and from zRy ∧ yRx, by transitivity of R, zRx; thus x ∼ z.
Exercise 5.73 Suppose that R is a symmetric and transitive relation on the set A
such that ∀x ∈ A∃y ∈ A(xRy). Show that R is reflexive on A.
Exercise 5.74 Let R be a relation on A. Show that R is an equivalence iff (i)∆A ⊆ R and (ii) R = R ◦ R−1 .

5.6 Equivalence Classes and Partitions
Equivalence relations on a set A enable us to partition the set A into equivalenceclasses.
Definition 5.75 Suppose R is an equivalence relation on A and that a ∈ A. The
set | a | = | a |R = { b ∈ A | bRa } is called the R-equivalence class of a, or theequivalence class of a modulo R.
Elements of an equivalence class are called representatives of that class.
Example 5.76 (continued from example 5.62) The equivalence class of a ∈ A
modulo ∆A is {a}.
The only equivalence class modulo A2 is A itself.

5.6. EQUIVALENCE CLASSES AND PARTITIONS

193

Example 5.77 (continued from example 5.63) The equivalence class of (1, 1, 1) ∈R3 modulo ∼ is the set {(r, r, r) | r > 0}: half a straight line starting at the origin(not including the origin). (A “direction”.)
Example 5.78 (continued from example 5.67) The equivalence class of 2 in Z
(mod 4) is the set{. . . , −6, −2, 2, 6, 10, 14, . . .} = {2 + 4n | n ∈ Z }.
The implementation yields:

REL> filter (modulo 4 2) [-15..15][-14,-10,-6,-2,2,6,10,14]

Example 5.79 (continued from example 5.68) The equivalence class of {0, 1, 2}
modulo the equivalence of having the same number of elements is the collectionof all three-element sets.
*(According to the Frege-Russell definition of natural number, this is the number three.)Lemma 5.80 Suppose that R is an equivalence on A. If a, b ∈ A, then:| a |R = | b |R ⇔ aRb.
Proof. ⇒ : Note that a ∈ | a |R (for, R is reflexive). Thus, if | a |R = | b |R , thena ∈ | b |R , i.e., aRb.
⇐ : Assume that aRb. Then also bRa (R is symmetric.)| a |R ⊆ | b |R : x ∈ | a |R signifies xRa, and hence xRb follows (R transitive);
therefore, x ∈ | b |R .
| b |R ⊆ | a |R : similarly.
Extensionality completes the proof.
Lemma 5.81 Let R be an equivalence on A.
1. Every equivalence class is non-empty,
2. every element of A belongs to some equivalence class,3. different equivalence classes are disjoint.
Proof. 1/2. Since R is reflexive on A, we have, for every a ∈ A: a ∈ | a |.
3. Suppose that | a | and | b | are not disjoint. Say, c ∈ | a | ∩ | b |. Then we have bothcRa and cRb. Since R is symmetric, it follows that aRc. Thus, aRb (R transitive).
Therefore, | a | = | b | follows, using Lemma 5.80.

194

CHAPTER 5. RELATIONS

Exercise 5.82 Use the implementation of relations Rel’ a as characteristic functions over type a to implement a function raccess that takes a relation r, an object
x, and a list L, and returns the list of all objects y from L such that rxy holds. The
type declaration should run:raccess :: Rel’ a -> a -> [a] -> [a]The concept of partitioning a set is made precise in the following definition.
Definition 5.83 A family A of subsets of a set A is called a partition of A if
• ∅∈
/ A,
S
• A = A,• for all X, Y ∈ A: if X 6= Y then X ∩ Y = ∅.

The elements of a partition are called its components.
This definition says that every element of A is in some member of A and that noelement of A is in more than one member of A.
Example 5.84 {{1, 2}, {3, 4}} is a partition of {1, 2, 3, 4}. ∅ (trivially) is a partition of ∅. Z can be partitioned into the negative integers, the positive integers, and{0}.
Exercise 5.85 Show the following: if {Ai | i ∈ I} is a partition of A and {Bj |j ∈ J} is a partition of B, then {Ai × Bj | (i, j) ∈ I × J} is a partition of A × B.
Definition 5.86 (Quotients) Assume that R is an equivalence on the set A. The
collection of equivalence classes of R, A/R = {| a | | a ∈ A}, is called thequotient of A modulo R.
The definition of partition was engineered in order to ensure the following:
Theorem 5.87 Every quotient (of a set, modulo an equivalence) is a partition (ofthat set).
Proof. This is nothing but a reformulation of Lemma 5.81.

5.6. EQUIVALENCE CLASSES AND PARTITIONS

195

Example 5.88 (continued from examples 5.62 and 5.76)
A/∆A = { {a} | a ∈ A }.

A/A2 = {A}.

Example 5.89 (continued from examples 5.63 and 5.77)
The partition R3 / ∼ of R3 has components {0} and all half-lines.
Example 5.90 (continued from examples 5.67 and 5.78)
The partition Z/mod(4) of Z induced by the equivalence mod(4) has four components: (i) the equivalence class {4n | n ∈ Z} of 0 (that also is the equivalenceclass of 4, 8, 12,. . . ,−4. . . ), (ii) the class {4n + 1 | n ∈ Z} of 1, (iii) the class{4n + 2 | n ∈ Z} of 2, and (iv) {4n + 3 | n ∈ Z}, the class of 3. The quotient
Z/mod(n) is usually written as Zn .
Example 5.91*(continued from examples 5.68 and 5.79)
The quotient of the collection of finite sets modulo the relation “same number of
elements” is —according to Frege and Russell— the set of natural numbers.
Exercise 5.92 Give the partition that the relation of example 5.59 induces on N.
Example 5.93 (continued from Example 5.72)
Let R be a pre-order on A. Then ∼ given by x ∼ y :≡ xRy ∧ yRx is an equivalence relation. Consider the relation R∼ on A/ ∼ given by |x|R∼ |y| :≡ xRy. The
relation R∼ is a partial order on A/ ∼ called the po-set reflection of R.

The definition is independent of the representatives, because assume x ∼ x 0 andy ∼ y 0 and xRy. Then xRx0 ∧ x0 Rx, yRy 0 ∧ y 0 Ry, and x0 Ry 0 . From x0 Rx, xRy
and yRy 0 by transitivity of R, x0 Ry 0 .
R∼ is reflexive, because R is. R∼ is anti-symmetric, for suppose |x|R∼ |y| and|y|R∼ |x|. Then xRy and yRx, so by definition of ∼, |x| = |y|. Finally R∼ istransitive because R is.
Thus, equivalences induce partitions. But the process can be inverted.
Theorem 5.94 Every partition (of a set) is a quotient (of that set, modulo a certain
equivalence).
Specifically: Suppose that A is a partition of A. Then the relation R on A definedby
xRy :≡ ∃K ∈ A (x, y ∈ K)

CHAPTER 5. RELATIONS

196(x and y member of the same component of A) is an equivalence on A, and A isthe collection of equivalence classes of R.
Proof. Exercise 5.121.
According to Theorems 5.87 and 5.94, equivalences and partitions are two sidesof the same coin.
Example 5.95 Consider the following relation on {0, 1, 2, 3, 4}:{(0, 0), (0, 1), (1, 0), (1, 1), (2, 2), (2, 4), (3, 3), (4, 2), (4, 4)}.
This is an equivalence, and the corresponding partition is{{0, 1}, {2, 4}, {3}}.

Exercise 5.96 Is the following relation an equivalence on {0, 1, 2, 3, 4}? If so,give the corresponding partition.
{(0, 0), (0, 3), (0, 4), (1, 1), (1, 2),(2, 1), (2, 2), (3, 0), (3, 3), (3, 4), (4, 0), (4, 3), (4, 4)}.

Example 5.97 The equivalence corresponding to the partition{{0}, {1, 2, 3}, {4}}of {0, 1, 2, 3, 4} is:{(0, 0), (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 4)}.

Exercise 5.98 What are the equivalences corresponding to the following partitions?:
1. {{0, 3}, {1, 2, 4}}, of {0, 1, 2, 3, 4},
2. {{n ∈ Z | n < 0}, {0}, {n ∈ Z | n > 0}}, of Z,
3. {{even numbers}, {odd numbers}}, of N.

5.6. EQUIVALENCE CLASSES AND PARTITIONS

197

Exercise 5.99 A = {1, 2, 3, 4, 5},
R = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4),(3, 3), (3, 5), (4, 1), (4, 2), (4, 4), (5, 3), (5, 5)}.
1. Is R an equivalence on A? If so, answer 2 and 3:
2. Determine |2|R .
3. Determine A/R.

Example 5.100 If ∼ on ℘(N) is given by
A ∼ B :≡ (A − B) ∪ (B − A) is finite,then ∼ is reflexive. If A ⊆ N is arbitrary, then (A − A) ∪ (A − A) = ∅ ∪ ∅ = ∅,and ∅ is finite.
Exercise 5.101* Define the relation ∼ on ℘(N) by:
A ∼ B :≡ (A − B) ∪ (B − A) is finite.
Thus, N 6∼ ∅, since (N − ∅) ∪ (∅ − N) = N ∪ ∅ = N is infinite.
Show that ∼ is an equivalence (reflexivity is shown in example 5.100).
Exercise 5.102 Define the relation R on all people by: aRb :≡ a and b have acommon ancestor. Is R transitive?
Same question for the relation S defined by: aSb :≡ a and b have a common
ancestor along the male line.
Example 5.103 For counting the partitions of a set A of size n, the key is to countthe numberof ways of partitioning a set A of size n into k non-empty classes.
Let us use nk for this number, and see if we can find a recurrence for it, i.e., anequation in terms of the function for smallerarguments. Distributing n objects
over 1 set can be done in only one way, so n1 = 1. Similarly, distributing n
objects over n non-empty sets can be done in only one way, so nn = 1.

To distribute n objects over k different sets, we can either (i) put the last object
into an equivalence class of its own, or (ii) put it in one of the existing classes.

CHAPTER 5. RELATIONS

198


to distribute n − 1(i) can be done in n−1
k−1 ways, for this is the number of ways
n−1
objects over k − 1 non-empty classes. (ii) can be done in k · k ways, for there
ways to distribute n − 1 objects over k
are k classes to choose from, and n−1
kclasses. Thus, the recurrence we need is:

 

 
n−1
n−1
n
.
+
=k·
k−1
k
k
In terms of this, the number of partitions b(n) is given by:
n  
X
n
b(n) =.
k
k=1

The numbers
bers.

n
k

are called Stirling set numbers. The b(n) are called Bell num-

Exercise 5.104 Implement functions bell and stirling to count the number ofdifferent partitions of a set of n elements.
Exercise 5.105 Use the result of Exercise 5.104 to fill out the last column in thetable of Exercise 5.71 on p. 191.
Exercise 5.106 Show: conditions 2 and 3 of Definition 5.83 taken together are
equivalent with: for every a ∈ A there exists exactly one K ∈ A such that a ∈ K.
Exercise 5.107 Is the intersection R ∩ S of two equivalences R and S on a set A
again an equivalence? And the union R ∪ S? Prove, or supply a simple counterexample. (See the table of Exercise 5.50.)
Exercise 5.108* Suppose that R and S are equivalences on A such that R ⊆ S.
Show that every S-equivalence class is a union of R-equivalence classes.
Exercise 5.109 A list partition is the list counterpart of a partition: list partitions
are of type Eq a => [[a]], and a list partition xss of xs has the following properties:• [] is not an element of xss,
• xs and concat xss have the same elements,• if ys and zs are distinct elements of xss, then ys and zs have no elements
in common.

5.6. EQUIVALENCE CLASSES AND PARTITIONS

199

Implement a function
listPartition :: Eq a => [a] -> [[a]] -> Boolthat maps every list xs to a check whether a given object of type [[a]] is a list
partition of xs.
Exercise 5.110 Implement a function
listpart2equiv :: Ord a => [a] -> [[a]] -> Rel athat generates an equivalence relation from a list partition (see Exercise 5.109).
The first argument gives the domain, the second argument the list partition. Generate an error if the second argument is not a list partition on the domain given by
the first argument.q
Exercise 5.111 R = {(0, 3), (1, 5), (2, 0)}; A = {0, 1, 2, 3, 4, 5}.
1. What is the smallest (in the sense of: number of elements) equivalence S ⊇R on A?
2. Determine A/S.
3. How many equivalences exist on A that include R?
4. Give the corresponding partitions.

Exercise 5.112 Implement a function
equiv2listpart :: Ord a => Set a -> Rel a -> [[a]]that maps an equivalence relation to the corresponding list partition (see Exercise
5.109). Generate an error if the input relation is not an equivalence.
Exercise 5.113 Use the function equiv2listpart to implement a function
equiv2part :: Ord a => Set a -> Rel a -> Set (Set a)that maps an equivalence relation to the corresponding partition.

200

CHAPTER 5. RELATIONS

Exercise 5.114* R is a relation on A. One of the following is the smallest equivalence on A that includes R. Which? Prove.
1. (∆A ∪ R)+ ∪ (∆A ∪ R−1 )+ .
2. ∆A ∪ R+ ∪ (R−1 )+ .
3. ∆A ∪ (R ∪ R−1 )+ .

Exercise 5.115 Let R be a relation on A. Show that S = R ∗ ∩ R−1∗ is anequivalence on A. Next, show that the relation T on the quotient A/S given by|a|S T |b|S :≡ aR∗ b is a partial order.
Note that the reasoning of Exercise 5.115 only uses reflexivity and transitivity ofR∗ . So, in general, if R is a reflexive transitive relation on A (i.e., R is a pre-order),then S = R ∩ R−1 is an equivalence on A, and the relation T on the quotient A/S
given by |a|S T |b|S :≡ aRb is a partial order.
Example 5.116 Consider the pre-order |= of Example 5.25. Note that the equivalence relation |= ∩ |=−1 is nothing other than the relation ≡ of logical equivalence.
Remark. In constructions with quotients, it often happens that a relation on thequotient is defined in terms of a relation on the underlying set. In such a case one
should always check that the definition is proper in the sense that it is independentof the representatives that are mentioned in it.
Take again the example of exercise 5.115, of a set A with a reflexive transitive
relation R on it. If S = R ∩ R−1 , then the definition of relation T on the quotient
A/S given by |a|S T |b|S :≡ aRb is proper, because it holds that aRb, a0 ∈ |a|Sand b0 ∈ |b|S together imply that a0 Rb0 .

To see that this is so, assume a0 ∈ |a|S and aRb. Then a0 Sa, so a0 Ra, and bytransitivity of R, a0 Rb. Suppose b0 ∈ |b|S . Then b0 Sb, so bRb0 . Together with
a0 Rb this gives, again by transitivity of R, that a0 Rb0 .

Exercise 5.117 Define the relations ∼ and ≈ on R by p ∼ q :≡ p × q ∈ Z,p ≈ q :≡ p − q ∈ Z. Are these equivalences? If so, describe the corresponding
partition(s).
Exercise 5.118 Define the relation R on R × R by (x, y)R(u, v) iff 3x − y =3u − v.

5.6. EQUIVALENCE CLASSES AND PARTITIONS

201

1. Show that R is an equivalence.
2. Describe the equivalence classes of (0, 0) and (1, 1).
3. Describe R × R/R in geometrical terms.

Exercise 5.119 Define an equivalence on R × R that partitions the plane in concentric circles with (0, 0) as centre.
Exercise 5.120 Q = {(0, 0), (0, 1), (0, 5), (2, 4), (5, 0)}. For an equivalence R
on {0, 1, 2, 3, 4, 5} it is given, that Q ⊆ R and (0, 2) ∈/ R.
1. Show that (1, 5) ∈ R and (4, 5) 6∈ R.
2. Give the partition corresponding to the smallest (in the sense of number of
elements) equivalence ⊇ Q.
3. How many equivalences S are there on {0, 1, 2, 3, 4, 5} such that Q ⊆ S
and (0, 2) 6∈ S? Give the corresponding partitions.

Exercise 5.121 Prove Theorem 5.94.
Hint. Make sure that you use all properties of partitions.
Exercise 5.122* On a certain planet there are 20 countries. Every two of themeither are at war with each other or have a peace treaty. Every two countries witha common enemy have such a peace treaty. What is the least possible number ofpeace treaties on this planet?
Hint: Note that the relations P and W , for being at peace and at war, respectively,
exclude one another, that they are both irreflexive and symmetric, and that theysatisfy the principle that there are no war-triangles. For if x is at war with both yand z, then y and z have x as common enemy, hence y and z have a peace treaty:∀xyz((xW y ∧ xW z) ⇒ yP z).
Find a recurrence (see page 212) for the maximum number of countries at war
among 2n countries. Next, use this to derive the minimum number of peace treaties
among 2n countries.

CHAPTER 5. RELATIONS

202

5.7 Integer Partitions
Integer partitions of n ∈ N+ are lists of non-zero natural numbers that add up to
exactly n. For example, the four integer partitions of 4 are[4], [1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1].
The integer partitions of n correspond to the sizes of the set partitions of a set A
with |A| = n. Here is an algorithm for generating integer partitions, in lexicographically decreasing order:• The first integer partition of n is [n].
• Let B be the last integer partition generated. If B consists of only 1’s, thendone. Otherwise, there is a smallest non-1 part m. To generate the next
partition, subtract 1 from m and collect all the units so as to match the newsmallest part m − 1.
Examples The partition after [1, 1, 3, 3] is [1, 2, 2, 3], for after subtracting 1 from3, we should pack the three units that result in parcels of size 2. The partition after[1, 1, 1, 1, 1, 1, 5] is [3, 4, 4], for after subtracting 1 from 5, we should pack theseven units that result in parcels with maximum size 4, which gives three units andone parcel of size 4, which in turn gives one parcel of size 3 and one of size 4. The
partition after [3, 3, 5] is [1, 2, 3, 5]. The partition after [1, 3, 3, 4] is [2, 2, 3, 4].

Implementation An integer partition is represented as a list of integers. For
convenience we count the number of 1’s, and remove them from the partition.
This gives a compressed representation (2, [3, 3]) of [1, 1, 3, 3]. These compressed
partitions have type CmprPart.

type Part = [Int]type CmprPart = (Int,Part)

Expansion of a compressed partition (n, p) is done by generating n 1’s followedby p:5.7. INTEGER PARTITIONS

203

expand :: CmprPart -> Part
expand (0,p) = p
expand (n,p) = 1:(expand ((n-1),p))In generating the next partition from (k,xs) we may assume that xs is non-empty,
and that its elements are listed in increasing order.
The partition that follows (k,x:xs) is generated by packing (k+x,x:xs) for
maximum size x − 1. This assumes that x is the smallest element in x:xs.
nextpartition :: CmprPart -> CmprPart
nextpartition (k,(x:xs)) = pack (x-1) ((k+x),xs)

To pack a partition (m,xs) for size 1, there is nothing to do. To pack a partition(m,xs) for maximum size k > 1 and k > m, decrease the parcel size to k − 1.
To pack a partition (m,xs) for size k > 1 and k 6 m, use k units to generate one
parcel of size k, and go on with (m-k,k:xs), for the same parcel size.

pack :: Int -> CmprPart -> CmprPart
pack 1 (m,xs) = (m,xs)
pack k (m,xs) = if k > m then pack (k-1) (m,xs)
else
pack k(m-k,k:xs)

To generate all partitions starting from a given partition (n,[]), just list the partition consisting of n units, for this is the last partition. To generate all partitions
starting from a given partition (n,x:xs), list this partition, and then generate fromthe successor of (n,x:xs).

generatePs :: CmprPart -> [Part]
generatePs p@(n,[])= [expand p]
generatePs p@(n,(x:xs)) =(expand p: generatePs(nextpartition p))

204

CHAPTER 5. RELATIONSGenerate all partitions starting from [n]. The case where n = 1 is special, for itis the only case where the compressed form of the first partition has a non-zero
number of units.

part :: Int -> [Part]
part n | n < 1= error "part: argument <= 0"| n == 1= [[1]]| otherwise = generatePs (0,[n])Here is what we get out:
REL> part 5[[5],[1,4],[2,3],[1,1,3],[1,2,2],[1,1,1,2],[1,1,1,1,1]]
REL> part 6[[6],[1,5],[2,4],[1,1,4],[3,3],[1,2,3],[1,1,1,3],[2,2,2],[1,1,2,2],[1,1,1,1,2],[1,1,1,1,1,1]]
REL> length (part 20)
627

Exercise 5.123 Write a program change :: Int -> [Int] that returns change
in EURO coins for any positive integer, in the least number of coins. Measure thevalues of the EURO coins 0.01, 0.02, 0.05, 0.10, 0.20, 0.50, 1, 2 in EURO cents,
as 1, 2, 5, 10, 20, 50, 100, 200. Use pack for inspiration.
Exercise 5.124 Modify the integer partition algorithm so that it generates all thepossible ways of giving coin change for amounts of money up to 10 EURO, using
all available EURO coins (0.01, 0.02, 0.05, 0.10, 0.20, 0.50, 1, 2). Measure thevalues of the EURO coins in EURO cents, as 1, 2, 5, 10, 20, 50, 100, 200.
Exercise 5.125 How many different ways are there to give change for one EURO?
Use the program from the previous exercise.

5.8 Further Reading
More on relations in the context of set theory in [DvDdS78]. Binary relations on
finite sets are studied in graph theory. See Chapters 4 and 5 of [Bal91]. Relationsin the context of database theory are the topic of [AHV95].

Chapter 6

Functions
Preview

In mathematics, the concept of a function is perhaps even more important than thatof a set. Also, functions are crucial in computer programming, as the functional
programming paradigm demonstrates. This chapter introduces basic notions andthen moves on to special functions, operations on functions, defining equivalencesby means of functions, and compatibility of equivalences with operations.
Many concepts from the abstract theory of functions translate directly into computational practice. Most of the example functions mentioned in this chapter can be
implemented directly in Haskell by just keying in their definitions, with domains
and co-domains specified as Haskell types.
Still, we have to bear in mind that an implementation of a function as a computer program is a concrete incarnation of an abstract object. The same function may be computed by vastly different computation procedures. If you key
in sum [2*k | k <- [1 .. 100] ] at the hugs prompt you get the answer
10100, and if you key in 100 * 101 you get the same answer, but the computation steps that are performed to get at the answers are different. We have alreadyseen that there is no mechanical test for checking whether two procedures performthe same task (Section 4.2), although in particular cases such results can be provedby mathematical induction (see Chapter 7). E.g., in Example 7.4 inPChapter 7 itn
is proved by mathematical induction that the computational recipes k=1 2k andn(n + 1) specify the same function.
205

CHAPTER 6. FUNCTIONS

206

module FCT
whereimport List

6.1 Basic Notions
A function is something that transforms an object given to it into another one. The
objects that can be given to a function are called its arguments, and the results ofthe transformation are called values. The set of arguments is called the domain of
the function. We say that a function is defined on its domain.
If f is a function and x one of its arguments, then the corresponding value isdenoted by f (x). A function value y = f (x) is called the image of x under f .
That f (x) = y can also be indicated by f : x 7−→ y. The domain of f is denotedby dom (f ). Its range is ran(f ) = {f (x) | x ∈ dom (f )}.
Example 6.1 A function can be given as a rule or a prescription how to carry out
the transformation.
• First square, next add one is the function that transforms a real x ∈ R intox2 + 1.
Letting f stand for this function, it is customary to describe it by the equationf (x) = x2 + 1.
The Haskell implementation uses the same equation:f x = x^2 + 1• The function described by
|x| =



x
−x

if x > 0
if x < 06.1. BASIC NOTIONS

207transforms a real into its absolute value. The Haskell implementation given
in Prelude.hs follows this definition to the letter:

absReal x | x >= 0= x| otherwise = -x

• The identity function 1A defined on A does not “transform” at all in theusual sense of the word: given an argument x ∈ A, it outputs x itself. A
polymorphic identity function is predefined in Haskell as follows:

id :: a -> a
id x = x

Set theory has the following simple definition of the concept of a function.
Definition 6.2 A function is a relation f that satisfies the following condition.
(x, y) ∈ f ∧ (x, z) ∈ f =⇒ y = z.
That is: for every x ∈ dom (f ) there is exactly one y ∈ ran(f ) such that (x, y) ∈f.
If x ∈ dom (f ), then f (x) is by definition the unique object y ∈ ran(f ) for which(x, y) ∈ f .
Note that we use dom here in the sense defined for relations, but that the relationand the function-sense of the notion coincide: the domain dom (f ) of the function
f is {x | ∃y((x, y) ∈ f )}; exactly as in the relation-case (cf. Definition 5.1 p. 162).
Similarly, the range of f coincides with the range of f as a relation.
The set-theoretic and the computational view on functions are worlds apart, forcomputationally a function is an algorithm for computing values. However, incases of functions with finite domains it is easy to switch back and forth between
the two perspectives, as the following conversions demonstrate.

208

CHAPTER 6. FUNCTIONS

list2fct :: Eq a => [(a,b)] -> a -> blist2fct [] _ = error "function not total"
list2fct ((u,v):uvs) x | x == u= v| otherwise = list2fct uvs x
fct2list :: (a -> b) -> [a] -> [(a,b)]fct2list f xs = [ (x, f x) | x <- xs ]The range of a function, implemented as a list of pairs, is given by:

ranPairs :: Eq b => [(a,b)] -> [b]ranPairs f = nub [ y | (_,y) <- f ]If a function is defined on an enumerable domain, we can list its (finite or infinite)range starting from a given element.

listValues :: Enum a => (a -> b) -> a -> [b]listValues f i = (f i) : listValues f (succ i)

If we also know that the domain is bounded, we can generate the whole range as afinite list.

listRange :: (Bounded a, Enum a) => (a -> b) -> [b]listRange f = [ f i | i <- [minBound..maxBound] ]Example 6.3• The function x 7→ x2 +1 defined on R is identified with the relation {(x, y) |x ∈ R ∧ y = x2 + 1} = {(x, x2 + 1) | x ∈ R}.

6.1. BASIC NOTIONS

209• f = {(1, 4), (2, 4), (3, 6)} is a function. dom (f ) = {1, 2, 3}, ran(f ) ={4, 6}.
• {(1, 4), (2, 4), (2, 6)} is not a function.
• ∆X , the identity-relation on X (Definition 5.7, p. 163), also is a function
from X to X. When viewed as a function, the identity relation on X isusually written as 1X .
• If f is a function, we clearly have that f = {(a, f (a)) | a ∈ dom (f )}, andran(f ) = {f (a) | a ∈ dom (f )}.
• The relation ∅ is a function. dom (∅) = ∅ = ran(∅).

As is the case for relations, functions are more often given in a context of two sets.
Compare 5.3 (p. 162).
Definition 6.4 (From . . . to, On, Co-domain) Suppose that X and Y are sets. A
function f is from X to Y ; notation:
f : X −→ Y,
if dom (f ) = X and ran(f ) ⊆ Y . (Note the difference in terminology comparedwith 5.3!)
In this situation, Y is called the co-domain of f .
A function f is said to be defined on X if dom (f ) = X.
Note that the set-theoretic way of identifying the function f with the relation R ={(x, f (x)) | x ∈ X} has no way of dealing with this situation: it is not possible torecover the intended co-domain Y from the relation R. As far as R is concerned,
the co-domain of f could be any set that extends ran(R).
More Than One Argument. Functions as introduced here are unary, i.e., they
apply to only one argument. But of course, functions with more than one argument(binary, ternary. . . ) do exist. E.g., addition and multiplication on N are of thiskind. However, such a binary (ternary. . . ) function can be viewed as a unary one
that applies to ordered pairs (resp., triples. . . ).
As we have seen, Haskell has predefined operations curry and uncurry to switch
back and forth between functions of types (a,b) -> c and a -> b -> c. We

CHAPTER 6. FUNCTIONS

210

can extend this to cases of functions that take triples, quadruples, etc. as arguments.
As an example, here is the case for currying functions that take triples to functions
that take three arguments, and for uncurrying functions that take three arguments
to functions that take triples.

curry3 :: ((a,b,c) -> d) -> a -> b -> c -> dcurry3 f x y z = f (x,y,z)
uncurry3 :: (a -> b -> c -> d) -> (a,b,c) -> duncurry3 f (x,y,z) = f x y z

Fact 6.5 (Function Equality) If f and g are functions that share their domain(dom (f ) = dom (g)) and, on it, carry out the same transformation (i.e., ∀x ∈dom (f )(f (x) = g(x))), then — according to Extensionality 4.1, p. 115 — wehave that f = g.
To establish that two functions f, g : X → Y are different we have to find anx ∈ X with f (x) 6= g(x). As we have seen in Section 4.2, there is no genericalgorithm for checking function equality. Therefore, to establish that two functions
f, g : X → Y are equal, we need a proof. The general form, spelled out in full, ofsuch a proof is:Given: f, g : X → Y .
To be proved: f = g.
Proof:
Let x be an arbitrary object in X.
To be proved: f (x) = g(x).
Proof:...
Thus f (x) = g(x).
Thus f = g.

Example 6.6 If f and g are defined on R by f (x) = x2 + 2x + 1, resp., g(x) =(x + 1)2 , then f = g. Thus, functions are not distinguished by the details of how
they actually transform (in this respect, f and g differ), but only with respect totheir output-behaviour.

6.1. BASIC NOTIONS

211

Warning. If functions f : X → Y and g : X → Z are given in the domainco-domain-context, and ∀x ∈ X(f (x) = g(x)), then f and g count as equal only
if we also have that Y = Z. The co-domains are taken as an integral part of thefunctions.
Function Definitions. If t(x) is an expression that describes an element of Y
in terms of an element x ∈ X, then we can define a function f : X −→ Y bywriting:
Let the function f : X → Y be defined by f (x) = t(x).
For completely specifying a function f three things are sufficient:• Specify dom (f ),• Specify the co-domain of f ,• Give an instruction for how to construct f (x) from x.
Examples of such instructions are x 7→ x + 1, x 7→ x2 , and in fact all the Haskell
definitions of functions that we have encountered in this book.
A very convenient notation for function construction is by means of lambda abstraction (page 58). In this notation, λx.x+1 encodes the specification x 7→ x+1.
The lambda operator is a variable binder, so λx.x+1 and λy.y +1 denote the samefunction. In fact, every time we specify a function foo in Haskell by means offoo x y z = t

we can think of this as a specification λxyz.t. If the types of x, y, z, t are known,
this also specifies a domain and a co-domain. For if x :: a, y :: b, z :: c,t :: d, then λxyz.t has type a -> b -> c -> d.
Example 6.7
1. t(x) is the expression 2x2 + 3:
Let the function g : R −→ R be defined by g(x) = 2x2 + 3.
Rx
2. t(x) is the expression 0 y sin(y)dy:Rx
Let the function h : R −→ R be defined by h(x) = 0 y sin(y) dy.

212

CHAPTER 6. FUNCTIONS

Example 6.8 The Haskell way of defining functions is very close to standard
mathematical practice, witness the following examples (note that all these equations define the same function):

f1 x = x^2 + 2 * x + 1
g1 x = (x + 1)^2
f1’ = \x -> x^2 + 2 * x + 1
g1’ = \x -> (x + 1)^2

Recurrences versus Closed Forms. A definition for a function f : N → A interms of algebraic operations is called a closed form definition. A function definition for f in terms of the values of f for smaller arguments is called a recurrence
for f . The advantage of a closed form definition over a recurrence is that it allows
for more efficient computation, since (in general) the computation time of a closed
form does not grow exponentially with the size of the argument.
Example 6.9 Consider the following recurrence.

g 0 = 0g n = g (n-1) + n

A closed form definition of the same function is:g’ n = ((n + 1) * n ) / 2

Exercise 6.10 Give a closed form implementation of the following function:

6.1. BASIC NOTIONS

213

h 0 = 0h n = h (n-1) + (2*n)

Exercise 6.11 Give a closed form implementation of the following function:

k 0 = 0k n = k (n-1) + (2*n-1)

To show that a particular closed form defines the same function as a given recurrence, we need a proof by induction: see Chapter 7. It is not always possible tofind useful definitions in closed form, and recurrences are in general much easier
to find than closed forms. E.g., a closed form for the factorial function n! would
be an expression that allows us to compute n! with at most a fixed number of‘standard’
Qn operations. The number of operations should be independent of n, so
n! = k=1 k = 1 × · · · (n − 1) × n does not count, for · · · hides a number of
productthat does depend on n. No closed form for n! is known, andQoperationsnn! = k=1 k performs essentially the same calculation as n! = (n − 1)! × n.
Thus, computationally, there is nothing to choose between the following two implementations of n!.

fac 0 = 1
fac n = fac (n-1) * nfac’ n = product [ k | k <- [1..n] ]Note that there is no need to add fac’ 0 = 1 to the second definition, because of
the convention that product [] gives the value 1.
A simple example of defining a function in terms of another function is the following (see Figure 6.1).

CHAPTER 6. FUNCTIONS

214
4

4

3

3

2

2

1

1

0

0
-2

-1

0

1

2

0

1

2

Figure 6.1: Restricting the function λx.x2 .
Definition 6.12 (Restrictions) Suppose that f : X −→ Y and A ⊆ X. The
restriction of f to A is the function h : A → Y defined by h(a) = f (a). The
notation for this function is fA.
Here is the implementation, for functions implemented as type a -> b:

restrict :: Eq a => (a -> b) -> [a] -> a -> b
restrict f xs x | elem x xs = f x| otherwise = error "argument not in domain"

And this is the implementation for functions implemented as lists of pairs:

restrictPairs :: Eq a => [(a,b)] -> [a] -> [(a,b)]
restrictPairs xys xs = [ (x,y) | (x,y) <- xys, elem x xs ]Definition 6.13 (Image, Co-image) Suppose that f : X −→ Y , A ⊆ X andB ⊆Y.
1. f [A] = {f (x) | x ∈ A} is called the image of A under f ;6.1. BASIC NOTIONS

215

2. f −1 [B] = {x ∈ X | f (x) ∈ B} is called the co-image of B under f .
From this definition we get:
1. f [X] = ran(f ),2. f −1 [Y ] = dom (f ),3. y ∈ f [A] ⇔ ∃x ∈ A(y = f (x)),
4. x ∈ f −1 [B] ⇔ f (x) ∈ B.
From 3. it follows that x ∈ A ⇒ f (x) ∈ f [A]. But note that we do notnecessarily have that f (x) ∈ f [A] ⇒ x ∈ A. (Example: f = {(0, 2), (1, 2)},x = 0, A = {1}.)
Two Types of Brackets. Distinguish f (a) and f [A]. The notation f (a) presupposes that a ∈ dom (f ). Then, f (a) is the f -value of a. The notation f [A]presupposes A ⊆ dom (f ). Then f [A] is the set of values f (x) where x ∈ A.
Remark. Many texts do not employ f [ ] but use f ( ) throughout. In that case,
you have to figure out from the context what is meant.
In the expression f −1 [B], the part f −1 has no meaning when taken by itself. The
notation f −1 will be used later on for the inverse function corresponding to f .
Such an inverse only exists if f happens to be a bijection. However, the notationf −1 [B] is always meaningful and does not presuppose bijectivity.
Here are the implementations of image and co-image:

image :: Eq b => (a -> b) -> [a] -> [b]image f xs = nub [ f x | x <- xs ]coImage :: Eq b => (a -> b) -> [a] -> [b] -> [a]coImage f xs ys = [ x | x <- xs, elem (f x) ys ]This gives:
FCT> image (*2) [1,2,3][2,4,6]
FCT> coImage (*2) [1,2,3] [2,3,4][1,2]

CHAPTER 6. FUNCTIONS216

And here are the versions for functions represented as lists of pairs:

imagePairs :: (Eq a, Eq b) => [(a,b)] -> [a] -> [b]imagePairs f xs = image (list2fct f) xs
coImagePairs :: (Eq a, Eq b) => [(a,b)] -> [a] -> [b] -> [a]coImagePairs f xs ys = coImage (list2fct f) xs ys

Exercise 6.14 Consider the relation R = {(0, 4), (1, 2), (1, 3)}.
1. Is R a function?
If so, determine dom (R) and ran(R).
2. Remember: R−1 = {(b, a) | (a, b) ∈ R} is the inverse of the relation R.
(Definition 5.7, p. 163).
Is R−1 a function?
If so, determine dom (R−1 ) and ran(R−1 ).

Exercise 6.15 Suppose that f : X −→ Y and A ⊆ X. Verify:
1. f [A] = ran(fA),
2. f [dom (f )] = ran(f ),3. f −1 [B] = dom (f ∩ (X × B)),
4. f −1 [ran(f )] = dom (f ),5. fA = f ∩ (A × Y ).

Exercise 6.16 Let X = {0, 1, 2, 3}, Y = {2, 3, 4, 5}, f = {(0, 3), (1, 2), (2, 4), (3, 2)}.
Determine f  {0, 3}, f [{1, 2, 3}] and f −1 [{2, 4, 5}]. Next, check your answers
with the implementation.

6.1. BASIC NOTIONS

217

Exercise 6.17 Suppose that f : A → Y , g : B → Y , and A ∩ B = ∅. Show thatf ∪ g : A ∪ B → Y.
What if A ∩ B 6= ∅?
Exercise 6.18* Let A be a partition
S of X. For every component A ∈ A a function
fA : A → Y is given. Show, that A∈A fA : X → Y .
Example 6.19 Suppose that f : X → Y and A, B ⊆ X. We show that f [A −B] ⊇ f [A] − f [B].
Assume that y ∈ f [A] − f [B]. Then y ∈ f [A] and y 6∈ f [B]. From the first we
obtain x ∈ A such that y = f (x). From the second we see that x 6∈ B (otherwise,y = f (x) ∈ f [B]). So, x ∈ A − B, and y = f (x) ∈ f [A − B].
To see that the inclusion cannot be replaced by an equality, take X = Y = A ={0, 1}, B = {1}, and let f be given by f (x) = 0. Then f [A − B] = f [{0}] = {0}and f [A] − f [B] = {0} − {0} = ∅.

Next, suppose that f : X → Y and C, D ⊆ Y . We show that f −1 [C − D] =f −1 [C] − f −1 [D]:x ∈ f −1 [C − D] ⇐⇒ f (x) ∈ C − D⇐⇒ (f (x) ∈ C) ∧ (f (x) 6∈ D)⇐⇒ x ∈ f −1 [C] ∧ x 6∈ f −1 [D]⇐⇒ x ∈ f −1 [C] − f −1 [D].
The required equality follows using Extensionality.

Exercise 6.20 Suppose that f : X → Y , A, B ⊆ X, and C, D ⊆ Y . See example
6.19. Show:
1. A ⊆ B ⇒ f [A] ⊆ f [B];C ⊆ D ⇒ f −1 [C] ⊆ f −1 [D],2. f [A ∪ B] = f [A] ∪ f [B];f [A ∩ B] ⊆ f [A] ∩ f [B],3. f −1 [C ∪ D] = f −1 [C] ∪ f −1 [D];f −1 [C ∩ D] = f −1 [C] ∩ f −1 [D],CHAPTER 6. FUNCTIONS

218
4. f [f −1 [C]] ⊆ C;
f −1 [f [A]] ⊇ A.

Give simple examples to show that the inclusions in 2 and 4 cannot be replaced byequalities.

6.2 Surjections, Injections, Bijections
If X is the domain of a function f , then for each x ∈ X there is only one ywith (x, y) ∈ f . However, there may be other elements z ∈ X with (z, y) ∈ f .
Functions for which this does not happen warrant a special name.
If f is a function from X to Y , then there may be elements of Y that are not inf [X]. Again, functions for which this does not happen warrant a special name.
Definition 6.21 (Surjections, Injections, Bijections)
A function f : X −→ Y is called
1. surjective, or a surjection, or onto Y, if every element b ∈ Y occurs as afunction value of at least one a ∈ X, i.e., if f [X] = Y ;2. injective, an injection, or one-to-one, if every b ∈ Y is value of at most onea ∈ X;
3. bijective or a bijection if it is both injective and surjective.
Example 6.22 Most functions are neither surjective, nor injective. For instance,• sin : R → R is not surjective (e.g., 2 ∈ R is not a value) and not injective(sin 0 = sin π).
• The identity function 1X : X → X is a bijection, whatever the set X.
• Let A be a set. According to Theorems 5.87 and 5.94, the function that transforms an equivalence R on A into its quotient A/R is a bijection between
the set of equivalences and the set of partitions on A.
• Consider f = {(0, 1), (1, 0), (2, 1)}. Thus, dom (f ) = {0, 1, 2}. The function
f : {0, 1, 2} −→ {0, 1}

6.2. SURJECTIONS, INJECTIONS, BIJECTIONS

219

is surjective, but
f : {0, 1, 2} −→ {0, 1, 2}

is not surjective. The concept of surjectivity presupposes that of a codomain. Cf. 6.5 (p. 210). However, whatever this co-domain, f clearly
is not injective, since 0 and 2 have the same image.

If the domain of a function is represented as a list, the injectivity test can be implemented as follows:

injective :: Eq b => (a -> b) -> [a] -> Bool
injective f [] = True
injective f (x:xs) =notElem (f x) (image f xs) && injective f xs

Similarly, if the domain and co-domain of a function are represented as lists, thesurjectivity test can be implemented as follows:

surjective :: Eq b => (a -> b) -> [a] -> [b] -> Boolsurjective f xs [] = True
surjective f xs (y:ys) =elem y (image f xs) && surjective f xs ys

Exercise 6.23 Implement a test for bijectivity.
Exercise 6.24 Implement tests
injectivePairs, surjectivePairs, bijectivePairs

for functions represented as lists of pairs.
Proving that a Function is Injective/Surjective. The following implication is a
useful way of expressing that f is injective:f (x) = f (y) =⇒ x = y.
The proof schema becomes:

CHAPTER 6. FUNCTIONS

220

To be proved: f is injective.
Proof:
Let x, y be arbitrary, and suppose f (x) = f (y).
..
.
Thus x = y.

The contraposition of f (x) = f (y) =⇒ x = y, i.e.,
x 6= y =⇒ f (x) 6= f (y),of course says the same thing differently, so an equivalent proof schema is:

To be proved: f is injective.
Proof:
Let x, y be arbitrary, and suppose x 6= y.
..
.
Thus f (x) 6= f (y).

That f : X → Y is surjective is expressed by:∀b ∈ Y ∃a ∈ X f (a) = b.
This gives the following pattern for a proof of surjectivity:

To be proved: f : X → Y is surjective.
Proof:
Let b be an arbitrary element of Y .
..
.
Thus there is an a ∈ X with f (a) = b.

Exercise 6.25 Are the following functions injective? surjective?

6.2. SURJECTIONS, INJECTIONS, BIJECTIONS

221

1. sin : R+ → R (N.B.: R+ = {x ∈ R | 0 < x}),
2. sin : R → [−1, +1],
3. sin : [−1, +1] → [−1, +1],
4. ex : R → R,
5. tan : R → R,
6. log : R+ → R,√
7.
: R+ → R + .

Remark. The functions of Exercise 6.25 are all predefined in Haskell: sin, exp,tan, log, sqrt. The base of the natural logarithm, Napier’s number e, is given by
exp 1.
Exercise 6.26 Give a formula for the number of injections from an n-element set
A to a k-element set B.
Exercise 6.27 Implement a function
injs :: [Int] -> [Int] -> [[(Int,Int)]]

that takes a finite domain and a finite codomain of type Int and produces the listof all injections from domain to codomain, given as lists of integer pairs.
Exercise 6.28 The bijections on a finite set A correspond exactly to the permutations of A. Implement a function
perms :: [a] -> [[a]]that gives all permutations of a finite list. The call perms [1,2,3] should yield:[[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]]

Hint: to get the permutations of (x:xs), take all the possible ways of inserting x
in a permutation of xs.

CHAPTER 6. FUNCTIONS

222

6.3 Function Composition
Definition 6.29 (Composition) Suppose that f : X −→ Y and g : Y −→ Z.
Thus, the co-domain of f coincides with the domain of g. The composition of f
and g is the function g ◦ f : X −→ Z defined by(g ◦ f )(x) = g(f (x)).
(“First, apply f , next, apply g” — thanks to the usual “prefix”-notation for functions, the f and the g are unfortunately in the reverse order in the notation g ◦ f .
To keep this reverse order in mind it is good practice to refer to g ◦ f as “g afterf ”.)
N.B.: The notation g ◦ f presupposes that the co-domain of f and the domain of
g are the same. Furthermore, g ◦ f has the domain of f and the co-domain of g.
Function composition is predefined in Haskell, as follows:(.)(f . g) x:: (b -> c) -> (a -> b) -> (a -> c)= f (g x)

Example 6.30 Haskell has a general procedure negate for negating a number.
The effect of first taking the absolute value, then negating can now be got bymeans of (negate . abs):
Prelude> (negate . abs) 5-5Prelude> (negate . abs) (-7)-7Prelude>Example 6.31 Another example from Prelude.hs is:even nodd= n ‘rem‘ 2 == 0
= not . even

6.3. FUNCTION COMPOSITION

223

Exercise 6.32 Implement an operation comp for composition of functions represented as lists of pairs.
Example 6.33 In analysis, compositions
abound. E.g., if f : R → R +√ is x 7→√2+2
x + 1, and g : R → R is x 7→ x,√ then g ◦ f : R → R is x 7→ x + 1:
2
2
(g ◦ f )(x) = g(f (x)) = g(x + 1) = x + 1.
The identity function behaves as a unit element for composition (see Example 6.1
for 1X ):Fact 6.34 If f : X → Y , then f ◦ 1X = 1Y ◦ f = f .
Suppose that f : X → Y , g : Y → Z and h : Z → U . There are now two ways
to define a function from X to U : (i) (h ◦ g) ◦ f , and (ii) = h ◦ (g ◦ f ). The next
lemma says that these functions coincide. Thus, composition is associative; andwe can safely write h ◦ g ◦ f if we mean either of these.
Lemma 6.35 If f : X → Y , g : Y → Z and h : Z → U , then (h ◦ g) ◦ f =h ◦ (g ◦ f ).
Proof. Suppose that x ∈ X. Then:((h◦g)◦f )(x) = (h◦g)(f (x)) = h(g(f (x))) = h((g ◦f )(x)) = (h◦(g ◦f ))(x).
The functions (h ◦ g) ◦ f and h ◦ (g ◦ f ) have the same domain (X) and co-domain(U ) and, on this domain, show the same action. Thus, they are (6.5, p. 210) equal.

Lemma 6.36 Suppose that f : X −→ Y , g : Y −→ Z. Then:
1. g ◦ f injective =⇒ f injective,
2. g ◦ f surjective =⇒ g surjective,
3. f and g injective =⇒ g ◦ f injective,
4. f and g surjective =⇒ g ◦ f surjective.
Proof. We prove 1, 2 and 4.
1. Given: g ◦ f injective, i.e., (g ◦ f )(a1 ) = (g ◦ f )(a2 ) ⇒ a1 = a2 .
To be proved: f injective, i.e., f (a1 ) = f (a2 ) ⇒ a1 = a2 .

CHAPTER 6. FUNCTIONS

224

Proof: Assume f (a1 ) = f (a2 ). Then of course, g(f (a1 )) = g(f (a2 )). (Applying
g twice to the same argument must produce the same value twice.) But then,(g ◦ f )(a1 ) = g(f (a1 )) = g(f (a2 )) = (g ◦ f )(a2 ). The given now shows thata1 = a 2 .
2. Given: g ◦ f surjective.
To be proved: g surjective. I.e., ∀c ∈ Z∃b ∈ Y (g(b) = c).
Proof: Assume c ∈ Z. Wanted: b ∈ Y such that g(b) = c. Since g ◦f is surjective,
there is a ∈ X such that (g ◦ f )(a) = c. But, g(f (a)) = (g ◦ f )(a). I.e., b = f (a)is the element looked for.
4. Given: f and g surjective.
To be proved: g ◦ f surjective. I.e., every c ∈ Z is a value of g ◦ f .
Proof: Assume c ∈ Z. Since g is surjective, b ∈ Y exists such that g(b) = c.
Since f is surjective, a ∈ X exists such that f (a) = b. It follows that (g ◦ f )(a) =g(f (a)) = g(b) = c. So, a is the element looked for.
Note that there are statements in the spirit of Lemma 6.36 that do not hold. In
particular, from the fact that g ◦ f is injective, it does not follow that g is injective.
Consider the case where f is the (empty) function from ∅ to Y , and g an arbitrary
non-injective function from Y → Z. Then g ◦ f is the (empty) function from ∅ toZ, which is surely injective. But g by assumption is not injective.
For an example in which g ◦ f is bijective, but g not injective and f not surjective,
take f : N → N given by f (n) = n + 1 and g : N → N given byn
if n = 0g(n) =
n − 1 if n > 1.
Clearly, g ◦f is the identity on N, but f is not a surjection, and g is not an injection.
Exercise 6.37 Can you think up an example with g ◦f bijective, but g not injective
and f not surjective, with the domain of f finite? If yes, give the example, if no,
prove that this is impossible.
Exercise 6.38 The function f : {0, 1, 2, 3, 4} → {0, 1, 2, 3, 4} is defined by thefollowing table:
xf (x)(f f )(x)(f f f )(x)(f f f f )(x)

0
1

1
2

2
0

3
0

4
3

6.3. FUNCTION COMPOSITION

225

1. Determine the compositions f ◦ f , f ◦ f ◦ f and f ◦ f ◦ f ◦ f by completingthe table.
2. How many elements has the set {f, f ◦ f, f ◦ f ◦ f, . . .}? (N.B.: the elements
are functions!)
3. Exhibit a function g : {0, 1, 2, 3, 4} → {0, 1, 2, 3, 4} such that {g, g ◦ g, g ◦g ◦ g, . . .} has 6 elements.

Exercise 6.39* Suppose that A is a finite set and f : A → A is a bijection. Thenf 1 = f , f 2 = f ◦ f , f 3 = f ◦ f ◦ f ,. . . all are bijections : A → A.
1. Show that, somewhere in this sequence, there occurs the bijection 1 A . I.e.,a number n exists such that f n = 1A .
2. Suppose that A has k elements. Can you give an upper bound for n?

Exercise 6.40 Suppose that h : X → X satisfies h ◦ h ◦ h = 1X . Show that h isa bijection.
Exhibit a simple example of a set X and a function h : X → X such that h◦h◦h =1X , whereas h 6= 1X .

Exercise 6.41 Prove Lemma 6.36.3.

Exercise 6.42 Suppose that f : X → Y and g : Y → Z are such that g ◦ f isbijective. Show that f is surjective iff g is injective.

Exercise 6.43 Suppose that limi→∞ ai = a, and that f : N → N is injective.
Show that limi→∞ af (i) = a. (Cf. also Exercise 8.22, p. 318.)

CHAPTER 6. FUNCTIONS

226

6.4 Inverse Function
If we consider f : X → Y as a relation, then we can consider its relational
inverse: the set of all (y, x) with (x, y) ∈ f . However, there is no guarantee thatthe relational inverse of a function is again a function. In case f is injective, weknow that the relational inverse of f is a partial function (some elements in thedomain may not have an image). If f is also surjective, we know that the relational
inverse of f is a function. Thus, an inverse function of f has to satisfy some special
requirements.

Figure 6.2: Inverse of the function λx.x2 (restricted to R+ ).
Definition 6.44 (Inverse Function) Suppose that f : X → Y . A function g :Y → X is an inverse of f if both (i) g ◦ f = 1X , and (ii) f ◦ g = 1Y .
The next theorem says all there is to know about inverses. Its proof describes how
to find an inverse if there is one.
Note that, by the first part of the theorem, we can safely talk about the inverse of afunction (provided there is one).
Theorem 6.45
1. A function has at most one inverse.
2. A function has an inverse iff it is bijective.
Proof. (1) Suppose that g and h are both inverses of f : X → Y . Then g =1X ◦ g = (h ◦ f ) ◦ g = h ◦ (f ◦ g) = h ◦ 1Y = h.
(2) (Only if.) Assume that g is inverse of f . Then since g ◦ f = 1X is injective, byLemma 6.36.1 f also is injective. And since f ◦ g = 1Y is surjective, by 6.36.2 f
also is surjective.

6.4. INVERSE FUNCTION

227(If.) Suppose that f is a bijection. I.e., for every y ∈ Y there is exactly one x ∈ Xsuch that f (x) = y. Thus, we can define a function g : Y → X by letting g(y) be
the unique x such that f (x) = y.
Then g is inverse of f : firstly, if x ∈ X, then g(f (x)) = x; secondly, if y ∈ Y ,then f (g(y)) = y.
Notation. If f : X → Y is a bijection, then its unique inverse is denoted by f −1 .
Remark. If f : X → Y is a bijection, then f −1 can denote either the inverse
function : Y → X, or the inverse of f considered as a relation. But from the proofof Theorem 6.45 it is clear that these are the same.
Example 6.46 The real function f that is given by f (x) = 59 x + 32 allows us
to convert degrees Celcius into degrees Fahrenheit. The inverse function f −1 isgiven by f −1 (x) = 95 (x − 32); it converts degrees Fahrenheit back into degrees
Celsius. Here are integer approximations:c2f, f2c :: Int -> Int
c2f x = div (9 * x) 5 + 32
f2c x = div (5 * (x - 32)) 9*Left and Right-inverse. Note that there are two requirements on inverse functions. If f : X → Y , g : Y → X, and we have that g ◦ f = 1X only, then g iscalled left-inverse of f and f right-inverse of g.
Example 6.47 The class Enum is (pre-)defined in Haskell as follows:class Enum a where
succ, pred
toEnumfromEnum:: a -> a:: Int -> a:: a -> Int

fromEnum should be a left-inverse of toEnum:CHAPTER 6. FUNCTIONS

228
fromEnum (toEnum x) = x

This requirement cannot be expressed in Haskell, so it is the responsibility of the
programmer to make sure that it is satisfied. Examples of use of toEnum andfromEnum from Prelude.hs:

ord
ord
chr
chr

:: Char -> Int= fromEnum:: Int -> Char= toEnum

Exercise 6.48 Show: if f : X → Y has left-inverse g and right-inverse h, then f
is a bijection and g = h = f −1 .
Exercise 6.49 Suppose that f : X → Y and g : Y → X. Show that the following
are equivalent.
1. g ◦ f = 1X ,2. {(f (x), x) | x ∈ X} ⊆ g.

Exercise 6.50 X = {0, 1}, Y = {2, 3, 4, 5}, f = {(0, 3), (1, 4)}. We have thatf : X → Y . How many functions g : Y → X have the property, that g ◦ f = 1 X ?

Exercise 6.51 Give an example of an injection f : X → Y for which there is nog : Y → X such that g ◦ f = 1X .

Exercise 6.52* Show that if f : X → Y is surjective, a function g : Y → X exists
such that f ◦ g = 1Y .

6.5. PARTIAL FUNCTIONS

229

Exercise 6.53 How many right-inverses are there to the function{(0, 5), (1, 5), (2, 5), (3, 6), (4, 6)}(domain: {0, 1, 2, 3, 4}, co-domain: {5, 6})?
Exercise 6.54
1. The surjection f : R → R+ is defined by f (x) = x2 . Givethree different right-inverses for f .
2. Same question for g : [0, π] → [0, 1] defined by g(x) = sin x.
Exercise 6.55 Suppose that f : X → Y is a surjection and h : Y → X. Show
that the following are equivalent.
1. h is right-inverse of f ,2. h ⊆ {(f (x), x) | x ∈ X}.

Exercise 6.56* Show:
1. Every function that has a surjective right-inverse is a bijection.
2. Every function that has an injective left-inverse is a bijection.

6.5 Partial Functions
A partial function from X to Y is a function with its domain included in X and
its range included in Y . If f is a partial function from X to Y we write this as f :X ,→ Y . It is immediate from this definition that f : X ,→ Y iff dom (f ) ⊆ X
and f  dom (f ) : dom (f ) → Y .
A way of defining a partial function (using ⊥ for ‘undefined’):⊥ if ...
f (x) =totherwise

230

CHAPTER 6. FUNCTIONS

The computational importance of partial functions is in the systematic perspective
they provide on exception handling. In Haskell, the crude way to deal with exceptions is by a call to the error abortion function error. The code below implements
partial functions succ0 and succ1. succ0 is partial because the pattern (x+1)
only matches positive integers. succ1 has an explicit call to error. The disadvantage of these implementations is that they are called by another program, theexecution of that other program may abort.

succ0 :: Integer -> Integer
succ0 (x+1) = x + 2
succ1 :: Integer -> Integersucc1 = \ x -> if x < 0then error "argument out of range"else x+1

This uses the reserved keywords if, then and else, with the obvious meanings.
A useful technique for implementing partial functions is to represent a partial function from type a to type b as a function of type a -> [b]. In case of an exception,
the empty list is returned. If a regular value is computed, the unit list with thecomputed value is returned.

succ2 :: Integer -> [Integer]succ2 = \ x -> if x < 0 then [] else [x+1]Composition of partial functions implemented with unit lists can be defined asfollows:

pcomp :: (b -> [c]) -> (a -> [b]) -> a -> [c]pcomp g f = \ x -> concat [ g y | y <- f x ]As an alternative to this trick with unit lists Haskell has a special data type for implementing partial functions, the data type Maybe, which is predefined as follows.

6.5. PARTIAL FUNCTIONS

231

data Maybe a = Nothing | Just aderiving (Eq, Ord, Read, Show)maybe:: b -> (a -> b) -> Maybe a -> bmaybe n f Nothing = nmaybe n f (Just x) = f x

Here is a third implementation of the partial successor function:

succ3 :: Integer -> Maybe Integer
succ3 = \ x -> if x < 0 then Nothing else Just (x+1)

The use of the predefined function maybe is demonstrated in the definition of composition for functions of type a -> Maybe b.

mcomp :: (b -> Maybe c) -> (a -> Maybe b) -> a -> Maybe c
mcomp g f = (maybe Nothing g) . f

Of course, the maybe function allows for all kinds of ways to deal with exceptions. E.g., a function of type a -> Maybe b can be turned into a function of typea -> b by the following part2error conversion.

part2error :: (a -> Maybe b) -> a -> b
part2error f = (maybe (error "value undefined") id) . f

Exercise 6.57 Define a partial function
stringCompare :: String -> String -> Maybe Ordering

232

CHAPTER 6. FUNCTIONS

for ordering strings consisting of alphabetic characters in the usual list order. If a
non-alphabetic symbol occurs, the ordering function should return Nothing. Use
isAlpha for the property of being an alphabetic character.

6.6 Functions as Partitions
In practice, equivalences are often defined by way of functions. Par abus de language functions sometimes are called partitions for that reason. Examples of suchfunctions on the class of all people: “the gender of x” (partitions in males and females), “the color of x” (partitions in races), “the age of x” (some hundred equivalence classes). The next exercise explains how this works and asks to show thatevery equivalence is obtained in this way.
Exercise 6.58 Suppose that f : A → I is a surjection. Define the relation R on A
by: aRb :≡ f (a) = f (b). Thus, R = {(a, b) ∈ A2 | f (a) = f (b)}. Show:
1. R is an equivalence on A,
2. A/R = {f −1 [{i}] | i ∈ I},
3. for every equivalence S on A there is a function g on A such that aSb ⇔g(a) = g(b).

Example 6.59 For any n ∈ Z with n 6= 0, let the function RMn :: Z → Z
be given by RMn (m) := r where 0 6 r < n and there is some a ∈ Z with
m = an + r. Then RMn induces the equivalence ≡n on Z.
Here is a Haskell implementation of a procedure that maps a function to the equivalence relation inducing the partition that corresponds with the function:

fct2equiv :: Eq a => (b -> a) -> b -> b -> Bool
fct2equiv f x y = (f x) == (f y)

You can use this to test equality modulo n, as follows:

6.6. FUNCTIONS AS PARTITIONS233

FCT> fct2equiv (‘rem‘ 3) 2 14
True

Exercise 6.60* Suppose that f : A → B.
1. Show: If f is an injection, then for all sets C and for every g : A → C there
is a function h : B → C such that g = h ◦ f .
2. Show: For all sets C, if to every g : A → C there is a function h : B → C
such that g = h ◦ f , then f is an injection.

Exercise 6.61* Suppose that R is an equivalence on the set A. Show: for every
equivalence S ⊇ R on A there exists a function g : A/R → A/S such that, fora ∈ A: | a |S = g(| a |R ).
Exercise 6.62* Suppose that ∼ is an equivalence on A, and that f : A 2 → A is abinary function such that for all a, b, x, y ∈ A:a ∼ x ∧ b ∼ y =⇒ f (a, b) ∼ f (x, y).
Show that a unique function f ∼ : (A/∼)2 → B exists such that, for a, b ∈ A:
f ∼ (|a|, |b|) = |f (a, b)|.
Exercise 6.63* Suppose that ∼ is an equivalence on A, and that R ⊆ A 2 is arelation such that for all a, b, x, y ∈ A:a ∼ x ∧ b ∼ y ∧ aRb ⇒ xRy.
Show that a unique relation R∼ ⊆ (A/ ∼)2 exists such that for all a, b ∈ A:
|a|R∼ |b| ⇔ aRb.
Exercise 6.64* A and B are sets, with B 6= ∅. Define ∼ on A × B by: (a, b) ∼(x, y) ≡ a = x.
1. Show that ∼ is an equivalence on A × B.
2. Exhibit a bijection : (A×B)/ ∼ −→ A from the quotient of A×B modulo∼ to A.

234

CHAPTER 6. FUNCTIONS

3. Exhibit, for every equivalence class, a bijection between the class and B.

Equivalence classes (restricted to a list) for an equivalence defined by a function
are generated by the following Haskell function:

block :: Eq b => (a -> b) -> a -> [a] -> [a]block f x list = [ y | y <- list, f x == f y ]This gives:
FCT> block (‘rem‘ 3) 2 [1..20][2,5,8,11,14,17,20]
FCT> block (‘rem‘ 7) 4 [1..20][4,11,18]

Exercise 6.65 Functions can be used to generate equivalences, or equivalently,
partitions. In an implementation we use list partitions; see Exercise 5.109 for adefinition. Implement an operation fct2listpart that takes a function and adomain and produces the list partition that the function generates on the domain.
Some example uses of the operation are:Main> fct2listpart even [1..20][[1,3,5,7,9,11,13,15,17,19],[2,4,6,8,10,12,14,16,18,20]]
Main> fct2listpart (\ n -> rem n 3) [1..20][[1,4,7,10,13,16,19],[2,5,8,11,14,17,20],[3,6,9,12,15,18]]

Exercise 6.66 Give an formula for the number of surjections from an n-element
set A to a k-element set B. (Hint: each surjection f : A → B induces a partition.
These partitions can be counted with the technique from Example 5.103.)

6.7 Products
Definition 6.67 (Product)
Q Suppose that, for every element i ∈ I a non-empty set
Xi is given. The product i∈I Xi is the set of all functions f for which dom (f ) =I and such that for all i ∈ I: f (i) ∈ Xi .

6.7. PRODUCTS

235

When I = {0, . . . , n − 1}, this product is also written as X0 × · · · × Xn−1 .

If all Xi (i ∈ I) are the same, Xi = X, the product is written as X I . Thus, X I is
the set of all functions f : I → X.
Exercise
6.68* There are now two ways to interpret the expression X 0 × X1 : (i)Q
as i∈{0,1} Xi , and (ii) as {(x, y) | x ∈ X0 ∧ y ∈ X1 }. Can you explain why
there is no harm in this?
In our implementation language, product types have the form (a,b), (a,b,c),
etcetera.
Exercise 6.69* Let A be any set. Exhibit two different bijections between ℘(A)and {0, 1}A.
Exercise 6.70* Suppose that X and Y are sets. On the set of functions Y X ={f | f : X → Y }, the relation ≈ is defined by: f ≈ g ≡ there are bijections
i : Y → Y and j : X → X such that i ◦ f = g ◦ j.
1. Show that ≈ is an equivalence.
2. Show: if f, g : X → Y are injective, then f ≈ g.
3. Suppose that Y = {0, 1, 2} and X = {0, 1, 2, 3}.
(a) Show that {(0, 0), (1, 0), (2, 1)} ≈ {(0, 1), (1, 3), (2, 3)}.

(b) How many equivalence classes has ≈? For every class, produce one
representative.

Exercise 6.71* Suppose that X, Y and Z are sets and that h : Y → Z. Define
F : Y X → Z X by F (g) := h ◦ g. Show:
1. if h is injective, then F is injective,
2. if h is surjective, then F is surjective.

Exercise 6.72* Suppose that X 6= ∅, Y , and Z are sets and that h : X → Y .
Define F : Z Y → Z X by F (g) := g ◦ h. Show:

CHAPTER 6. FUNCTIONS

236
1. if h is injective, then F is surjective,
2. if h is surjective, then F is injective.

6.8 Congruences
A function f : X n → X is called an n-ary operation on X. Addition and multiplication are binary operations on N (on Z, on Q, on R, on C).
If one wants to define new structures from old, an important method is taking
quotients for equivalences that are compatible with certain operations.
Definition 6.73 (Congruence) If f be an n-ary operation on A, and R an equivalence on A, then R is a congruence for f (or: R is compatible with f ) if forall
x1 , . . . , xn , y1 , . . . , yn ∈ A : x1 Ry1 , . . . , xn Ryn
imply thatf (x1 , . . . , xn )Rf (y1 , . . . , yn ).
If R is a congruence for f , then the operation induced by f on A/R is the operationfR : (A/R)n → A/R given byfR (|a1 |R , . . . , |an |R ) := |f (a1 , . . . , an )|R .
If (A, f ) is a set with an operation f on it and R is a congruence for f , then
(A/R, fR ) is the quotient structure defined by R.
Example 6.74 Consider the modulo n relation on Z. Suppose m ≡ n m0 andk ≡n k 0 . Then (Proposition 5.65) there are a, b ∈ Z with m0 = m + an andk 0 = k + bn. Thus m0 + k 0 = m + k + (a + b)n, i.e., m + k ≡n m0 + k 0 . Thisshows that ≡n is a congruence for addition. Similarly, it can be shown that ≡n is acongruence for subtraction. It follows that we can define [m]n + [k]n := [m + k]nand [m]n − [k]n := [m − k]n .
Exercise 6.75 Show that ≡n on Z is a congruence for multiplication, for any n ∈Z with n 6= 0.

6.8. CONGRUENCES

237

Example 6.76 Is ≡n on Z (n 6= 0) a congruence for exponentiation, i.e., is it
possible to define exponentiation of classes in Zn by means of:([k]n )([m]n ):= [k m ]n , for k ∈ Z, m ∈ N?

No, for consider the following example: ([2]3 )([1]3 ) = [21 ]3 = [2]3 . Since 1 ≡3 4
we also have: ([2]3 )([1]3 ) = ([2]3 )([4]3 ) = [24 ]3 = [16]3 = [1]3 6= [2]3 . What
this shows is that the definition is not independent of the class representatives.
Therefore, ≡n is not a congruence for exponentiation.
Example 6.77 The definition of the integers from the natural numbers, in Section
7.2 below, uses the fact that the relation R on N2 given by(m1 , m2 )R(n1 , n2 ) :≡ m1 + n2 = m2 + n1
is an equivalence relation, that the relation <R on N2 /R given by|(m1 , m2 )|R <R |(n1 , n2 )|R :≡ m1 + n2 < m2 + n1
is properly defined, and moreover, that R is a congruence for addition and multiplication on N2 .
To check that R is a congruence for addition on N2 , where addition on N2 is givenby(m1 , m2 ) + (n1 , n2 ) := (m1 + n1 , m2 + n2 ),we have to show that (m1 , m2 )R(p1 , p2 ) and (n1 , n2 )R(q1 , q2 ) together imply
that (m1 , m2 ) + (n1 , n2 )R(p1 , p2 ) + (q1 , q2 ).
Assume (m1 , m2 )R(p1 , p2 ) and (n1 , n2 )R(q1 , q2 ). Applying the definition of R,
this gives
m1 + p2 = p1 + m2 and n1 + q2 = q1 + n2 ,whencem1 + n 1 + p 2 + q 2 = n 1 + p 1 + m 2 + n 2 .
We now have:(m1 , m2 ) + (n1 , n2 ) = (m1 + n1 , m2 + n2 ),(p1 , p2 ) + (q1 , q2 ) = (p1 + q1 , p2 + q2 ),and by the definition of R we get from (*) that
(m1 , m2 ) + (n1 , n2 )R(p1 , p2 ) + (q1 , q2 ).
This proves that R is a congruence for addition.

(*)

CHAPTER 6. FUNCTIONS

238

Exercise 6.78* Show that the relation R on N2 from example 6.77 is a congruence
for the multiplication operation on N2 given by:(m1 , m2 ) · (n1 , n2 ) := (m1 n1 + m2 n2 , m1 n2 + n1 m2 ).

6.9 Further Reading
More on functions in the context of set theory in [DvDdS78]. A logical theory offunctions is developed in the lambda calculus. See [Bar84].

Chapter 7

Induction and Recursion
Preview
A very important proof method that is not covered by the recipes from Chapter 3 isthe method of proof by Mathematical Induction. Roughly speaking, mathematicalinduction is a method to prove things about objects that can be built from a finite
number of ingredients in a finite number of steps. Such objects can be thought ofas construed by means of recursive definitions. Thus, as we will see in this chapter,recursive definitions and inductive proofs are two sides of one coin.

module IAR
whereimport List
import STAL (display)

7.1 Mathematical Induction
Mathematical induction is a proof method that can be used to establish the truth ofa statement for an infinite sequence of cases 0, 1, 2, . . . . Let P (n) be a property of
239

240

CHAPTER 7. INDUCTION AND RECURSION

natural numbers. To prove a goal of the form ∀n ∈ N : P (n) one can proceed asfollows:
1. Basis. Prove that 0 has the property P .
2. Induction step. Assume the induction hypothesis that n has property P .
Prove on the basis of this that n + 1 has property P .
That’s all. The goal ∀n ∈ N : P (n) follows from this by the principle of mathematical induction.
By the principle of mathematical induction we mean the following fact:
Fact 7.1 For every set X ⊆ N, we have that:
if 0 ∈ X and ∀n ∈ N(n ∈ X ⇒ n + 1 ∈ X), then X = N.
This fact is obviously true.
The best way to further explain mathematical induction is by way of examples.
Example 7.2 Sum of the Angles of a Convex Polygon.
Suppose we want to prove that the sum of the angles of a convex polygon of n + 3
sides is (n + 1)π radians.
@
@
@
@

@
@

We can show this by mathematical induction with respect to n, as follows:
Basis For n = 0, the statement runs: the sum of the angles of a convex polygon of
3 sides, i.e., of a triangle, is π radians. We know from elementary geometrythat this is true.
Induction step Assume that the sum of the angles of a convex polygon of n + 3
sides is (n + 1)π radians. Take a convex polygon P of n + 4 sides. Then,since P is convex, we can decompose P into a triangle T and a convex
polygon P 0 of n + 3 sides (just connect edges 1 and 3 of P ). The sum of theangles of P equals the sum of the angles of T , i.e. π radians, plus the sumof the angles of P 0 , i.e., by the induction hypothesis, (n + 1)π radians. Thisshows that the sum of the angles of P is (n + 2)π radians.

7.1. MATHEMATICAL INDUCTION

241

From 1. and 2. and the principle of mathematical induction the statement follows.

Notation The next examples all involve sums of the general form a 1 + a2 +n
· · · + an , written in summation notation (cf. Example
P 2.29 above) as Σ k=1 ak .
Note that sum is the computational counterpart to . We agree that the “emptysum” Σ0k=1 ak yields 0. The same convention is followed in computations withsum, for sum [] has value 0.
Example 7.3 The Sum of the First n Odd Numbers.
The sum of the first n odd numbers equals n2 . More formally:nX
k=1(2k − 1) = n2 .

Proof by induction:
P
Basis For n = 1, we have 1k=1 (2k − 1) = 1 = 12 .
Pn
Pn+1
Induction step Assume k=1 (2k −1) = n2 . We have to show k=1 (2k −1) =PPn(n + 1)2 . Indeed, n+1
k=1 (2k − 1) =k=1 (2k − 1) + 2(n + 1) − 1. Using
the induction hypothesis, this gives:
n+1
X
k=1

(2k − 1) = n2 + 2n + 1 = (n + 1)2 .

The closed formula gives us an improved computation procedure for summing theodd numbers: sumOdds performs better on large input than sumOdds’.

sumOdds’ :: Integer -> Integer
sumOdds’ n = sum [ 2*k - 1 | k <- [1..n] ]sumOdds :: Integer -> Integer
sumOdds n = n^2Note that the method of proof by mathematical induction may obscure the processof finding the relation in the first place. To see why the sum of the first n odd
numbers equals n2 , it is instructive to look at the following picture of the sum
1 + 3 + 5 + 7 + 9.

242

CHAPTER 7. INDUCTION AND RECURSION
q
q
q
q
q

q
q
q
q
q

q
q
q
q
q

q
q
q
q
q

q
q
q
q
q

Example 7.4 The Sum of the First n Even Numbers.
What about the sum of the first n even natural numbers? Again, a picture suggests
the answer. Look at the following representation of 2 + 4 + 6 + 8 + 10:

q
q
q
q
q

q
q
q
q
q

q
q
q
q
q

q
q
q
q
q

q
q
q
q
q

q
q
q
q
q

Again, a picture
P is not (quite) a proof. Here is proof by mathematical induction of
the fact that nk=1 2k = n(n + 1).

Basis Putting k = 1 gives 2 = 1 · 2, which is correct.

Pn
Pn+1
Pn
Induction step Assume k=1 2k = n(n + 1). Then k=1 2k = k=1 2k +2(n + 1). Using the induction hypothesis we see that this is equal to n(n +1) + 2(n + 1) = n2 + 3n + 2 = (n + 1)(n + 2), and we are done.
Notice that we left the termPfor k = 0 out of the sum. We might as well haven
included it, for it holds that k=0 2k = n(n + 1). By the convention about empty
sums, the two versions are equivalent.
From what we found for the sum of the first n even natural numbers the formulafor the sum of the first n positive natural numbers follows immediately:
n
X
k=1

k=

n(n + 1).
2

Again, we get improved computation procedures from the closed forms that we
found:

7.1. MATHEMATICAL INDUCTION

243

sumEvens’ :: Integer -> Integer
sumEvens’ n = sum [ 2*k | k <- [1..n] ]sumEvens :: Integer -> Integer
sumEvens n = n * (n+1)
sumInts :: Integer -> Integer
sumInts n = (n * (n+1)) ‘div‘ 2

Example 7.5 Summing Squares.
Consider the problem of finding a closed formula for the sum of the first n squares(a closed formula, as opposed to a recurrence f (0) = 0, f (n) = f (n−1)+n 2 , n >0). By direct trial one might find the following:
12 + 2 2 = 5 =

2·3·5
.
6

3·4·7
.
6
4·5·9
12 + 22 + 32 + 42 = 14 + 16 = 30 =.
6
5 · 6 · 11
12 + 22 + 32 + 42 + 52 = 30 + 25 = 55 =.
6
12 + 22 + 32 = 5 + 9 = 14 =This suggests a general rule:
12 + . . . + n 2 =n(n + 1)(2n + 1).
6

But the trial procedure that we used to guess the rule is different from the procedurethat is needed to prove it.
Note that the fact that one can use mathematical induction to prove a rule gives no
indication about how the rule was found in the first place. We will return to theissue of guessing closed forms for polynomial sequences in Sections 9.1 and 9.2.
In Haskell, you can compute sums of squares in a naive way or in a sophisticatedway, as follows:

CHAPTER 7. INDUCTION AND RECURSION

244

sumSquares’ :: Integer -> Integer
sumSquares’ n = sum [ k^2 | k <- [1..n] ]sumSquares :: Integer -> Integer
sumSquares n = (n*(n+1)*(2*n+1)) ‘div‘ 6

Again, the insight that the two computation procedures will always give the same
result can be proved by mathematical induction:
Exercise 7.6 Summing Squares. Prove by mathematical induction:
n
X

k2 =

k=1

n(n + 1)(2n + 1).
6

Example 7.7 Summing Cubes.
Let us move on to the problem of summing cubes. By direct trial one finds:
13 + 23 = 9 = (1 + 2)2 .
13 + 23 + 33 = 9 + 27 = 36 = (1 + 2 + 3)2 .
13 + 23 + 33 + 43 = 36 + 64 = 100 = (1 + 2 + 3 + 4)2 .
13 + 23 + 33 + 43 + 53 = 100 + 125 = 225 = (1 + 2 + 3 + 4 + 5)2 .
This suggests a general rule:
13 + · · · + n3 = (1 + · · · + n)2 .
We saw in Example 7.4 that
n
X
k=1

k=

n(n + 1),
2

so the general rule reduces to:
13 + · · · + n 3 =n(n + 1)
2

2.

So much about finding a rule for the sum of the first n cubes. In Sections 9.1
and 9.2 we will give an algorithm for generating closed forms for polynomial
sequences.

7.1. MATHEMATICAL INDUCTION

245

What we found is that sumCubes defines the same function as the naive procedure
sumCubes’ for summing cubes:

sumCubes’ :: Integer -> Integer
sumCubes’ n = sum [ k^3 | k <- [1..n] ]sumCubes :: Integer -> Integer
sumCubes n = (n*(n+1) ‘div‘ 2)^2

Again, the relation we found suggests a more sophisticated computation procedure, and proving the general relationship between the two procedures is anotherexercise in mathematical induction.
Exercise 7.8 Summing Cubes.
Prove by mathematical induction:
n
X

k=1

k3 =



n(n + 1)
2

2.

Exercise 7.9 Prove that for all n ∈ N: 32n+3 + 2n is divisible by 7.
Remark. If one compares the proof strategy needed to establish a principle of theform ∀n ∈ N : P (n) with that for an ordinary universal statement ∀x ∈ A : P (x),where A is some domain of discourse, then the difference is that in the formercase we can make use of what we know about the structure of N. In case weknow nothing about A, and we have to prove P for an arbitrary element from A,
we have to take our cue from P . In case we have to prove something about anarbitrary element n from N we know a lot more: we know that either n = 0 or
n can be reached from 0 in a finite number of steps. The key property of N thatmakes mathematical induction work is the fact that the relation < on N is wellfounded: any sequence m0 > m1 > m2 > · · · terminates. This guarantees theexistence of a starting point for the induction process.
For any A that is well-founded by a relation ≺ the following principle holds. Let
X ⊆ A. If∀a ∈ A(∀b ≺ a(b ∈ X) ⇒ a ∈ X),

then X = A. In Section 11.1 we will say more about the use of well-foundedness
as an induction principle.

246

CHAPTER 7. INDUCTION AND RECURSION

7.2 Recursion over the Natural Numbers
Why does induction over the natural numbers work? Because we can think of any
natural number n as the the result of starting from 0 and applying the successor operation +1 a finite number of times. Let us use this fact about the natural numbers
to give our own Haskell implementation, as follows:data Natural = Z | S Naturalderiving (Eq, Show)

Here Z is our representation of 0, while S n is a representation of n + 1. The
number 4 looks in our representation like S (S (S (S Z))).
The symbol | is used to specify alternatives for the data construction.
With deriving (Eq, Show) one declares Natural as a type in the classes Eq
and Show. This ensures that objects of the type can be compared for equality anddisplayed on the screen without further ado.
We can define the operation of addition on the natural numbers recursively in termsof the successor operation +1 and addition for smaller numbers:
m + 0 := m
m + (n + 1) := (m + n) + 1
This definition of the operation of addition is called recursive because the operation + that is being defined is used in the defining clause, but for a smaller second
argument. Recursive definitions always have a base case (in the example: the firstline of the definition, where the second argument of the addition operation equals0) and a recursive case (in the example: the second line of the definition, where
the second argument of the addition operation is greater than 0, and the operationthat is being defined appears in the right hand side of the definition).
In proving things about recursively defined objects the idea is to use mathematicalinduction with the basic step justified by the base case of the recursive definition,
the induction step justified by the recursive case of the recursive definition. Thiswe will now demonstrate for properties of the recursively defined operation of
addition for natural numbers.
Here is the Haskell version of the definition of +, in prefix notation:

7.2. RECURSION OVER THE NATURAL NUMBERS

247

plus m Z = mplus m (S n) = S (plus m n)

If you prefer infix notation, just write m ‘plus‘ n instead of plus m n. The
back quotes around plus transform the two placed prefix operator into an infix
operator. This gives the following equivalent version of the definition:m ‘plus‘ Z = mm ‘plus‘ (S n) = S (m ‘plus‘ n)Now, with diligence, we can prove the following list of fundamental laws of addition from the definition.
m+0
m+n
m + (n + k)

=
m
=
n+m
= (m + n) + k(0 is identity element for +)(commutativity of +)(associativity of +)The first fact follows immediately from the definition of +.
In proving things about a recursively defined operator ⊕ it is convenient to be able
to refer to clauses in the recursive definition, as follows: ⊕.1 refers to the firstclause in the definition of ⊕, ⊕.2 to the second clause, and so on.
Here is a proof by mathematical induction of the associativity of +:Proposition 7.10 ∀m, n, k ∈ N : (m + n) + k = m + (n + k).
Proof. Induction on k.
Basis (m + n) + 0

+.1

=

+.1

m + n = m + (n + 0).

Induction step Assume (m+n)+k = m+(n+k). We show (m+n)+(k+1) =m + (n + (k + 1)):(m + n) + (k + 1)

+.2

=

i.h.

((m + n) + k) + 1= (m + (n + k)) + 1+.2= m + ((n + k) + 1)+.2
= m + (n + (k + 1)).

CHAPTER 7. INDUCTION AND RECURSION

248

The inductive proof of commutativity of + uses the associativity of + that we justestablished.
Proposition 7.11 ∀m, n ∈ N : m + n = n + m.
Proof. Induction on n.
Basis Induction on m.
Basis 0 + 0 = 0 + 0.
Induction Step Assume m+0 = 0+m. We show (m+1)+0 = 0+(m+1):+1(m + 1) + 0

=

m+1

+1

=

(m + 0) + 1

ih(0 + m) + 1=
prop 7.10=

0 + (m + 1).

Induction step Assume m + n = n + m. We show m + (n + 1) = (n + 1) + m:
m + (n + 1)+.2=

(m + n) + 1

ih=(n + m) + 1+.2=

n + (m + 1)

ihn + (1 + m)=
prop 7.10=(n + 1) + m.

Once we have addition, we can define multiplication in terms of it, again following
a recursive definition:m · 0 := 0m · (n + 1) := (m · n) + m
We call · the multiplication operator. It is common to use mn as shorthand form · n, or, in other words, one usually does not write the multiplication operator.

7.2. RECURSION OVER THE NATURAL NUMBERS

249

Here is a Haskell implementation for our chosen representation (this time we give
just the infix version):m ‘mult‘ Z = Zm ‘mult‘ (S n) = (m ‘mult‘ n) ‘plus‘ m

Let us try this out:
IAR> (S (S Z)) ‘mult‘ (S (S (S Z)))
S (S (S (S (S (S Z)))))

The following laws hold for ·, and for the interaction of · and +:m·1=
mm · (n + k) = m · n + m · k
m · (n · k) =(m · n) · k
m·n
=
n·m

(1 is identity element for ·)(distribution of · over +)(associativity of ·)(commutativity of ·)Exercise 7.12 Prove these laws from the recursive definitions of + and ·, plus thelaws that were established about +.
If we now wish to implement an operation expn for exponentiation on naturals,
the only thing we have to do is find a recursive definition of exponentiation, andimplement that. Here is the definition:
m0
m

n+1

:= 1
:= (mn ) · m

This leads immediately to the following implementation:

expn m Z = (S Z)
expn m (S n) = (expn m n) ‘mult‘ mThis gives:
IAR> expn (S (S Z)) (S (S (S Z)))
S (S (S (S (S (S (S (S Z)))))))

CHAPTER 7. INDUCTION AND RECURSION

250

Exercise 7.13 Prove by mathematical induction that k m+n = k m · k n .
We can define the relation 6 on N as follows:

m 6 n :≡

there is a k ∈ N : m + k = n

Instead of m 6 n we also write n > m, with the same meaning. We use m < nfor m 6 n and m 6= n. Instead of m < n we also write n > m.
This shows that we can define < or 6 in terms of addition. On further reflection,
successor (+1) is enough to define 6, witness the following recursive definition:
0 6 m,
m+16n+1

if

m 6 n.

This is translated into Haskell as follows:

leq Z _ = Trueleq (S _) Z = False
leq (S m) (S n) = leq m n

Note the use of _ for an anonymous variable: leq Z _ = True means that leq,
applied to the zero element Z and to anything else, gives the value True. The
expression (S _) specifies a pattern: it matches any natural that is a successor
natural.
To fully grasp the Haskell definition of leq one should recall that the three equations that make up the definition are read as a list of three mutually exclusive
cases. The Haskell system tries the equations one by one, from top to bottom, until it finds one that applies. Thus, leq Z _ applies to any pair of naturals with thefirst member equal to Z. leq (S _) Z applies to any pair of naturals with the firstone starting with an S, and the second one equal to Z. The third equation, finally,
applies to those pairs of naturals where both the first and the second member ofthe pair start with an S
We can now define geq, gt, lt, in terms of leq and negation:

7.3. THE NATURE OF RECURSIVE DEFINITIONS

251

geq m n = leq n m
gt m n = not (leq m n)
lt m n = gt n m

Exercise 7.14 Implement an operation for cut-off subtraction subtr on naturals:
the call subtr (S (S (S Z))) (S (S (S (S Z)))) should yield Z.
Exercise 7.15 Implement operations quotient and remainder on naturals. Dividing a by b yields quotient q and remainder r with 0 6 r < b, according to theformula a = q · b + r. (Hint: you will need the procedure subtr from Exercise
7.14.)

7.3 The Nature of Recursive Definitions
Not any set of equations over natural numbers can serve as a definition of an operation on natural numbers. Consider
f (0) := 1f (n + 1) := f (n + 2).
This does not define unique values for f (1), f (2), f (3), . . . , for the equations only
require that all these values should be the same, not what the value should be.
The following format does guarantee a proper definition:f (0) := cf (n + 1) := h(f (n)).
Here c is a description of a value (say of type A), and h is a function of type
A → A. A definition in this format is said to be a definition by structural recursion
over the natural numbers. The function f defined by this will be of type N → A.
This format is a particular instance of a slightly more general one called primitive
recursion over the natural numbers. Primitive recursion allows c to depend on anumber of parameters, so the function f will also depend on these parameters. But
we will leave those parameters aside for now.
Definition by structural recursion of f from c and h works like this: take a natural
number n, view it as· + 1+} 0,|1 + · ·{zn times

252

CHAPTER 7. INDUCTION AND RECURSION

replace 0 by c, replace each successor step 1+ by h, and evaluate the result:h( · · · (h(c))··).
| {z }
n timesThis general procedure is easily implemented in an operation foldn, defined asfollows:

foldn :: (a -> a) -> a -> Natural -> afoldn h c Z = cfoldn h c (S n) = h (foldn h c n)

Here is a first example application; please make sure you understand how and why
this works.

exclaim :: Natural -> String
exclaim = foldn (’!’:) []Now a function ‘adding m’ can be defined by taking m for c, and successor for h.
For this we should be able to refer to the successor function as an object in its own
right. Well, it is easily defined by lambda abstraction as (\ n -> S n). This isused in our alternative definition of plus, in terms of foldn. Note that there is noneed to mention the two arguments of plus in the definition.

plus :: Natural -> Natural -> Naturalplus = foldn (\ n -> S n)

Similarly, we can define an alternative for mult in terms of foldn. The recipe forthe definition of mult m (‘multiply by m’) is to take Z for c and plus m for h:mult :: Natural -> Natural -> Natural
mult m = foldn (plus m) Z

7.3. THE NATURE OF RECURSIVE DEFINITIONS

253

Finally, for exponentiation expn m (‘raise m to power . . . ’) we take (S Z) for c
and mult m for h:expn :: Natural -> Natural -> Natural
expn m = foldn (mult m) (S Z)

Exercise 7.16 Implement the operation of cut-off subtraction ( subtr m for ‘subtract from m’; Exercise (7.14)) in terms of foldn and a function for predecessor,
on the basis of the following definition:x −̇ 0 := xx −̇ (n + 1) := p(x −̇ n),
where p is the function for predecessor given by p(0) := 0, p(n + 1) := n. Callthe predecessor function pre.
Exercise 7.17 The Fibonacci numbers are given by the following recursion:
F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn for n > 0.
This gives:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, . . .
Prove with induction that for all n > 1:
Fn+1 Fn−1 − Fn2 = (−1)n .
Exercise 7.18 A bitlist is a list of zeros and ones. Consider the following code
bittest for selecting the bitlists without consecutive zeros.

bittest :: [Int] -> Boolbittest []= Truebittest [0]= Truebittest (1:xs)= bittest xs
bittest (0:1:xs) = bittest xs
bittest _= FalseCHAPTER 7. INDUCTION AND RECURSION254

1. How many bitlists of length 0 satisfy bittest? How many bitlists of length
1 satisfy bittest? How many bitlists of length 2 satisfy bittest? How
many bitlists of length 3 satisfy bittest?
2. Let an be the number of bitlists of length n without consecutive zeros. Givean induction proof to show that for every n > 0 it holds that an = Fn+2 ,where Fn is the n-th Fibonacci number. Take care: you will need two basecases (n = 0, n = 1), and an induction hypothesis of the form: “assume thatthe formula holds for n and for n + 1.” A further hint: the code for bittest
points the way to the solution.

Exercise 7.19* Consider the following two definitions of the Fibonacci numbers(Exercise 7.17):

fib 0 = 0
fib 1 = 1
fib n = fib (n-1) + fib (n-2)
fib’ n = fib2 0 1 n where
fib2 a b 0 = a
fib2 a b n = fib2 b (a+b) (n-1)

Use an induction argument to establish that fib and fib’ define the same function.
Hint: establish the more general claim that for all i,n it holds thatfib2 (fib i) (fib (i+1)) n = fib (i+n)
by induction on n.
Exercise 7.20 The Catalan numbers are given by the following recursion:
C0 = 1, Cn+1 = C0 Cn + C1 Cn−1 + · · · + Cn−1 C1 + Cn C0 .
This gives:[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, . . .
Use this recursion to give a Haskell implementation. Can you see why this is not
a very efficient way to compute the Catalan numbers?

7.4. INDUCTION AND RECURSION OVER TREES

255

Exercise 7.21 Let x0 , . . . , xn be a sequence of n + 1 variables. Suppose theirproduct is to be computed by doing n multiplications. The number of ways todo the multiplications corresponds to the number of bracketings for the sequence.
For instance, if n = 3 there are four variables x0 , x1 , x2 , x3 , and five possible
bracketings:(x1 x2 )(x3 x4 ), ((x1 x2 )x3 )x4 , (x1 (x2 x3 ))x4 , x1 ((x2 x3 )x4 ), x1 (x2 (x3 x4 )).
Show that the number of bracketings for n + 1 variables is given by the Catalan number Cn . (Hint: you will need strong induction, so-called because of its
strengthened induction hypothesis. Your induction hypothesis should run: “Forany i with 0 6 i 6 n, for any sequence of i + 1 variables x0 · · · xi it holds that Ci
gives the number of bracketings for that sequence.”
Example 7.22 Balanced sequences of parentheses of length 2n are defined recursively as follows: the empty sequence is balanced; if sequence w is balanced then(w) is balanced; if sequences w and v are balanced then wv is balanced. Thus,())(() is not balanced. The balanced sequences involving 3 left and 3 right parentheses are:()()(), (())(), ()(()), (()()), ((())).
There is a one-to-one mapping between bracketings for sequences of n + 1 variables and balanced sequences of parentheses with 2n parentheses. Let a bracketingfor x0 · · · xn be given. This can be changed into a balanced sequence of parentheses as follows: We illustrate with the example x0 ((x2 x3 )x4 ).
1. Put one extra pair of parentheses around the bracketing: (x0 ((x2 x3 )x4 )).
2. Insert multiplication operators at the appropriate places: (x0 ·((x2 ·x3 )·x4 )).
3. Erase the variables and the left-brackets: ··)·)).
4. Replace the ·’s with left-brackets: (()()).
This mapping gives a one-to-one correspondence between variable bracketings andbalanced parentheses strings, so we get from the previous exercise that there are
Cn different balanced parentheses strings of length 2n.

7.4 Induction and Recursion over Trees
Here is a recursive definition of binary trees:CHAPTER 7. INDUCTION AND RECURSION256• A single leaf node • is a binary tree.
• If t1 and t2 are binary trees, then the result of joining t1 and t2 under a single
node (called the root node) is a binary tree. A notation for this is: (• t 1 t2 )• Nothing else is a binary tree.
The depth of a binary tree is given by:• The depth of • is 0.
• The depth of (• t1 t2 ) is 1+ the maximum of the depths of t1 and t2 .
A binary tree is balanced if it either is a single leaf node •, or it has the form(• t1 t2 ), with both t1 and t2 balanced, and having the same depth.
We see the following: A balanced binary tree of depth 0 is just a single leaf node•, so its number of nodes is 1. A balanced binary tree of depth 1 has one internal
node and two leaves, so it has 3 nodes. A balanced binary tree of depth 2 has 3internal nodes and 4 leaf nodes, so it has 7 nodes. A binary tree of depth 3 has 7internal nodes plus 8 leaf nodes, so it has 15 nodes.
Recursion and induction over binary trees are based on two cases t = • and t =(• t1 t2 ).
Example 7.23 Suppose we want to find a formula for the number of nodes in abalanced binary tree of depth d.
q

q
q

q
q

q

q
q

q q q qqq
q q q qqq qq qq qq

Suppose we want to show in general that the number of nodes of a binary tree of
depth n is 2n+1 − 1. Then a proof by mathematical induction is in order. We
proceed as follows:
Basis If n = 0, then 2n+1 − 1 = 21 − 1 = 1. This is indeed the number of nodesof a binary tree of depth 0.
Induction step Assume the number of nodes of a binary tree of depth n is 2 n+1 −1. We have to show that the number of nodes of a binary tree of depth n + 1
equals 2n+2 − 1.

7.4. INDUCTION AND RECURSION OVER TREES

257

A binary tree of depth n + 1 can be viewed as a set of internal nodes constituting a binary tree of depth n, plus a set of leaf nodes, consisting of two
new leaf nodes for every old leaf node from the tree of depth n. By induction hypothesis, we know that a tree of depth n has 2n+1 − 1 nodes, so atree of depth n + 1 has 2n+1 − 1 internal nodes. It is easy to see that a treeof depth n + 1 has 2n+1 leaf nodes. The total number of nodes of a tree of
depth n + 2 is therefore 2n+1 − 1 + 2n+1 = 2 · 2n+1 − 1 = 2n+2 − 1, andwe have proved our induction step.

To illustrate trees and tree handling a bit further, here is a Haskell definition ofbinary trees, with a procedure for making balanced trees of an arbitrary depth n,
and a procedure for counting their nodes in a naive way.
We use L for a single leaf •. The data declaration specifies that a BinTree either
is an object L (a single leaf), or an object constructed by applying the constructor
N to two BinTree objects (the result of constructing a new binary tree (• t 1 t2 )from two binary trees t1 and t2 ).
The addition deriving Show ensures that data of this type can be displayed on
the screen.

data BinTree = L | N BinTree BinTree deriving Show
makeBinTree :: Integer -> BinTree
makeBinTree 0 = L
makeBinTree (n + 1) = N (makeBinTree n) (makeBinTree n)count :: BinTree -> Integer
count L = 1
count (N t1 t2) = 1 + count t1 + count t2

With this code you can produce binary trees as follows:
IAR> makeBinTree 6
N (N (N (N (N (N L L) (N L L)) (N (N L L) (N L L))) (N (N (N L L) (N
L L)) (N (N L L) (N L L)))) (N (N (N (N L L) (N L L)) (N (N L L) (N LL))) (N (N (N L L) (N L L)) (N (N L L) (N L L))))) (N (N (N (N (N L L)
(N L L)) (N (N L L) (N L L))) (N (N (N L L) (N L L)) (N (N L L) (N LL)))) (N (N (N (N L L) (N L L)) (N (N L L) (N L L))) (N (N (N L L) (N258

CHAPTER 7. INDUCTION AND RECURSIONL L)) (N (N L L) (N L L)))))IAR>

If you want to check that the depth of a the result of maketree 6 is indeed 6,
or that maketree 6 is indeed balanced, here are some procedures. Note that theprocedures follow the definitions of depth and balanced to the letter:

depth :: BinTree -> Integer
depth L = 0
depth (N t1 t2) = (max (depth t1) (depth t2)) + 1

balanced :: BinTree -> Boolbalanced L = Truebalanced (N t1 t2) = (balanced t1)&& (balanced t2)&& depth t1 == depth t2

The programs allow us to check the relation between count (makeBinTree n)and 2^(n+1) - 1 for individual values of n:
IAR>
True

count (makeBinTree 6) ==2^7 - 1

What the proof by mathematical induction provides is an insight that the relation
holds in general. Mathematical induction does not give as clue as to how to find aformula for the number of nodes in a tree. It only serves as a method of proof oncesuch a formula is found.
So how does one find a formula for the number of nodes in a binary tree in thefirst place? By noticing how such trees grow. A binary tree of depth 0 has 1 node,and this node is a leaf node. This leaf grows two new nodes, so a binary tree of
depth 1 has 1 + 2 = 3 nodes. In the next step the 2 leaf nodes grow two new nodes
each, so we get 22 = 4 new leaf nodes, and the number of nodes of a binary treeof depth 2 equals 1 + 2 + 4 = 7. In general, a tree of depth n − 1 is transformed
into one of depth n by growing 2n new leaves, and the total number of leaves of
the new tree is given by 20 + 21 + · · · + 2n . In other words, the number of nodes

7.4. INDUCTION AND RECURSION OVER TREES
of a balanced binary tree of depth n is given by :here is a simple trick:nX
k=0

2k = 2 ·

n
X
k=0

2k −

n
X

k=0

259

Pn

k
k=0 2 . To get a value for this,2k = (2 · 2n + 2n · · · + 2) − (2n + · · · + 1) == 2 · 2n − 1 = 2n+1 − 1.
Example 7.24 Counting the Nodes of a Balanced Ternary Tree.
Now suppose we want to find a formula for the number of nodes in a balanced
ternary tree of depth n. The number of leaves of a balanced ternary tree of depthn isP
3n , so the total number of nodes of a balanced ternary tree of depth n is givennby k=0 3k .
Pnn+1
We prove by induction that k=0 3k = 3 2 −1 .
Basis A ternary tree of depth n = 0 consists of just a single node. And indeed,
0X

3k = 1 =k=0

31 − 1.
2

Induction step Assume that the number of nodes of a ternary tree of depth n is
3n+1 −1. The number of leaf nodes is 3n , and each of these leaves grows 3
2
new leaves to produce the ternary tree of depth n + 1. Thus, the number ofleaves of the tree of depth n + 1 is given by
n+1
X
k=0

3k =

n
X

3k + 3n+1 .

k=0

Using the induction hypothesis, we see that this is equal to
3n+1 − 1 2 · 3n+1
3n+2 − 1
3n+1 − 1+ 3n+1 =
+
=
.
2
2
2
2
Pn
n+1
But how did we get at k=0 3k = 3 2 −1 in the first place? In the same way asin the case of the binary trees:
2·

n
X

k=0

3k = 3 ·

n
X

k=0

3k −

n
X
k=0

3k = (3 · 3n + 3n · · · + 3) − (3n + · · · + 1) == 3 · 3n − 1 = 3n+1 − 1.

CHAPTER 7. INDUCTION AND RECURSION

260
Therefore

n
X

3k =

k=0

3n+1 − 1.
2

Exercise 7.25 Write a Haskell definition of ternary trees, plus procedures for generating balanced ternary trees and counting their node numbers.
Example 7.26 Counting the Nodes of a Balanced m-ary Tree. The
Pnumber ofnodes of a balanced m-ary tree of depth n (with m > 1) is given by nk=0 mk =mn+1 −1m−1 . Here is a proof by mathematical induction.
Basis An m-ary tree of depth 0 consists of just a single node. In fact,
1 = m−1
m−1 .

P0

k=0 m

k

=

Induction step Assume that the number of nodes of an m-ary tree of depth n is
mn+1 −1nm−1 . The number of leaf nodes is m , and each of these leaves grows
m new leaves to produce the ternary tree of depth n + 1. Thus, the numberof leaves of the tree of depth n + 1 is given by
n+1
X

k

m =

n
X

mk + mn+1 .

k=0

k=0

Using the induction hypothesis, we see that this is equal tomn+1 − 1
mn+1 − 1 mn+2 − mn+1
mn+2 − 1+ mn+1 =
+
=
.
m−1
m−1
m−1
m−1
Note that the proofs by mathematical induction do not tell you how to find the formulas for which the induction proof works in the first place. This is an illustrationof the fact that mathematical induction is a method of verification, not a method of
invention. Indeed, mathematical induction is no replacement for the use of creative
intuition in the process of finding meaningful relationships in mathematics.
Exercise 7.27 Geometric Progressions.
Prove by mathematical induction (assuming q 6= 1, q 6= 0):
n
X
k=0

qk =

q n+1 − 1.
q−1

Note that this exercise is a further generalization of Example 7.26.

7.4. INDUCTION AND RECURSION OVER TREES

261

To get some further experience with tree processing, consider the following definition of binary trees with integer numbers at the internal nodes:

data Tree = Lf | Nd Int Tree Tree deriving ShowWe say that such a tree is ordered if it holds for each node N of the tree that the
integer numbers in the left subtree below N are all smaller than the number at nodeN , and the number in the right subtree below N are all greater than the number at
N.
Exercise 7.28 Write a function that inserts a number n in an ordered tree in sucha way that the tree remains ordered.
Exercise 7.29 Write a function list2tree that converts a list of integers to anordered tree, with the integers at the tree nodes. The type is [Int] -> Tree.
Also, write a function tree2list for conversion in the other direction.
Exercise 7.30 Write a function that checks whether a given integer i occurs in anordered tree.
Exercise 7.31 Write a function that merges two ordered trees into a new ordered
tree containing all the numbers of the input trees.
Exercise 7.32 Write a function that counts the number of steps that are needed toreach a number i in an ordered tree. The function should give 0 if i is at the top
node, and −1 if i does not occur in the tree at all.
A general data type for binary trees with information at the internal nodes is givenby:data Tr a = Nil | T a (Tr a) (Tr a) deriving (Eq,Show)

Exercise 7.33 Write a function mapT :: (a -> b) -> Tr a -> Tr b that doesfor binary trees what map does for lists.

262

CHAPTER 7. INDUCTION AND RECURSION

Exercise 7.34 Write a function
foldT :: (a -> b -> b -> b) -> b -> (Tr a) -> bthat does for binary trees what foldn does for naturals.
Exercise 7.35 Conversion of a tree into a list can be done in various ways, depending on when the node is visited:
Preorder traversal of a tree is the result of first visiting the node, next visiting
the left subtree, and finally visiting the right subtree.
Inorder traversal of a tree is the result of first visiting the left subtree, next visiting the node, and finally visiting the right subtree.
Postorder traversal of a tree is the result of first visiting the left subtree, next
visiting the right subtree, and finally visiting the node.
Define these three conversion functions from trees to lists in terms of the foldT
function from Exercise 7.34.
Exercise 7.36 An ordered tree is a tree with information at the nodes given insuch manner that the item at a node must be bigger than all items in the left subtree and smaller than all items in the right subtree. A tree is ordered iff the list
resulting from its inorder traversal is ordered and contains no duplicates. Give an
implementation of this check.
Exercise 7.37 An ordered tree (Exercise 7.36) can be used as a dictionary by
putting items of of type (String,String) at the internal nodes, and defining
the ordering as: (v, w) 6 (v 0 , w0 ) iff v 6 v 0 . Dictionaries get the following type:

type Dict = Tr (String,String)

Give code for looking up a word definition in a dictionary. The type declaration is:lookupD :: String -> Dict -> [String]7.4. INDUCTION AND RECURSION OVER TREES

263

If (v, w) occurs in the dictionary, the call lookupD v d should yield [w], otherwise []. Use the order on the dictionary tree.
Exercise 7.38 For efficient search in an ordered tree (Exercises 7.36 and 7.37) itis crucial that the tree is balanced: the left and right subtree should have (nearly)
the same depth and should themselves be balanced.
The following auxiliary function splits non-empty lists into parts of (roughly)
equal lengths.

split :: [a] -> ([a],a,[a])split xs = (ys1,y,ys2)
where
ys1
= take n xs
(y:ys2) = drop n xs
n= length xs ‘div‘ 2

Use this to implement a function buildTree :: [a] -> Tr a for transforming
an ordered list into an ordered and balanced binary tree.
Here is a data type LeafTree for binary leaf trees (binary trees with information
at the leaf nodes):data LeafTree a = Leaf a| Node (LeafTree a) (LeafTree a) deriving Show

Here is an example leaf tree:

ltree :: LeafTree String
ltree = Node(Leaf "I")(Node
(Leaf "love")(Leaf "you"))

264

CHAPTER 7. INDUCTION AND RECURSION

Exercise 7.39 Repeat Exercise 7.33 for leaf trees. Call the new map function
mapLT.
Exercise 7.40 Give code for mirroring a leaf tree on its vertical axis. Call the function reflect. In the mirroring process, the left- and right branches are swapped,
and the same swap takes place recursively within the branches. The reflection of
Node (Leaf 1) (Node (Leaf 2) (Leaf 3))

isNode (Node (Leaf 3) (Leaf 2)) (Leaf 1).

Exercise 7.41 Let reflect be the function from Exercise 7.40. Prove with induction on tree structure that reflect (reflect t) == t. holds for every leaf
tree t.
A data type for trees with arbitrary numbers of branches (rose trees), with information of type a at the buds, is given by:data Rose a = Bud a | Br [Rose a] deriving (Eq,Show)

Here is an example rose:

rose = Br [Bud 1, Br [Bud 2, Bud 3, Br [Bud 4, Bud 5, Bud 6]]]

Exercise 7.42 Write a function mapR :: (a -> b) -> Rose a -> Rose b that
does for rose trees what map does for lists. For the example rose, we should get:
IAR> rose
Br [Bud 1, Br [Bud 2, Bud 3, Br [Bud 4, Bud 5, Bud 6]]]
IAR> mapR succ rose
Br [Bud 2, Br [Bud 3, Bud 4, Br [Bud 5, Bud 6, Bud 7]]]

7.5. INDUCTION AND RECURSION OVER LISTS

265

7.5 Induction and Recursion over Lists
Induction and recursion over natural numbers are based on the two cases n = 0
and n = k + 1. Induction and recursion over lists are based on the two casesl = [] and l = x:xs, where x is an item and xs is the tail of the list. An example
is the definition of a function len that gives the length of a list. In fact, Haskell
has a predefined function length for this purpose; our definition of len is just for
purposes of illustration.

len []= 0len (x:xs) = 1 + len xs

Similarly, Haskell has a predefined operation ++ for concatenation of lists. For
purposes of illustration we repeat our version from Section (4.8).

cat :: [a] -> [a] -> [a]cat []
ys = yscat (x:xs) ys = x : (cat xs ys)

As an example of an inductive proof over lists, we show that concatenation of lists
is associative.
Proposition 7.43 For all lists xs, ys and zs over the same type a:cat (cat xs ys) zs = cat xs (cat ys zs).
Proof. Induction on xs.
Basiscat (cat [] ys) zs

cat.1

=

cat.1

=

cat ys zs
cat [] (cat ys zs).

CHAPTER 7. INDUCTION AND RECURSION

266
Induction step

cat x:xs (cat ys zs)

cat.2

=

i.h.

=

cat.2

=

cat.2

=

x:(cat xs (cat ys zs)
x:(cat (cat xs ys) zs)
cat x:(cat xs ys) zs
cat (cat x:xs ys) zs.

Exercise 7.44 Prove by induction that cat xs [] = cat [] xs.
Exercise 7.45 Prove by induction:len (cat xs ys) = (len xs) + (len ys).

A general scheme for structural recursion over lists (without extra parameters) is
given by:f [] := zf (x : xs) := h x (f xs)
For example, the function s that computes the sum of the elements in a list of
numbers is defined by:s [] := 0s(n : xs) := n + s xs
Here 0 is taken for z, and + for h.
As in the case for natural numbers, it is useful to implement an operation forthis general procedure. In fact, this is predefined in Haskell, as follows (fromthe Haskell file Prelude.hs):foldr:: (a -> b -> b) -> b -> [a] -> bfoldr f z []= zfoldr f z (x:xs) = f x (foldr f z xs)

7.5. INDUCTION AND RECURSION OVER LISTS

267

In general, z is the identity element of the operation f, i.e., the value you would
start out with in the base case of a recursive definition of the operation. The identity
element for addition is 0, for multiplication it is 1 (see Section 7.3).
The base clause of the definition of foldr says that if you want to fold an empty
list for operation f, the result is the identity element for f. The recursive clausesays that to fold a non-empty list for f, you perform f to its first element and to
the result of folding the remainder of the list for f.
The following informal version of the definition of foldr may further clarify its
meaning:
foldr (⊕) z [x1 , x2 , . . . , xn ] := x1 ⊕ (x2 ⊕ (· · · (xn ⊕ z) · · · )).
The function add :: [Natural] -> Natural can now be defined as:

add = foldr plus Z

The function mlt :: [Natural] -> Natural is given by:

mlt = foldr mult (S Z)

And here is an alternative definition of list length, with values of type Natural.
Note that the function (\ _ n -> S n) ignores its first argument (you don’t haveto look at the items in a list in order to count them) and returns the successor ofits second argument (for this second argument represents the number of items thatwere counted so far).

ln :: [a] -> Naturalln = foldr (\ _ n -> S n) Z

It is also possible to use foldr on the standard type for integers. Computing theresult of adding or multiplying all elements of a list of integers can be done asfollows:

CHAPTER 7. INDUCTION AND RECURSION

268

Prelude> foldr (+) 0 [1..10]
55
Prelude> foldr (*) 1 [1..10]3628800

Exercise 7.46 Use foldr to give a new implementation of generalized union andfoldr1 to give a new implementation of generalized intersection for lists. (Lookup the code for foldr1 in the Haskell prelude. Compare Exercise 4.53.)
Consider the following definitions of generalized conjunction and disjunction:

or :: [Bool] -> Boolor [] = Falseor (x:xs) = x || or xs
and :: [Bool] -> Booland [] = Trueand (x:xs) = x && and xs

The function or takes a list of truth values and returns True if at least one memberof the list equals True, while and takes a list of truth values and returns True if
all members of the list equal True. (We have encountered and before, in Section
2.2.)
In fact, Haskell has predefined these operations, in terms of foldr. To see how
we can use foldr to implement generalized conjunction and disjunction, we only
need to know what the appropriate identity elements of the operations are. Should
the conjunction of all elements of [] count as true or false? As true, for it is indeed(trivially) the case that all elements of [] are true. So the identity element for
conjunction is True. Should the disjunction of all elements of [] count as true or
false? As false, for it is false that [] contains an element which is true. Therefore,
the identity element for disjunction is False. This explains the following Haskell
definition in Prelude.hs:

and, orandor:: [Bool] -> Bool
= foldr (&&) True
= foldr (||) False

7.5. INDUCTION AND RECURSION OVER LISTS

269

Exercise 7.47 Define a function srt that sorts a list of items in class Ord a byfolding the list with a function insrt.
The operation foldr folds ‘from the right’. Folding ‘from the left’ can be donewith its cousin foldl, predefined in Prelude.hs as follows:foldl:: (a -> b -> a) -> a -> [b] -> afoldl f z []= zfoldl f z (x:xs) = foldl f (f z x) xs

An informal version may further clarify this:foldl (⊕) z [x1 , x2 , . . . , xn ] := (· · · ((z ⊕ x1 ) ⊕ x2 ) ⊕ · · · ) ⊕ xn .
This can be used to flesh out the following recursion scheme:
f z [] := z
f z (x : xs) := f (h z x) xs
This boils down to recursion over lists with an extra parameter, and in fact foldl
can be used to speed up list processing. The case of reversing a list is an example:r zs [] := zs
r zs (x : xs) := r (prefix zs x) xs,
where prefix is given by prefix ys y = y:ys. Here is a definition in terms offoldl:

rev = foldl (\ xs x -> x:xs) []The list [1,2,3] is reversed as follows:
rev [1,2,3] = foldl (\ xs x -> x:xs) [] [1,2,3]= foldl (\ xs x -> x:xs) ((\ xs x -> x:xs) [] 1) [2,3]= foldl (\ xs x -> x:xs) [1] [2,3]= foldl (\ xs x -> x:xs) ((\ xs x -> x:xs) [1] 2) [3]270

CHAPTER 7. INDUCTION AND RECURSION= foldl (\ xs x -> x:xs) [2,1] [3]= foldl (\ xs x -> x:xs) ((\ xs x -> x:xs) [2,1] 3) []= foldl (\ xs x -> x:xs) [3,2,1] []= [3,2,1]

Note that (\ xs x -> x:xs) has type [a] -> a -> [a]. An alternative definition of rev, in terms of foldr, would need a swap function\ x xs -> xs ++ [x] :: a -> [a] -> [a]and would be much less efficient:rev’ = foldr (\ x xs -> xs ++ [x]) []The inefficiency resides in the fact that ++ itself is defined recursively, as follows:[]++ ys = ys
(x:xs) ++ ys = x : (xs ++ ys)

To see why rev’ is less efficient than rev, look at the following, where we write
postfix for (\ x xs -> xs ++[x]).
rev’ [1,2,3] = foldr postfix [] [1,2,3]= postfix 1 (foldr postfix [] [2,3])= (foldr postfix [] [2,3]) ++ [1]= (postfix 2 (foldr postfix [] [3])) ++ [1]= (foldr postfix [] [3]) ++ [2] ++ [1]= (postfix 3 (foldr postfix [] [])) ++ [2] ++ [1]= (foldr postfix [] []) ++ [3] ++ [2] ++ [1]= [] ++ [3] ++ [2] ++ [1]= ([3] ++ [2]) ++ [1]= 3:([] ++ [2]) ++ [1]= [3,2] ++ [1]= 3:([2] ++ [1])= 3:2:([] ++ [1])= [3,2,1]

7.5. INDUCTION AND RECURSION OVER LISTS

271

If we compare the two list recursion schemes that are covered by foldr andfoldl, then we see that the two folding operations h and h0 are close cousins:
f 0 [] := zf (x : xs) := h0 (f 0 xs) x

f [] := zf (x : xs) := h x (f xs)

0

An obvious question to ask now is the following: what requirements should h andh0 satisfy in order to guarantee that f and f 0 define the same function? Exercise
7.48 and Example 7.49 invite you to further investigate this issue of the relation
between the schemes.
Exercise 7.48 Let h :: a -> b -> b and h’ :: b -> a -> b, and z :: b.
Assume that for all x :: a it holds that h x (h’ y z) = h’ z (h x y). Show
that for every x,y :: a and every finite xs :: [a] the following holds:
h x (foldl h’ y xs) = foldl h’ (h x y) xs

Use induction on xs.
Example 7.49 Let h :: a -> b -> b and h’ :: b -> a -> b, and z :: b.
Assume we have for all x,y :: a and all xs :: [a] the following:h x z = h’ z x
h x (h’ y z) = h’ z (h x y)

We show that we have for all finite xs :: [a] thatfoldr h z xs = foldl h’ z xs
We use induction on the structure of xs.
Basis Immediate from the definitions of foldr and foldl we have that:foldr h z [] = z = foldl h’ z []Induction step Assume the induction hypothesis foldr h z xs = foldl h’ z xs.
We have to show that foldr h z x:xs = foldl h’ z x:xs. Here is thereasoning:
foldr h z x:xs

foldr

=

h x (foldr h z xs)

IH=

h x (foldl h z xs)

7.48

=

foldl h0 (h x z) xs

given

=

foldl h0 (h0 z x) xs

foldl

foldl h0 z x:xs

=

272

CHAPTER 7. INDUCTION AND RECURSION

For an application of Example (7.49), note that the functions postfix and prefix
are related by:
postfix x [] = [] ++ [x] = [x] == prefix [] x
postfix x (prefixxs y) = (prefix xs y) ++ [x]= y:(xs ++ [x])= y:(postfix x xs)= prefix (postfix x xs) y

It follows from this and the result of the example that rev and rev’ indeed compute the same function.
Exercise 7.50 Consider the following version of rev.

rev1 :: [a] -> [a]rev1 xs = rev2 xs []where
rev2 []
ys = ysrev2 (x:xs) ys = rev2 xs (x:ys)

Which version is more efficient, the original rev, the version rev’, or this version?
Why?
Exercise 7.51 Define an alternative version ln’ of ln using foldl instead offoldr.
In Section 1.8 you got acquainted with the map and filter functions. The two
operations map and filter can be combined:
Prelude> filter (>4) (map (+1) [1..10])[5,6,7,8,9,10,11]
Prelude> map (+1) (filter (>4) [1..10])[6,7,8,9,10,11]

These outcomes are different. This is because the test (>4) yields a different result
after all numbers are increased by 1. When we make sure that the test used in thefilter takes this change into account, we get the same answers:

7.6. SOME VARIATIONS ON THE TOWER OF HANOI

273

Prelude> filter (>4) (map (+1) [1..10])[5,6,7,8,9,10,11]
Prelude> map (+1) (filter ((>4).(+1)) [1..10])[5,6,7,8,9,10,11]

Here (f . g) denotes the result of first applying g and next f. Note: ((>4).(+1))
defines the same property as (>3). Exercise 7.52 gives you an opportunity to showthat these identical answers are no accident.
Exercise 7.52 Let xs :: [a], let f :: a -> b, and let p :: b -> Bool be a
total predicate. Show that the following holds:
filter p (map f xs) = map f (filter (p · f ) xs).
Note: a predicate p :: b -> Bool is total if for every object x :: b, the application p x gives either true or false. In particular, for no x :: b does p x raise
an error.

7.6 Some Variations on the Tower of Hanoi

Figure 7.1: The Tower of Hanoi.
The Tower of Hanoi is a tower of 8 disks of different sizes, stacked in order ofdecreasing size on a peg. Next to the tower, there are two more pegs. The task is
to transfer the whole stack of disks to one of the other pegs (using the third peg
as an auxiliary) while keeping to the following rules: (i) move only one disk at atime, (ii) never place a larger disk on top of a smaller one.
Exercise 7.53 In this exercise, you are required to invent your own solution, andnext prove it by mathematical induction. Make sure that the reasoning also establishes that the formula you find for the number of moves is the best one can
do.

274

CHAPTER 7. INDUCTION AND RECURSION

1. How many moves does it take to completely transfer a tower consisting of ndisks?
2. Prove by mathematical induction that your answer to the previous question
is correct.
3. How many moves does it take to completely transfer the tower of Hanoi?

Exercise 7.54 Can you also find a formula for the number of moves of the disk ofsize k during the transfer of a tower with disks of sizes 1, . . . , n, and 1 6 k 6 n?
Again, you should prove by mathematical induction that your formula is correct.

According to legend, there exists a much larger tower than the tower of Hanoi, thetower of Brahma, with 64 disks. Monks are engaged in transferring the disks of
the Tower of Brahma, from the beginning of the universe to the present day. As
soon as they will have completed their task the tower will crumble and the world
will end.

Exercise 7.55 How long will the universe last, given that the monks move one
disk per day?
For an implementation of the disk transfer procedure, an obvious way to represent
the starting configuration of the tower of Hanoi is ([1,2,3,4,5,6,7,8],[],[]).
For clarity, we give the three pegs names A, B and C. and we declare a type Tower:

data Peg= A | B | C
type Tower = ([Int], [Int], [Int])

There are six possible single moves from one peg to another:

7.6. SOME VARIATIONS ON THE TOWER OF HANOI

275

IAR> (display 88 . show . take 200 . hanoi) 8[([1,2,3,4,5,6,7,8],[],[]),([2,3,4,5,6,7,8],[1],[]),([3,4,5,6,7,8],[1],[2]),([3,4,5,6,7,8],[],[1,2]),([4,5,6,7,8],[3],[1,2]),([1,4,5,6,7,8],[3],[2]),([1,4,5,6,7,8],[2,3],[]),([
4,5,6,7,8],[1,2,3],[]),([5,6,7,8],[1,2,3],[4]),([5,6,7,8],[2,3],[1,4]),([2,5,6,7,8],[3],[1,4]),([1,2,5,6,7,8],[3],[4]),([1,2,5,6,7,8],[],[3,4]),([2,5,6,7,8],[1],[3,4]),([5,6,7,
8],[1],[2,3,4]),([5,6,7,8],[],[1,2,3,4]),([6,7,8],[5],[1,2,3,4]),([1,6,7,8],[5],[2,3,4]),([1,6,7,8],[2,5],[3,4]),([6,7,8],[1,2,5],[3,4]),([3,6,7,8],[1,2,5],[4]),([3,6,7,8],[2,5],[1,4]),([2,3,6,7,8],[5],[1,4]),([1,2,3,6,7,8],[5],[4]),([1,2,3,6,7,8],[4,5],[]),([2,3,6,7,8],[1,4,5],[]),([3,6,7,8],[1,4,5],[2]),([3,6,7,8],[4,5],[1,2]),([6,7,8],[3,4,5],[1,2]),([1,6,7,8],[3,4,5],[2]),([1,6,7,8],[2,3,4,5],[]),([6,7,8],[1,2,3,4,5],[]),([7,8],[1,2,3,4,5],[6]),([7,8],[2,3,4,5],[1,6]),([2,7,8],[3,4,5],[1,6]),([1,2,7,8],[3,4,5],[6]),([1,2,7,8],[4,5],[3,6]),([2,7,8],[1,4,5],[3,6]),([7,8],[1,4,5],[2,3,6]),([7,8],[4,5],[1,2,3,6]),([4,7,8],[5],[1,2,3,6]),([1,4,7,8],[5],[2,3,6]),([1,4,7,8],[2,5],[3,6]),([4,7,8],[1,2,5],[3,6]),([3,4,7,8],[1,2,5],[6]),([3,4,7,8],[2,5],[1,6]),([2,3,4,7,8],[5],[1,6]),([1,2,3,4,7,8],[5],[6]),([1,2,3,4,7,8],[],[5,6]),([2,3,4,7,8],[1],[5,6]),([3,4,7,8],[1],[2,5,6]),([3,4,7,8],[],[1,2,5,6]),([4,7,8],[3],[1,2,5,6]),([1,4,7,8],[3],[2,5,6]),([1,4,7,8],[2,3],[5,6]),([4,7,8],[1,2,3],[5,6]),([7,8],[1,2,3],[4,5,6]),([7,8],[2,3],[1,4,5,6]),([2,7,8],[3],[1,4,5,6]),([1,2,7,8],[3],[4,5,6]),([1,2,7,8],[],[3,4,5,6]),([2,7,8],[1],[3,4,5,6]),([7,8],[1],[2,3,4,5,6]),([7,8],[],[1,2,3,4,5,6]),([8],[7],[1,2,3,4,5,6]),([1,8],[7],[2,3,4,5,6]),([1,8],[2,7],[3,4,5,6]),([8],[1,2,7],[3,4,5,6]),([3,8],[1,2,7],[4,5,6]),([3,8],[2,7],[1,4,5,6]),([2,3,8],[7],[1,4,5,6]),([1,2,3,8],[7],[4,5,6]),([1,2,3,8],[4,7],[5,6]),([2,3,8],[1,4,7],[5,6]),([3,8],[1,4,7],[2,5,6]),([3,8],[4,7],[1,2,5,6]),([8],[3,4,7],[1,2,5,6]),([1,8],[3,4,7],[2,5,6]),([1,8],[2,3,4,7],[5,6]),([8],[1,2,3,4,7],[5,6]),([5,8],[1,2,3,4,7],[6]),([5,8],[2,3,4,7],[1,6]),([2,5,8],[3,4,7],[1,6]),([1,2,5,8],[3,4,7],[6]),([1,2,5,8],[4,7],[3,6]),([2,5,8],[1,4,7],[3,6]),([5,8],[1,4,7],[2,3,6]),([5,8],[4,7],[1,2,3,6]),([4,5,8],[7],[1,2,3,6]),([1,4,5,8],[7],[2,3,6]),([1,4,5,8],[2,7],[3,6]),([4,5,8],[1,2,7],[3,6]),([3,4,5,8],[1,2,7],[6]),([3,4,5,8],[2,7],[1,6]),([2,3,4,5,8],[7],[1,6]),([1,2,3,4,5,8],[7],[6]),([1,2,3,4,5,8],[6,7],[]),([2,3,4,5,8],[1,6,7],[]),([3,4,5,8],[1,6,7],[2]),([3,4,5,8],[6,7],[1,2]),([4,5,8],[3,6,7],[1,2]),([1,4,5,8],[3,6,7],[2]),([1,4,5,8],[2,3,6,7],[]),([4,5,8],[1,2,3,6,7],[]),([5,8],[1,2,3,6,7],[4]),([5,8],[2,3,6,7],[1,4]),([2,5,8],[3,6,7],[1,4]),([1,2,5,8],[3,6,7],[4]),([1,2,5,8],[6,7],[3,4]),([2,5,8],[1,6,7],[3,4]),([5,8],[1,6,7],[2,3,4]),([5,8],[6,7],[1,2,3,4]),([8],[5,6,7],[1,2,3,4]),([1,8],[5,6,7],[2,3,4]),([1,8],[2,5,6,7],[3,4]),([8],[1,2,5,6,7],[3,4]),([3,8],[1,2,5,6,7],[4]),([3,8],[2,5,6,7],[1,4]),([2,3,8],[5,6,7],[1,4]),([1,2,3,8],[5,6,7],[4]),([1,2,3,8],[4,5,6,7],[]),([2,3,8],[1,4,5,6,7],[]),([3,8],[1,4,5,6,7],[2]),([3,8],[4,5,6,7],[1,2]),([8],[3,4,5,6,7],[1,2]),([1,8],[3,4,5,6,7],[2]),([1,8],[2,3,4,5,6,7],[]),([8],[1,2,3,4,5,6,7],[]),([],[1,2,3,4,5,6,7],[8]),([],[2,3,4,5,6,7],[1,8]),([2],[3,4,5,6,7],[1,8]),([1,2],[3,4,5,6,7],[8]),([1,2],[4,5,6,7],[3,8]),([2],[1,4,5,6,7],[3,8]),([],[1,4,5,6,
7],[2,3,8]),([],[4,5,6,7],[1,2,3,8]),([4],[5,6,7],[1,2,3,8]),([1,4],[5,6,7],[2,3,8]),([1,4],[2,5,6,7],[3,8]),([4],[1,2,5,6,7],[3,8]),([3,4],[1,2,5,6,7],[8]),([3,4],[2,5,6,7],[1,8]),([2,3,4],[5,6,7],[1,8]),([1,2,3,4],[5,6,7],[8]),([1,2,3,4],[6,7],[5,8]),([2,3,4],[1,6,7],[5,8]),([3,4],[1,6,7],[2,5,8]),([3,4],[6,7],[1,2,5,8]),([4],[3,6,7],[1,2,5,8]),([1,4],[3,6,7],[2,5,8]),([1,4],[2,3,6,7],[5,8]),([4],[1,2,3,6,7],[5,8]),([],[1,2,3,6,7],[4,5,8]),([],[2,3,6,7],[1,4,5,8]),([2],[3,6,7],[1,4,5,8]),([1,2],[3,6,7],[4,5,8]),([1,2],[6,7],[3,4,5,8]),([2],[1,6,7],[3,4,5,8]),([],[1,6,7],[2,3,4,5,8]),([],[6,7],[1,2,3,4,5,8]),([6],[7],[1,2,3,4,5,8]),([1,6],[7],[2,3,4,5,8]),([1,6],[2,7],[3,4,5,8]),([6],[1,2,7],[3,4,5,8]),([3,6],[1,2,7],[4,5,8]),([3,6],[2,7],[1,4,5,8]),([2,3,6],[7],[1,4,5,8]),([1,2,3,6],[7],[4,5,8]),([1,2,3,6],[4,7],[5,8]),([2,3,6],[1,4,7],[5,8]),([3,6],[1,4,7],[2,5,8]),([3,6],[4,7],[1,2,5,8]),([6],[3,4,7],[1,2,5,8]),([1,6],[3,4,7],[2,5,8]),([1,6],[2,3,4,7],[5,8]),([6],[1,2,3,4,7],[5,8]),([5,6],[1,2,3,4,7],[8]),([5,6],[2,3,4,7],[1,8]),([2,5,6],[3,4,7],[1,8]),([1,2,5,6],[3,4,7],[8]),([1,2,5,6],[4,7],[3,8]),([2,5,6],[1,4,7],[3,8]),([5,6],[1,4,7],[2,3,8]),([5,6],[4,7],[1,2,3,8]),([4,5,6],[7],[1,2,3,8]),([1,4,5,6],[7],[2,3,8]),([1,4,5,6],[2,7],[3,8]),([4,5,6],[1,2,7],[3,8]),([3,4,5,6],[1,2,7],[8]),([3,4,5,6],[2,7],[1,8]),([2,3,4,5,6],[7],[1,8]),([1,2,3,4,5,6],[7],[8]),([1,2,3,4,5,6],[],[7,8]),([2,3,4,5,6],[1],[7,8]),([3,4,5,6],[1],[2,7,8]),([3,4,5,6],[],[1,2,7,8]),([4,5,6],[3],[1,2,7,8]),([1,4,5,6],[3],[2,7,8]),([1,4,5,6],[2,3],[7,8]),([4,5,6],[1,2,3],[7,8])]Figure 7.2: Tower of Hanoi (first 200 configurations)

276

CHAPTER 7. INDUCTION AND RECURSION

move :: Peg -> Peg -> Tower -> Tower
move A B (x:xs,ys,zs) = (xs,x:ys,zs)
move B A (xs,y:ys,zs) = (y:xs,ys,zs)
move A C (x:xs,ys,zs) = (xs,ys,x:zs)
move C A (xs,ys,z:zs) = (z:xs,ys,zs)
move B C (xs,y:ys,zs) = (xs,ys,y:zs)
move C B (xs,ys,z:zs) = (xs,z:ys,zs)

The procedure transfer takes three arguments for the pegs, an argument for thenumber of disks to move, and an argument for the tower configuration to move.
The output is a list of tower configurations. The procedure hanoi, finally, takes asize argument and outputs the list of configurations to move a tower of that size.

transfer :: Peg -> Peg -> Peg -> Int -> Tower -> [Tower]transfer _ _ _ 0 tower = [tower]transfer p q r n tower = transfer p r q (n-1) tower++transfer r q p (n-1) (move p q tower’)
where tower’ = last (transfer p r q (n-1) tower)
hanoi :: Int -> [Tower]
hanoi n = transfer A C B n ([1..n],[],[])The output for hanoi 8 is given in Figure 7.2. Here is the output for hanoi 4:
IAR> hanoi 4[([1,2,3,4],[],[]),([2,3,4],[1],[]),([3,4],[1],[2]),([3,4],[],[1,2]),([4],[3],[1,2]),([1,4],[3],[2]),([1,4],[2,3],[]),([4],[1,2,3],[]),([],[1,2,3],[4]),([],[2,3],[1,4]),([2],[3],[1,4]),([1,2],[3],[4]),([1,2],[],[3,4]),([2],[1],[3,4]),([],[1],[2,3,4]),([],[],[1,2,3,4])]

If you key in hanoi 64 and expect you can start meditating until the end of theworld, you will discover that the program suffers from what functional programmers call a space leak or black hole: as the execution progresses, the list of tower
configurations that is kept in memory grows longer and longer, and execution willabort with an ‘out of memory’ error. All computation ceases before the worldends.

7.6. SOME VARIATIONS ON THE TOWER OF HANOI

277

Now consider the following. If there is a best way to transfer the tower of Hanoi(or the tower of Brahma), then in any given configuration it should be clear what
the next move is. If they only have to look at the present configuration, the monks
in charge of the tower of Brahma can go about their task in a fully enlightenedmanner, with complete focus on the here and now.
Observe that there are 3n ways to stack n disks of decreasing sizes on 3 pegs insuch a way that no disk is on top of a smaller disk. Since the number of moves toget from ([1..n],[],[]) to ([],[],[1..n]) is less than this, not all of these
configurations are correct. How can we implement a check for correct configurations? Here is a checking procedure, with an argument for the size of the largest
disk.

check :: Int -> Tower -> Boolcheck 0 t = t == ([],[],[])
check n (xs,ys,zs)| xs /= [] && last xs == n = check (n-1) (init xs, zs, ys)| zs /= [] && last zs == n = check (n-1) (ys, xs, init zs)| otherwise= FalseExercise 7.56 To see that the implementation of check is correct, we need an
inductive proof. Give that proof.
Here is a function for finding the largest disk in a configuration, and a function
checkT for checking a configuration of any size.

maxT :: Tower -> Int
maxT (xs, ys, zs) = foldr max 0 (xs ++ ys ++ zs)
checkT :: Tower -> BoolcheckT t = check (maxT t) t

The following exercise gives a more direct criterion.
Exercise 7.57 Show that configuration (xs,ys,zs) with largest disk n is correct
iff it holds that every disk m is either on top of a disk k with k − m odd, or at thebottom of the source or destination peg, with n − m even, or at the bottom of theauxiliary peg, with n − k odd.

278

CHAPTER 7. INDUCTION AND RECURSION

The previous exercise immediately gives a procedure for building correct configurations: put each disk according to the rule, starting with the largest disk. This
will give exactly two possibilities for every disk k: the largest disk can either go atsource or at destination; if k + 1 is already in place, k can either go on top of k + 1
or to the only other place with the same parity as k + 1.
For the implementation of a parity check, for the case where the largest disk has
size n, we can take any configuration (xs,ys,zs), extend it to(xs ++ [n+1], ys ++ [n], zs ++ [n+1])

and then define the parities by means of par (x:xs) = x mod 2. Here is theimplementation:

parity :: Tower -> (Int,Int,Int)
parity (xs,ys,zs) = par (xs ++ [n+1], ys ++ [n],zs ++ [n+1])
where
n = maxT (xs, ys, zs)par (x:xs,y:ys,z:zs) = (mod x 2, mod y 2, mod z 2)

Exercise 7.58 Show that if (xs,ys,zs) is a correct configuration, thenparity (xs,ys,zs) ∈ {(1, 1, 0), (1, 0, 1), (0, 1, 1)}.

A little reflection shows that there are only two kinds of moves.
• Moves of the first kind move disk 1 (the smallest disk).
• Moves of the second kind move a disk other than 1.
Moves of the first kind are fully determined, as can be seen by the fact that disk 1
should always be moved to the place with parity 0.
Exercise 7.59 Give an argument for this.
Moves of the second kind are also fully determined, for if there is one empty peg,then there are disks 1, k on top of the other pegs, and we can only move k to the7.6. SOME VARIATIONS ON THE TOWER OF HANOI

279

empty peg, otherwise there are disks 1 < k < m on top of the pegs, and we canonly move k on top of m.
This gives a new algorithm for tower transfer, without space leak. We determine
the target of a move of disk 1 by means of a parity check:target :: Tower -> Peg
target t@(xs,ys,zs) | parity t == (0,1,1) = A| parity t == (1,0,1) = B| parity t == (1,1,0) = C

The moves of disk 1 are now given by:

move1 :: Tower -> Tower
move1 t@(1:_,ys,zs) = move A (target t) t
move1 t@(xs,1:_,zs) = move B (target t) t
move1 t@(xs,ys,1:_) = move C (target t) t

The moves of the middle disk are given by:

move2 :: Tower -> Tower
move2 t@(1:xs,[],zs) = move C B t
move2 t@(1:xs,ys,[]) = move B C t
move2 t@(1:xs,ys,zs) = if ys < zs then move B C t else move C B t
move2 t@([],1:ys,zs) = move C A t
move2 t@(xs,1:ys,[]) = move A C t
move2 t@(xs,1:ys,zs) = if xs < zs then move A C t else move C A t
move2 t@([],ys,1:zs) = move B A t
move2 t@(xs,[],1:zs) = move A B t
move2 t@(xs,ys,1:zs) = if xs < ys then move A B t else move B A t

The check for completion is:

280

CHAPTER 7. INDUCTION AND RECURSION

done :: Tower -> Booldone ([],[], _) = True
done (xs,ys,zs) = False

Transfer of a tower takes place by alternation between the two kinds of moves,
until the tower is completely transferred. Since the last move has to be a move1,
we only have to check for complete transfer right after such moves, i.e., in thefunction transfer2:transfer1, transfer2 :: Tower -> [Tower]transfer1 t = t : transfer2 (move1 t)transfer2 t = if done t then [t] else t : transfer1 (move2 t)

And here is our new Hanoi procedure:

hanoi’ :: Int -> [Tower]
hanoi’ n = transfer1 ([1..n],[],[])
zazen :: [Tower]
zazen = hanoi’ 64

By now we know enough about correct tower configurations to be able to orderthem.
Exercise 7.60 Define and implement a total ordering on the list of all correct tower
configurations.
Exercise 7.60 makes clear that it should be possible to define a bijection between
the natural numbers and the correct tower configurations, in their natural order.
For this, we first define a function for finding the k-th correct configuration in thelist of transitions for ([1..n],[],[]).

7.7. INDUCTION AND RECURSION OVER OTHER DATA STRUCTURES281

hanoiCount :: Int -> Integer -> Tower
hanoiCount n k | k < 0= error "argument negative"| k > 2^n - 1 = error "argument not in range"| k == 0= ([1..n],[],[])| k == 2^n - 1 = ([],[],[1..n])| k < 2^(n-1) = (xs ++ [n], zs, ys)| k >= 2^(n-1) = (ys’, xs’, zs’ ++ [n])where(xs,ys,zs)= hanoiCount (n-1) k(xs’,ys’,zs’) = hanoiCount (n-1) (k - 2^(n-1))

In terms of this we define the bijection. Note that for the definition we need
the inverse of the function λn.2n , i.e., the function λn. log2 n. The predefined
Haskell function logBase gives logarithms, and logarithms to base 2 are given by
logBase 2. Since the results are in class Floating (the class of floating point
numbers), we need conversion to get back to class Integral (the class consistingof Int and Integer). For this we use truncate.

toTower :: Integer -> Tower
toTower n = hanoiCount k m
wheren’ = fromInteger (n+1)
k = truncate (logBase 2 n’)
m = truncate (n’ - 2^k)

Exercise 7.61 The function hanoiCount gives us yet another approach to thetower transfer problem. Implement this as hanoi’’ :: Int -> [Tower].
Exercise 7.62 Implement the function fromTower :: Tower -> Integer thatis the inverse of toTower.

7.7 Induction and Recursion over Other Data Structures
A standard way to prove properties of logical formulas is by induction on theirsyntactic structure. Consider e.g. the following Haskell data type for propositional

282

CHAPTER 7. INDUCTION AND RECURSION

formulas.

data Form = P Int | Conj Form Form | Disj Form Form | Neg Form
instance Show Form whereshow (P i) = ’P’:show ishow (Conj f1 f2) = "(" ++ show f1 ++ " & " ++ show f2 ++ ")"show (Disj f1 f2) = "(" ++ show f1 ++ " v " ++ show f2 ++ ")"show (Neg f)= "~" ++ show f

The instance Show Form ensures that the data type is in the class Show, and thefunction show indicates how the formulas are displayed.
It is assumed that all proposition letters are from a list P0 , P1 , . . .. Then ¬(P1 ∨¬P2 ) is represented as Neg (Disj (P 1) (Neg (P 2))), and shown on thescreen as ~(P1 v ~P2), and so on.
We define the list of sub formulas of a formula as follows:

subforms :: Form -> [Form]
subforms (P n) = [(P n)]
subforms (Conj f1 f2) = (Conj f1 f2):(subforms f1 ++ subforms f2)
subforms (Disj f1 f2) = (Disj f1 f2):(subforms f1 ++ subforms f2)
subforms (Neg f) = (Neg f):(subforms f)

This gives, e.g.:
IAR> subforms (Neg (Disj (P 1) (Neg (P 2))))[~(P1 v ~P2),(P1 v ~P2),P1,~P2,P2]

The following definitions count the number of connectives and the number of
atomic formulas occurring in a given formula:7.7. INDUCTION AND RECURSION OVER OTHER DATA STRUCTURES283

ccount :: Form -> Int
ccount (P n) = 0
ccount (Conj f1 f2) = 1 + (ccount f1) + (ccount f2)
ccount (Disj f1 f2) = 1 + (ccount f1) + (ccount f2)
ccount (Neg f) = 1 + (ccount f)

acount :: Form -> Intacount (P n) = 1
acount (Conj f1 f2) = (acount f1) + (acount f2)
acount (Disj f1 f2) = (acount f1) + (acount f2)
acount (Neg f) = acount f

Now we can prove that the number of sub formulas of a formula equals the sum of
its connectives and its atoms:
Proposition 7.63 For every member f of Form:length (subforms f) = (ccount f) + (acount f).
Proof.
Basis If f is an atom, then subforms f = [f], so this list has length 1. Also,
ccount f = 0 and acount f = 1.
Induction step If f is a conjunction or a disjunction, we have:• length (subforms f) = 1 + (subforms f1) + (subforms f2),
• ccount f = 1 + (ccount f1) + (ccount f2),
• acount f = (acount f1) + (acount f2),

where f1 and f2 are the two conjuncts or disjuncts. By induction hypothesis:
length (subforms f1) = (ccount f1) + (acount f1).
length (subforms f2) = (ccount f2) + (acount f2).

284

CHAPTER 7. INDUCTION AND RECURSION
The required equality follows immediately from this.
If f is a negation, we have:• length (subforms f) = 1 + (subforms f1),
• ccount f = 1 + (ccount f1),

• acount f = (acount f1),

and again the required equality follows immediately from this and the induction hypothesis.

If one proves a property of formulas by induction on the structure of the formula,then the fact is used that every formula can be mapped to a natural number thatindicates its constructive complexity: 0 for the atomic formulas, the maximum ofrank(Φ) and rank(Ψ) plus 1 for a conjunction Φ ∧ Ψ, and so on.

7.8 Further Reading
Induction and recursion are at the core of discrete mathematics. See [Bal91]. A
splendid textbook that teaches concrete (and very useful) mathematical skills inthis area is Graham, Knuth and Patashnik [GKP89]. Recursion is also crucial foralgorithm design. If you are specifically interested in the design and analysis ofalgorithms you should definitely read Harel [Har87]. Algorithm design in Haskell
is the topic of [RL99].

Chapter 8

Working with Numbers
Preview
When reasoning about mathematical objects we make certain assumptions about
the existence of things to reason about. In the course of this chapter we will takea look at integers, rational numbers, real numbers, and complex numbers. We
will recall what they all look like, and we will demonstrate that reasoning aboutmathematical objects can be put to the practical test by an implementation of thedefinitions that we reason about. The implementations turn the definitions into
procedures for handling representations of the mathematical objects. This chapter
will also present some illuminating examples of the art of mathematical reasoning.

module WWN
whereimport List
import Nats

285

286

CHAPTER 8. WORKING WITH NUMBERS

8.1 A Module for Natural Numbers
The natural numbers are the conceptual basis of more involved number systems.
The implementation of Natural from the previous chapter has the advantage thatthe unary representation makes the two cases for inductive proofs and recursive
definitions, Z and S n, very clear. We repeat the code of our implementation of
Natural, wrapped up this time in the form of a module, and integrated into the
Haskell type system: see Figure 8.1. Note that an implementation of quotRem
was added: quotRem n m returns a pair (q,r) consisting of a quotient and aremainder of the process of dividing n by m, i.e., q and r satisfy 0 6 r < m andq × m + r = n.
The integration of Natural in the system of Haskell types is achieved by meansof instance declarations. E.g., the declaration instance Ord Natural makes
the type Natural an instance of the class Ord. From an inspection of Prelude.hs
we get that (i) a type in this class is assumed to also be in the class Eq, and (ii)a minimal complete definition of a type in this class assumes definitions of <= orcompare. Condition (i) is fulfilled for Natural by the deriving Eq statement,
while (ii) is taken care of by the code for compare in the module. The generalcode for class Ord provides methods for the relations <=, <, >=, >, and for thecomparison operators max and min, so putting Natural in class Ord provides us
with all of these.
Similarly, the declaration instance Enum Natural ensures that Natural is in
the class Enum. Definitions for the following functions for types in this class areprovided in the module:
succ :: Natural -> Natural
pred :: Natural -> Natural
toEnum :: Int -> NaturalfromEnum :: Natural -> Int
enumFrom :: Natural -> [Natural]

The general code for the class provides, e.g., enumFromTo, so we get:Nats> enumFromTo Z (toEnum 5)[Z,S Z,S (S Z),S (S (S Z)),S (S (S (S Z))),S (S (S (S (S Z))))]

Next, the declaration instance Num Natural ensures that Natural is in theclass Num. This is where the functions for addition, multiplication, sign, absolute value belong, as well as a type conversion function fromInteger. Putting in
the appropriate definitions allows the use of the standard names for these operators. In terms of these, Prelude.hs defines subtraction, the type conversion function

8.1. A MODULE FOR NATURAL NUMBERS

module Nats wheredata Natural = Z | S Natural deriving (Eq, Show)instance Ord Natural wherecompare Z Z = EQ
compare Z _ = LTcompare _ Z = GTcompare (S m) (S n) = compare m ninstance Enum Natural where
succ = \ n -> S npred Z = Z
pred (S n) = n
toEnum= fromInt
fromEnum = toInt
enumFrom n = map toEnum [(fromEnum n)..]instance Num Natural where(+) = foldn succ(*) = \m -> foldn (+m) Z(-) = foldn pred
abs = idsignum Z = Z
signum n = (S Z)
fromInteger n | n < 0= error "no negative naturals"| n == 0= Z| otherwise = S (fromInteger (n-1))
foldn :: (a -> a) -> a -> Natural -> afoldn h c Z = cfoldn h c (S n) = h (foldn h c n)
instance Real Natural where toRational x = toInteger x % 1
instance Integral Natural where
quotRem n d | d > n= (Z,n)| otherwise = (S q, r)
where (q,r) = quotRem (n-d) d
toInteger = foldn succ 0

Figure 8.1: A Module for Natural Numbers.

287

288

CHAPTER 8. WORKING WITH NUMBERS

fromInt, and negate. Since negate is defined in terms of subtraction, and subtraction for naturals is cut-off subtraction, we get:Nats> negate (S (S (S (S Z))))Z

The next instance declaration puts Natural in the class Real. The only code
we have to provide is that for the type conversion function toRational. Putting
Natural in this class is necessary for the next move: the instance declarationthat puts Natural in the class Integral: types in this class are constrained by
Prelude.hs to be in the classes Real and Enum. The benefits of making Natural
an integral type are many: we now get implementations of even, odd, quot, rem,
div, mod, toInt and many other functions and operations for free (see the contentsof Prelude.hs for full details).
The implicit type conversion also allows us to introduce natural numbers in shorthand notation:Nats> 12 :: Natural
S (S (S (S (S (S (S (S (S (S (S (S Z)))))))))))The standard representation for natural numbers that we are accustomed to, decimal string notation, is built into the Haskell system for the display of integers. To
display naturals in decimal string notation, all we have to do is use type conversion:Nats> toInt (S (S (S Z)))
3Nats> (S (S Z)) ^ (S (S (S Z)))
S (S (S (S (S (S (S (S Z)))))))
Nats> toInt (S (S Z)) ^ (S (S (S Z)))
8

For other representations, i.e., binary string representation of an integral numbern, divide by 2 repeatedly, and collect the remainders in a list. The reversal of thislist gives the binary representation bk . . . b0 satisfying n = bk · 2k + bk−1 · 2k−1 +· · · + b0 . The nice thing is that we can provide this code for the whole class ofIntegral types:

8.2. GCD AND THE FUNDAMENTAL THEOREM OF ARITHMETIC

289

binary :: Integral a => a -> [Int]binary x = reverse (bits x)
where bits 0 = [0]bits 1 = [1]bits n = toInt (rem n 2) : bits (quot n 2)

To display this on the screen, we need intToDigit for converting integers into
digital characters:

showDigits :: [Int] -> String
showDigits = map intToDigit
bin :: Integral a => a -> Stringbin = showDigits . binary

Exercise 8.1 Give a function hex for displaying numbers in type class Integral
in hexadecimal form, i.e., in base 16 representation. The extra digits a, b, c, d,e, f for 10, 11, 12, 13, 14, 15 that you need are provided by intToDigit. The
call hex 31 should yield "1f". (Hint: it is worthwhile to provide a more generalfunction toBase for converting to a list of digits in any base in {2, . . . , 16}, andthen define hex in terms of that.)

8.2 GCD and the Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic, stating that every n ∈ N with n > 1 hasa unique prime factorization, was known in Antiquity. Let us reflect a bit to see
why it is true. First note that the restriction on n is necessary, for m · 0 = 0 for allm ∈ N, so a factorization of 0 can never be unique, and since we have ruled out 1
as a prime, 1 does not have a prime factorization at all. The fundamental theoremof arithmetic constitutes the reason, by the way, for ruling out 1 as a prime number.
We have n = 1m · n for any n, m ∈ N, so a factorization of n that admits 1m as afactor can never be unique.

CHAPTER 8. WORKING WITH NUMBERS

290

From the prime factorization algorithm (1.7) we know that prime factorizations ofevery natural number > 1 exist. To show uniqueness, we still have to establish thatno number has more than one prime factorization.

Euclid’s GCD algorithm The greatest common divisor of two natural numbers
a, b, notation GCD(a, b), is the natural number d with the property that d divides
both a and b, and for all natural numbers d0 that divide both a and b it holds thatd0 divides d. For example, GCD(30, 84) = 6, for 6 divides 30 and 84, and every
other common divisor of 30 and 84 divides 6 (these other common divisors being
1, 2 and 3).
Clearly, if such d exists, it is unique, for suppose that d0 also qualifies. Thenbecause d0 divides a and b, it follows from the definition of d that d0 divides d.
Similarly, because d divides a and b, it follows from the definition of d 0 that ddivides d0 . But if d divides d0 and d0 divides d then it follows that d = d0 .
The greatest common divisor of a and b can be found from the prime factorizations
β1
βk
αk
1
of a and b as follows. Let a = pα
1 · · · pk and b = p1 · · · pk be prime factorizations of a and b, with p1 , . . . , pk distinct primes, and α1 , . . . , αk , β1 , . . . , βknatural numbers. Then the greatest common divisor of a and b equals the natural
number pγ11 · · · pγkk , where each γi is the minimum of αi and βi .

For example, the greatest common divisor of 30 = 21 · 31 · 51 · 70 and 84 =22 · 31 · 50 · 71 is given by 21 · 31 · 50 · 70 = 6.
But there is an easier way to find GCD(a, b). Here is Euclid’s famous algorithm(assume neither of a, b equals 0):
WHILE a 6= b DO IF a > b THEN a := a − b ELSE b := b − a.
Let us run this for the example case a = 30, b = 84. We get the following(conventions about variable names as in Section 1.7).

a0 < b 0
a1 < b 1
a2 > b 2
a3 < b 3
a4 < b 4
a5 < b 5
a6 = b 6 = 6

a0 = 30
a1 = 30
a2 = 30
a3 = 30 − 24 = 6
a4 = 6a5 = 6a6 = 6b0 = 84
b1 = 84 − 30 = 54b2 = 54 − 30 = 24
b3 = 24b4 = 24 − 6 = 18
b5 = 18 − 6 = 12
b6 = 12 − 6 = 6

Now why does this work? The key observation is the following.

8.2. GCD AND THE FUNDAMENTAL THEOREM OF ARITHMETIC

291

If d divides a and b and a > b then d divides a − b (for then there are natural
numbers m, n with m > n and a = md , b = nd, and therefore a−b = md−nd =d(m − n)), and similarly, if d divides a and b and a < b then d divides b − a.
Conversely, if a > b and d divides a − b and b, then d divides a (for then there
are natural numbers m, n with a − b = md and b = nd, hence a = md + nd =d(m + n)), and similarly, if a > b and d divides b − a and a, then d divides a.
Thus, if a > b then the set of common divisors of a and b equals the set of common
divisors of a − b and b, and if b > a then the set of common divisors of a and bequals the set of common divisors of a and b − a.
Since the sets of all common divisors are equal, the greatest common divisors must
be equal as well. Therefore we have: if a > b then GCD(a, b) = GCD(a − b, b),
and if b > a then GCD(a, b) = GCD(a, b − a).
Using this we see that every iteration through the loop preserves the greatest common divisor in the following sense: GCD(ai , bi ) = GCD(ai+1 , bi+1 ). Since weknow that the algorithm terminates we have: there is some k with ak = bk . Therefore ak = GCD(ak , bk ) = GCD(a, b).
Haskell contains a standard function gcd for the greatest common divisor of a pairof objects of type Integral:

gcd
gcd 0 0gcd x y:: Integral a => a -> a -> a
= error "Prelude.gcd: gcd 0 0 is undefined"= gcd’ (abs x) (abs y)where gcd’ x 0 = x
gcd’ x y = gcd’ y (x ‘rem‘ y)

Exercise 8.2 If you compare the code for gcd to Euclid’s algorithm, you see that
Euclid uses repeated subtraction where the Haskell code uses rem. Explain as
precisely as you can how the Haskell version of the GCD algorithm is related to
Euclid’s method for finding the GCD.
We can use the GCD to define an interesting relation. Two natural numbers n andm are said to be co-prime or relatively prime if GCD(m, n) = 1. Here is an
implementation:

292

CHAPTER 8. WORKING WITH NUMBERS

coprime :: Integer -> Integer -> Bool
coprime m n = (gcd m n) == 1

Exercise 8.3 Consider the following experiment:
WWN> coprime 12 25
True
WWN> 12 + 25
37
WWN> coprime 25 37
True
WWN> 25 + 37
62
WWN> coprime 37 62
True
WWN> 37 + 62
99
WWN> coprime 62 99
True
WWN> 62 + 99
161
WWN> coprime 99 161
True
WWN>

This experiment suggests a general rule, for you to consider . . .
Does it follow from the fact that a and b are co-prime with a < b that b and a+b are
co-prime? Give a proof if your answer is ‘yes’ and a counterexample otherwise.
Theorem 8.4 For all positive a, b ∈ N there are integers m, n with ma + nb =GCD(a, b).
Proof. Consider the pairs (a0 , b0 ), (a1 , b1 ), . . . , (ak , bk ) generated by Euclid’s
algorithm. We know that (a0 , b0 ) = (a, b) and that ak = bk = GCD(a, b).
a0 satisfies a0 = ma + nb for m = 1, n = 0, b0 satisfies ma + nb = 1 for
m = 0, n = 1.
Suppose ai satisfies ai = m1 a + n1 b and bi satisfies bi = m2 a + n2 b. If ai > bi ,then ai+1 satisfies ai+1 = (m1 − m2 )a + (n1 − n2 )b and bi+1 satisfies bi+1 =8.3. INTEGERS

293

m2 a + n2 b. If ai < bi , then ai+1 satisfies ai+1 = m1 a + n1 b and bi+1 satisfies
bi+1 = (m2 −m1 )a+(n2 −n1 )b. Thus, every iteration through the loop of Euclid’s
algorithm preserves the fact that ai and bi are integral linear combinations ma+nb
of a and b.
This shows that there are integers m, n with ak = ma + nb, hence that ma + nb =GCD(a, b).
Theorem 8.5 If p is a prime number that divides ab then p divides a or b.
Proof. Suppose p divides ab and p does not divide a. Then GCD(a, p) = 1. By
the previous theorem there are integers m, n with ma + np = 1. Multiplying bothsides by b gives:
mab + nbp = b.
By the fact that p divides ab we know that p divides both mab and nbp. Hence p
divides mab + nbp. Hence p divides b.
Theorem 8.5 is the tool for proving the fundamental theorem of arithmetic.
Theorem 8.6 (Fundamental Theorem of Arithmetic)
Every natural number greater than 1 has a unique prime factorization.
Proof. We established in 1.7 that every natural number greater than 1 has at least
one prime factorization.
To show that every natural number has at most one prime factorization, assume tothe contrary that there is a natural number N > 1 with at least two different prime
factorizations. Thus,
N = p 1 · · · pr = q 1 · · · qs ,with all of p1 , . . . , pr , q1 , . . . , qs prime. Divide out common factors if necessary.
This gives a pi that is not among the q’s. But this is a contradiction with theorem
8.5, because pi divides N = q1 · · · qs but pi does not divide any of q1 , . . . , qs ,since these are all prime numbers different from pi .

8.3 Integers
Suppose n, m, k are natural numbers. When n = m + k we can view k as thedifference of n and m, and it is tempting to write this as k = n − m for theoperation of subtraction. Addition and subtraction are called inverse operations,

294

CHAPTER 8. WORKING WITH NUMBERS

for if the addition of n to m is followed by the subtraction of n, the end result isthe original natural number m. In other words, we have:(m + n) − n = m.
But we have to be careful here. The operation of subtracting a natural number nfrom a natural number m will only result in a natural number if m > n. To makesubtraction possible between any pair of numbers, we have to introduce negative
numbers. This gives the domain of integers:
Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}.
The symbol Z derives from Zahl, the German word for number.
In fact, we need not consider the integers as given by God, but we can view themas constructed from the natural numbers. This can be done in several ways. The
following Haskell code illustrates one possibility:

data Sgn = P | N deriving (Eq,Show)
type MyInt = (Sgn,Natural)
myplus :: MyInt -> MyInt -> MyInt
myplus (s1,m) (s2,n) | s1 == s2= (s1,m+n)| s1 == P && n <= m = (P,m-n)| s1 == P && n > m = (N,n-m)| otherwise= myplus (s2,n) (s1,m)Another way is as follows. We represent every integer as a ‘difference pair’ of twonatural numbers. For example, (0, 3) represents −3, but the same number −3 isalso represented by (1, 4) or (2, 5).
In general, for all m1 , m2 ∈ N, if m1 > m2 then there is a k ∈ N with m2 + k =m1 , and (m1 , m2 ) represents k. If, on the other hand, m1 < m2 then there is ak ∈ N with m1 + k = m2 , and (m1 , m2 ) represents −k. Thus, let R ⊆ N2 be
defined as follows:(m1 , m2 )R(n1 , n2 ) :≡ m1 + n2 = m2 + n1 .
It is easy to see that R is an equivalence on N2 . If m ∈ N, then the integer −m isrepresented by (0, m), but also by (k, k + m), for any k ∈ N.
Intuitively (m1 , m2 ) and (n1 , n2 ) are equivalent modulo R when their differences
are the same. This is the case precisely when m1 + n2 = m2 + n1 . In this case we8.3. INTEGERS

295

say that (m1 , m2 ) and (n1 , n2 ) represent the same number. In Section 6.8 we sawthat the equivalence relation R is a congruence for addition on N 2 , where addition
on N2 is given by:(m1 , m2 ) + (n1 , n2 ) := (m1 + n1 , m2 + n2 ),Denoting the equivalence class of (m1 , m2 ) as [m1 − m2 ], we get:[m1 − m2 ] := {(n1 , n2 ) ∈ N2 | m1 + n2 = m2 + n1 }.
We identify the integers with the equivalence classes [m1 − m2 ], for m1 , m2 ∈N. Note that in this notation − is not an operator. Call the equivalence classes
difference classes.
If an integer m = [m1 − m2 ], then we can swap the sign by swapping the order of
m1 and m2 , and put −m = [m2 − m1 ]. Swapping the order twice gets us back to
the original equivalence class. This reflects the familiar rule of sign m = −(−m).
We define addition and multiplication on difference classes as follows:[m1 − m2 ] + [n1 − n2 ] := [(m1 + n1 ) − (m2 + n2 )][m1 − m2 ] · [n1 − n2 ] := [(m1 n1 + m2 n2 ) − (m1 n2 + n1 m2 )].

The purpose of this definition is to extend the fundamental laws of arithmetic to
the new domain. To see whether we have succeeded in this we have to perform averification, as a justification of the definition.
In the verification that integers satisfy the law of commutativity for addition, we
make use of the definition of addition for difference classes, and of laws of commutativity for addition on N:
Proposition 8.7 For all m, n ∈ Z: m + n = n + m.
Proof. Representing m, n as difference classes we get:[m1 − m2 ] + [n1 − n2 ][(m1 + n1 ) − (m2 + n2 )][(n1 + m1 ) − (n2 + m2 )]=[definition of + for difference classes]=[commutativity of + for N]=[definition of + for difference classes][(n1 − n2 ] + [m1 − m2 )].

Note that the proof uses just the definition of + for difference classes and the commutativity of + on N. Note also that every equality statement is justified in the296

CHAPTER 8. WORKING WITH NUMBERS

proof by a reference to a definition or to a fundamental law of arithmetic, wherethese laws are in turn justified by inductive proofs based on the recursive definitions of the natural number operations. Thus, the fundamental laws of arithmetic
and the definition of a new kind of object (difference classes) in terms of a familiar
one (natural numbers) are our starting point for the investigation of that new kindof object.

Exercise 8.8 Show (using the definition of integers as difference classes and thedefinition of addition for difference classes) that the associative law for addition
holds for the domain of integers.

In a similar way it can be shown from the definition of integers by means of difference classes and the definition of multiplication for difference classes that theassociative and commutative laws for multiplication continue to hold for the domain of integers.
As a further example of reasoning about this representation for the integers, weshow that the distributive law continues to hold in the domain of integers (viewed
as difference classes). In the verification we make use of the definition of addition
and multiplication for difference classes, and of the fundamental laws for addition
and multiplication on the natural numbers. Again, we justify every step by meansof a reference to the definition of an operation on difference classes or to one of
the laws of arithmetic for the natural numbers.

Proposition 8.9 For all m, n, k ∈ Z : m(n + k) = mn + mk.
Proof. Representing m, n, k as difference classes we get:===[m1 − m2 ]([n1 − n2 ] + [k1 − k2 ])[definition of + for difference classes][m1 − m2 ][(n1 + k1 ) − (n2 + k2 )][definition of · for difference classes][(m1 (n1 + k1 ) + m2 (n2 + k2 )) − (m1 (n2 + k2 ) + (n1 + k1 )m2 )][distribution of · over + for N][(m1 n1 + m1 k1 + m2 n2 + m2 k2 ) − (m1 n2 + m1 k2 + m2 n1 + m2 k1 )]8.4. IMPLEMENTING INTEGER ARITHMETIC
=
=

=

297

[commutativity of + for N][(m1 n1 + m2 n2 + m1 k1 + m2 k2 ) − (m1 n2 + m2 n1 + m1 k2 + m2 k1 )][definition of + for difference classes][(m1 n1 + m2 n2 ) − (m1 n2 + m2 n1 )]+ [(m1 k1 + m2 k2 ) − (m1 k2 + m2 k1 )][definition of · for difference classes][m1 − m2 ] · [n1 − n2 ] + [m1 − m2 ] · [k1 − k2 ].
Once the integers and the operations of addition and multiplication on it are defined, and we have checked that the definitions are correct, we can forget of course
about the representation by means of difference classes. It simply takes too muchmental energy to keep such details of representation in mind. Mathematicians
make friends with mathematical objects by forgetting about irrelevant details of
their definition.
The natural numbers have representations as difference classes too: a natural number m is represented by [m − 0]. In fact, the original numbers and their representations in the new domain Z behave exactly the same, in a very precise sense: thefunction that maps m ∈ N to [m − 0] is one-to-one (it never maps different numbers to the same pair) and it preserves the structure of addition and multiplication,in the following sense (see also Section 6.8). If m + n = k then[m − 0] + [n − 0] = [(m + n) − 0] = [k − 0],and if mn = k then[m − 0] · [n − 0] = [mn − 0] = [k − 0].

Again we forget about the differences in representation and we say: N ⊆ Z.

8.4 Implementing Integer Arithmetic
If we represent natural numbers as type Natural, then the moral of the above isthat integers can be represented as pairs of naturals. E.g., minus five is represented
by the pair (S Z,S(S(S(S(S(S Z)))))), or by the pair (Z,S(S(S(S(S Z))))),and so on. Here is an appropriate data type declaration:

type NatPair = (Natural,Natural)

298

CHAPTER 8. WORKING WITH NUMBERS

The gist of the previous section can be nicely illustrated by means of implementations of the integer operations. For suppose we have the natural number operations
plus for addition and times for multiplication available for the type Natural).
Then addition for integer pairs is implemented in Haskell as:

plus1 :: NatPair -> NatPair -> NatPair
plus1 (m1, m2) (n1, n2) = (m1+n1, m2+n2)

Subtraction is just addition, but with the sign of the second operand reversed. Sign
reversal is done by swapping the elements of a pair, as we have seen. Thus, theimplementation of subtraction can look like this:subtr1 :: NatPair -> NatPair -> NatPair
subtr1 (m1, m2) (n1, n2) = plus1 (m1, m2) (n2, n1)

Here is the implementation of multiplication:

mult1 :: NatPair -> NatPair -> NatPair
mult1 (m1, m2) (n1, n2) = (m1*n1 + m2*n2, m1*n2 + m2*n1)

The implementation of equality for pairs of naturals is also straightforward:

eq1 :: NatPair -> NatPair -> Booleq1 (m1, m2) (n1, n2) = (m1+n2) == (m2+n1)

Finally, it is useful to be able to reduce a naturals pair to its simplest form (sucha simplest form is often called a canonical representation). The simplest form ofa naturals pair is a pair which has either its first or its second member equal toZ. E.g., the simplest form of the pair (S(S Z),S(S(S(S Z)))) is (Z,S(S Z)).
Reduction to simplest form can again be done by recursion.

8.5. RATIONAL NUMBERS

299

reduce1 :: NatPair -> NatPair
reduce1 (m1,Z) = (m1,Z)
reduce1 (Z,m2) = (Z,m2)
reduce1 (S m1, S m2) = reduce1 (m1, m2)

Exercise 8.10 Define and implement relations leq1 for 6 and gt1 for > for difference classes of naturals.

8.5 Rational Numbers
A further assumption that we would like to make is that any integer can be divided
by any non-zero integer to form a fraction, or rational number, and that any rational
number m/n can be ‘canceled down’ to its lowest form by dividing m and n by
the same number. E.g., 12/39 cancels down to 4/13.
Again we can view the rational numbers as constructed by means of pairs, in thiscase pairs (m, n) with m, n ∈ Z, n 6= 0. In this case the pairs are ‘ratio pairs’.
One and the same rational number can be represented in many different ways:(1, 2), (2, 4), (13, 26) all represent the rational number 1/2. Or in other words:
the rational number m/n is nothing but the class of all (p, q) with p, q ∈ Z, q 6= 0,
and the property that mq = np. In such a case we say that m/n and p/q represent
the same number.
As in the case of the representation of integers by means of difference classes
we have an underlying notion of equivalence. Let S be the equivalence relation on
Z×(Z−{0}) given by (m, n)S(p, q) :≡ mq = np. Note that (m, n) and (km, kn)
are equivalent modulo S (provided k 6= 0), which justifies the simplification of
km/kn to m/n by means of canceling down. The set Q of rational numbers (orfractional numbers) can be defined as:
Q := (Z × (Z − {0}))/S.
We write a class [(m/n)]S as [m/n]. Addition, multiplication and equality of rational numbers are now defined in terms of addition and multiplication of integers,
as follows:[m/n] + [p/q] := [(mq + pn)/nq][m/n] · [p/q] := [mp/nq][m/n] = [p/q] :≡ mq = np

CHAPTER 8. WORKING WITH NUMBERS

300

Again, we have to check that these definitions make sense. It is easy to see thatthe sum, the difference, and the product of two ratio classes are again ratio classes.
For instance, if x ∈ Q and y ∈ Q, then there are integers m, n, p, q ∈ Z with
n 6= 0, q 6= 0, and x = [m/n] and y = [p/q]. Then x + y = [m/n] + [p/q] =[(mq + pn)/nq], and from mq + pn ∈ Z, nq ∈ Z, and nq 6= 0 it follows thatx + y ∈ Q.
Proposition 8.11 The law of associativity for + holds for the rationals.
Proof. If x, y, z ∈ Q then there are m, n, p, q, r, s ∈ Z with x = [m/n], y = [p/q],z = [r/s], n 6= 0, q 6= 0 and s 6= 0. Now:
x + (y + z)=

[definitions of x, y, z]pr[mn ] + ([ q ] + [ s ])=[definition of + for Q]
ps+rq[mn ] + [ qs ]=[definition of + for Q]][ mqs+(ps+rq)nnqs=[distribution law for Z][ mqs+psn+rqn]
nqs[assoc of · for Z, dist law for Z]=][ (mq+pn)s+rqn
nqs= [definition of + for Q]r[ mq+pnnq ] + [ s ]= [definition of + for Q]pr([ mn ] + [ q ]) + [ s ]=[definitions of x, y, z](x + y) + z.

In a similar way, it is not difficult (but admittedly a bit tedious) to check that thelaw of commutativity for + holds for the rationals, that the laws of associativity

8.5. RATIONAL NUMBERS

301

and commutativity for · hold for the rationals, and that the law of distribution holds
for the rationals.
A thing to note about Proposition 8.11 is that the statement about rational numbers that it makes also tells us something about our ways of doing things in an
implementation. It tells us that any program that uses the procedure add which
implements addition for the rationals need not distinguish between(add (add (m,n) (p,q)) (r,s)) and (add (m,n) (add (p,q) (r,s))).

Wherever one expression fits the other will fit.
Again forgetting about irrelevant differences in representation, we see that Z ⊆ Q,
for every integer n has a representation in the rationals, namely n/1.
If x, y ∈ Q, then x + y ∈ Q, x − y ∈ Q, xy ∈ Q. If, in addition y 6= 0,
we have x/y ∈ Q. We say that the domain of rationals is closed under addition,
subtraction, and multiplication and ‘almost closed’ under division. If y 6= 0, then
1/y exists, which shows that every rational number y except 0 has an inverse: arational number that yields 1 when multiplied with y.
A domain of numbers that is closed under the four operations of addition, subtraction, multiplication and division (with the extra condition on division that thedivisor should be 6= 0) is called a field. The domain of rational numbers is a field.
It is easy to see that each rational number m/n can be written in decimal form, byperforming the process of long division. There are two possible outcomes:
1. the division process stops: m/n can be written as a finite decimal expansion.
2. the division process repeats: m/n can be written as a infinite decimal expansion with a tail part consisting of a digit or group of digits which repeats
infinitely often.
Examples of the first outcome are 1/4 = 0.25, −1/4 = −0.25, 22/5 = 4.4.
Examples of the second outcome are 1/3 = 0.3333 . . ., 1/6 = 0.166666 . . .,29/7 = 4.142857142857142857 . . ., 23/14 = 1.6428571428571428571 . . .. A
handy notation for this is 1/3 = 0.3, 1/6 = 0.6, 29/7 = 4.142857, 23/14 =1.6428571. The part under the line is the part that repeats infinitely often.
If the division process repeats for m/n then at each stage of the process there must
be a non-zero remainder. This means that at each stage, the remainder must be in
the range 1, . . . , n − 1. But then after at most n steps, some remainder k is boundto reoccur, and after the reappearance of k the process runs through exactly thesame stages as after the first appearance of k. The same remainders turn up, in thesame order. We are in a loop.

CHAPTER 8. WORKING WITH NUMBERS

302

Conversely, it is not very difficult to see that every repeating decimal corresponds
to a rational number. First an example. Suppose M = 0.133133133 . . . = 0.133.
Then: 1000 ∗ M = 133.133, and (1000 ∗ M ) − M = 133.133 − 0.133 = 133, so
133.
M = 999
Now for the general case.
Theorem 8.12 Every repeating decimal corresponds to a rational number.
Proof. A repeating decimal can always be written in our over-line notation in theform
M ± 0.a1 a2 · · · am b1 b2 · · · bn ,where M is an integer and the ai and bj are decimal digits. For example, −37/14 =−2.6428571 = −2 − 0.6428571. If we can prove that the partp = 0.a1 a2 · · · am b1 b2 · · · bn
is rational, then we are done, for if p is rational and M is an integer then there
are integers n, k with p = n/k, so M + p = M + n/k = kMk+n , i.e., M + p is
rational, and similarly for M − p. Setting
A = 0.a1 a2 · · · am , B = 0.b1 b2 · · · bn , C = 0.b1 b2 · · · bn ,we getp = 0.a1 a2 · · · am b1 b2 · · · bn = A + 10−m C.
If we can write this in terms of rational operations on A and B we are done.
C = 0.b1 b2 · · · bn = 0.b1 b2 · · · bn b1 b2 · · · bn .
Thus,
C − 10−n C= 0.b1 b2 · · · bn b1 b2 · · · bn − 0.01 · · · 0n b1 b2 · · · bn == 0.b1 b2 · · · bn = B.

B
Therefore, C = 1−10
−n , and we get:p=A+
so we have proved that p is rational.

10−m B,1 − 10−n8.5. RATIONAL NUMBERS

3 r
J

1 r
J

J

303

J

J

J

J
Jr

J

J

2
3

J
Jr
2

Figure 8.2: Constructing the fraction 2/3.
There is a nice geometrical interpretation of the process of constructing rational
numbers. Assume that we have been given the line of integers:
...

−3

•

−2

•

−1

•

0

•

1

•

2

•

3

•

...

Place these on a (horizontal) x-axis and a (vertical) y-axis:

x

···

• •

..
.
•
•
• • • ···
•
•
..
.
y

Now construct further points on the x-axis by first drawing a straight line l througha point m on the x-axis and a point n on the y-axis, and next drawing a line l 0parallel to l through the point (0, 1). The intersection of l 0 and the x-axis is therational point m/n. (Use congruence reasoning for triangles to establish that theratio between m and n equals the ratio between the intersection point m/n and 1.)
Figure 8.2 gives the construction of the fraction 2/3 on the x-axis.
Figures 8.3, 8.4, 8.5 and 8.6 give the geometrical interpretations of addition, negation, multiplication and reciprocal. Subtraction is addition of a negated number,division is multiplication with a reciprocal, so these give all the rational operations.
Note that these constructions can all be performed by a process of (i) connecting
previously constructed points by a straight line, and (ii) drawing lines parallel toCHAPTER 8. WORKING WITH NUMBERS

304

a

1
0

a

1

a+b

b

Figure 8.3: Geometrical Interpretation of Addition.

a

−a

−1

0

1

a

Figure 8.4: Geometrical Interpretation of Negation.

1
1
b

0

1

a

b

Figure 8.5: Geometrical Interpretation of Multiplication.

ab8.6. IMPLEMENTING RATIONAL ARITHMETIC

305

a

1
0 a1

1

a

Figure 8.6: Geometrical Interpretation of Reciprocal.
previously constructed lines. These are the so-called linear constructions: the constructions that can be performed by using a ruler but no compass. In particular, useof a compass to construct line segments of equal lengths is forbidden.

8.6 Implementing Rational Arithmetic
Haskell has a standard implementation of the above, a type Rational, predefined
as follows:data Integral a => Ratio a = a :% a deriving (Eq)type Rational= Ratio Integer

To reduce a fraction to its simplest form, we first make sure that the denominator
is non-negative. If (x,y) represents a fraction with numerator x and denominatory, then(x * signum y) (abs y)is an equivalent representation with positive denominator. Here abs gives the absolute value, and signum the sign (1 for positive integers, 0 for 0, −1 for negative
integers). The reduction to canonical form is performed by:

CHAPTER 8. WORKING WITH NUMBERS

306

(%)
x % y

:: Integral a => a -> a -> Ratio a= reduce (x * signum y) (abs y)

reducereduce x y | y == 0
| otherwise:: Integral a => a -> a -> Ratio a= error "Ratio.%: zero denominator"= (x ‘quot‘ d) :% (y ‘quot‘ d)where d = gcd x yFunctions for extracting the numerator and the denominator are provided:

numerator, denominator
numerator (x :% y)denominator (x :% y):: Integral a => Ratio a -> a= x
= yNote that the numerator of (x % y) need not be equal to x and the denominatorneed not be equal to y:
Prelude> numerator (2 % 4)1
Prelude> denominator (2 % 10)
5A total order on the rationals is implemented by:instance Integral a => Ord (Ratio a) wherecompare (x:%y) (x’:%y’) = compare (x*y’) (x’*y)

The standard numerical operations from the class Num are implemented by:8.6. IMPLEMENTING RATIONAL ARITHMETIC

307

instance Integral a => Num (Ratio a) where(x:%y) + (x’:%y’) = reduce (x*y’ + x’*y) (y*y’)(x:%y) * (x’:%y’) = reduce (x*x’) (y*y’)
negate (x :% y)= negate x :% yabs (x :% y)= abs x :% ysignum (x :% y)= signum x :% 1

The rationals are also closed under the division operation λxy. xy and the reciprocal
operation λx. x1 . These are implemented as follows:

instance Integral a => Fractional (Ratio a) where(x:%y) / (x’:%y’)= (x*y’) % (y*x’)
recip (x:%y)= if x < 0 then (-y) :% (-x) else y :% x

If you want to try out decimal expansions of fractions on a computer, here is a
Haskell program that generates the decimal expansion of a fraction.

decExpand :: Rational -> [Integer]
decExpand x | x < 0= error "negative argument"| r == 0= [q]| otherwise = q : decExpand ((r*10) % d)
where(q,r) = quotRem n dn= numerator xd= denominator x

If the decimal expansion repeats, you will have to interrupt the process by typing
control c:
WWN> decExpand (1 % 7)
[0, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8,
5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2,
8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4,
2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1,
4, 2, 8, 5,{Interrupted!}

308

CHAPTER 8. WORKING WITH NUMBERS

This problem can be remedied by checking every new quotient remainder pairagainst a list of quotient-remainder pairs, to spot a repetition. The main function
decForm produces the integer part, and relegates the task of calculating the lists of
non-repeating and repeating decimals to an auxiliary function decF.

decForm :: Rational -> (Integer,[Int],[Int])decForm x | x < 0= error "negative argument"| otherwise = (q,ys,zs)
where
(q,r)
= quotRem n dn= numerator xd= denominator x(ys,zs) = decF (r*10) d []The function decF has a parameter for the list of quotient remainder pairs thathave to be checked for repetition. The code for dForm uses elemIndex from themodule List.hs to find the index of the first repeating digit, and splitAt to split
a list at an index.

decF :: Integer -> Integer -> [(Int,Integer)] -> ([Int],[Int])decF n d xs | r == 0
= (reverse (q: (map fst xs)),[])| elem (q,r) xs = (ys,zs)| otherwise= decF (r*10) d ((q,r):xs)
where(q’,r) = quotRem n dq= toInt q’
xs’= reverse xs
Just k = elemIndex (q,r) xs’(ys,zs) = splitAt k (map fst xs’)

Here are a few examples:
WWN> decForm (133 % 999)
(0,[],[1,3,3])
WWN> decForm (1 % 7)(0,[],[1,4,2,8,5,7])
WWN> decForm (2 % 7)

8.7. IRRATIONAL NUMBERS

309(0,[],[2,8,5,7,1,4])
WWN> decForm (3 % 7)(0,[],[4,2,8,5,7,1])
WWN> decForm (4 % 7)(0,[],[5,7,1,4,2,8])
WWN> decForm (5 % 7)(0,[],[7,1,4,2,8,5])
WWN> decForm (6 % 7)(0,[],[8,5,7,1,4,2])

There is no upper limit to the length of the period. Here is an example with aperiod length of 99:
WWN> decForm (468 % 199)
(2,[],[3,5,1,7,5,8,7,9,3,9,6,9,8,4,9,2,4,6,2,3,1,1,5,5,7,7,8,8,9,4,4,
7,2,3,6,1,8,0,9,0,4,5,2,2,6,1,3,0,6,5,3,2,6,6,3,3,1,6,5,8,2,9,1,4,5,
7,2,8,6,4,3,2,1,6,0,8,0,4,0,2,0,1,0,0,5,0,2,5,1,2,5,6,2,8,1,4,0,7,0])

Exercise 8.13 Write a Haskell program to find the longest period that occurs indecimal expansions of fractions with numerator and denominator taken from theset {1, . . . , 999}.

8.7 Irrational Numbers
The Ancients discovered to their dismay that with just ruler and compass it ispossible to construct line segments whose lengths do not form rational fractions.
In other words, it is possible to construct a length q such that for no m, n ∈ N
it holds that q = m
n . Here is the famous theorem from Antiquity stating this
disturbing fact, with its proof.

@

@

@

@

1

1

x
@

@

@

@
@

310

CHAPTER 8. WORKING WITH NUMBERS

Theorem 8.14 There is no rational number x with x2 = 2.
Proof. Assume there is a number x ∈ Q with x2 = 2. Then there are m, n ∈ Z,
n 6= 0 with (m/n)2 = 2. We can further assume that m/n is canceled down to its
lowest form, i.e., there are no k, p, q ∈ Z with k 6= 1, m = kp and n = kq.

We have: 2 = (m/n)2 = m2 /n2 , and multiplying both sides by n2 we find
2n2 = m2 . In other words, m2 is even, and since squares of odd numbers arealways odd, m must be even, i.e., there is a p with m = 2p. Substitution in
2n2 = m2 gives 2n2 = (2p)2 = 4p2 , and we find that n2 = 2p2 , which leads tothe conclusion that n is also even. But this means that there is a q with n = 2q,
and we have a contradiction with the assumption that m/n was in lowest form. It
follows that there is no number x ∈ Q with x2 = 2. The square root of 2 is notrational.
√√
Of course, we all use 2 for the square/√ root of 2. The theorem tells us that 2 ∈Q. The collection of numbers that 2 does belong to is called the collection ofreal numbers, R. It is possible to give a formal construction of the reals from therationals, but we will not do so here. Instead, we just mention that Q ⊆ R, andwe informally introduce the reals as the set of all signed (finite or infinite) decimal
expansions. The domain of real numbers is closed under the four operations ofaddition, subtraction, multiplication and division, i.e., just like the rationals, thereals form a field.
Exercise 8.15 Use the method from the proof of Theorem 8.14 to show thatis irrational.
Exercise 8.16 Show that if p is prime, then√√3

p is irrational.

Exercise√ 8.17 Show that if n is a natural number withthen n is irrational.

√

n not a natural number,

Using the fundamental theorem of arithmetic, we can give the following alternative
proof of Theorem 8.14:√
Proof. If 2 = (p/q), then 2q 2 = p2 . In the representation of p2 as a productof prime factors, every prime factor has an even number of occurrences (for thesquare of p equals the product of the squares of p’s prime factors). In the representation of 2q 2 as a product of prime factors, the prime factor 2 has an odd numberof occurrences. Contradiction with Theorem 8.6.

8.7. IRRATIONAL NUMBERS

311

In Section 8.8 we will discuss an algorithm, the mechanic’s rule, for approaching
the square root of any positive fraction p with arbitrary precision.
The irrational numbers are defined here informally as the infinite non-periodic
decimal expansions. This is not a very neat definition, for its relies on the use ofdecimals, and there is nothing special about using the number ten as a basis fordoing calculations. Other bases would serve just as well. In any case, looking at Q
and R in terms of expansions provides a neat perspective on the relation between
rational and irrational numbers. What, e.g., are the odds against constructing arational by means of a process of tossing coins to get an infinite binary expansion?
Exercise 8.18 Does it follow from the fact that x + y is rational that x is rationalor y is rational? If so, give a proof, if not, give a refutation.
√
We have seen that writing out the decimal expansion of a real number like 2 does
not give a finite representation. In fact, since there are uncountably many reals (seeChapter 11), no finite representation scheme for arbitrary reals is possible.
In implementations it is customary to use floating point representation (or: scientific representation) of approximationsof reals. E.g., the decimal fraction 1424213.56√+ 5. Theis an approximation of 2 × 106 , and gets represented as 1.42421356E√√
2
decimal fraction 0.00000142421356 is an approximation of 106 = 2 × 10−6 ,and gets represented as 1.42421356E − 6. The general form is x.xxxEm, where
x.xxx is a decimal fraction called the mantissa and m is an integer called theexponent. Here are the Hugs versions:Prelude> sqrt 2 * 10^61.41421e+06
Prelude> sqrt 2 / 10^61.41421e-06

Haskell has a predefined type Float for single precision floating point numbers,
and a type Double for double precision floating point numbers. Together these
two types form the class Floating:Prelude> :t sqrt 2
sqrt 2 :: Floating a => aFloating point numbers are stored as pairs (m, n), where m is the matrix and n
the exponent of the base used for the encoding. If x is a floating point number, thebase of its representation is given by floatRadix and its matrix and exponent, as avalue of type (Integer,Int), by decodeFloat. Thus, if floatRadix x equalsb and decodeFloat x equals (m, n), then x is the number m · b n . floatDigits
gives the number of digits of m in base b representation. In particular, if (m, n)

312

CHAPTER 8. WORKING WITH NUMBERS

is the value of decodeFloat x, and d the value of floatDigits, then either mand n are both zero, or bd−1 6 m < bd . Here is an example:
Prelude> floatRadix (sqrt 2)
2
Prelude> decodeFloat (sqrt 2)(11863283,-23)
Prelude> 11863283 * 2^^(-23)
1.41421
Prelude> floatDigits (sqrt 2)
24
Prelude> 2^23 <= 11863283 && 11863283 < 2^24
True

The inverse to the function decodeFloat is encodeFloat:
Prelude> sqrt 2
1.41421
Prelude> encodeFloat 11863283 (-23)
1.41421

Scaling a floating point number is done by ‘moving the point’:Prelude> scaleFloat 4 (sqrt 2)
22.6274
Prelude> 2^4 * sqrt 2
22.6274

A floating point number can always be scaled in such a way that its matrix is in
the interval (−1, 1). The matrix in this representation is given by the function
significand, the exponent by exponent:
Prelude> significand (sqrt 2)
0.707107
Prelude> exponent (sqrt 2)
1
Prelude> 0.707107 * 2^1
1.41421
Prelude> scaleFloat 1 0.707107
1.41421

The definitions of exponent, significand and scaleFloat (from the Prelude):8.8. THE MECHANIC’S RULE

exponent x
significand x
scaleFloat k x

313= if m==0 then 0 else n + floatDigits x
where (m,n) = decodeFloat x= encodeFloat m (- floatDigits x)
where (m,_) = decodeFloat x= encodeFloat m (n+k)
where (m,n) = decodeFloat x

8.8 The Mechanic’s Rule
Sequences of fractions can be used to find approximations to real numbers thatthemselves are not fractions (see Section 8.7). A well known algorithm for generating such sequences is the so-called mechanic’s rule (also known as Newton’s
method, a bit misleadingly, for the algorithm was already in use centuries before
Newton):1
pp > 0, a0 > 0, an+1 = (an +).
2an
In Exercise 8.19 you are asked to prove that this can be used to approximate thesquare root of any positive fraction p to any degree of accuracy. The Haskell implementation uses some fresh ingredients. The function recip takes the reciprocalof a fraction, the operation iterate iterates a function by applying it again to theresult of the previous application, and takeWhile takes a property and a list andconstructs the largest prefix of the list consisting of objects satisfying the property.
In the present case, the property is (\ m -> m^2 <= p), having a square 6 p,
and the list is the list of positive naturals. You should look up the implementationsin Prelude.hs and make sure you understand.

mechanicsRule :: Rational -> Rational -> Rational
mechanicsRule p x = (1 % 2) * (x + (p * (recip x)))mechanics :: Rational -> Rational -> [Rational]
mechanics p x = iterate (mechanicsRule p) x

CHAPTER 8. WORKING WITH NUMBERS

314

sqrtM :: Rational -> [Rational]
sqrtM p | p < 0= error "negative argument"| otherwise = mechanics p s
where
s = if xs == [] then 1 else last xs
xs = takeWhile (\ m -> m^2 <= p) [1..]√As a demonstration, here are the first seven√ steps in the approximation to 2, thefirst seven steps in the√approximation to 4, and the first, second, . . . , sixth step in
the approximation to 50, respectively. This already gives greater accuracy thanis needed in most applications; the algorithm converges very fast.
WWN> take 5 (sqrtM 2)[1 % 1,3 % 2,17 % 12,577 % 408,665857 % 470832]
WWN> take 7 (sqrtM 4)[2 % 1,2 % 1,2 % 1,2 % 1,2 % 1,2 % 1,2 % 1]
WWN> sqrtM 50 !! 0
7 % 1
WWN> sqrtM 50 !! 1
99 % 14
WWN> sqrtM 50 !! 2
19601 % 2772
WWN> sqrtM 50 !! 3768398401 % 108667944
WWN> sqrtM 50 !! 41180872205318713601 % 167000548819115088

Exercise 8.19

1. Prove that for every n,√√an+1 − p(an − p)2=√
√ .
an+1 + p(an + p)2

2. From the first item it follows (by induction on n) that

√
√  n
an+1 − pa0 − p 2.
√ =√an+1 + p
a0 + pDerive from this that sqrtM p converges tonumber p.

√
p for any positive rational

8.9. REASONING ABOUT REALS

315

3. Show that the approximation is from above, i.e., show that n > 1 implies√
that an > p.

Exercise 8.20
1. Find a rule to estimate the number of correct decimal places√
in approximation an of p. Use the result of Exercise 8.19. (Hint: try√
to find an inequality of the form an − p 6 t, with t an expression that
employs a0 and a1 .)2. Use the previous item to give an estimate of√the number of correct decimal
places in the successive approximations to 2.

8.9 Reasoning about Reals
Suppose one wants to describe the behaviour of moving objects by plotting theirposition or speed as a function of time. Moving objects do not suddenly disappear
and reappear somewhere else (outside the Bermuda triangle, at least), so the pathof a moving object does not have holes or gaps in it. This is where the notions ofcontinuity and limit arise naturally. Analysis, with its emphasis on continuity andlimit, is a rich source of examples where skill in quantifier reasoning is called for.
The following Lemma illustrates that many common functions are continuous. Its
proof is given as an example: it is a nice illustration of the use of the logical rules.
In fact, logic is all there is to this proof: properties of real numbers are not neededat all.
Lemma. The composition of two continuous functions is continuous.
I.e., if f and g are continuous functions from reals to reals, then the function hdefined by h(x) = g(f (x)) (cf. Definition 6.29 p. 222) is continuous as well.
Proof. Given: f and g are continuous, i.e., (“ε-δ-definition” p. 66; for clarity weuse different variables for the arguments):∀x ∀ε > 0 ∃δ > 0 ∀y (|x − y| < δ =⇒ |f (x) − f (y)| < ε),(8.1)∀a ∀ε > 0 ∃δ > 0 ∀b (|a − b| < δ =⇒ |g(a) − g(b)| < ε).

(8.2)

To be proved: ∀x ∀ε > 0 ∃δ > 0 ∀y (|x − y| < δ =⇒ |g(f (x)) − g(f (y))| < ε).

CHAPTER 8. WORKING WITH NUMBERS

316

Proof (detailed version): Note that what is To be proved begins with two universal
quantifiers ∀x and ∀ε > 0. Therefore, the proof has to start (recall the obligatory
opening that goes with ∀-introduction!) with choosing two arbitrary values x ∈ R
and ε > 0. We now have to show that∃δ > 0 ∀y (|x − y| < δ =⇒ |g(f (x)) − g(f (y))| < ε).

(8.3)

The next quantifier asks us to supply some example-δ that is > 0. There appears
to be no immediate way to reduce the proof problem further, so we start looking at
the givens in order to obtain such an example.
It turns out that we have to use the second given (8.2) first. It can be used (∀elimination) if we specify values for a and ε. Later on it will show that a = f (x)and the earlier ε will turn out to be useful.
Thus, (8.2) delivers (∃-elimination) some δ1 > 0 such that∀b (|f (x) − b| < δ1 =⇒ |g(f (x)) − g(b)| < ε).

(8.4)

Applying the given (8.1) to our x and ε = δ1 (∀-elimination) we obtain (∃elimination) δ > 0 such that∀y (|x − y| < δ =⇒ |f (x) − f (y)| < δ1 ).

(8.5)

And this is the example-δ we are looking for, i.e.:
Claim: ∀y (|x − y| < δ =⇒ |g(f (x)) − g(f (y))| < ε).
(From this follows what we have to show using ∃-introduction.)
Proof: Suppose that (to prepare for ∀-introduction and Deduction Rule — cf. Exercise 3.18 p. 93) y is such that |x − y| < δ.
From (8.5) (using this y — ∀-elimination and Modus Ponens) you find, that |f (x)−f (y)| < δ1 .
Finally, from (8.4) (with b = f (y) — ∀-elimination) you find that |g(f (x)) −g(f (y))| < ε.
The Claim follows.
Making up the score: the proof applies one rule for each of the fifteen (!) occurrences of logical symbols in Given and To be proved. Make sure you understandevery detail.
Of course, the amount of detail is excessive and there is a more common concise
version as well. The following version is the kind of argument you will find in8.9. REASONING ABOUT REALS

317

analysis textbooks. As is usual, it leaves to the reader to fill in which rules havebeen applied, and where.
Proof. Assume that x ∈ R and ε > 0.
From (8.2), obtain δ1 > 0 such that∀b (|f (x) − b| < δ1 =⇒ |g(f (x)) − g(b)| < ε).

(8.6)

Applying (8.1) to x and δ1 , obtain δ > 0 such that∀y (|x − y| < δ =⇒ |f (x) − f (y)| < δ1 ).

(8.7)

Then if |x − y| < δ, by (8.7) we get that |f (x) − f (y)| < δ1 , and from (8.6) it
follows that |g(f (x)) − g(f (y))| < ε.
This version of the proof will be considered very complete by every mathematician. Nevertheless, the compression attained is approximately 4 : 1.
The remaining examples of this section are about sequences of reals.
Limits. Assume that a0 , a1 , a2 , . . . is a sequence of reals and that a ∈ R. The
expression limi→∞ ai = a (“the sequence converges to a”, “a is limit of the sequence”) by definition means that∀ε > 0 ∃n ∀i > n (|a − ai | < ε).
Generating concrete examples of converging sequences in Haskell is easy, as wehave seen in Section 8.8. We have seen that the sequences produced by the mechanic’s rule converge. We will prove some results about convergence.
Theorem. Every sequence of reals has at most one limit.
Proof. The situation should be analyzed as follows.
Given: limi→∞ ai = a, limi→∞ ai = b.
To be proved: a = b.
Proof: Proof by Contradiction is not a bad idea here, since the new given it provides, nl., that a 6= b, is equivalent with the positive |a − b| > 0. Thus, assume
this. Proof by Contradiction now asks to look for something false.
In order to use the old given, you need to choose (∀-elimination!) values for ε. As
you’ll see later, it is useful to choose ε = 21 |a − b|. Note that ε > 0.

CHAPTER 8. WORKING WITH NUMBERS

318

From the given limi→∞ ai = a we obtain now, that ∃n ∀i > n (|a − ai | < ε).
Thus (∃-elimination) some n1 exists such that ∀i > n1 (|a − ai | < ε).
From the given limi→∞ ai = b we obtain, similarly, some n2 such that ∀i >n2 (|b − ai | < ε).
Define n = max(n1 , n2 ). Since n > n1 , n2 , by ∀-elimination we now get fromthese facts that both |a − an | < ε and |b − an | < ε.
Lastly, using the triangle-inequality|x + y| 6 |x| + |y|,it follows that |a − b| = |a − an + an − b| 6 |a − an | + |b − an | < 2ε = |a − b|— and this is the falsity looked for.
Exercise 8.21 Write a concise version of the above proof.
Exercise 8.22 Assume that limi→∞ ai = a.
1. Show that limi→∞ a2i = a.
2. Assume that f : N → N is a function such that ∀n∃m∀i > m f (i) > n.
Show that limi→∞ af (i) = a.

∞
Exercise 8.23 Assume that the sequences of reals {an }∞n=0 and {bn }n=0 have
limits a resp. b, and that a < b. Show that a number n exists such that ∀m >n (am < bm ).

Exercise 8.24 Assume that limi→∞ ai = a and that limi→∞ bi = b.
Show that limi→∞ (ai + bi ) = a + b.

Exercise 8.25 Show that a function f : R → R is continuous iff limi→∞ f (ai ) =f (a) whenever limi→∞ ai = a.
Cauchy. A sequence of reals {an }∞n=0 is called Cauchy if∀ε > 0 ∃n ∀i, j > n (|ai − aj | < ε).

8.10. COMPLEX NUMBERS

319

Exercise 8.26 Assume that the sequence {an }∞n=0 is Cauchy.
1. Show that the sequence is bounded. I.e., that numbers b and c exist such that∀i (b < ai < c).
2. Assume that a ∈ R is such that ∀ε > 0∀n∃i > n (|a − ai | < ε). (Theexistence of such an a follows from the sequence being bounded, but you
are not asked to prove this.) Show that lim i→∞ ai = a.

It follows immediately from Exercise 8.19 that the sequences of rationals producedby the Mechanic’s rule are Cauchy. Thanks to that we can implement a programfor calculating square roots as follows (for still greater precision, change the valueof ε in apprx):

approximate :: Rational -> [Rational] -> Rational
approximate eps (x:y:zs)| abs (y-x) < eps = y| otherwise= approximate eps (y:zs)
apprx :: [Rational] -> Rational
apprx = approximate (1/10^6)
mySqrt :: Rational -> Rational
mySqrt p = apprx (sqrtM p)

Exercise 8.27 Just as we defined the integers from the naturals and the rationals
from the integers by means of quotient sets generated from suitable equivalenceclasses, we can do so now, by defining the set of real numbers R as the set of all
Cauchy sequences in Q modulo an appropriate equivalence relation. Define thatequivalence relation and show that it is indeed an equivalence.

8.10 Complex Numbers
In the domain of rational numbers we cannot solve the equationx 2 − 2√= 0, but√
in the domain of real numbers we can: its roots are x = 2 and x = − 2. What

320

CHAPTER 8. WORKING WITH NUMBERS

about solving x2 + 1 = 0? There are no real number solutions, for the square rootof −1 does not exist in the realm of real numbers. The field of real numbers is not
closed under the operation of taking square roots. To remedy this, we follow theby now familiar recipe of extending the number domain, by introducing numbersof a new kind. We extend the domain of reals to a domain of complex numbers,
C, by introducing an entity i called ‘the imaginary unit’, and postulating i 2 = −1.
We do not want to lose closure under addition, subtraction,
multiplication and√division, so we should be able to make sense of 2 + i, 2i, −i, and so on. In
general, we want to be able to make sense of x + iy, where x and y are arbitrary
real numbers, and we need rules for adding and multiplying complex numbers, insuch a way that the laws of commutativity and associativity of + and ·, and thelaw of distribution of · over + continue to hold. x is called the real part of x + iy,
iy its imaginary part. Adding complex numbers boils down to adding real andimaginary parts separately:(x + iy) + (u + iw) = (x + u) + i(y + w).
This also gives a recipe for subtraction:(x + iy) − (u + iw) = (x + iy) + (−u + −iw) = (x − u) + i(y − w).
For multiplication, we use the fact that i2 = −1:(x + iy)(u + iw) = xu + iyu + ixw + i2 yw = (xu − yw) + i(yu + xw).
Division uses the fact that (x + iy)(x − iy) = x2 + y 2 . It is given by:
x + iy u − iw
xu + yw
yu − xw
x + iy
=
·
= 2
+i 2
.
u + iw
u + iw u − iw
u + w2u + w2We see that like the rationals and the reals, the complex numbers are closed under
the four operations addition, subtraction, multiplication and division, so C is afield. Moreover, any real number a can be viewed as a complex number of theform a + 0i, so we have that R ⊆ C.

Solving the equation x2 + 1 = 0 in the domain C gives two roots, viz., x = i andx = −i. In general, solving the equation xn + an−1 xn−1 + . . . + a1 x + a0 = 0,
where a0 , . . . , an−1 may be either real or complex, gives n complex roots, for thefundamental theorem of algebra (which we will not prove here) states that every
polynomial of degree n,f (x) = xn + an−1 xn−1 + . . . + a1 x + a0 ,can be factored into a product of exactly n factors,(x − b1 )(x − b2 ) · · · (x − bn ).

8.10. COMPLEX NUMBERS

321

Exercise 8.28 Check that the commutative and associative laws and the distributive law hold for C.
There is a standard Haskell module Complex.hs with an implementation of complex numbers.

infix

6

:+

data (RealFloat a) => Complex a = !a :+ !aderiving (Eq,Read,Show)

The exclamation marks in the typing !a :+ !a indicate that the real and imaginary parts of type RealFloat are evaluated in a strict way.
The real part of a complex number x + iy is the real number x, the imaginary
part the real number y. Notation for the real part of z: Re(z). Notation for theimaginary part of z: Im(z). The Haskell implementations are:realPart, imagPart :: (RealFloat a) => Complex a -> a
realPart (x:+y)= x
imagPart (x:+y)= yThe complex number z = x + iy can be represented geometrically in either of twoways:
1. Associate with z the point with coordinates (x, y) in the plane R2 . In thisway, we view the plane R2 as the complex plane.
2. Associate with z the two-dimensional vector with components x and y.
Think of this vector as a free vector, i.e., a vector that may be moved aroundfreely as long as its direction remains unchanged.
The two representations can be combined by attaching the vector z to the origin of
the plane. We then get the picture of Figure 8.7. Call the horizontal axis throughthe origin the real axis and the vertical axis through the origin the imaginary axis.
The conjugate of a complex number z = x + iy is the number z̄ = x − iy. Its
geometrical representation is the reflection of z in the real axis (see again Figure8.7). Its implementation is given by:

CHAPTER 8. WORKING WITH NUMBERS

322

z

iy
i

x

θ
−1

1
−i
z̄

Figure 8.7: Geometrical Representation of Complex Numbers.

conjugate :: (RealFloat a) => Complex a -> Complex aconjugate (x:+y)= x :+ (-y)

The magnitude or modulus or absolute value of a complex number z is thep length rof the z vector. Notation |z|. The magnitude of z = x + iy is given by x2 + y 2 .
Its Haskell implementation:

magnitude :: (RealFloat a) => Complex a -> amagnitude (x:+y) =scaleFloat k (sqrt ((scaleFloat mk x)^2 + (scaleFloat mk y)^2))
where k = max (exponent x) (exponent y)
mk = - k

The phase or argument of a complex number z is the angle θ of the vector z.
Notation: arg(z). The phase θ of a vector of magnitude 1 is given by the vector cos(θ) + i sin(θ) (see Figure 8.8), so the phase of z = x + iy is given byarctan xy for x > 0, and arctan xy + π for x < 0. In case x = 0, the phase is8.10. COMPLEX NUMBERS

323

z

iy

i sin(θ)θ

xcos(θ)

Figure 8.8: ∠θ = cos(θ) + i sin(θ).

CHAPTER 8. WORKING WITH NUMBERS

324

0. This computation is taken care of by the Haskell function atan2. Here is theimplementation:

phase
phase (0:+0)
phase (x:+y):: (RealFloat a) => Complex a -> a
= 0
= atan2 y x

The polar representation of a complex number z = x + iy is r∠θ, where r = |z|and θ = arg(z). The advantage of polar representation is that complex multiplication looks much more natural: just multiply the magnitudes and add the phases.
(R∠ϕ)(r∠θ) = Rr∠(ϕ + θ).
The polar representation of a complex number is given by:polar
polar z:: (RealFloat a) => Complex a -> (a,a)= (magnitude z, phase z)

Note that polar representations are not unique, for we have:. . . = r∠(θ − 2π) = r∠θ = r∠(θ + 2π) = r∠(θ + 4π) = . . .
To get from representation r∠θ to the representation as a vector sum of real andimaginary parts, use r∠θ = r(cos(θ) + i sin(θ)) (see again Figure 8.8). Implementation:

mkPolarmkPolar r theta:: (RealFloat a) => a -> a -> Complex a= r * cos theta :+ r * sin theta

Converting a phase θ to a vector in the unit circle is done by:cis
cis theta:: (RealFloat a) => a -> Complex a= cos theta :+ sin theta

8.10. COMPLEX NUMBERS325

The implementations of the arithmetical operations of addition, subtraction, multiplication, negation, absolute value and signum are given by:

instance (RealFloat a) => Num (Complex a) where(x:+y) + (x’:+y’) = (x+x’) :+ (y+y’)(x:+y) - (x’:+y’) = (x-x’) :+ (y-y’)(x:+y) * (x’:+y’) = (x*x’-y*y’) :+ (x*y’+y*x’)
negate (x:+y)= negate x :+ negate y
abs z= magnitude z :+ 0
signum 0= 0
signum z@(x:+y)= x/r :+ y/r where r = magnitude z

Note that the signum of a complex number z is in fact the vector representationof arg(z). This suggests that arg(z) is a generalization of the + or − sign forreal numbers. That this is indeed the case can be seen when we perform complex
multiplication on real numbers: positive real numbers, represented in the complex
plane, have phase 0, or in general, phase 2kπ. Negative real number, represented
in the complex plane, have phase π, or in general, (2k + 1)π. Multiplying two
negative real numbers means multiplying their values and adding their phases, so
we get phase 2π, which is the same as phase 0, modulo 2kπ.
Complex division is the inverse of multiplication, so it boils down to performing
division on the magnitudes, and subtraction on the phases. The implementation interms of the vector representations looks slightly more involved than this, however:instance (RealFloat a) => Fractional (Complex a) where(x:+y) / (x’:+y’) = (x*x’’+y*y’’) / d :+ (y*x’’-x*y’’) / d
where x’’ = scaleFloat k x’y’’ = scaleFloat k y’k= - max (exponent x’) (exponent y’)d= x’*x’’ + y’*y’’

For the definition of further numerical operations on complex numbers we refer tothe library file Complex.hs.
Complex numbers lose their mystery when one gets well acquainted with theirgeometrical representations. Here are√some examples, for getting the feel of them.
The number 1 + i has magnitude 2 and phase π4 : see Figure 8.9. Squaring

326

CHAPTER 8. WORKING WITH NUMBERS

√

i

2

π
4

1

Figure 8.9: The number 1 + i.

iπ2Figure 8.10: The number (1 + i)2 .

8.10. COMPLEX NUMBERS

327

√
2 2

3π
4

Figure 8.11: The number (1 + i)3 .

CHAPTER 8. WORKING WITH NUMBERS

328

this number involves squaring the magnitude and doubling the phase, so (1 + i) 2
has magnitude 2 and phase π2 : see Figure 8.10. Raising 1 + i to the third power√
involves multiplying the magnitudes 2 and 2 and adding the phases π4 and π2 .
√This gives magnitude 2 2 and phase 3π
4 . See Figure 8.11 for a picture of thenumber (1+i)3 . Raising 1+i to the fourth power involves squaring (1+i)2, so themagnitude 2 is squared and the phase π2 is doubled, which gives magnitude 4 andphase π. Translating all of this back into vector sum notation, we get (1+i) 2 = 2i,(1 + i)3 = −2 + 2i, and (1 + i)4 = −4. And sure enough, the Haskell library
Complex.hs confirms these findings:
Complex> (1 :+ 1)^2
0.0 :+ 2.0
Complex> (1 :+ 1)^3(-2.0) :+ 2.0
Complex> (1 :+ 1)^4(-4.0) :+ 0.0

Similarly, we see that multiplying i and −i involves multiplying the magnitudes
1 and adding the phases π2 and 3π
2 , with result the number with magnitude 1 andphase 2π = 0, i.e., the number 1. Here is the Haskell confirmation:
Complex> (0 :+ 1) * (0 :+ (-1))
1.0 :+ 0.0

Exercise 8.29 You are encouraged to familiarize yourself further with complex
numbers by checking the following by means of pictures:
1. Re(z) = 12 (z + z̄).
1(z − z̄).
2. Im(z) = 2i

3. tan(arg(z)) = Im(z)Re(z) .
R
4. R∠ϕr∠θ = r ∠(ϕ − θ).

5. arg( zz12 ) = arg(z1 ) − arg(z2 ).

6. arg( 1z ) = − arg(z).

Exercise 8.30

1. Use induction on n to prove De Moivre’s formula for n ∈ N:(cos(ϕ) + i sin(ϕ))n = cos(nϕ) + i sin(nϕ).

Draw a picture to see what is happening!

8.11. FURTHER READING

329

2. Prove De Moivre’s formula for exponents in Z by using the previous item,plus:
1(cos(ϕ) + i sin(ϕ))−m =.
(cos(ϕ) + i sin(ϕ))m8.11 Further Reading
A classic overview of the ideas and methods of mathematics, beautifully written,
and a book everyone with an interest in mathematics should possess and read is
Courant and Robbins [CR78, CrbIS96]. Here is praise for this book by Albert
Einstein:
A lucid representation of the fundamental concepts and methods ofthe whole field of mathematics. It is an easily understandable introduction for the layman and helps to give the mathematical student a
general view of the basic principles and methods.
Another beautiful book of numbers is [CG96]. For number theory and its history
see [Ore88].

330

CHAPTER 8. WORKING WITH NUMBERS

Chapter 9

Polynomials
Preview
Polynomials or integral rational functions are functions that can be represented bya finite number of additions, subtractions, and multiplications with one independent variable. The closed forms that we found and proved by induction for thesums of evens, sums of odds, sums of squares, sums of cubes, and so on, in Chapter 7, are all polynomial functions. In this chapter we will first study the process
of automating the search for polynomial functions that produce given sequencesof integers. Next, we establish the connection with the binomial theorem, we implement a datatype for the polynomials themselves, and we use this datatype forthe study of combinatorial problems.

module POL
whereimport Polynomials

331

CHAPTER 9. POLYNOMIALS

332

9.1 Difference Analysis of Polynomial Sequences
Suppose {an } is a sequence of natural numbers, i.e., f = λn.an is a function in
N → N. The function f is a polynomial function of degree k if f can be presented
in the form
ck nk + ck−1 nk−1 + · · · + c1 n + c0 ,with ci ∈ Q and ck 6= 0.
Example 9.1 The sequence[1, 4, 11, 22, 37, 56, 79, 106, 137, 172, 211, 254, 301, 352, . . .]is given by the polynomial function f = λn.(2n2 + n + 1). This is a function of
the second degree.
Here is the Haskell check:Prelude> take 15 (map (\ n -> 2*n^2 + n + 1) [0..])[1,4,11,22,37,56,79,106,137,172,211,254,301,352,407]

Consider the difference sequence given by the functiond(f ) = λn.an+1 − an .
The Haskell implementation looks like this:difs :: [Integer] -> [Integer]
difs [] = []difs [n] = []difs (n:m:ks) = m-n : difs (m:ks)

This gives:
POL> difs [1,4,11,22,37,56,79,106,137,172,211,254,301][3,7,11,15,19,23,27,31,35,39,43,47]

9.1. DIFFERENCE ANALYSIS OF POLYNOMIAL SEQUENCES

333

The difference function d(f ) of a polynomial function f is itself a polynomial
function. E.g., if f = λn.(2n2 + n + 1), then:d(f ) = λn.(2(n + 1)2 + (n + 1) + 1 − (2n2 + n + 1)
= λn.4n + 3.
The Haskell check:
POL> take 15 (map (\n -> 4*n + 3) [0..])[3,7,11,15,19,23,27,31,35,39,43,47,51,55,59]
POL> take 15 (difs (map (\ n -> 2*n^2 + n + 1) [0..]))[3,7,11,15,19,23,27,31,35,39,43,47,51,55,59]

Proposition 9.2 If f is a polynomial function of degree k then d(f ) is a polynomial function of degree k − 1.
Proof. Suppose f (n) is given by
ck nk + ck−1 nk−1 + · · · + c1 n + c0 .
Then d(f )(n) is given by
ck (n + 1)k +ck−1 (n + 1)k−1 + · · · + c1 (n + 1) + c0

− (ck nk + ck−1 nk−1 + · · · + c1 n + c0 ).

It is not hard to see that f (n + 1) has the form ck nk + g(n), with g a polynomialof degree k − 1. Since f (n) also is of the form ck nk + h(n), with h a polynomialof degree k − 1, d(f )(n) has the form g(n) − h(n), so d(f ) is itself a polynomialof degree k − 1.
It follows from Proposition 9.2 that if f is a polynomial function of degree k, thendk (f ) will be a constant function (a polynomial function of degree 0).
Here is a concrete example of computing difference sequences until we hit at aconstant sequence:
-12

-11
1

6
17

16

45
39

22
6

112
67

28
6

213
101

34
6

354
141

40
6

541
187

46
6

We find that the sequence of third differences is constant, which means that theform of the original sequence is a polynomial of degree 3. To find the next number

334

CHAPTER 9. POLYNOMIALS

in the sequence, just take the sum of the last elements of the rows. This gives
6 + 46 + 187 + 541 = 780.
Charles Babbage (1791–1871), one of the founding fathers of computer science,used these observations in the design of his difference engine. We will give a
Haskell version of the machine.
According to Proposition 9.2, if a given input list has a polynomial form of degreek, then after k steps of taking differences the list is reduced to a constant list:
POL> difs [-12,-11,6,45,112,213,354,541,780,1077][1,17,39,67,101,141,187,239,297]
POL> difs [1,17,39,67,101,141,187,239,297][16,22,28,34,40,46,52,58]
POL> difs [16,22,28,34,40,46,52,58][6,6,6,6,6,6,6]

The following function keeps generating difference lists until the differences getconstant:

difLists :: [[Integer]]->[[Integer]]
difLists [] = []difLists lists@(xs:xss) =if constant xs then lists else difLists ((difs xs):lists)
whereconstant (n:m:ms) = all (==n) (m:ms)
constant _= error "lack of data or not a polynomial fct"

This gives the lists of all the difference lists that were generated from the initial
sequence, with the constant list upfront.
POL> difLists [[-12,-11,6,45,112,213,354,541,780,1077]][[6,6,6,6,6,6,6],[16,22,28,34,40,46,52,58],[1,17,39,67,101,141,187,239,297],[-12,-11,6,45,112,213,354,541,780,1077]]

The list of differences can be used to generate the next element of the original
sequence: just add the last elements of all the difference lists to the last element of
the original sequence. In our example case, to get the next element of the list[−12, −11, 6, 45, 112, 213, 354, 541, 780, 1077]

9.1. DIFFERENCE ANALYSIS OF POLYNOMIAL SEQUENCES

335

add the list of last elements of the difference lists (including the original list):
6 + 58 + 297 + 1077 = 1438. To see that this is indeed the next element, note thatthe difference of 1438 and 1077 is 361, the difference of 361 and 297 is 64, andthe difference of 64 and 58 is 6, so the number 1438 ‘fits’ the difference analysis.
The following function gets the list of last elements that we need (in our example
case, the list [6,58,297,1077]):

genDifs :: [Integer] -> [Integer]
genDifs xs = map last (difLists [xs])A new list of last elements of difference lists is computed from the current one bykeeping the constant element d1 , and replacing each di+1 by di + di+1 .

nextD :: [Integer] -> [Integer]
nextD [] = error "no data"
nextD [n] = [n]nextD (n:m:ks) = n : nextD (n+m : ks)

The next element of the original sequence is given by the last element of the newlist of last elements of difference lists:

next :: [Integer] -> Integer
next = last . nextD . genDifs

In our example case, this gives:
POL> next [-12,-11,6,45,112,213,354,541,780,1077]
1438

All this can now be wrapped up in a function that continues any list of polynomial
form, provided that enough initial elements are given as data:

CHAPTER 9. POLYNOMIALS

336

continue :: [Integer] -> [Integer]
continue xs = map last (iterate nextD differences)where
differences = nextD (genDifs xs)

This uses the predefined Haskell function iterate, that is given by:

iterateiterate f x:: (a -> a) -> a -> [a]= x : iterate f (f x)This is what we get:
POL> take 20 (continue [-12,-11,6,45,112,213,354,541,780,1077])[1438,1869,2376,2965,3642,4413,5284,6261,7350,8557,9888,11349,
12946,14685,16572,18613,20814,23181,25720,28437]

If a given list is generated by a polynomial, then the degree of the polynomial canbe computed by difference analysis, as follows:degree :: [Integer] -> Int
degree xs = length (difLists [xs]) - 1

The difference engine is smart enough to be able to continue a list of sums ofsquares, or a list of sums of cubes:
POL> take 10 (continue [1,5,14,30,55])[91,140,204,285,385,506,650,819,1015,1240]
POL> take 10 (continue [1,9,36,100,225,441])[784,1296,2025,3025,4356,6084,8281,11025,14400,18496]

Exercise 9.3 What continuation do you get for [3,7,17,39,79,143]? Can you
reconstruct the polynomial function that was used to generate the sequence?

9.2. GAUSSIAN ELIMINATION

337

Difference analysis yields an algorithm for continuing any finite sequence with apolynomial form. Is it also possible to give an algorithm for finding the form? This
would solve the problem of how to guess the closed forms for the functions that
calculate sums of squares, sums of cubes, and so on. The answer is ‘yes’, and themethod is Gaussian elimination.

9.2 Gaussian Elimination
If we know that a sequence a0 , a1 , a2 , a3 , . . . has a polynomial form of degree
3, then we know that the form is a + bx + cx2 + dx3 (listing the coefficientsin increasing order). This means that we can find the form of the polynomial bysolving the following quadruple of linear equations in a, b, c, d:a = a0a + b + c + d = a1
a + 2b + 4c + 8d = a2
a + 3b + 9c + 27d = a3
Since this is a set of four linear equations in four unknowns, where the equations
are linearly independent (none of them can be written as a multiple of any of theothers), this can be solved by eliminating the unknowns one by one.
Example 9.4 Find the appropriate set of equations for the sequence[−7, −2, 15, 50, 109, 198, 323]
and solve it.
Difference analysis yields that the sequence is generated by a polynomial of thethird degree, so the sequence leads to the following set of equations:a = −7a + b + c + d = −2

a + 2b + 4c + 8d = 15
a + 3b + 9c + 27d = 50
Eliminating a and rewriting gives:
b+c+d = 5
2b + 4c + 8d = 22
3b + 9c + 27d = 57

338

CHAPTER 9. POLYNOMIALS

Next, eliminate the summand with factor d from the second and third equation.
Elimination from the second equation is done by subtracting the second equationfrom the 8-fold product of the first equation. This gives 6b + 4c = 18, which
can be simplified to 3b + 2c = 9. Elimination from the third equation is doneby subtracting the third equation from the 27-fold product of the first, with result
24b + 18c = 78. We get the following pair of equations:
3b + 2c = 924b + 18c = 78
Elimination of c from this pair is done by subtracting the second equation fromthe 9-fold product of the first. This gives 3b = 3, whence b = 1. Together with
3b + 2c = 9 we get c = 3. Together with b + c + d = 5 we get d = 1. Thus, thepolynomial we are looking for has the form λn.(n3 + 3n2 + n − 7).
Exercise 9.5 Find the appropriate set of equations for the sequence[13, 21, 35, 55, 81, 113, 151]
and solve it.
Solving sets of linear equations can be viewed as manipulation of matrices ofcoefficients. E.g., the quadruple of linear equations in a, b, c, d for a polynomial of
the third degree gives the following matrix:
1 0 0 0 a0
 1 1 1 1 a1 
 1 2 4 8 a2 
1 3 9 27 a3

To solve this, we transform it to an equivalent matrix in so-called echelon form orleft triangular form, i.e., a matrix of the form:
a00 a01 a02 a03 b0
 0 a11 a12 a13 b1  0
0 a22 a23 b2 00
0 a33 b3

From this form, compute the value of variable d from the last row, next eliminate
this variable from the third row, and find the value of c. Then use the values ofd and c to find the value of b from the second row, and finally, use the values ofb, c, d to find the value of a from the first row.

To handle matrices, the following type declarations are convenient.

9.2. GAUSSIAN ELIMINATION

339

type Matrix = [Row]type Row= [Integer]

It is also convenient to be able to have functions for the numbers of rows andcolumns of a matrix.

rows, cols :: Matrix -> Int
rows m = length mcols m | m == []= 0| otherwise = length (head m)

The function genMatrix produces the appropriate matrix for a list generated by apolynomial:

genMatrix :: [Integer] -> Matrix
genMatrix xs = zipWith (++) (genM d) [ [x] | x <- xs ]whered= degree xs
genM n = [ [ (toInteger x^m) | m <- [0..n] ] | x <- [0..n] ]zipWith is predefined in the Haskell prelude as follows:

zipWithzipWith z (a:as) (b:bs)
zipWith _ __:: (a->b->c) -> [a]->[b]->[c]= z a b : zipWith z as bs= []In a picture:[(z x0 y0 ), (z x1 y1 ), . . . , (z xn yn ),*[xn+1 , xn+2 , . . .
[yn+1 , yn+2 , . . .

CHAPTER 9. POLYNOMIALS

340
genMatrix gives, e.g.:

POL> genMatrix [-7,-2,15,50,109,198,323][[1,0,0,0,-7],[1,1,1,1,-2],[1,2,4,8,15],[1,3,9,27,50]]

The process of transforming the matrix to echelon form is done by so-called forward elimination: use one row to eliminate the first coefficient from the other rows
by means of the following process of adjustment (the first row is used to adjust thesecond one):

adjustWith :: Row -> Row -> Row
adjustWith (m:ms) (n:ns) = zipWith (-) (map (n*) ms) (map (m*) ns)

To transform a matrix into echelon form, proceed as follows:
1. If the number of rows or the number of columns of the matrix is 0, then thematrix is already in echelon form.
2. If every row of rs begins with a 0 then the echelon form of rs can be foundby putting 0’s in front of the echelon form of map tail rs.
3. If rs has rows that do not start with a 0, then take the first one of these, piv,and use it to eliminate the leading coefficients from the other rows. This
gives a matrix of the form
a00 a01 a02 a03 · · · b0
 0 a11 a12 a13 · · · b1 
 0 a21 a22 a23 · · · b2 
 0 a31 a32 a33 · · · b3 


 ..
.. 
..
..
..
..
 .
. 
.
.
.
.
0 an1 an2 an3 · · · bn
where the first row is the pivot row. All that remains to be done in this caseis to put the following sub matrix in echelon form:
a11 a12 a13 · · · b1
 a21 a22 a23 · · · b2  a31 a32 a33 · · · b3 


 ..
..
..
..
.. 
 .
.
.
.
. 
an1 an2 an3 · · · bn

9.2. GAUSSIAN ELIMINATION341

The code for this can be found in the Haskell demo file Matrix.hs (part of the Hugs
system):echelon:: Matrix -> Matrix
echelon rs| null rs || null (head rs) = rs
| null rs2= map (0:) (echelon (map tail rs))
| otherwise= piv : map (0:) (echelon rs’)
where rs’
= map (adjustWith piv) (rs1++rs3)(rs1,rs2)= span leadZero rs
leadZero (n:_) = n==0
(piv:rs3)= rs2

Here is an example:

POL> echelon [[1,0,0,0,-7],[1,1,1,1,-2],[1,2,4,8,15],[1,3,9,27,50]][[1,0,0,0,-7],[0,-1,-1,-1,-5],[0,0,-2,-6,-12],[0,0,0,-12,-12]]

Backward Gaussian elimination, or computing the values of the variables from amatrix in echelon form, is done by computing the value of the variable in the lastrow, eliminate that variable from the other rows to get a smaller matrix in echelon
form, and repeating that process until the values of all variables are found. If
we know that ax = c (we may assume a 6= 0), and the coordinate of x is thecoordinate ain = b of ai1 | · · · | ain−1 | ain | d, then we can eliminate thiscoordinate by replacing ai1 | · · · | ain−1 | ain | d with the following:a · ai1 | · · · | a · ain−1 | ad − bc.
It does make sense to first reduce x = c/a to its simplest form by dividing out
common factors of c and a. The implementation of rational numbers does thisfor us if we express x as a number of type Rational (see Section 8.6 for further
details). Note that an elimination step transforms a matrix in echelon form, minus
its last row, into a smaller matrix in echelon form. Here is the implementation:

342

CHAPTER 9. POLYNOMIALS

eliminate :: Rational -> Matrix -> Matrix
eliminate p rs = map (simplify c a) rs
wherec = numeratorpa = denominator p
simplify c a row = init (init row’) ++ [a*d - b*c]whered= last rowb= last (init row)row’ = map (*a) row

The implementation of backward substitution runs like this:backsubst :: Matrix -> [Rational]
backsubst rs = backsubst’ rs []wherebacksubst’ [] ps = psbacksubst’ rs ps = backsubst’ rs’ (p:ps)wherea= (last rs) !! ((cols rs) - 2)c= (last rs) !! ((cols rs) - 1)p
= c % ars’
= eliminate p (init rs)

We get:
POL> backsubst [[1,0,0,0,-7],[0,-1,-1,-1,-5],[0,0,-2,-6,-12],[0,0,0,-12,-12]][-7 % 1,1 % 1,3 % 1,1 % 1]

To use all this to analyze a polynomial sequence, generate the appropriate matrix(appropriate for the degree of the polynomial that we get from difference analysisof the sequence), put it in echelon form, and compute the values of the unknowns
by backward substitution.

solveSeq :: [Integer] -> [Rational]
solveSeq = backsubst . echelon . genMatrix

9.2. GAUSSIAN ELIMINATION

343

Recall that the sequence of sums of squares starts as follows: 0, 1, 5, 14, 30, . . ..
Solving this sequence with solveSeq gives:
POL> solveSeq [0,1,5,14,30][0 % 1,1 % 6,1 % 2,1 % 3]

This gives the form
2n3 + 3n2 + nn(n + 1)(2n + 1)
1 3 1 2 1n + n + n=
=
.
3
2
6
6
6
Here is a Haskell check (the use of the / operator creates a list of Fractionals):
POL> map (\ n -> (1/3)*n^3 + (1/2)*n^2 + (1/6)*n) [0..4][0.0,1.0,5.0,14.0,30.0]

Similarly, 0, 1, 9, 36, 100, 225 is the start of the sequence of sums of cubes. Solving
this with solveSeq gives:
POL> solveSeq [0, 1, 9, 36, 100, 225]
[0 % 1,0 % 1,1 % 4,1 % 2,1 % 4]

This gives the form
1 4 1 3 1 2
n4 + 2n3 + n2
n2 (n + 1)2
n + n + n =
=
=
4
2
4
4
4



n(n + 1)2

2.

The running example from the previous section is solved as follows:
POL> solveSeq [-12,-11,6,45,112,213,354,541,780,1077][-12 % 1,-5 % 1,5 % 1,1 % 1]

Thus, the sequence has the form n3 + 5n2 − 5n − 12.
Before we look at the confirmation, let us note that we are now in fact using representations of polynomial functions as lists of their coefficients, starting from theconstant coefficient. It is easy to implement a conversion from these representations to the polynomial functions that they represent:

p2fct :: Num a => [a] -> a -> ap2fct [] x = 0p2fct (a:as) x = a + (x * p2fct as x)

We can use this in the confirmations, as follows:

CHAPTER 9. POLYNOMIALS

344

POL> [ n^3 + 5 * n^2 - 5 * n - 12 | n <- [0..9] ][-12,-11,6,45,112,213,354,541,780,1077]
POL> map (p2fct [-12,-5,5,1]) [0..9][-12,-11,6,45,112,213,354,541,780,1077]

Finally, here is the automated solution of Exercise 9.3:
POL> solveSeq [3,7,17,39,79][3 % 1,3 % 1,0 % 1,1 % 1]

This represents the form n3 + 3n + 3.
Exercise 9.6 Suppose you want to find a closed form for the number of pieces
you can cut a pie into by making n straight cuts. After some experimentation
it becomes clear that to obtain the maximum number of pieces no cut should beparallel to a previous cut, and no cut should pass through an intersection point ofprevious cuts. Under these conditions you find that the n-th cut can be made tointersect all the n − 1 previous cuts, and can thereby made to split n of the old
regions. This gives the recurrence C0 = 1 and Cn = Cn−1 + n, which yields thesequence 1, 2, 4, 7, 11, . . . Next, you use solveSeq to obtain:
POL> solveSeq [1,2,4,7,11][1 % 1,1 % 2,1 % 2]

You conclude that n cuts can divide a pie into 21 n2 + 12 n + 1 = n(n+1)+ 1 pieces.
2
Is there still need for an inductive proof to show that this answer is correct?

9.3 Polynomials and the Binomial Theorem
In this section we will establish a connection between polynomials and lists, namely
lists of coefficients of a polynomial. Let f (x) be a polynomial of degree n, i.e., let
f (x) be a function
f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 ,with ai constants, and an 6= 0. Let c be a constant and consider the case x = y + c.
Substitution of y + c for x in f (x) gives a new polynomial of degree n in y, say
f (x) = f (y + c) = bn y n + bn−1 y n−1 + · · · + b1 y + b0 .

9.3. POLYNOMIALS AND THE BINOMIAL THEOREM

a0

=

0,

0

=

n 0

=

a · f 0 (x),(f (x) ± g(x))0

=

0

=

(f (g(x)))0=(a · f (x))(x )(f (x) · g(x))

345nxn−1 ,

f 0 (x) ± g 0 (x),f 0 (x)g(x) + f (x)g 0 (x)b,g 0 (x) · f 0 (g(x)).

Figure 9.1: Differentiation Rules.
To consider an example, take f (x) = 3x4 − x + 2, and let c = −1. Substituting
y + c for x we get:f (x) = f (y + c) = f (y − 1)
= 3(y − 1)4 − (y − 1) + 2= (3y 4 − 12y 3 + 18y 2 − 12y + 3) − (y − 1) + 2= 3y 4 − 12y 3 + 18y 2 − 13y + 6.

We will see shortly that the coefficients bi can be computed in a very simple way.
Substitution of x − c for y in in f (y + c) gives
f (x) = bn (x − c)n + bn−1 (x − c)n−1 + · · · + b1 (x − c) + b0 .
Calculation of f 0 (x), f 00 (x), . . . , f (n) (x) (the first, second, . . . , n-th derivative off ), is done with the familiar rules of Figure 9.1.1 In particular, this gives (b(x −c)k )0 = kb(x − c)k−1 , and we get:f 0 (x)f 00 (x)f (n) (x)= b1 + 2b2 (x − c) + · · · + nbn (x − c)n−1= 2b2 + 3 · 2b3 (x − c) + · · · + n(n − 1)bn (x − c)n−2..
.
= n(n − 1)(n − 2) · · · 3 · 2bn .

1 If these rules are unfamiliar to you, or if you need to brush up your knowledge of analysis, you
should consult a book like [Bry93].

CHAPTER 9. POLYNOMIALS

346
Substitution of x = c gives:f (c) = b0 , f 0 (c) = b1 , f 00 (c) = 2b2 , . . . , f (n) (c) = n!bn .
This yields the following instruction for calculating the bk :b0 = f (c), b1 = f 0 (c), b2 =f 00 (c)f (n) (c), . . . , bn =.
2n!

In general:f (k) (c).
k!
Applying this to the example f (x) = 3x4 − x + 2, with c = −1, we see thatb0 = f (−1) = 3(−1)4 + 1 + 2 = 6, b1 = f 0 (−1) = 12(−1)3 − 1 = −13,
bk =

2

00

= 36(−1)= 18, b3 = f
b2 = f (−1)
2
2

(3)

(−1)
f (4) (−1)
= −72
= 3.
6
6 = −12, b4 =24nAnother example is the expansion of (z + 1) . Using the calculation method with
derivatives, we get, for c = 0:
f (z) = (z + 1)n = bn z n + bn−1 z n−1 + · · · + b1 z + b0 ,with the following derivatives:
f 0 (z) = n(z + 1)n−1 , f 00 (z) = n(n − 1)(z + 1)n−2 , . . . , f (n) (z) = n!.
Substituting z = 0 this gives:n!
, · · · , bk = n(n−1)···(n−k+1), · · · , bn = n!
= 1.
b0 = 1, b1 = n, b2 = n(n−1)
2k!

The general coefficient bk has the formbk==

Define:n(n − 1) · · · (n − k + 1) (n − k)!
n(n − 1) · · · (n − k + 1)
=
·
k!
k!
(n − k)!
n!
.
k! (n − k)!
 
n
n!
:=
.
k
k! (n − k)!

Pronounce nk as ‘n choose k’ or ‘n over k’. We have derived:

Theorem 9.7 (Newton’s binomial theorem)
n  
X
n k
(z + 1)n =
z .
k
k=0

9.3. POLYNOMIALS AND THE BINOMIAL THEOREM

347


If A is a set of n objects,
then there are nk ways to pick a subset B from A with
|B| = k. Thus, nk is also the number of k-sized subsets of an n-sized set. To
see why this is so, note that the number of k-sequences picked from A, without
repetitions, equalsn · (n − 1) · · · · · (n − k + 1),

for there are n ways to pick the first element in the sequence, n − 1 ways to pickthe second element in the sequence, . . . , and n − (k − 1) ways to pick the k-th
element in the sequence. The number n · (n − 1) · · · · · (n − k + 1) is equal ton!
(n−k)! .

For picking k-sized subsets from A, order does not matter. There are k! waysof arranging sequences of size k without repetition. These are all equivalent. Thisn!
for the number of k-sized subsets of a set of size n. This connection
gives k! (n−k)!
explains the phrasing ‘n choose k’.
Note the following:
 
 
 
 
n(n − 1)(n − 2)nnnn(n − 1)
n
,
.
=
= 1,
= n,
=
2
6
3
0
1
2
Here is a straightforward implementation of



n
k .

choose n k = (product [(n-k+1)..n]) ‘div‘ (product [1..k])

The more general version of Newton’s binomial theorem runs:
Theorem 9.8 (Newton’s binomial theorem, general version)n  Xn k n−kn(x + y) =
x y
.
k
k=0

Note: a binomial is the sum of two terms, so (x + y) is a binomial.

Pn
Proof: To get this from the special case (z + 1)n = k=0 nk z k derived above, kPn
set z = xy to get ( xy + 1)n = k=0 nk xyk , and multiply by y n :nxn+ 1 · yn(x + y)=y!
n   kn   kn  XXXn x · ynn k n−k
n x
n
=
·y =
=
x y
.
k
yk
k yk
k
k=0

k=0

k=0

CHAPTER 9. POLYNOMIALS

348

Because of their use in the binomial theorem, the numbers
coefficients. What the binomial theorem gives us is:nkare called binomial(x + y)0(x + y)1= 1x0 y 0
= 1x1 y 0 + 1x0 y 1(x + y)2(x + y)3= 1x2 y 0 + 2x1 y 1 + 1x0 y 2= 1x3 y 0 + 3x2 y 1 + 3x1 y 2 + 1x0 y 3(x + y)4(x + y)5= 1x4 y 0 + 4x3 y 1 + 6x2 y 2 + 4x1 y 3 + 1x0 y 4= 1x5 y 0 + 5x4 y 1 + 10x3 y 2 + 10x2 y 3 + 5x1 y 4 + 1x0 y 5..
.

To see how this pattern arises, look at what happens when we raise x + y to then-th power by performing the n multiplication steps to work out the product of(x + y)(x + y) · · · (x + y):|{z}
n factors

x+y
x+y ×
x2 + xy + xy + y 2
x+y ×x3 + x2 y + x2 y + x2 y + xy 2 + xy 2 + xy 2 + y 3
x+y ×
..
.

Every term in this expansion is itself the product of x-factors and y-factors, witha total number of factors always n, so that each term has the form xk y n−k . Everybinomial (x+y) in (x+n)n either contributes an x-factor or a y-factor to xk y n−k .
The number of ways to get at the term xk y n−k equals the number of k-sized subsets from a set of size n (pick any subset of k binomials from the set of n binomial
factors). Thus, this term occurs exactly nk times.
We can arrange the binomial coefficients in the well-known triangle of Pascal.


4
0


3
0


2
0


4
1


1
0


3
1

0
0




2
1


4
2

..
.

1
1




3
2

2
2




4
3

3
3



4
4



9.3. POLYNOMIALS AND THE BINOMIAL THEOREM

349

Working this out, we get:
1
1
1
1
1

1
2

3
4

1
3

6
..
.

1
4

1

Studying the pattern of Pascal’s triangle, we see that that it is built according to thefollowing law:
  
 

n
n−1
n−1
=
+
.
k
k−1
k

This is called the addition law for binomial coefficients. To see that this law iscorrect, consider a set A of size n, and single out one of its objects, a. To countthe number of ways of picking a k-sized subset B from A, consider the two cases(i) a ∈ B and (ii) a ∈/ B. The number of ways of picking a k-sized subset B from
A with a ∈ B is equal
 to the number of ways of picking a k − 1-sized subset fromA − {a}, i.e., n−1
k−1 . The number of ways of picking a k-sized subset B from
A with a ∈/ B is equalways
pickinga k-sized subset from to the number of ofn−1
n−1
A − {a}, i.e., n−1
.
Thus,
there
are
+
ways
of picking a k-sizedkk−1ksubset from an n-sized set.
It is of course also possible to prove the addition law directly from the definitionof nk . Assume k > 0. Then: (n − 1)!
n−1n−1(n − 1)!
++=(k − 1)! (n − k)! k! (n − 1 − k)!
k−1k(n − 1)! (n − k)(n − 1)! k+=k! (n − k)!
k! (n − k)!
 n!
n(n − 1)! n
=
=
.
=
k
k! (n − k)!
k! (n − k)!
We can use the addition law for an implementation of nk . In accordance with
the interpretationof nk as the number of k-sized subset of an n-sized set, wenwill put 0 = 1 (there is just one way to pick a 0-sized subset from any set) andnk = 0 for n < k (no ways to pick a k-sized subset from a set that has lessthan k elements). A further look at Pascal’s triangle reveals the following law ofsymmetry:
  

n
n
=
.
k
n−k

CHAPTER 9. POLYNOMIALS

350


This makes sense under the interpretation of nk as the number of k-sized subsets
of a set of size n, for the number of k-sized subsets equals the number of theircomplements. There is just one way to pick an n-sized subset from an n-sized set(pick the whole set), so nn = 1. This leads to the following implementation:choose’ n 0 = 1
choose’ n k | n < k= 0| n == k= 1| otherwise =choose’ (n-1) (k-1) + (choose’ (n-1) (k))

Exercise 9.9 Which implementation is more efficient, choose or choose’? Why?

Exercise 9.10 Derive the symmetry law for binomial coefficients directly fromthe definition.
We will now give an inductive proof of Newton’s binomial theorem.
 The proofnuses the addition law for binomials, in the form k−1+ nk = n+1
k .
Theorem 9.11 (Newton’s binomial theorem again)n(x + y) =

n  
X
n

k=0

k

xk y n−k .

Proof. Induction on n.
Basis:(x + y)0 = 1 =

 
0  
0 0 0 X 0 k 0−k
x y =
x y
.
0
k
k=0

Induction step: Assumen(x + y) =

n  
X
n

k=0

k

xk y n−k .

9.3. POLYNOMIALS AND THE BINOMIAL THEOREM

351

Then:
(x + y)n+1 =
ih

=
=
=
=
=
=
add

=

=

(x + y)(x + y)nn  Xn k n−kx y(x + y)
k
k=0
n  
n  
X
X
n k n−kn k n−k
x
x y
+y
x y
k
k
k=0
k=0
n  
n  
X
n k+1 n−k X n k (n+1)−k
x
y
+
x y
k
k
k=0
k=0
n−1
n  
X n
X
n k (n+1)−kxn+1 +
xk+1 y n−k +x y+ y n+1
k
k
k=0
k=1

n 
n  
X
X
n
n k (n+1)−kn+1
k (n+1)−k
x
+
x y
+
x y
+ y n+1
k−1
k
k=1
k=1

 

n 
X
n
n k (n+1)−kxn+1 +
xk y (n+1)−k +x y+ y n+1
k−1
k
k=1

n 
X
n + 1 k (n+1)−k
n+1
x
+
x y
+ y n+1
k
k=1
n+1
n+1
X n + 1
X n + 1
xk y (n+1)−k + y n+1 =xk y (n+1)−k .
k
k
k=1

k=0

Exercise 9.12 Show from the definition that if 0 < k 6 n then:
 


n
n
n−1
= ·
.
k
k
k−1

The law from exercise 9.12 is the so-called absorption law for binomial coefficients. It allows for an alternative implementation of a function for binomial coefficients, for we have the following recursion:


 
 
 
n−1
n
n
n
n
for 0 < k 6 n.
= 1,
= 0 for n < k,
= ·
k−1
0
k
k
k

Thus we get a more efficient function for nk :

352

CHAPTER 9. POLYNOMIALS

binom n 0 = 1
binom n k | n < k= 0| otherwise = (n * binom (n-1) (k-1)) ‘div‘ k

Exercise 9.13 Use a combinatorial argument (an argument in terms of sizes thesubsets of a set) to prove Newton’s law:      

n
m
n
n−k
·
=
·
.
m
k
k
m−k

Exercise 9.14 Prove:
  
 


 

n
n+1
n+2
n+k
n+k+1
+
+
+···+
=
.
n
n
n
n
n+1

9.4 Polynomials for Combinatorial Reasoning
To implement the polynomial functions in a variable z, we will represent a polynomial
f (z) = f0 + f1 z + f2 z 2 + · · · + fn−1 z n−1 + fn z n
as a list of its coefficients:[f0 , f1 , . . . , fn ].
As we have seen, the function p2fct maps such lists to the corresponding functions.
The constant zero polynomial has the form f (z) = 0. In general we will avoid
trailing zeros in the coefficient list, i.e., we will assume that if n > 0 then f n 6= 0.
The constant function λz.c will get represented as [c], so there is a map from integers to polynomial representations, given by λc.[c]. We will also allow rationals
as coefficients, to there is also a map from rationals to polynomial representations,
given by λr.[r].
We need some conventions for switching back and forth between a polynomial andits list of coefficients. If f (z) is a polynomial, then we use f for its coefficient list.
If this list of coefficients is non-empty then, as before, we will indicate the tail of f

9.4. POLYNOMIALS FOR COMBINATORIAL REASONING

353

as f . Thus, if f = [f0 , f1 , . . . , fn ], then f = [f1 , . . . , fn ], and we have the identity
f = f0 : f . Moreover, if f (z) = f0 + f1 z + f2 z 2 + · · · + fn−1 z n−1 + fn z n , thenwe use f (z) for f1 + f2 z + · · · + fn−1 z n−2 + fn z n−1 . This convention yields thefollowing important equality:
f (z) = f0 + zf(z).
The identity function λz.z will get represented as [0, 1], for this function is of theform λz.f0 + f1 z, with f0 = 0 and f1 = 1. This gives:

z :: Num a => [a]z = [0,1]

To negate a polynomial, simply negate each term in its term expansion. For if
f (z) = f0 + f1 z + f2 z 2 + · · · , then −f (z) = −f0 − f1 z − f2 z 2 − · · · . This gives:

negate []negate (f:fs)= []= (negate f) : (negate fs)

To add two polynomials f (z) and g(z), just add their coefficients, for clearly, if
f (z) = f0 + f1 z + f2 z 2 + · · · + fk z k and g(z) = b0 + b1 z + g2 z 2 + · · · = gm z m ,thenf (z) + g(z) = (f0 + g0 ) + (f1 + g1 )z + (f2 + g2 )z 2 + · · ·This translates into Haskell as follows:

fs
+ []
= fs
[]
+ gs
= gs
(f:fs) + (g:gs) = f+g : fs+gs

Note that this uses overloading of the + sign: in f+g we have addition of numbers,
in fs+gs addition of polynomial coefficient sequences.

CHAPTER 9. POLYNOMIALS354

The product of f (z) = f0 + f1 z + f2 z 2 + · · · + fk z k and g(z) = g0 + g1 z +g2 z 2 + · · · + gm z m looks like this:f (z) · g(z) = (f0 + f1 z + f2 z 2 + · · · + fk z k ) · (g0 + g1 z + g2 z 2 + · · · + gm z m )= f0 g0 + (f0 g1 + f1 g0 )z + (f0 g2 + f1 g1 + f2 g0 )z 2 + · · ·= f0 g0 + z(f0 g(z) + g0 f(z) + zf(z)g(z))= f0 g0 + z(f0 g(z) + f (z)g(z))
Here f(z) = f1 + f2 z + · · · + fk z k−1 and g(z) = g1 + g2 z + · · · + gm z m−1 ,i.e., f (z) and g(z) are the polynomials that get represented as [f 1 , . . . , fk ] and[g1 , . . . , gm ], respectively.
If f (z) and g(z) are polynomials of degree k, then for all n 6 k, in the list ofcoefficients for f (z)g(z), the n-th coefficient has the form f0 gn + f1 gn−1 + · · · +fn−1 g1 + fn g0 . The list of coefficients of the product is called the convolution of
the lists of coefficients f and g.
Multiplying a polynomial by z boils down to shifting its sequence of coefficients
one place to the right. This leads to the following Haskell implementation, where(.*) is an auxiliary multiplication operator for multiplying a polynomial by a numerical constant. Note that (*) is overloaded: f*g multiplies two numbers, but
fs * (g:gs) multiplies two lists of coefficients. We cannot extend this overloading to multiplication of numbers with coefficient sequences, since Haskell insists
on operands of the same type for (*). Hence the use of (.*).

infixl 7 .*(.*) :: Num a => a -> [a] -> [a]c .* []= []c .* (f:fs) = c*f : c .* fs
fs
* []
= []
[]
* gs
= []
(f:fs) * (g:gs) = f*g : (f .* gs + fs * (g:gs))

Example 9.15 In Figure 9.2 the polynomials are declared as a data type in class
Num. This entails that all Haskell operations for types in this class are available.
We get:
POL> (z + 1)^0

9.4. POLYNOMIALS FOR COMBINATORIAL REASONING

355

[1]
POL> (z + 1)
[1,1]
POL> (z + 1)^2
[1,2,1]
POL> (z + 1)^3
[1,3,3,1]
POL> (z + 1)^4
[1,4,6,4,1]
POL> (z + 1)^5
[1,5,10,10,5,1]
POL> (z + 1)^6
[1,6,15,20,15,6,1]

This gives yet another way to get at the binomial coefficients.
Now suppose we have a polynomial f (z). We are interested in the difference listof its coefficients [f0 , f1 − f0 , f2 − f1 , . . .]. It is easy to see that this difference list
is the list of coefficients of the polynomial (1 − z)f (z):f (z) ;
−zf (z) ;(1 − z)f (z) ;[ f0 ,
f1 ,f2 ,f3 , · · ·
[ 0,
−f0 ,
−f1 ,
−f2 , · · ·[ f 0 , f1 − f 0 , f2 − f 1 , f3 − f 2 , · · ·]]]This is implemented by the following function:delta :: Num a => [a] -> [a]delta = ([1,-1] *)This gives, for example:
POL> delta [2,4,6][2,2,2,-6]Note that the coefficient of z 4 in [2, 4, 6] is 0, so this is correct. Note also that weare now looking at difference lists of coefficients, not at different lists of the resultof mapping the polynomial function to [0..], as in Section 9.1.

CHAPTER 9. POLYNOMIALS

356

module Polynomials
whereinfixl 7 .*(.*) :: Num a => a -> [a] -> [a]c .* []= []c .* (f:fs) = c*f : c .* fs
z :: Num a => [a]z = [0,1]
instance Num a => Num [a] where
fromInteger c= [fromInteger c]negate []= []negate (f:fs)= (negate f) : (negate fs)
fs
+ []
= fs
[]
+ gs
= gs
(f:fs) + (g:gs) = f+g : fs+gs
fs
* []
= []
[]
* gs
= []
(f:fs) * (g:gs) = f*g : (f .* gs + fs * (g:gs))
delta :: Num a => [a] -> [a]delta = ([1,-1] *)shift :: [a] -> [a]shift = tailp2fct :: Num a => [a] -> a -> ap2fct [] x = 0p2fct (a:as) x = a + (x * p2fct as x)comp :: Num a => [a] -> [a] -> [a]comp _[]= error ".."
comp []_= []comp (f:fs) (0:gs) = f : gs * (comp fs (0:gs))
comp (f:fs) (g:gs) = ([f] + [g] * (comp fs (g:gs)))+ (0 : gs * (comp fs (g:gs)))
deriv :: Num a => [a] -> [a]deriv []
= []deriv (f:fs) = deriv1 fs 1 where
deriv1 []_ = []
deriv1 (g:gs) n = n*g : deriv1 gs (n+1)

Figure 9.2: A Module for Polynomials.

9.4. POLYNOMIALS FOR COMBINATORIAL REASONING357

The composition of two polynomials f (z) and g(z) is again a polynomial f (g(z)).
It is given by:
f (z) = f0
f (g(z)) = f0+
f1 z +
f2 z 2+ f1 g(z) + f2 (g(z))2+f3 z 3+ f3 (g(z))3

+ ···
+ ···

We see from this:f (g(z)) = f0 + g(z) · f (g(z))
This leads immediately to the following implementation (the module of Figure 9.2
has a slightly more involved implementation comp that gets explained in the next
chapter, on page 389):

comp1 :: Num a => [a] -> [a] -> [a]comp1 _ [] = error ".."
comp1 [] _ = []comp1 (f:fs) gs = [f] + (gs * comp1 fs gs)

Example 9.16 We can use this to pick an arbitrary layer in Pascal’s triangle:
POL> comp1 (z^2) (z+1)[1,2,1]
POL> comp1 (z^3) (z+1)[1,3,3,1]
POL> comp1 (z^12) (z+1)[1,12,66,220,495,792,924,792,495,220,66,12,1]

We can also use it to generate Pascal’s triangle up to arbitrary depth:
POL> comp1 [1,1,1,1,1,1] [[0],[1,1]][[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1]]

Note that this uses the composition of f (z) = 1 + z + z 2 + z 3 + z 4 + z 5 + z 6 withg(z) = (y + 1)z + 0. The result of this is f (g(z)) = 1 + (y + 1)z + (y + 1) 2 z 2 +· · · + (y + 1)6 z 6 .
If f (z) = f0 + f1 z + f2 z 2 + · · · + fk z k , the derivative of f (z) is given by (Figure
9.1):
f 0 (z) = f1 + 2f2 z + · · · + kfk z k−1 .

CHAPTER 9. POLYNOMIALS

358

This has a straightforward implementation, as follows:deriv :: Num a => [a] -> [a]deriv []
= []deriv (f:fs) = deriv1 fs 1 where
deriv1 []_ = []
deriv1 (g:gs) n = n*g : deriv1 gs (n+1)

The close link between binomial coefficients and combinatorial notions makes
polynomial reasoning a very useful tool for finding solutions to combinatorial
problems.
Example 9.17 How many ways are there of selecting ten red, blue or white marbles from a vase, in such a way that there are at least two of each color and at most
five marbles have the same colour? The answer is given by the coefficient of z 10
in the following polynomial:(z 2 + z 3 + z 4 + z 5 )3 .
This is easily encoded into a query in our implementation:
POL> ([0,0,1,1,1,1]^3) !! 10
12

How many ways are there of selecting ten red, blue or white marbles from a vase,in such manner that there is even number of marbles of each colour:
POL> ([1,0,1,0,1,0,1,0,1,0,1]^3) !! 1021

We associate coefficient lists with combinatorial problems by saying that
[f0 , f1 , f2 , . . . , fn ]solves a combinatorial problem if fr gives the number of solutions for that problem.
Example 9.18 The polynomial (1+z)10 solves the problem of picking r elementsfrom a set of 10. The finite list [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1] solves
the problem. It is implemented by:9.5. FURTHER READING

359

POL> (1 + z)^10[1,10,45,120,210,252,210,120,45,10,1]

Example 9.19 The list [1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1] is a solution forthe problem of picking r marbles from a vase containing red, white or blue marbles, with a maximum of five of each colour. A polynomial for this problem is(1 + z + z 2 + z 3 + z 4 + z 5 )3 . In the implementation:
POL> (1 + z + z^2 + z^3 + z^4 + z^5)^3[1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1]

Exercise 9.20 Use polynomials to find out how many ways there are of selecting
ten red, blue or white marbles from a vase, in such manner that the number ofmarbles from each colour is prime.

9.5 Further Reading
Charles Babbage’s difference engine is described in [Lar34] (reprinted in [Bab61]),and by Babbage himself in his memoirs [Bab94]. The memoirs are very amusing:Among the various questions which have been asked respecting theDifference Engine, I will mention a few of the most remarkable: one
gentleman addressed me thus: ‘Pray, Mr Babbage, can you explain
to me in two words what is the principle of your machine?’ Had thequerist possessed a moderate acquaintance with mathematics I mightin four words have conveyed to him the required information by answering, ‘The Method of Differences.’[. . . ] On two occasions I have been asked - ‘Pray, Mr Babbage, if you
put into the machine wrong figures, will the right answers come out?’
In one case a member of the Upper, and in the other a member of theLower, House put this question.
There are many good textbooks on calculus, but [Bry93] is particularly enlightening. An excellent book on discrete mathematics and combinatorial reasoning is[Bal91].

360

CHAPTER 9. POLYNOMIALSChapter 10

Corecursion
Preview
In this chapter we will look the construction of infinite objects and at proof methods suited to reasoning with infinite data structures. The most important kind of
infinite data structures are streams (infinite lists), so the main topic of this chapter
is the logic of stream processing. We will show how non-deterministic processes
can be viewed as functions from random integer streams to streams. For the implementation of this we will use two functions from Random.hs, a module for
random number generation and processing from the Haskell library. At the endof the chapter we will connect combinatorial reasoning with stream processing,
via the study of power series and generating functions. Our Haskell treatment of
power series is modeled after the beautiful [McI99, McI00].

module COR
whereimport Random (mkStdGen,randomRs)
import Polynomials
import PowerSeries

The default for the display of fractional numbers in Haskell is floating point nota361

362

CHAPTER 10. CORECURSION

tion. As we are going to develop streams of integers and streams of fractions inthis chapter, it is convenient to have them displayed with unlimited precision in
integer or rational notation. The default command takes care of that.

default (Integer, Rational, Double)

10.1 Corecursive Definitions
As we have seen, it is easy to generate infinite lists in Haskell. Infinite lists areoften called streams. Here is the code again for generating an infinite list (or astream) of ones:

ones = 1 : ones

This looks like a recursive definition, but there is no base case. Here is a definitionof a function that generates an infinite list of all natural numbers:

nats = 0 : map (+1) nats

Again, the definition of nats looks like a recursive definition, but there is no basecase. Definitions like this are called corecursive definitions. Corecursive definitions always yield infinite objects. When you come to think of it, the funny
explanation of the acronym GNU as GNU’s Not Unix is also an example.
As we have seen in Section 3.7, generating the odd natural numbers can be doneby corecursion.

odds = 1 : map (+2) odds

10.1. CORECURSIVE DEFINITIONS

363

Exercise 10.1 Write a corecursive definition that generates the even natural numbers.
We can make the corecursive definitions more explicit with the use of iterate.
The definition of iterate in the Haskell prelude is itself an example of corecursion:

iterateiterate f x:: (a -> a) -> a -> [a]= x : iterate f (f x)

Here are versions of the infinite lists above in terms of iterate:

theOnes = iterate id 1
theNats = iterate (+1) 0
theOdds = iterate (+2) 1

Exercise 10.2 Use iterate to define the infinite stream of even natural numbers.

The list [0..] can be defined corecursively from ones with zipWith.
Suppose n is a natural number. Then its successor can be got by adding 1 to n. 0
is the first natural number. The second natural number, 1, is got by adding 1 to 0.
The third natural number, 2, is got by adding 1 to the second natural number, andso on:

theNats1 = 0 : zipWith (+) ones theNats1

The technique that produced theNats1 can be used for generating the Fibonacci
numbers:

theFibs = 0 : 1 : zipWith (+) theFibs (tail theFibs)

364

CHAPTER 10. CORECURSION

The process on Fibonacci numbers that was defined in Exercise 7.17 can be definedwith corecursion, as follows:

pr (x1:x2:x3:xs) = x1*x3 - x2*x2 : pr (x2:x3:xs)

As we proved in Exercise 7.17, applying this process to theFibs gives the list
λn.(−1)n+1 :COR> take 20 (pr theFibs)[-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1]

The definition of the sieve of Eratosthenes (page 106) also uses corecursion:

sieve :: [Integer] -> [Integer]
sieve (0 : xs) = sieve xs
sieve (n : xs) = n : sieve (mark xs 1 n)
wheremark (y:ys) k m | k == m
= 0 : (mark ys 1m)| otherwise = y : (mark ys (k+1) m)

What these definitions have in common is that they generate infinite objects, andthat they look like recursive definitions, except for the fact that there is no base
case.
Here is a faster way to implement the Sieve of Eratosthenes. This time, we actually
remove multiples of x from the list on encountering x in the sieve. The counting
procedure now has to be replaced by a calculation, for the removals affect the
distances in the list. The property of not being a multiple of n is implemented
by the function (\ m -> (rem m n) /= 0). Removing all numbers that do nothave this property is done by filtering the list with the property.

sieve’ :: [Integer] -> [Integer]
sieve’ (n:xs) = n : sieve’ (filter (\ m -> (rem m n) /= 0) xs)
primes’ :: [Integer]
primes’ = sieve’ [2..]10.2. PROCESSES AND LABELED TRANSITION SYSTEMS365

How does one prove things about corecursive programs? E.g., how does one provethat sieve and sieve’ compute the same stream result for every stream argument? Proof by induction does not work here, for there is no base case.
Exercise 10.3* The Thue-Morse sequence is a stream of 0’s and 1’s that is produced as follows. First produce 0. Next, at any stage, swap everything that was
produced so far (by interchanging 0’s and 1’s) and append that. The first few stagesof producing this sequence look like this:
0
01
0110
01101001
0110100110010110
Thus, if Ak denotes the first 2k symbols of the sequence, then Ak+1 equals Ak ++Bk , where Bk is obtained from Ak by interchanging 0’s and 1’s. Give a corecursive program for producing the Thue-Morse sequence as a stream.

10.2 Processes and Labeled Transition Systems
The notion of a nondeterministic sequential process is so general that it is impossible to give a rigorous definition. Informally we can say that processes are interacting procedures. Typical examples are (models of) mechanical devices such as
clocks, protocols for traffic control, vending machines, operating systems, clientserver computer systems, and so on. A formal notion for modeling processes thathas turned out to be extremely fruitful is the following.
A labeled transition system (Q, A, T ) consists of a set of states Q, a set ofaction labels A, and a ternary relation T ⊆ Q × A × Q, the transition relation.
aIf (q, a, q 0 ) ∈ T we write this as q −→ q 0 .
Example 10.4 Perhaps the simplest example of a labeled transition system is the
tick
crack
system given by the two states c and c0 and the two transitions c −→ c and c −→c0 (Figure 10.1). This is a model of a clock that ticks until it gets unhinged.
Note that the process of the ticking clock is nondeterministic. The clock keeps
ticking, until at some point, for no reason, it gets stuck.
To implement nondeterministic processes like the clock process from Example
10.4, we have to find a way of dealing with the nondeterminism. Nondeterministic

CHAPTER 10. CORECURSION

366

tick

crack
Figure 10.1: Ticking clock.
behaviour is behaviour determined by random factors, so a simple way of modelingnondeterminism is by modeling a process as a map from a randomly generated listof integers to a stream of actions. The following function creates random streams
of integers, within a specified bound [0, .., b], and starting from a particular seed s.
It uses randomRs and mkStdGen from the library module Random.hs.

randomInts :: Int -> Int -> [Int]randomInts bound seed =tail (randomRs (0,bound) (mkStdGen seed))

Exercise 10.5 Note that randomInts 1 seed generates a random stream of 0’s
and 1’s. In the long run, the proportion of 0’s and 1’s in such a stream will be 1 to
1. How would you implement a generator for streams of 0’s and 1’s with, in thelong run, a proportion of 0’s and 1’s of 2 to 1?
We define a process as a map from streams of integers to streams of action labels.
To start a process, create an appropriate random integer stream and feed it to theprocess.

type Process = [Int] -> [String]
start :: Process -> Int -> Int -> [String]
start process bound seed = process (randomInts bound seed)10.2. PROCESSES AND LABELED TRANSITION SYSTEMS

367

The clock process can now be modeled by means of the following corecursion:clock :: Processclock (0:xs) = "tick" : clock xs
clock (1:xs) = "crack" : []This gives:
COR> start clock 1 1["tick","crack"]
COR> start clock 1 2["crack"]
COR> start clock 1 25["tick","tick","tick","tick","crack"]

The parameter for the integer bound in the start function (the second argumentof start function) should be set to 1, to ensure that we start out from a list of 0’s
and 1’s.

Example 10.6 Consider a very simple vending machine that sells mineral waterand beer. Water costs one euro, beer two euros. The machine has a coin slot anda button. It only accepts 1 euro coins. If a coin is inserted and the dispense button
is pushed, it dispenses a can of mineral water. If instead of pushing the dispense
button, another one euro coin is inserted and next the dispense button is pushed,
it dispenses a can of beer. If, instead of pushing the button for beer, a third coin
is inserted, the machine returns the inserted money (three 1 euro coins) and goesback to its initial state. This time we need four states, and the following transitions
coin

water

coin

beer

coin

(Figure 10.2): q −→ q1 , q1 −→ q, q1 −→ q2 , q2 −→ q, q2 −→ q3 , q3

moneyback−→q.

Again, this is easily modeled. This time, the random stream is not needed for thecoinmoneybacktransitions q −→ q1 and q3 −→ q, for insertion of a coin is the only possibility
for the first of these, and return of all the inserted money for the second.

CHAPTER 10. CORECURSION

368

moneyback
beer

water

coin

coin

Figure 10.2: A simple vending machine.

coin10.2. PROCESSES AND LABELED TRANSITION SYSTEMS

369

vending, vending1, vending2, vending3 :: Process
vending (0:xs) = "coin": vending1 xs
vending (1:xs) =vending xs
vending1 (0:xs) = "coin": vending2 xs
vending1 (1:xs) = "water": vending xs
vending2 (0:xs) = "coin": vending3 xs
vending2 (1:xs) = "beer": vending xs
vending3 (0:xs) = "moneyback": vending xs
vending3 (1:xs) =vending3 xs

This gives:

COR> take 9 (start vending 1 1)["coin","water","coin","water","coin","water","coin","coin","beer"]
COR> take 8 (start vending 1 3)["coin","water","coin","coin","coin","moneyback","coin","water"]
COR> take 8 (start vending 1 22)["coin","water","coin","water","coin","coin","coin","moneyback"]

Example 10.7 A parking ticket dispenser works as follows. As long as pieces of
1 or 2 euro are inserted, the parking time is incremented by 20 minutes per euro.
If the red button is pressed, all the inserted money is returned, and the machine
returns to its initial state. If the green button is pressed, a parking ticket is printed
indicating the amount of parking time, and the machine returns to its initial state.
return(i)
1euro
2euro
There are the following transitions: q(i) −→ q(0), q(i) −→ q(i + 1), q(i) −→no timetime i*20 min
q(i + 2), q(0) −→ q(0), q(i)−→q(0). Note that the number of states isinfinite.

Here is an implementation of the parking ticket dispenser:

CHAPTER 10. CORECURSION

370

ptd :: Process
ptd = ptd0 0
ptd0 :: Int -> Process
ptd0 0 (0:xs) = ptd0 0 xs
ptd0 i (0:xs) = ("return " ++ show i ++ " euro") : ptd0 0 xs
ptd0 i (1:xs) = "1 euro" : ptd0 (i+1) xs
ptd0 i (2:xs) = "2 euro" : ptd0 (i+2) xs
ptd0 0 (3:xs) = ptd0 0 xs
ptd0 i (3:xs) = ("ticket " ++ show (i * 20) ++ " min") : ptd0 0 xs

This yields:
COR> take 6 (start ptd 3 457)["1 euro","2 euro","2 euro","ticket 100 min","1 euro","ticket 20 min"]

tick

tick

crack

crack

Figure 10.3: Another ticking clock.

Example 10.8 Intuitively, the clock process of Example 10.4 is the same as the
tick
crack
clock process of the following example (Figure 10.3): c1 −→ c2 , c1 −→ c3 ,
tick
crack
c2 −→ c2 and c2 −→ c4 . It is clear that this is also a clock that ticks until it getsstuck.

10.2. PROCESSES AND LABELED TRANSITION SYSTEMS

moneyback
beer

coin

coin

coin

water

Figure 10.4: Another simple vending machine.

coin

371

CHAPTER 10. CORECURSION

372

Exercise 10.9 Consider the vending machine given by the following transitions
coin

coin

coin

beer

coin

moneyback(Figure 10.4): q −→ q1 , q −→ q4 , q1 −→ q2 , q2 −→ q, q2 −→ q3 , q3 −→ q,
water
q4 −→ q. Taking the machine from Example 10.6 and this machine to be blackboxes, how can a user find out which of the two machines she is operating?
Exercise 10.10 Give a Haskell implementation of the vending machine from Exercise 10.9.
The key question about processes is the question of identity: How does one provethat two processes are the same? How does one prove that they are different? Proof
methods for this will be developed in the course of this chapter.
Before we end this brief introduction to processes we give a simple example ofprocess interaction.
Example 10.11 A user who continues to buy beer from the vending machine in
coin
coin
beer
Example 10.6 can be modeled by: u −→ u1 , u1 −→ u2 , u2 −→ u. It is clear how
this should be implemented.
How about a beer drinker who interacts with a vending machine? It turns out that
we can model this very elegantly as follows. We let the user start with buying his(her?) first beer. Next, we feed the stream of actions produced by the vending
machine as input to the user, and the stream of actions produced by the user as
input to the vending machine, and they keep each other busy. This interaction canbe modeled corecursively, as follows:

actions= user [0,0,1] responses
responses = vending actions
user acts ~(r:s:p:resps) = acts ++ user (proc [r,s,p]) resps
proc ["coin","coin","beer"] = [0,0,1]

This gives:
COR> take 8 actions[0,0,1,0,0,1,0,0]
COR> take 8 responses["coin","coin","beer","coin","coin","beer","coin","coin"]10.3. PROOF BY APPROXIMATION373

The user starts by inserting two coins and pressing the button, the machine responds with collecting the coins and issuing a can of beer, the user responds tothis by inserting two more coins and pressing the button once more, and so on.
One hairy detail: the pattern ~(r:s:p:resps) is a so-called lazy pattern. Lazypatterns always match, they are irrefutable. This allows the initial request to besubmitted ‘before’ the list (r:s:p:resps) comes into existence by the response
from the vending machine.

10.3 Proof by Approximation
One of the proof methods that work for corecursive programs is proof by approximation. For this, we have to extend each data type to a so-called domain with apartial ordering v (the approximation order). Every data type gets extended with
an element ⊥. This is the lowest element in the approximation order.
Let (D, v) be a set D with a partial order v on it, and let A ⊆ D.
An element x ∈ A is the greatest element of A if a v x for all a ∈ A. x ∈ A isthe least element of A if x v a for all a ∈ A. Note that there are D with A ⊆ D
for which such least and greatest elements do not exist.
Exercise 10.12 Give an example of a set D, a partial order v on D, and a subset
A of D such that A has no greatest and no least element.
An element x ∈ D is an upper bound of A if a v x for all a ∈ A. Use Au for theset of all upper bounds of A, i.e., Au := {x ∈ D | ∀a ∈ A a v x}.

E.g., consider N with the usual order 6. Take {2, 4, 6} ⊆ N. Then {2, 4, 6} u ={n ∈ N | 6 6 n}. But the set of even numbers has no upper bounds in N.

An element x ∈ D is a lower bound of A if x v a for all a ∈ A. Use Al for theset of all upper bounds of A, i.e., Al := {x ∈ D | ∀a ∈ A x v a}.
An element x ∈ D is the lub or least upper bound of A if x is the least elementof Au . The lub of A is also called the supremum of A. Notation tA.

nE.g., consider R with the usual order 6. Take A = { n+1| n ∈ N} ⊆ R. Then
1 2 3u
A = {0, 2 , 3 , 4 , . . .}, A = {r ∈ R | r > 1}, and tA = 1.

An element x ∈ D is the glb or greatest lower bound of A if x is the greatest
element of Al . The glb of A is also called the infimum of A. Notation uA.
Caution: there may be A ⊆ D for which tA and uA do not exist.

CHAPTER 10. CORECURSION

374

Exercise 10.13 Give an example of a set D, a partial order v on D, and a subset
A of D such that A has no lub and no glb.
A subset A of D is called a chain in D if the ordering on A is total, i.e., if for all
x, y ∈ A either x v y or y v x. E.g., the set A = {{k ∈ N | k < n} | n ∈ N} isa chain in ℘(N) under ⊆.
A set D with a partial order v is called a domain if D has a least element ⊥ andtA exists for every chain A inS
D. E.g., ℘(N) with ⊆ is a domain: the least element
is given by ∅, tA is given by A.
Exercise 10.14 Show that N, with the usual ordering 6, is not a domain. Can you
extend N to a domain?

We will now define approximation orders that make each data type into a domain.
For basic data types A the approximation order is given by:
xvy:≡ x = ⊥ ∨ x = y.

To see that the result is indeed a domain, observe that ⊥ is the least element of thedata type, and that the only chains in basic data types are {⊥}, {x} and {⊥, x}.
Obviously, we have t{⊥} = ⊥, t{x} = x, t{⊥, x} = x. Thus, every chain has
a lub.
For pair data types A×B — represented in Haskell as (a,b) — the approximation
order is given by:⊥ v (x, y)(x, y) v (u, v) :≡ x v u ∧ y v v.
Again, it is not difficult to see that every chain in a pair data type has a lub.
For functional data types A → B the approximation order is given by:
f vg:≡ ∀x ∈ A (f x v gx).

Here it is assumed that A and B are domains.
If A is a basic data type, f, g can be viewed as partial functions, with f v g
indicating that g is defined wherever f is defined, and f and g agreeing on every x
for which they are both defined.
Exercise 10.15 Show that functional data types A → B under the given approximation order form a domain. Assume that A and B are domains.

10.3. PROOF BY APPROXIMATION375

For list data types [A] the approximation order is given by:⊥ v xs[] v xs :≡ xs = []x : xs v y : ys :≡ x v y ∧ xs v ys
Exercise 10.16 Show that list data types [a] under the given approximation orderform a domain. Assume that a is a domain.
Using ⊥ one can create partial lists, where a partial list is a list of the form x0 :x1 : · · · : ⊥. The Haskell guise of ⊥ is a program execution error or a programthat diverges. The value ⊥ shows up in the Haskell prelude in the following weird
definition of the undefined object.

undefined:: aundefined | False = undefined

A call to undefined always gives rise to an error due to case exhaustion. An
example of a partial list would be ’1’:’2’:undefined. Partial lists can be used
to approximate finite and infinite lists. It is easy to check that⊥ v x0 : ⊥ v x0 : x1 : ⊥ v · · · v x0 : x1 : · · · : xn : ⊥ v x0 : x1 : · · · : xn : [].
This finite set of approximations is a chain, and we have:G{⊥, x0 : ⊥, x0 : x1 : ⊥, . . . , x0 : x1 : · · · : xn : ⊥, x0 : x1 : · · · : xn : []}= x0 : x1 : · · · : xn : [].

Also, for infinite lists, we can form infinite sets of approximations:⊥ v x 0 : ⊥ v x 0 : x1 : ⊥ v x 0 : x1 : x2 : ⊥ v · · ·This infinite set of approximations is a chain. We will show (in Lemma 10.17) thatfor infinite lists xs = x0 : x1 : x2 : · · · we have:G{⊥, x0 : ⊥, x0 : x1 : ⊥, x0 : x1 : x2 : ⊥, . . .} = xs.
The function approx can be used to give partial approximations to any list:

CHAPTER 10. CORECURSION

376

approx :: Integer -> [a] -> [a]approx (n+1) []= []approx (n+1) (x:xs) = x : approx n xs

Since n + 1 matches only positive integers, the call approx n xs, with n lessthan or equal to the length of the list, will cause an error (by case exhaustion) aftergeneration of n elements of the list, i.e., it will generate x0 : x1 : · · · : xn−1 : ⊥.
If n is greater than the length of xs, the call approx n xs will generate the whole
list xs. We can now writeG{⊥, x0 : ⊥, x0 : x1 : ⊥, x0 : x1 : x2 : ⊥, . . .}as∞G

approx n xs .

n=0

Lemma 10.17 (Approximation Lemma) For any list xs:∞
Gapprox n xs = xs.

n=0

Proof. We have to show two things:
1. xs ∈ {approx n xs | n ∈ N}u .

2. xs is the least element of {approx n xs | n ∈ N}u .
To show (1), we have to establish that for every n and every xs, approx n xs v xs.
We prove this with induction on n, for arbitrary lists.
Basis: approx 0 xs = ⊥ v xs for any list xs.
Induction step: Assume that for every xs, approx n xs v xs. We have to showthat for any list xs, approx (n + 1) xs v xs.
If xs = ⊥ or xs = [], then the result trivially holds by the definition of approx, so
we assume xs = x:xs’. Then:approx (n + 1) x : xs’= { def approx }x : approx n xs’v { induction hypothesis }x : xs’.

10.3. PROOF BY APPROXIMATION

377

To show (2), we have to show that for any list ys, if ys ∈ {approx n xs | n ∈ N} u ,then xs v ys, i.e., we have to show that xs is the least element of {approx n xs |n ∈ N}u .

Assume ys ∈ {approx n xs | n ∈ N}u . This means that for all n ∈ N, approx n xs v
ys. We have to show that xs v ys.
Suppose xs 6v ys. Then there has to be a k with (approx (k + 1) xs) !! k 6v ys !! k.
But then approx (k + 1) xs 6v ys , and contradiction with ys ∈ {approx n xs | n ∈N}u .
Theorem 10.18 (Approximation Theorem)
xs = ys ⇔ ∀n (approx n xs = approx n ys).

Proof. ⇒: Immediate from the fact that xs !! n = ys !! n, for every n ∈ N.
⇐:=⇒∀n(approx n xs = approx n ys){ property of lub }∞∞GG(approx n xs) =(approx n ys)
n=0

⇐⇒

n=0

{ Lemma 10.17 }xs = ys.

The approximation theorem is one of our tools for proving properties of streams.
Example 10.19 Suppose we want to prove the following:
map f (iterate f x) = iterate f (f x).
This equality cannot be proved by list induction, as the lists on both sides of theequality sign are infinite. We can prove an equivalent property by induction on n,
as follows.
∀n(approx n (map f (iterate f x)) = approx n (iterate f (f x))).
Proof by induction on n.
Basis. For n = 0, the property holds by the fact that approx 0 xs = ⊥ for all lists
xs.

CHAPTER 10. CORECURSION

378
Induction step. Assume (for arbitrary x):

approx n (map f (iterate f x)) = approx n (iterate f (f x)).
We have:approx (n + 1) (map f (iterate f x))= { definition of iterate }approx (n + 1) (map f (x : iterate f (f x)))= { definition of map }approx (n + 1) (f x : map f (iterate f (f x)))= { definition of approx }f x : approx n (map f (iterate f (f x)))= { induction hypothesis }f x : approx n (iterate f (f (f x)))= { definition of approx }approx (n + 1) (f x : iterate f (f (f x)))= { definition of iterate }approx (n + 1) (iterate f (f x)).

Exercise 10.20 In Exercise 7.52 you showed that for every finite list xs :: [a],every function f :: a -> b, every total predicate p :: b -> Bool the following holds:
filter p (map f xs) = map f (filter (p · f ) xs).
Use proof by approximation to show that this also holds for infinite lists.
Example 10.21 To reason about sieve, we would like to show that mark n k
has the same effect on a list of integers as mapping with the following function:
λm.if rem m n 6= 0 then m else 0.
This will only hold when mark n k is applied to sequences that derive from asequence [q..] by replacement of certain numbers by zeros. Let us use [q..] •for the general form of such a sequence. Suppose xs equals [q..] • for some q,i.e., suppose that xs is the result of replacing some of the items in the infinite list
[q, q + 1, q + 2, . . .] by zeros. We prove by approximation that if q = an + k, with
1 6 k 6 n, then:mark n k xs = map (λm.if rem m n 6= 0 then m else 0) xs.

10.4. PROOF BY COINDUCTION

379

Basis. For n0 = 0, the property holds by the fact that approx 0 xs = ⊥ for all lists
xs.
Induction step. Assume (for arbitrary xs of the form [q..]• with q = an + k and1 6 k 6 n):approx n0 (mark n k xs) = approx n0 (map (λm.if rem m n 6= 0 then m else 0) xs).
Two cases: (i) k = n, and (ii) k < n.
Case (i) is the case where n|q, i.e., rem x n = 0. We get:approx (n0 + 1) (mark n k x:xs)= { definition of mark }approx (n0 + 1) (0 : mark n 1 xs)= { definition of approx }0 : approx n0 mark n 1 xs
= { induction hypothesis }0 : approx n0 (λm.if rem m n 6= 0 then m else 0) xs= { definition of approx }approx (n0 + 1) (0 : (λm.if rem m n 6= 0 then m else 0) xs)= { definition of map, plus the fact that rem x n = 0 }approx (n0 + 1) map (λm.if rem m n 6= 0 then m else 0) x:xs.

Case (ii) is the case where n 6 | q. The reasoning for this case uses the other caseof the definition of mark.
A proof by approximation of the fact that sieve and sieve’ define the samefunction on streams can now use the result from Example 10.21 as a lemma.

10.4 Proof by Coinduction
To compare two streams xs and ys, intuitively it is enough to compare their observational behaviour. The key observation on a stream is to inspect its head. If two
streams have the same head, and their tails have the same observational behaviour,then they are equal. Similarly, to compare two infinite binary trees it is enoughto observe that they have the same information at their root nodes, that their left
daughters have the same observational behaviour, and that their right daughters
have the same observational behaviour. And so on, for other infinite data structures.

CHAPTER 10. CORECURSION380

To make this talk about observational behaviour rigorous, it is fruitful to consider
the infinite data structures we are interested in as labeled transition systems. The
observations are the action labels that mark a transition.
A bisimulation on a labeled transition system (Q, A, T ) is a binary relation R on
Q with the following properties. If qRp and a ∈ A then:aa

1. If q −→ q 0 then there is a p0 ∈ Q with p −→ p0 and q 0 Rp0 .
aa

2. If p −→ p0 then there is an q 0 ∈ Q with p −→ p0 and q 0 Rp0 .
Example 10.22 Take decimal representations of rational numbers with over-line
notation to indicate repeating digits (Section 8.5). The following representations
all denote the same number 71 :0.142857, 0.1428571, 0.14285714, 0.142857142857142.
Why? Because if you check them against each other decimal by decimal, youwill never hit at a difference. The relation ‘having the same infinite expansion’is a bisimulation on decimal representations, and the representations for 17 are allbisimilar.
Exercise 10.23 Show that the relation of equality on a set A is a bisimulation on
A.
Example 10.24 Collect the transition for the two clock processes of Examples
10.4 and 10.8 together in one transition system. Then it is easy to check that therelation R given by cRc1 , cRc2 , c0 Rc3 , c0 Rc4 is a bisimulation on this transition
system (Figure 10.5). The bisimulation connects the states c and c 1 . The two clock
processes are indeed indistinguishable.
Exercise 10.25 Show that there is no bisimulation that connects the starting statesof the two vending machines of Examples 10.6 and 10.9.
Exercise 10.26 Show that the union of two bisimulations is again a bisimulation.

Exercise 10.27 Show that there is always a greatest bisimulation on a set A under
the inclusion ordering. (Hint: show that the union of all bisimulations on A is itself
a bisimulation.)
Exercise 10.28 The bisimulation relation given in Example 10.24 is not the greatest bisimulation on the transition system. Find the greatest bisimulation.

10.4. PROOF BY COINDUCTION

381

tick

crack

crack

tick
tick

crack

Figure 10.5: Bisimulation between ticking clocks.

CHAPTER 10. CORECURSION

382

Use ∼ for the greatest bisimulation on a given set A (Exercise 10.27). Call twoelements of A bisimilar when they are related by a bisimulation on A. Being
bisimilar then coincides with being related by the greatest bisimulation:a ∼ b ⇔ ∃R(R is a bisimulation, and aRb).
To show that two infinite objects x and y are equal, we show that they exhibit
the same behaviour, i.e. we show that x ∼ y. Such a proof is called a proof bycoinduction.
In the particular case of comparing streams we can make the notion of a bisimulation more specific. There are only two kinds of observations on a stream: observing its head and observing its tail. Comparison of the heads should exhibit objects
that are the same. Comparison of the tails should exhibit bisimilar streams.
When talking about streams f = [f0 , f1 , f2 , f3 , . . .] it is convenient to refer to thetail of the stream as f. Thus, a stream f = [f0 , f1 , f2 , f3 , . . .] always has the formf = f0 : f , where f0 is the head and f is the tail.

f

f0

f

Figure 10.6: A stream viewed as a transition system.

f



g



g0

f0

f




g

Figure 10.7: Bisimulation between streams.
f0

Viewing a stream f as a transition system, we get transitions f −→ f , and soon (Figure 10.6). A bisimulation between streams f and g that connects f and10.4. PROOF BY COINDUCTION

383

g is given in Figure 10.7. It is immediate from the picture that the notion of abisimulation between streams boils down to the following.
A stream bisimulation on a set A is a relation R on [A] with the following property. If f, g ∈ [A] and f Rg then both f0 = g0 and f Rg.
The general pattern of a proof by coinduction using stream bisimulations of f ∼ g,
where f, g ∈ [A], is simply this. Define a relation R on [A]. Next, show that R isa stream bisimulation, with f Rg.
Example 10.29 As an example, we will show:
map f (iterate f x) ∼ iterate f (f x).
Let R be the following relation on [A].
{(map f (iterate f x), iterate f (f x)) | f :: a → a, x :: a}.
Suppose:(map f (iterate f x)) R (iterate f (f x)).
We show that R is a bisimulation. We have:
map f (iterate f x)

iterate=map=

iterate f (f x)

iterate=map f x : (iterate f (f x))(f x) : map f (iterate f (f x))(f x) : iterate f (f (f x)).

This shows:head

map f (iterate f x) −→ (f x)(1)

map f (iterate f x) −→ map f x : (iterate f (f x))

(2)

tail

head

iterate f (f x) −→ (f x)(3)iterate f (f x) −→ iterate f (f (f x))

(4)

tail

Clearly, map f (iterate f x) and iterate f (f x) have the same heads, and also,
their tails are R-related. This shows that R is a bisimimulation that connects
map f (iterate f x) and iterate f (f x). Hence(map f (iterate f x)) ∼ (iterate f (f x)).

384

CHAPTER 10. CORECURSION

Given: A labeled transition system (Q, A, T ) . . .
To be proved: q ∼ p
Proof:Let R be given by . . . and suppose qRp.
To be proved: R is a bisimulation.
aSuppose q −→ q 0a
To be proved: There is a p0 ∈ Q with p −→ p0 and q 0 Rp0 .
Proof: . . .
aSuppose p −→ p0aTo be proved: There is a q 0 ∈ Q with q −→ q 0 and q 0 Rp0 .
Proof: . . .
Thus R is a bisimulation with qRp.
Thus q ∼ p.

Given: . . .
To be proved: f ∼ g
Proof:Let R be given by . . . and suppose f Rg.
To be proved: R is a stream bisimulation.
To be proved: f0 = g0 .
Proof: . . .
To be proved: f Rg.
Proof: . . .
Thus R is a stream bisimulation with f Rg.
Thus f ∼ g.

Figure 10.8: Proof Recipes for Proofs by Coinduction.

10.5. POWER SERIES AND GENERATING FUNCTIONS

385

Exercise 10.30 Show by means of a coinduction argument that for every infinite
list xs :: [a], every f :: a -> b, and every total predicate p :: b -> Boolthe following holds:
filter p (map f xs) = map f (filter (p.f ) xs).

10.5 Power Series and Generating Functions

module PowerSeries
whereimport Polynomials
instance Fractional a => Fractional [a] where
fromRational c = [fromRational c]fs/ []= error "division by 0 attempted"
[]
/ gs
= []
(0:fs) / (0:gs) = fs / gs(_:fs) / (0:gs) = error "division by 0 attempted"(f:fs) / (g:gs) = let q = f/g inq : (fs - q.*gs) / (g:gs)
int :: Fractional a => [a] -> [a]int fs = 0 : int1 fs 1 whereint1 []_ = []
int1 (g:gs) n = g/n : (int1 gs (n+1))
expz = 1 + (int expz)

Figure 10.9: A Module for Power Series.
In chapter 9 we have seen how polynomials can get represented as finite lists of
their coefficients, and how operations on polynomials can be given directly onthese list representations. We will now generalize this to infinite sequences. A

CHAPTER 10. CORECURSION

386

possible motivation for this (one of many) is the wish to define a division operation
on polynomials.
Suppose we want to divide
f (z) = f0 + f1 z + f2 z 2 + · · · + fk z kbyg(z) = g0 + g1 z + g2 z 2 + · · · + gm z m .

The outcome h(z) = h0 + h1 z + h2 z 2 + · · · has to satisfy:
f (z) = h(z) · g(z).

This gives:
f0 + zf(z) = (h0 + zh(z))g(z)= h0 g(z) + zh(z)g(z)= h0 (g0 + zg(z)) + zh(z)g(z)= h0 g0 + z(h0 g(z) + h(z)g(z)).
Thus, f0 = h0 g0 , hence h0 = fg00 , and f (z) = h0 g(z) + h(z)g(z), so from this0 g(z)
h(z) = f (z)−h. We see from this that computing fractions can be done by ag(z)process of long division:f(z) − (f0 /g0 )g(z)
f0
f (z)
=
+z
.
g(z)
g0
g(z)

(div)

1
An example case is 1−z. Long division gives:

1−z
1
1
+z
=1+z
1−z
1−z
1−z
1
1
) = 1 + z(1 + z(1 + z)) = · · ·= 1 + z(1 + z
1−z
1−z
The representation of 1 − z is [1, −1], so in terms of computation on sequences we
get:
1
1−z

=

[1]
[1, −1]

=
=
=
=
=

1
[1]
:
1
[1, −1]
1
[1]
1
:
:
1
1
[1, −1]
1
1
1
[1]
:
:
:
1
1
1
[1, −1]
1
1
1
1
[1]
:
:
:
:
1
1
1
1
[1, −1]...

10.5. POWER SERIES AND GENERATING FUNCTIONS

387

1This shows that 1−z
does not have finite degree. It is not a polynomial.

To get at a class of functions closed under division we define power series to befunctions of the form∞Xf (z) = f0 + f1 z + f2 z 2 + · · · =fk z k .
k=0

A power series is represented by an infinite stream [f0 , f1 , f2 , . . .]. Power series ina variable z can in fact be viewed as approximations to complex numbers (Section
8.10), but we will not pursue that connection here.
The implementation of division follows the formula (div). Since trailing zeros
are suppressed, we have to take the fact into account that [] represents an infinite
stream of zeros.

fs/ []= error "division by 0 attempted"
[]
/ gs
= []
(0:fs) / (0:gs) = fs / gs(_:fs) / (0:gs) = error "division by 0 attempted"(f:fs) / (g:gs) = let q = f/g in q : (fs - q.*gs)/(g:gs)

Here are some Haskell computations involving division:
COR> take 10 (1/(1-z))[1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1]
COR> take 10 (1/(1+z))[1 % 1,-1 % 1,1 % 1,-1 % 1,1 % 1,-1 % 1,1 % 1,-1 % 1,1 % 1,-1 % 1]
3,
Example 10.31 To get a feel for ‘long division with sequences’, we compute 3−z
as a calculation on sequences. The sequence representation of 3 − z is [3, −1], so
we get:[3][3, −1]

=
=
=
=
=

3
[1]
:
3
[3, −1]
1
[1/3]
3
:
:
3
3
[3, −1]
3
1
1
[1/9]
:
:
:
3
3
9
[3, −1]3
1
1
1[1/27]
:
:
:
:
3
3
9
27
[3, −1]...

CHAPTER 10. CORECURSION

388
This is indeed what Haskell calculates for us:

COR> take 9 (3 /(3-z))[1 % 1,1 % 3,1 % 9,1 % 27,1 % 81,1 % 243,1 % 729,1 % 2187,1 % 6561]

Integration involves division, forRz0f (t)dt is given by

1
1
0 + f 0 z + f1 z 2 + f2 z 3 + · · ·2
3Rz
1
To give an example, if g(z) = 1−z , then 0 g(t)dt equals:
1
1
0 + z + z2 + z3 + · · ·2
3Integration is implemented by:int :: Fractional a => [a] -> [a]int [] = []int fs = 0 : int1 fs 1 whereint1 []_ = []
int1 (g:gs) n = g/n : (int1 gs (n+1))

We get:
COR> take 10 (int (1/(1-z)))[0 % 1,1 % 1,1 % 2,1 % 3,1 % 4,1 % 5,1 % 6,1 % 7,1 % 8,1 % 9]

To extend composition to power series, we have to be careful, for the equationf (g(z)) = f0 + g(z) · f(g(z)) has a snag: the first coefficient depends on all of f ,which is a problem if f is a power series. Here is an example:
COR> comp1 ones [0,2] where ones = 1 : ones
ERROR - Garbage collection fails to reclaim sufficient space

To solve this, we must develop the equation for composition a bit further:f (g(z)) = f0 + g(z) · f(g(z))= f0 + (g0 + zg(z)) · f(g(z))= (f0 + g0 · f (g(z))) + zg(z) · f (g(z))

10.5. POWER SERIES AND GENERATING FUNCTIONS

389

In the special case where g0 = 0 we can simplify this further:f (zg(z)) = f0 + zg(z) · f(zg(z)).
This leads immediately to the following implementation (part of the module for
Polynomials):

comp :: Num a => [a] -> [a] -> [a]comp _[]= error ".."
comp []_= []comp (f:fs) (0:gs) = f : gs * (comp fs (0:gs))
comp (f:fs) (g:gs) = ([f] + [g] * (comp fs (g:gs)))+ (0 : gs * (comp fs (g:gs)))

This gives:
COR> take 15 (comp ones [0,2]) where ones = 1 : ones
[1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384]

Figure 10.9 gives a module for power series. We will now show how this is canbe used as a tool for combinatorial calculations. We are interested in finite lists[f0 , . . . , fn ] or infinite lists [f0 , f1 , f2 , . . .] that can be viewed as solutions to combinatorial problems.
We associate sequences with combinatorial problems by saying that [f 0 , f1 , f2 , . . .]solves a combinatorial problem if fr gives the number of solutions for that problem (cf. the use of polynomials for solving combinatorial problems in section 9.4,
which is now generalized to power series).
We call a power series f (z) a generating function for a combinatorial problem
if the list of coefficients of f (z) solves that problem. If f (z) has power series
expansion f0 + f1 z + f2 z 2 + f3 z 3 + · · · , then f (z) is a generating function for[f0 , f1 , f2 , f3 , . . .].
Example 10.32 Find a generating function for [1, 1, 1, 1, . . .]. Solution: as we saw
1above 1−z
has power series expansion 1 + z + z 2 + z 3 + z 4 + · · · . The list ofcoefficients is [1, 1, 1, 1, . . .].
1
The generating function 1−z
is an important building block for constructing othergenerating functions, so we should give it a name. In fact, we already did: ones
names the list [1,1,1,1,..]. As long as we don’t use division, this is OK.

CHAPTER 10. CORECURSION390

Example 10.33 Find a generating function for the list of natural numbers. Solution: recall the corecursive definition of the natural numbers in terms of zipWith (+).
This can now be expressed succinctly in terms of addition of power series:
COR> take 20 nats where nats = 0 : (nats + ones)[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19]

This gives the following specification for the generating function:g(z) = z(g(z) +g(z) − zg(z) =g(z) =z1−z
z(1 − z)21)
1−zHere is the check:
COR> take 20 (z * ones^2)[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19]

Another way of looking at this: differentiation of 1 + z + z 2 + z 3 + z 4 + · · · gives01z
1 + 2z + 3z 2 + 4z 3 + · · · . This means that z 1−z= (1−z)2 is a generating
function for the naturals.
Example 10.34 Find the generating function for [0, 0, 0, 1, 2, 3, 4, . . .]. Solution:
multiplication by z has the effect of shifting the sequence of coefficients one place
to the right and inserting a 0 at the front. Thus the generating function can be
got from that of the previous example through multiplying by z 2 . The generatingz3
function is: (1−z)
2 . Here is the check:
COR> take 20 (z^3 * ones^2)[0,0,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17]

Example 10.35 Find the generating function for the sequence of powers of two(the sequence λn.2n ). Solution: start out from a corecursive program for the powers of two:
COR> take 15 powers2 where powers2 = 1 : 2 * powers2[1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384]

10.5. POWER SERIES AND GENERATING FUNCTIONS

391

This immediately leads to the specification g(z) = 1 + 2zg(z) (the factor z shifts
one place to the right, the summand 1 puts 1 in first position). It follows that1g(z) = 1−2z
. Here is the confirmation:
COR> take 10 (1/(1-2*z))[1 % 1,2 % 1,4 % 1,8 % 1,16 % 1,32 % 1,64 % 1,128 % 1,256 % 1,512 % 1]

Example 10.36 If g(z) is the generating function for λn.gn , then λn.gn+1 is gen1
erated by g(z)z , for multiplication by z shifts the coefficients of the power series
one place to the left. Thus, the generating function for the sequence λn.n + 1 is1(1−z)2 . This sequence can also be got by adding the sequence of naturals and thez
1
sequence of ones, i.e., it is also generated by (1−z)
2 + 1−z . This identification
makes algebraic sense:
1
z
1−z
1
z
+
=
+
=
.
(1 − z)2
1−z(1 − z)2(1 − z)2(1 − z)2
Exercise 10.37 Find the generating functions for[0, 0, 0, 1, 1, 1, . . .],for[1, 1, 1, 0, 0, 0, . . .],and for

1 1 1[1, , , , . . .].
2 4 8Check your answers by computer.

Example 10.38 Find the generating function for [0, 1, 0, 1, 0, 1, . . .]. Solution: thesequence is generated by the following corecursive program.
COR> take 20 things where things = 0 : 1 : things[0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1]

This leads to the specification g(z) = 0 + z(1 + zg(z)), which reduces to g(z) =z
z + z 2 g(z). From this, g(z) = 1−z
2 . In a similar way we can derive the generat1
ing function for [1, 0, 1, 0, 1, 0, . . .], which turns out to be g(z) = 1−z
2 . Alternatively, observe that [1, 0, 1, 0, 1, 0, . . .] is the result of shifting [0, 1, 0, 1, 0, 1, . . .]one place to the left, so division by z does the trick.

392

CHAPTER 10. CORECURSION

Theorem 10.39 summarizes a number of recipes for finding generating functions.
Theorem 10.39 Suppose f (z) is a generating function for λn.fn and g(z) is thegenerating function for λn.gn . Then:c
1. 1−zis the generating function for λn.c.
z
2. (1−z)2 is the generating function for λn.n.

3. f (z)z is the generating function for λn.fn+1 .
4. cf (z) + dg(z) is the generating function for λn.cfn + dgn .
5. (1 − z)f (z) is the generating function for the difference sequence
λn.fn − fn−1 .
6. 1−zz f (z) is the generating function for the difference sequence
λn.fn+1 − fn .
1
7. 1−zf (z) is the generating function for λn.f0 + f1 + · · · + fn .

8. f (z)g(z) is the generating function for λn.f0 gn + f1 gn−1 + · · · + fn−1 g1 +fn g0 (the convolution of f and g).
9. zf 0 (z) is the generating function for λn.nfn .
Rz
fn.
10. z1 0 f (t)dt is the generating function for λn. n+1

Exercise 10.40 Prove the statements of theorem 10.39 that were not proved in theexamples.
Example 10.41 Find the generating function for the sequence λn.2n + 1. Solution: the odd natural numbers are generated by the following corecursive program:
COR> take 20 (ones + 2 * nats)[1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39]
1
2This immediately gives g(z) = 1−z+ (1−z)2.

Example 10.42 Find the generating function for the sequence λn.(n + 1) 2 . Solution: the generating function is the sum of the generating functions for λn.n 2 ,λn.2n and λn.1. One way to proceed would be to find these generating functions
and take their sum. But there is an easier method. Recall that (n + 1) 2 is the sumof the first n odd numbers, and use the recipe from Theorem 10.39 for forming10.5. POWER SERIES AND GENERATING FUNCTIONS

393

the sequence λn.a0 + · · · + an , by means of dividing the generating function from
1
2z
Example 10.41 by 1 − z. This immediately gives (1−z)
2 + (1−z)3 .
Alternatively, Theorem 10.39 tells us that λn.(n + 1)2 can also be defined fromthe naturals with the help of differentiation:
COR> take 20 (deriv nats)[1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400]

It follows that the generating function g(z), for g(z) should satisfy:g(z) =

z(1 − z)2

0

.

1
2z
Working this out we find that g(z) = (1−z)
2 + (1−z)3 .

Finally, here is computational confirmation:
COR> take 20 (ones^2 + 2 * z * ones^3)[1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400]

Example 10.43 Find the generating function for λn.n2 . Solution: shift the solution for the previous example to the right by multiplication by z. This gives:
COR> take 20 (z * ones^2 + 2 * z^2 * ones^3)[0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361]
2z
2z
So the generating function is g(z) = (1−z)
2 + (1−z)3 . Sure enough, we can also
get the desired sequence from the naturals by shift and differentiation:

COR> take 20 (z * deriv nats)[0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361]

Example 10.44 Find the generating function for the sequence of Fibonacci numbers. Solution: consider again the corecursive definition of the Fibonacci numbers.
We can express this succinctly in terms of addition of power series, as follows:
COR> take 20 fibs where fibs = 0 : 1 : (fibs + (tail fibs))[0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181]

CHAPTER 10. CORECURSION

394

This is easily translated into an instruction for a generating function:g(z)g(z) = z 2 g(z) ++ z.
zExplanation: multiplying by z 2 inserts two 0’s in front of the fibs sequence, andadding z changes the second of these to 1 (the meaning is: the second coefficientis obtained by add 1 to the previous coefficient). g(z)z gives the tail of the fibs
sequence. From this we get:g(z) − zg(z) − z 2 g(z) = z
g(z) =z
1 − z − z2

And lo and behold:
COR> take 10 (z/(1-z-z^2))[0 % 1,1 % 1,1 % 1,2 % 1,3 % 1,5 % 1,8 % 1,13 % 1,21 % 1,34 % 1]

Exercise 10.45 The Lucas numbers are given by the following recursion:
L0 = 2, L1 = 1, Ln+2 = Ln + Ln+1 .
This gives:[2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, . . .]The recipe is that for the Fibonacci numbers, but the initial element of the list is 2
instead of 0. Find a generating function for the Lucas numbers, starting out froman appropriate corecursive definition.
Example 10.46 Recall the definition of the Catalan numbers from Exercise 7.20:
C0 = 1, Cn+1 = C0 Cn + C1 Cn−1 + · · · + Cn−1 C1 + Cn C0 .
Clearly, Cn+1 is obtained from [C0 , . . . , Cn ] by convolution of that list with itself.
This leads to the following corecursive definition of the Catalan numbers:
COR> take 15 cats where cats = 1 : cats * cats
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]

10.5. POWER SERIES AND GENERATING FUNCTIONS

395

From this, we get at the following specification for the generating function:2g(z) = 1 + z · (g(z))2z(g(z)) − g(z) + 1 = 0

Considering g(z) as the unknown, we cansolve this as a quadratic equation ax 2 +√−b± b2 −4acbx + c = 0, using the formula x =. We get the following generating
2afunctions:√
1 ± 1 − 4z
g(z) =.
2z

Consider the following power series.
∞
X
z2
z3
zn
g(z) = 1 + z +++··· =
2!
3!
n!
n=0

Note that by the definition of the derivative operation for power series, g(z) 0 =g(z). The function with this property is the function ez , where e is the base of thenatural logarithm, Napier’s number e. The number e is defined by the following
infinite sum:
∞
X
1
1
1
1
1+ + + +··· =1! 2! 3!
n!
n=0Therefore we call g(z) the exponential function ez . Note that by the rules for
integration for power series,
Z z
z2
z3
et dt = 0 + z +++ · · · = ez − 1,
2!
3!
0
Rz
and therefore, ez = 1 + 0 et dt. This gives us a means of defining ez by corecursion, as follows:

expz = 1 + (int expz)

We get:
COR> take 9 expz
[1 % 1,1 % 1,1 % 2,1 % 6,1 % 24,1 % 120,1 % 720,1 % 5040,1 % 40320]

CHAPTER 10. CORECURSION

396

1
Since we know (Theorem 10.39) that 1−z
f (z) gives incremental sums of the sequence generated by f (z), we now have an easy means to approximate Napier’s
number (the number e):

COR> take 20 (1/(1-z) * expz)[1 % 1,2 % 1,5 % 2,8 % 3,65 % 24,163 % 60,1957 % 720,685 % 252,
109601 % 40320,98641 % 36288,9864101 % 3628800,13563139 % 4989600,
260412269 % 95800320,8463398743 % 3113510400,
47395032961 % 17435658240,888656868019 % 326918592000,
56874039553217 % 20922789888000,7437374403113 % 2736057139200,
17403456103284421 % 6402373705728000,82666416490601 % 30411275102208]

10.6 Exponential Generating Functions
Up until now, the generating functions were used for solving combinatorial problems in which order was irrelevant. These generating functions are sometimes
called ordinary generating functions. To tackle combinatorial problems whereorder of selection plays a role, it is convenient to use generating functions in aslightly different way. The exponential generating function for a sequence λn.f n
is the function
f (z) = f0 + f1 z +∞X
f2 z 2 f3 z 3
fn z n++··· =
2!
3!
n!
n=0

1
Example 10.47 The (ordinary) generating function of [1, 1, 1, . . .] is 1−z, the exzzponential generating function of [1, 1, 1, . . .] is e . If e is taken as an ordinary
1 1 1 1
generating function, it generates the sequence [1, 1!
, 2! , 3! , 4! , . . .].

Example 10.48 (1 + z)r is the ordinary generating function for the problem ofpicking an n-sized subset from an r-sized set, and it is the exponential generating
function for the problem of picking a sequence of n distinct objects from an r-sizedset.
We see from Examples 10.47 and 10.48 that the same function g(z) can be used forgenerating different sequences, depending on whether g(z) is used as an ordinary
or an exponential generating function.
It makes sense, therefore, to define an operation that maps ordinarily generated
sequences to exponentially generated sequences.

10.6. EXPONENTIAL GENERATING FUNCTIONS

397

o2e :: Num a => [a] -> [a]o2e []= []o2e (f:fs) = f : o2e (deriv (f:fs))

With this we get:
COR> take 10 (o2e expz)[1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1]

A function for converting from exponentially generated sequences to ordinarily
generated sequences is the converse of this, so it uses integration:

e2o :: Fractional a => [a] -> [a]e2o []= []e2o (f:fs) = [f] + (int (e2o (fs)))

This gives:
COR> take 9 (e2o (1/(1-z)))[1 % 1,1 % 1,1 % 2,1 % 6,1 % 24,1 % 120,1 % 720,1 % 5040,1 % 40320]

Example 10.49 Here is how (z + 1)10 is used to solve the problem of finding thenumber of ways of picking subsets from a set of 10 elements:
COR> (1+z)^10[1,10,45,120,210,252,210,120,45,10,1]

And here is how the same function is used to solve the problem of finding thenumber of ways to pick sequences from a set of 10 elements:
COR> o2e ((1+z)^10)[1,10,90,720,5040,30240,151200,604800,1814400,3628800,3628800]

Example 10.50 Suppose a vase contains red, white and blue marbles, at least fourof each kind. How many ways are there of arranging sequences of marbles from398

CHAPTER 10. CORECURSION

the vase, with at most four marbles of the same colour? Solution: the exponential
2
3
4
generating function is (1 + z + z2 + z6 + z24 )3 . The idea: if you pick n marbles ofthe same colour, then n! of the marble orderings become indistinguishable. Hereis the Haskell computation of the solution.
COR> o2e ([1,1,1/2,1/6,1/24]^3)[1 % 1,3 % 1,9 % 1,27 % 1,81 % 1,240 % 1,690 % 1,1890 % 1,4830 % 1,11130 % 1,22050 % 1,34650 % 1,34650 % 1]

Example 10.51 Suppose a vase contains red, white and blue marbles, an unlimited supply of each colour. How many ways are there of arranging sequences ofmarbles from the vase, assume that at least two marbles are of the same colour?
32
Solution: a suitable exponential generating function is (ez − z − 1)3 = ( z2 + z6 +z4
3
24 + · · · ) :COR> take 10 (o2e ((expz - z - 1)^3))[0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,90 % 1,630 % 1,2940 % 1,11508 % 1]

This gives the number of solutions for up to 9 marbles. For up to 5 marbles there
are no solutions. There are 90 = 26!3 sequences containing two marbles of eachcolour. And so on.
Exercise 10.52 Suppose a vase contains red, white and blue marbles, at least fourof each kind. How many ways are there of arranging sequences of marbles fromthe vase, with at least two and at most four marbles of the same colour?

10.7 Further Reading
Domain theory and proof by approximation receive fuller treatment in [DP02].
Generating functions are a main focus in [GKP89]. The connection between generating functions and lazy semantics is the topic of [Kar97, McI99, McI00]. A
coinductive calculus of streams and power series is presented in [Rut00]. Corecursion is intimately connected with circularity in definitions [BM96] and with thetheory of processes and process communication [Fok00, Mil99].

Chapter 11

Finite and Infinite Sets
Preview
Some sets are bigger than others. For instance, finite sets such as ∅ and {0, 1, 2},
are smaller than infinite ones such as N and R. But there are varieties of infinity:
the infinite set R is bigger than the infinite set N, in a sense to be made precise inthis chapter.
This final chapter starts with a further discussion of the Principle of Mathematical
Induction as the main instrument to fathom the infinite with, and explains why
some infinite sets are incomparably larger than some others.

module FAIS
where

11.1 More on Mathematical Induction
The natural numbers form the set N = {0, 1, 2, . . .}. The Principle of Mathematical Induction allows you to prove universal statements about the natural numbers,that is, statements of the form ∀n ∈ N E(n).
399

CHAPTER 11. FINITE AND INFINITE SETS

400

Fact 11.1 (Mathematical Induction) For every set X ⊆ N, we have that:
if 0 ∈ X and ∀n ∈ N(n ∈ X ⇒ n + 1 ∈ X), then X = N.
This looks not impressive at all, since it is so obviously true. For, suppose thatX ⊆ N satisfies both 0 ∈ X and ∀n ∈ N(n ∈ X ⇒ n + 1 ∈ X). Then
we see that N ⊆ X (and, hence, that X = N) as follows. First, we have as afirst Given, that 0 ∈ X. Then the second Given (∀-elimination, n = 0) yields
that 0 ∈ X ⇒ 1 ∈ X. Therefore, 1 ∈ X. Applying the second Given again(∀-elimination, n = 1) we obtain 1 ∈ X ⇒ 2 ∈ X; therefore, 2 ∈ X. And soon, for all natural numbers.
Nevertheless, despite its being so overly true, it is difficult to overestimate theimportance of induction. It is essentially the only means by with we acquire information about the members of the infinite set N.
Sets versus Properties. With every property E of natural numbers there corresponds a set {n ∈ N | E(n)} of natural numbers. Thus, induction can also be
formulated using properties instead of sets. It then looks as follows:
If E is a property of natural numbers such that(i) E(0),(ii) ∀n ∈ N [E(n) ⇒ E(n + 1)],then ∀n ∈ N E(n).
Induction: Basis, Step, Hypothesis. According to Induction, in order that E(n)is true for all n ∈ N, it suffices that(i) this holds for n = 0, i.e., E(0) is true, and(ii) this holds for a number n + 1, provided it holds for n.
As we have seen, the proof of (i) is called basis of a proof by induction, the proofof (ii) is called the induction step. By the introduction rules for ∀ and ⇒, in orderto carry out the induction step, you are granted the Given E(n) for an arbitrary
number n, whereas you have to show that E(n + 1). The assumption that E(n) is
called the induction hypothesis. The nice thing about induction is that this induction hypothesis comes entirely free.
Fact 11.2 (Scheme for Inductions) Viewed as a rule of proof, induction can be
schematized as follows.

11.1. MORE ON MATHEMATICAL INDUCTION

401

E(n)
..
.
E(0) E(n + 1)
∀nE(n)A proof using the Principle of Induction of a statement of the form ∀n E(n) is
called a proof with induction with respect to n. It can be useful to mention theparameter n when other natural numbers surface in the property E.
According to the above schema, the conclusion ∀n E(n) can be drawn on the basisof (i) the premiss that E(0), and (ii) a proof deriving E(n + 1) on the basis of theinduction hypothesis E(n).
The following strong variant of induction is able to perform tasks that are beyond
the reach of the ordinary version. This time, we formulate the principle in terms
of properties.
Fact 11.3 (Strong Induction) For every property E of natural numbers:
if ∀n ∈ N [ (∀m < n E(m)) ⇒ E(n) ], then ∀n ∈ N E(n).
Using strong induction, the condition ∀m < n E(m) acts as the induction hypothesis from which you have to derive that E(n).
This is (at least, for n > 1) stronger than in the ordinary form. Here, to establishthat E(n), you are given that E(m) holds for all m < n. In ordinary induction,
you are required to show that E(n + 1) on the basis of the Given that it holds forn only.
Note: using ordinary induction you have to accomplish two things: basis, andinduction step. Using strong induction, just one thing suffices. Strong induction
has no basis.
Proof. (of 11.3.) Suppose that ∀n ∈ N [ (∀m < n E(m)) ⇒ E(n) ].

(1)

Define the set X by X = {n | ∀m < n E(m)}. Assumption (1) says that every
element of X has property E. Therefore, it suffices showing that every natural
number is in X. For this, we employ ordinary induction.
Basis. 0 ∈ X.
I.e.: ∀m < 0 E(m); written differently: ∀m[m < 0 ⇒ E(m)]. This holds triviallysince there are no natural numbers < 0. The implication m < 0 ⇒ E(m) is
trivially satisfied.
Induction step.

CHAPTER 11. FINITE AND INFINITE SETS

402

Assume that (induction hypothesis) n ∈ X. That means: ∀m < n E(m). Assumption (1) implies, that E(n) is true. Combined with the induction hypothesis,
we obtain that ∀m < n + 1 E(m). This means, that n + 1 ∈ X. The induction
step is completed.
Induction now implies, that X = N. According to the remark above, we have that∀n ∈ N E(n).
It is useful to compare the proof schemes for induction and strong induction. Herethey are, side by side:
E(n)
..
.
E(0) E(n + 1)
∀nE(n)∀m < n E(m)
..
.
E(n)
∀nE(n)An alternative to strong induction that is often used is the minimality principle.
Fact 11.4 (Minimality Principle) Every non-empty set of natural numbers has aleast element.
Proof. Assume that A is an arbitrary set of natural numbers. It clearly suffices toshow that, for each n ∈ N,n ∈ A ⇒ A has a least element.
For, if also is given that A 6= ∅, then some n ∈ A must exist. Thus the implication
applied to such an element yields the required conclusion.
Here follows a proof of the statement using strong induction w.r.t. n. Thus, assume
that n is an arbitrary number for which:
Induction hypothesis: for every m < n: m ∈ A ⇒ A has a least element.
To be proved: n ∈ A ⇒ A has a least element.
Proof: Assume that n ∈ A. There are two cases.
(i) n is (by coincidence) the least element of A. Thus, A has a least element.
(ii) n is not least element of A. Then some m ∈ A exists such that m < n. So we
can apply the induction hypothesis and again find that A has a least element.
Remark. The more usual proof follows the logic from Chapter 2. This is much
more complicated, but has the advantage of showing the Minimality Principle tobe equivalent with Strong Induction.

11.1. MORE ON MATHEMATICAL INDUCTION

403

Strong Induction is the following schema:∀n[∀m < n E(m) ⇒ E(n)] ⇒ ∀n E(n).
Since E can be any property, it can also be a negative one. Thus:∀n[∀m < n¬E(m) ⇒ ¬E(n)] ⇒ ∀n¬E(n).
The contrapositive of this is (Theorem 2.10 p. 45):¬∀n¬E(n) ⇒ ¬∀n[∀m < n¬E(m) ⇒ ¬E(n)].
Applying Theorem 2.40 (p. 65):∃n E(n) ⇒ ∃n¬[∀m < n¬E(m) ⇒ ¬E(n)].
Again Theorem 2.10:∃n E(n) ⇒ ∃n[∀m < n¬E(m) ∧ ¬¬E(n)].
Some final transformations eventually yield:∃n E(n) ⇒ ∃n[E(n) ∧ ¬∃m < n E(m)]— which is Minimality formulated using properties.
Exercise 11.5 Prove Induction starting at m, that is: for every X ⊆ N, if m ∈ X
and ∀n > m(n ∈ X ⇒ n + 1 ∈ X), then ∀n ∈ N(m 6 n ⇒ n ∈ X).
Hint. Apply induction to the set Y = {n | m + n ∈ X}.
Exercise 11.6 Suppose that X ⊆ N is such that 1 ∈ X and ∀n ∈ N(n ∈ X ⇒n + 2 ∈ X). Show that every odd number is in X.
Definition 11.7 A relation ≺ on A is called well-founded if no infinite sequence· · · ≺ a2 ≺ a1 ≺ a0 exists in A. Formulated differently: every sequence a0 
a1  a2  · · · in A eventually terminates.
Example 11.8 According to Exercise 4.34 (p. 136), the relation ≺ on sets definedby: a ≺ b iff ℘(a) = b, is well-founded.
An easier example is in the next exercise.
Exercise 11.9 Show that < is well-founded on N. That is: there is no infinite
sequence n0 > n1 > n2 > · · · .
Hint. Show, using strong induction w.r.t. n, that for all n ∈ N, E(n); where
E(n) signifies that no infinite sequence n0 = n > n1 > n2 > · · · starts at n.
Alternatively, use Exercise 11.10.

CHAPTER 11. FINITE AND INFINITE SETS

404

Exercise 11.10* Well-foundedness as an induction principle. Let ≺ be a relation on a set A.
1. Suppose that ≺ is well-founded. Assume that X ⊆ A satisfies∀a ∈ A(∀b ≺ a(b ∈ X) ⇒ a ∈ X).
Show that X = A.
Hint. Show that any a0 ∈ A − X can be used to start an infinite sequencea0  a1  a2  · · · in A.
2. Conversely:
Suppose that every X ⊆ A satisfying ∀a ∈ A(∀b ≺ a(b ∈ X) ⇒ a ∈ X)
coincides with A. Show that ≺ is well-founded.

Hint. Suppose that a0  a1  a2  · · · , and consider the set X = A −{a0 , a1 , a2 , . . .}.

Exercise 11.11* Let R be a relation on a set A. Recall that R ∗ denotes the reflexive
transitive closure of R, cf. Exercise 5.46 p. 173.
1. Assume that for all a, b1 , b2 ∈ A, if aRb1 and aRb2 , then c ∈ A exists suchthat b1 Rc and b2 Rc.
Show that R is confluent, that is: for all a, b1 , b2 ∈ A, if aR∗ b1 and aR∗ b2 ,then c ∈ A exists such that b1 R∗ c and b2 R∗ c.
2. Assume that R is weakly confluent, that is: for all a, b1 , b2 ∈ A, if aRb1 andaRb2 , then c ∈ A exists such that b1 R∗ c and b2 R∗ c.
A counter-example to confluence is called bad. That is: a is bad iff there
are b1 , b2 ∈ A such that aR∗ b1 and aR∗ b2 , and for no c ∈ A we have thatb1 R∗ c and b2 R∗ c.
Show: if a is bad, then a bad b exists such that aRb.
3. (In abstract term rewriting theory, the following result is known as Newman’s Lemma.)
Assume that R is weakly confluent.
Furthermore, assume that R−1 is well-founded.
Show that R is confluent.

11.1. MORE ON MATHEMATICAL INDUCTION

405

Hint. Use part 2; alternatively, use Exercise 11.10.
Remark. That R−1 is well-founded is necessary here.
For example, R = {(1, 0), (1, 2), (2, 1), (2, 3)} is weakly confluent but notconfluent.

Exercise 11.12 Suppose that ∅ 6= X ⊆ N, and that X is bounded, i.e.: that m ∈ Nexists such that for all n ∈ X, n 6 m. Show that X has a maximum, that is: anelement m ∈ X such that for all n ∈ X, n 6 m.
Hint. Induction w.r.t. m. Thus, E(m) is: every non-empty X ⊆ N such that∀n ∈ X (n 6 m), has a maximum.
Exercise 11.13 Suppose that f : N → N is such that n < m ⇒ f (n) < f (m).
Show by means of an induction argument that for all n ∈ N: n 6 f (n).
Exercise 11.14 Suppose that a0 , a1 , a2 , . . . is an infinite sequence of natural numbers. Prove that i, j ∈ N exist such that both i < j and ai 6 aj .
Exercise 11.15 The function g : N+ × N+ → N+ has the following properties.
1. If n < m, then g(n, m) = g(n, m − n).
2. g(n, m) = g(m, n).
3. g(n, n) = n.
Show that g(n, m) is the gcd (greatest common divisor) of n and m.
Exercise 11.16* You play Smullyan’s ball game on your own. Before you is abox containing finitely many balls. Every ball carries a natural number. Next to thebox, you have a supply of as many numbered balls as you possibly need. A movein the game consists in replacing one of the balls in the box by arbitrarily (possibly
zero, but finitely) many new balls that carry a number smaller than that on the9
one you replace. (Thus, 99 balls numbered 7 can replace one ball numbered 8,
but a ball numbered 0 can only be taken out of the box since there are no natural
numbers smaller than 0.)
Repeat this move.
Show that, no matter how you play the game, you’ll end up with an empty box
eventually. (Thus, your last moves necessarily consist in throwing away balls numbered 0.)

406

CHAPTER 11. FINITE AND INFINITE SETS

Hint. Proof by Contradiction. Suppose that you can play ad infinitum, and that B k
is how the box looks after your k-th move. Derive a contradiction applying strong
induction w.r.t. the greatest number n present on one of the balls in the box B 0
you start from. If m is the number of balls in B0 carrying n, apply a second strong
induction, now w.r.t. m.
Exercise 11.17 Implement a simplified version of Smullyan’s ball game from theprevious exercise, where (i) the box of balls is represented as a list of integers, (ii)
it is always the first integer n > 1 that gets replaced, (iii) an integer n > 1 gets
replaced by two copies of n − 1, (iv) the game terminates with a list consisting ofjust ones. The type declaration should run:ballgame :: [Integer] -> [[Integer]].

How long will it take before ballgame [50] terminates? Minutes? Hours?
Days? Years?
Exercise 11.18 The following theorem implies that, e.g., all inhabitants of Amsterdam have the same number of hairs.
Theorem: in every set of n people, everyone has the same number of hairs.
Proof. Induction w.r.t. n.
Basis. n = 0 (or n = 1): trivial.
Induction step. Induction hypothesis: the statement holds for sets of n people.
Now assume that A is an (n + 1)-element set of humans. Arbitrarily choose different p and q in A; we show that p and q have the same number of hairs. A − {p}and A − {q} have n elements, thus the induction hypothesis applies to these sets.
Choose r ∈ A − {p, q}. Then r and q have the same number of hairs (they are bothin the set A − {p}); and r and p have the same number of hairs (similar reason).
Thus, p and q have the same number of hairs as well.
Explain this apparent contradiction with common sense observation.

11.2 Equipollence
In order to check whether two finite sets have the same number of elements, it is
not necessary at all to count them. For, these numbers are the same iff there is abijection between the two sets.
Sometimes, it is much easier to construct a bijection than to count elements. Imagine a large room full of people and chairs, and you want to know whether there are
as many people as there are chairs. In order to answer this question, it suffices to
ask everyone to sit down, and have a look at the resulting situation.

11.2. EQUIPOLLENCE

407

This observation motivates the following definition.
Definition 11.19 (Equipollence) Two sets A and B are called equipollent if there
is a bijection from A to B.
Notation: A ∼ B.
Example 11.20 (Trivial but Important) The set N is equipollent with its proper
subset N+ = N−{0}. For, this is witnessed by the successor function n 7−→ n+1.
We can generate the graph of this function in Haskell by means of:
succs = [ (n, succ n) | n <- [0..] ].

Of course, that a set is equipollent with one of its proper subsets can only happen
in the case of infinite sets. The example shows that the notion of equipollence canhave surprising properties when applied to infinite sets.
The following theorem shows that ∼ is an equivalence.
Theorem 11.21 For all sets A, B, C:
1. A ∼ A(reflexivity),

2. A ∼ B =⇒ B ∼ A(symmetry),

3. A ∼ B ∧ B ∼ C =⇒ A ∼ C(transitivity).

Proof.
1. 1A is a bijection from A to itself.
2. If f : A → B is a bijection, then f −1 is a bijection : B → A.
3. If f : A → B and g : B → C are bijections, then g ◦ f : A → C is a bijection
as well. (Cf. Lemma 6.36, p. 223)
The standard example of an n-element set is {0, . . . , n−1}. (Of course, {1, . . . , n}serves as well.) This motivates part 1 of the following definition.
Definition 11.22 (Finite, Infinite)
1. A set has n elements if it is equipollent with {0, . . . , n − 1}.
2. It is finite if n ∈ N exists such that it has n elements.

CHAPTER 11. FINITE AND INFINITE SETS

408

3. It is infinite if it is not finite.
Example 11.23
1. The set ∅ has 0 elements, hence ∅ is finite.
2. If A has n elements and x is arbitrarily chosen, then A ∪ {x} has n or n + 1
elements (depending on whether x ∈ A). Thus, if A is finite then so is A ∪ {x}.
The proof of the following unsurprising theorem illustrates the definitions and theuse of induction.
Theorem 11.24 N is infinite.
Proof. We show, using induction w.r.t. n, that for all n ∈ N,
N 6∼ {0, . . . , n − 1}.
The induction step applies Exercise 11.25.
Basis.
If n = 0, then {0, . . . , n − 1} = ∅. And a non-empty set like N cannot be
equipollent with ∅.
Induction step.
Induction hypothesis: N 6∼ {0, . . . , n − 1}.
To be proved: N 6∼ {0, . . . , n}.
Proof: Assume that, nevertheless, a bijection from N to {0, . . . , n}exists. According to Exercise 11.25 we may assume there is a bijection f : N → {0, . . . , n} such that f (0) = n. Its restrictionf  (N−{0}) (Definition 6.12 p. 214) then is a bijection from N−{0}to {0, . . . , n − 1}. We have that N ∼ (N − {0}) (This is the Trivial but Important Example 11.20). Conclusion (Theorem 11.21.3):
N ∼ {0, . . . , n − 1}. But this contradicts the induction hypothesis.

Exercise 11.25 Suppose that A ∼ B, a ∈ A and b ∈ B. Show that a bijection
f : A → B exists such that f (a) = b.
Hint. By A ∼ B, we have a bijection g : A → B. If, by coincidence, g(a) = b,
we let f be g. Thus, assume g(a) = b0 6= b. Since g is surjective, a0 ∈ A exists
such that g(a0 ) = b. Make a picture of the situation and look whether you can findf by suitably modifying g.

11.2. EQUIPOLLENCE

409

Exercise 11.26* Suppose that A ∼ B. Show that ℘(A) ∼ ℘(B). Write out thebijection, using f ∗ : ℘(A) → ℘(B), with f ∗ (X) = f [X], for f a bijection thatwitnesses A ∼ B.
Exercise 11.27 Show that, for every set A: ℘(A) ∼ {0, 1}A.
Hint. Associate with X ⊆ A its characteristic function, that is, the function χ X :A → {0, 1} defined by: χX (a) = 1 iff a ∈ X. The function X 7→ χX (that sends
sets to functions) is the bijection you are looking for.
Exercise 11.28 Suppose that A ∼ B. Show:
1. if A has n elements,then so has B,
2. if A is finite, then so is B,
3. if A is infinite, then so is B.

Exercise 11.29 f is a function. Show that f ∼ dom (f ).
Exercise 11.30* Suppose that R is an equivalence on A. V = A/R is the corresponding quotient. Show that the set of all partitions of V is equipollent with theset of all equivalences Q on A for which R ⊆ Q.
Exercise 11.31* Suppose that n, m ∈ N and n < m. Show that {0, . . . , n − 1} 6∼{0, . . . , m − 1}.
Hint. Use induction to show that ∀n < m {0, . . . , n − 1} 6∼ {0, . . . , m − 1}.
Employ the method of proof of Theorem 11.24.
Exercise 11.32 Suppose that X ⊆ N and m ∈ N are give such that ∀n ∈ X(n <m). Show that X is finite.
The following exercise explains how to apply induction to finite sets.
Exercise 11.33* Prove the following induction principle for finite sets.
If E is a property of sets such that
1. E(∅),
2. for every set A and every object x 6∈ A: if E(A), then also E(A ∪ {x}),410

CHAPTER 11. FINITE AND INFINITE SETS

then E holds for every finite set.
Hint. Apply induction to the property E 0 of numbers, defined by E 0 (n) ≡ ∀A[Ahas n elements ⇒ E(A)].
Exercise 11.34* Show that a subset of a finite set is finite.
Exercise 11.35* Show that the union of two finite sets is finite.
Exercise 11.36* Suppose that h is a finite injection with dom (h) ⊆ A andrng (h) ⊆ B. Suppose that A ∼ B. Show that a bijection f : A → B exists
such that f ⊇ h. (And what, if h is infinite?)Hint. Induction on the number n of elements of h. (The case n = 1 is Exercise 11.25.)
Exercise 11.37* Show that a proper subset of a finite set never is equipollent withthat set.
Exercise 11.38 Suppose that A and B are finite sets and f : A → B a bijection.
Show:
1. (B − A) ∼ (A − B),
2. there exists a bijection g : A ∪ B → A ∪ B such that f ⊆ g.

Exercise 11.39* Show: a set A is finite iff the following condition is satisfied forevery collection E ⊆ ℘(A): if ∅ ∈ E and ∀B ∈ E ∀a ∈ A (B ∪ {a} ∈ E), thenA ∈ E.

11.3 Infinite Sets
One of the great discoveries of Cantor is that for any set A, the set ℘(A) is ‘larger’(in a sense to made precise below) than A. This shows the existence of ‘Cantor’s
paradise’ of an abundance of sets with higher and higher grades of infinity.

11.3. INFINITE SETS

411

At Most Equipollent To
Recall Definition 11.19: sets are equipollent means that there is a bijection between the two.
Definition 11.40 (At Most Equipollent To) The set A is at most equipollent to B
if an injection exists from A into B.
Notation: A  B.
Example 11.41 Z  R+ .
Theorem 11.42 For every infinite set A: N  A.
Proof. Suppose that A is infinite. An injection h : N → A as required can beproduced in the following way.
Note: A 6= ∅ (since ∅ is finite). Thus, it i s possible to choose an element h(0) ∈ A.
Now, A 6= {h(0)} (for, {h(0)} is finite). Thus, another element h(1) ∈ A−{h(0)}exist.
Again, A 6= {h(0), h(1)}, etc.
Going on, this argument produces different elements h(0), h(1), h(2), . . . in A;
thus, the corresponding function h : N → A is an injection.
Thus, N is the “simplest” infinite set.
Exercise 11.43 Show:
1. A  A,
2. A ∼ B =⇒ A  B,
3. A  B ∧ B  C =⇒ A  C,
4. A ⊆ B =⇒ A  B,

Exercise 11.44* Show the reverse of Theorem 11.42: if N  A, then A is infinite.
Hint. Cf. the proof of Theorem 11.24, p. 408. A slight modification (that uses a
modification of Exercise 11.25) shows that for all n, N 6 {0, . . . , n − 1}; this
implies what is required.

412

CHAPTER 11. FINITE AND INFINITE SETS

Exercise 11.45 Suppose that h : A → A is an injection that is not surjective. Say,b ∈ A − rng (h). Define f : N → A by: f (0) = b, f (n + 1) = h(f (n)). E.g.,
f (3) = h(f (2)) = h(h(f (1))) = h(h(h(f (0)))) = h(h(h(b))).
Show that f (n) is different from f (0), . . . , f (n − 1) (n ∈ N). (Induction w.r.t. n.)
Conclusion: f is a injection, and N  A.
Exercise 11.46 Show: if N  A, then a non-surjective injection h : A → A
exists.
Exercise 11.47 Show: a set is infinite iff it is equipollent with one of its proper
subsets.
Hint. Use Theorem 11.42 and Exercises 11.44, 11.45 and 11.46.
Exercise 11.48 Suppose that A is finite and that f : A → A. Show: f is surjective
iff f is injective.
Hint. ⇐ : use Exercises 11.44 and 11.45.

Countably Infinite
The prototype of a countably infinite set is N.
Definition 11.49 (Countable) A set A is countably infinite (or: denumerable) if
N ∼ A.
Exercise 11.50* Show: a subset of a countably infinite set is countably infinite or
finite.
Exercise 11.51 Show:
1. Z (the set of integers) is countably infinite,
2. a union of two countably infinite sets is countably infinite.

Theorem 11.52 N2 = {(n, m) | n, m ∈ N} is countably infinite.
Proof. Define S(p) = {(n, m) | n + m = p}.
The sets S(p) are pairwise disjoint, and N2 = S(0) ∪ S(1) ∪ S(2) ∪ . . . Moreover,S(p) has exactly p+1 elements: (0, p),(1, p−1),. . . ,(p, 0). Verify that the function

11.3. INFINITE SETS

413

j : N2 → N that is defined by j(n, m) = 12 (n + m)(n + m + 1) + n enumerates
the pairs of the subsequent sets S(p).
Note that the function j enumerates N2 as follows:(0, 0), (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (0, 3), (1, 2), (2, 1), (3, 0), . . .
Look at all pairs (n, m) as the corresponding points in two-dimensional space.
Visualize j as a walk along these points.
Of course, there are many other “walks” along these points that prove this theorem.
E.g., (visualize!)(0, 0), (1, 0), (1, 1), (0, 1), (0, 2), (1, 2), (2, 2), (2, 1), (2, 0), (3, 0), (3, 1), . . .
Theorem 11.53 The set of positive rationals Q+ is countably infinite.
Proof. Identify a positive rational q ∈ Q+ with the pair (n, m) ∈ N2 for which q =nm and for which n and m are co-prime. Use Theorem 11.52 and Exercise 11.50.
Exercise 11.54 Show that Q is countably infinite.
Exercise 11.55 Show that N∗ is countably infinite.
Exercise 11.56 Show that a union of countably infinitely many countably infinite
sets is countably infinite.

Uncountable
Definition 11.57 (Less Power Than) A set A has power less than B if both A 
B and A 6∼ B.
Notation: A ≺ B.
Thus:
A ≺ B ⇐⇒ A  B ∧ A 6∼ B.
Example 11.58 {0, . . . , n − 1} ≺ N (Theorem 11.24);
N ≺ R (Theorem 11.60).
Warning. That A ≺ B implies but is not equivalent with: there exists a nonsurjective injection from A into B.

414

CHAPTER 11. FINITE AND INFINITE SETS

That A ≺ B means by definition: A  B and A 6∼ B. That A  B means that an
injection f : A → B exists. If, moreover, A 6∼ B, then no bijection exists between
A and B; in particular, the injection f cannot be surjective.
Counter-examples to the converse: (Example 11.20) the successor-function n 7−→
n + 1 is a non-surjective injection : N → N, but of course it is false that N ≺ N;
the identity function 1N is a non-surjective injection from N into Q, but we do nothave (Exercise 11.54), that N ≺ Q.
Definition 11.59 (Uncountable) A set A is uncountable in case N ≺ A.
The following is Cantor’s discovery.
Theorem 11.60 R is uncountable.
Proof. (i) That N  R is clear (Exercise 11.43.3, N ⊆ R). (ii) It must be shown,that no bijection h : N → R exists. In fact, there is no surjection from N to R.
That is:Claim. To every function h : N → R there exists a real r such hat r 6∈ ran(h).
Proof. Suppose that h : N → R. Write down the decimal expansion for every
real h(n): h(n) = pn + 0.r0n r1n r2n · · · , wherepn ∈ Z, pn 6 h(n) < pn + 1, and√decimals rin ∈ {0, 1, 2, . . . , 9}. (E.g., − 2 = −2 + 0, 15 · · · )For every n, choose a digit rn ∈ {0, 1, 2, . . . , 9} such that rn 6= rnn . The realr = 0, r0 r1 r2 · · · then differs from h(n) in its n-th decimal digit (n = 0, 1, 2, . . .).
However, even if pn = 0, this does not imply r 6= h(n). For, a real can have two
decimal notations. E.g., 0.5000 · · · = 0.4999 · · · . But, a tail of zeros vs. a tail ofnines is the only case for which this happens. So, this problem can be avoided if
we require rn to be different from 0 and 9.
Recall, that ℘(A) = {X | X ⊆ A} is the collection of all subsets of A (Definition 4.22, p. 132). The powerset operation produces in a direct way sets of greater
power.
In particular, not every uncountable set is equipollent with R.
Theorem 11.61 A ≺ ℘(A).
Proof. (i) The injection a 7→ {a} from A into ℘(A) shows that A  ℘(A).
(ii) To show that A 6∼ ℘(A), we exhibit, as in the previous proof, for every function
h : A → ℘(A), an element D ∈ ℘(A) − rng (h).

11.3. INFINITE SETS

415

Such an element can be simply described in this context: we take D = {a ∈ A |a 6∈ h(a)}.
If we would have that D ∈ rng (h), then D would be the value of some d ∈ A.
Now there are two cases: either d ∈ D, or d 6∈ D.
If d ∈ D, then, by definition of D, we would have that d 6∈ h(d) = D: contradiction. And if d 6∈ D, then, since D = h(d), we would have that d 6∈ h(d), andhence d ∈ D; again a contradiction.
Conclusion: D 6∈ rng (h).
Corollary 11.62

1. N ≺ ℘(N) ≺ ℘(℘(N)) ≺ · · · ,2. for every set A there exists a set B such that A ≺ B.
Exercise 11.63 Show:
1. A 6≺ A,
2. A  B ⇐⇒ A ≺ B ∨ A ∼ B,
3. A ≺ B ∧ B ∼ C =⇒ A ≺ C.
4. What is wrong with the following “proof ” for 2 ⇒ ?:Given is that A  B. Suppose that f : A → B is an injection.
(a) f is (by accident) surjective. Then A ∼ B.
(b) f is not surjective. Then A 6∼ B, hence A ≺ B.

Exercise 11.64 Show: if A is finite, then A ≺ N.
(The converse of this is true as well)
Exercise 11.65 Show: the real interval (0, 29 ] = {r ∈ R | 0 < r 6 92 } is uncountable.
Exercise 11.66 A is a set. Define h : A → ℘(A) by h(a) = {a}. Determine{a ∈ A | a 6∈ h(a)}.
Exercise 11.67 Show that N ≺ {0, 1}N.
Exercise 11.68* Show that N ≺ NN . (NN is the set of all functions : N → N.)

Hint. Produce, for every function ϕ : N → NN , a function f ∈ NN − rng (ϕ).

416

CHAPTER 11. FINITE AND INFINITE SETS

Exercise 11.69 Suppose that h : N → Q is surjective and that we apply theprocedure from the proof of Theorem 11.60 to h to produce a real r. Is r a rationalor not?

*Cantor-Bernstein Theorem
The following result is of fundamental importance in the theory of equipollence.
Theorem 11.70 (Cantor-Bernstein) A  B ∧ B  A =⇒ A ∼ B.
The proof of this theorem is delegated to the exercises.
Note that we have two examples of uncountable sets: R (Theorem 11.60) and ℘(N)
(Theorem 11.61). It turns out that these sets are of the same magnitude.
Theorem 11.71 R ∼ ℘(N).
Proof. We show that R  ℘(Q) ∼ ℘(N) ∼ {0, 1}N  R. From this, the CantorBernstein Theorem produces the required conclusion.
1. R  ℘(Q). The function r 7−→ {q ∈ Q | q < r} is an injection from R into℘(Q).
2. ℘(Q) ∼ ℘(N). Choose a bijection h : Q → N (Exercise 11.54, p. 413). Now
X 7−→ h[X] is a bijection between ℘(Q) and ℘(N) (Exercise 11.26, p. 409).
3. ℘(N) ∼ {0, 1}N. Cf. Exercise 11.27.

4. {0, 1}N  R. Associate with h : N → {0, 1} the real (in the interval [0, 19 ]) thathas decimal expansion 0, h(0)h(1)h(2) · · · .
Continuum Problem and Hypothesis.
Since N ≺ R, it is tempting to ask whether sets A exist such that N ≺ A ≺ R.
(If so, such an A exists for which A ⊆ R.) This question is Cantor’s Continuum
Problem. (The continuum is the set R.)
Cantor’s Continuum Hypothesis asserts that such a set does not exist.
The usual set-theoretic axioms cannot answer the question. Gödel proved in 1938
that the axioms cannot show that the Continuum Hypothesis is false. Cohen
proved in 1963 that the axioms cannot show that the Continuum Hypothesis istrue. Indeed, as far as the axioms are concerned, the power of R can be unimaginably big and there can be arbitrarily many sets A, B, C, . . . ⊆ R such thatN ≺ A ≺ B ≺ C ≺ · · · ≺ R.

11.3. INFINITE SETS

417

Example 11.72 We show that [0, 1] ∼ [0, 1). Although these intervals differ in
only one element, establishing a bijection between them is far from trivial. (Thereader who doubts this should give it a try before reading on.) Let f : [0, 1] →[0, 1) be an arbitrary injection; say, f (x) = 12 x. Consider the following function
h : [0, 1] → [0, 1): h sends 1 to f (1) = 21 , 12 to f ( 21 ) = 14 , 14 to f ( 14 ) = 81 etc.; onother arguments r 6= 2−n in [0, 1], you let h(r) = r.
Verify that h is bijective. To check injectivity, let r 6= s ∈ [0, 1]. If neither of r, s
is of the form 2−n , then by definition of h we have h(r) = r 6= s = h(s). If bothare of the form 2−n , say r = 2−i and s = 2−j with i 6= j, then, by definitionof h, h(r) = 2−i−1 6= 2−j−1 = h(s). If one of r, s is of the form 2−n and theother is not, then one of h(r), h(s) is of the form 2−n and the other is not, so againh(r) 6= h(s). This shows that h is injective. For surjectivity, let r ∈ [0, 1). If r is
not of the form 2−n then h(r) = r, so there is an s ∈ [0, 1] with h(s) = r. If r isof the form 2−n , say r = 2−i with i > 0, then h(2−i+1 ) = 2−i = r. Again, there
is an s ∈ [0, 1] with h(s) = r.
Lemma 11.73 If A  B ⊆ A, then A ∼ B.
Exercise 11.74* Prove the Lemma.
Hint. Generalize the solution for Example 11.72.
Exercise 11.75* Prove Theorem 11.70.
Hint. Apply Lemma 11.73 to the composition of the two functions that are given.

Exercise 11.76 Show the following variations on the fact that [0, 1] ∼ [0, 1):
1. [0, 1] ∼ [0, 32 ),2. {(x, y) | x2 + y 2 6 1} ∼ {(x, y) | x2 + y 2 < 1},
3. {(x, y) | x2 + y 2 6 1} ∼ {(x, y) | |x|, |y| < 21 }.

Exercise 11.77* Suppose that A ⊆ R is finite or countably infinite. Show thatR − A is uncountable. Can you show that (R − A) ∼ R?
Exercise 11.78 Show that (R − Q) ∼ R.

418

CHAPTER 11. FINITE AND INFINITE SETS

11.4 Cantor’s World Implemented
The following program illustrates that N2 is denumerable:

natpairs = [(x,z-x) | z <- [0..], x <- [0..z]]This gives:
FAIS> natpairs[(0,0),(0,1),(1,0),(0,2),(1,1),(2,0),(0,3),(1,2),(2,1),(3,0),(0,4),(1,3),(2,2),(3,1),(4,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0){Interrupted!}

Exercise 11.79 Implement the function pair :: (Int,Int) -> Int that is theinverse of natpairs. It should hold for all natural numbers n thatpair (natpairs !! n) = n.

Exercise 11.80 Implement a function natstar :: [[Int]] to enumerate N ∗(cf. Exercise 11.55).
The following code illustrates that Q is denumerable:

rationals = [ (n,m) | (n,m) <- natpairs, m /= 0, gcd n m == 1 ]This gives:
FAIS> rationals[(0,1),(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(2,3),(3,2),(4,1),(1,5),(5,1),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(1,7),(3,5),(5,3),(7,1),(1,8),(2,7),(4,5),(5,4),(7,2),(8,1),(1,9),(3,7),(7,3),(9,1),(1,10),(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2),(10,1),(1,11),(5,7),(7,5),(11,1),(1,12),(2,11),(3,10),(4,9),(5,8),(6,7),(7,6),(8,5),(9,4),(10,3),(11,2),(12,1),(1,13),(3,11),(5,9),(9,5),(11,3),(13,1),(1,14),

11.4. CANTOR’S WORLD IMPLEMENTED

419(2,13),(4,11),(7,8),(8,7),(11,4),(13,2),(14,1),(1,15),(3,13),(5,11),(7,9),(9,7),(11,5),(13,3),(15,1),(1,16),(2,15),(3,14),(4,13),(5,12),(6,11),(7,10),(8,9),(9,8),(10,7),(11,6),(12,5),(13,4),(14,3),(15,2),(16,1),(1,17),(5,13),(7,11),(11,7),(13,5),(17,1),(1,18),(2,17),(3,16),(4,15),(5,14),(6,13),(7,12),(8,11),(9,10),(10,9),(11,8),(12,7),(13,6),(14,5),(15,4),(16,3),(17,2),(18,1),(1,19),(3,17),(7,13),(9,11),(11,9),(13,7),(17,3),(19,1),(1,20),(2,19),(4,17),(5,16),(8,13),(10,11),(11,10),(13,8),(16,5){Interrupted!}

The following code illustrates that {True, False}N is not denumerable:

diagonal :: (Integer -> [Bool]) -> Integer -> Booldiagonal f n = not ((f n)!!(fromInteger n))
f :: Integer -> [Bool]f 0 = cycle [False]f (n+1) = True : f n

Now [ f n | n <- [0..] ] is a list of streams of booleans, all different, anddiagonal f is a new stream of booleans, different from all members of[ f n | n <- [0..] ].

Here is an illustration for initial segments of the lists:
FAIS> [ take 11 (f n) | n <- [0..10] ][[False,False,False,False,False,False,False,False,False,False,False],[True,False,False,False,False,False,False,False,False,False,False],[True,True,False,False,False,False,False,False,False,False,False],[True,True,True,False,False,False,False,False,False,False,False],[True,True,True,True,False,False,False,False,False,False,False],[True,True,True,True,True,False,False,False,False,False,False],[True,True,True,True,True,True,False,False,False,False,False],[True,True,True,True,True,True,True,False,False,False,False],[True,True,True,True,True,True,True,True,False,False,False],[True,True,True,True,True,True,True,True,True,False,False],[True,True,True,True,True,True,True,True,True,True,False]]
FAIS> [ diagonal f n | n <- [0..10] ][True,True,True,True,True,True,True,True,True,True,True]

420

CHAPTER 11. FINITE AND INFINITE SETS

11.5 *Cardinal Numbers
By Theorem 11.21, equipollence is an equivalence on the collection of all sets.
Definition 11.81 (Cardinal Number) A cardinal number is an equivalence class
w.r.t. equipollence.
|A| denotes the cardinal number of the set A modulo ∼.
The following is immediate (cf. Lemma 5.80, p. 193):
Lemma 11.82 |A| = |B| ⇐⇒ A ∼ B.
Usually, the cardinal number |{0, . . . , n − 1}| of the n-element sets is identified
with the natural number n.

Aleph-zero. ℵ0 = |N|.1
The concept of a cardinal number can be considered as a generalisation of that ofa natural number.
It is possible to generalize the definitions of addition, multiplication and exponentiation to cardinal numbers, and to prove natural laws that generalize those for thenatural numbers.
The operations are defined as follows. |A|+|B| = |A∪B| (provided that A∩B =∅: only if A and B are disjoint sets of n resp. m elements does it follow that theirunion has n + m elements), |A| × |B| = |A × B| and |A||B| = |AB |.

Thus, by Theorem 11.71, we have that |R| = 2ℵ0 .

The cardinal ℵ0 is the starting point of an infinite series of cardinal numbers called
alephs: ℵ0 < ℵ1 < ℵ2 < · · · < ℵω < · · · ; (ℵω is the smallest cardinal bigger
than every ℵn ).
Using cardinals sometimes makes for amazingly compact proofs. An example isthe following theorem.
Theorem 11.83 R × R ∼ R, i.e.: there are as many points in the plane as on aline.
1 ℵ (aleph) is the first letter of the Hebrew alphabet.

11.5. CARDINAL NUMBERS

421

Proof.
|R × R| = |R| × |R|= 2 ℵ0 × 2 ℵ0
= 2ℵ0 +ℵ0
= 2 ℵ0= |R|.
The third equality uses the rule np × nq = np+q , the fourth that ℵ0 + ℵ0 = ℵ0 , cf.
Exercise 11.51, p. 412.

*Further Exercises
Exercise 11.84 Suppose that A1 ∼ A2 and B1 ∼ B2 . Show:
1. if A1 ∩ B1 = A2 ∩ B2 = ∅, then A1 ∪ B1 ∼ A2 ∪ B2 ,2. A1 × B1 ∼ A2 × B2 ,

|B |

|B |

B2
B1
1
3.* AB| = |A1 | 1 = |A2 | 2 =1 ∼ A2 (Hint: it does not do to say |A1
B2|A2 |, for we don’t have a rule of exponentiation for cardinal numbers as
B2
1
yet. Instead, show how to establish a bijection between AB
1 and A2 .)Exercise 11.85 Suppose that A1  A2 and B1  B2 . Show:
1. if A2 ∩ B2 = ∅, then A1 ∪ B1  A2 ∪ B2 ,2. A1 × B1  A2 × B2 ,3. ℘(A1 )  ℘(A2 ),B2
1
4.* if A2 6= ∅, then AB
1  A2 (Hint: Use the fact that, since A2 6= ∅, you
can pick a ∈ A2 for the definition of the injection that you need.)

Exercise 11.86 Give counter-examples to the following implications:
1. A1 ≺ A2 ⇒ A1 ∪ B ≺ A2 ∪ B (A1 ∩ B = A2 ∩ B = ∅),422

CHAPTER 11. FINITE AND INFINITE SETS

2. A1 ≺ A2 ⇒ A1 × B ≺ A2 × B,
B
3. A1 ≺ A2 ⇒ AB
1 ≺ A2 ,4. A1 ≺ A2 ⇒ B A1 ≺ B A2 .

Exercise 11.87 Show:
1. if B ∩ C = ∅, then AB∪C ∼ AB × AC ,2. (A × B)C ∼ AC × B C
3.* (AB )C ∼ AB×C . (Hint: use the currying operation.)

Exercise 11.88 Show (n > 1):
1. {0, 1}N ∼ {0, . . . , n}N ∼ NN ∼ RN ∼ R,
2. {0, 1}R ∼ {0, . . . , n}R ∼ NR ∼ RR ∼ (℘(R))R ∼ (RR )R .

Exercise 11.89 Show, for all n ∈ N+ : Nn ∼ N (n.b.: NN 6∼ N) and Rn ∼ R (n.b.:
RN ∼ R).
Exercise 11.90 Show: {(x, y) ∈ R2 | x2 + y 2 6 1 ∧ y > 0} ∼ {(x, y) ∈ R2 |1 < y 6 2}.
Exercise 11.91 Show: if A is infinite and B finite, then (A − B) ∪ (B − A) ∼ A.

The Greek Alphabet
Mathematicians are in constant need of symbols, and most of them are very fondof Greek letters. Since this book might be your first encounter with this new set ofsymbols, we list the Greek alphabet below.
name
alpha
beta
gamma
delta
epsilon
zeta
eta
theta
iota
kappa
lambda
mu
nu
xi
pi
rho
sigma
tau
upsilon
phi
chi
psi
omega

lower case
α
β
γ
δ
ε
ζ
η
θ
ι
κ
λ
µ
ν
ξ
π
ρ
σ
τ
υ
ϕ
χ
ψ
ω

423

upper case

Γ
∆

Θ

Λ

Ξ
Π
Σ
Υ
Φ
Ψ
Ω

424

Bibliography[AHV95]

S Abiteboul, R. Hull, and V. Vianu. Foundations of Databases. Addison Wesley, 1995.

[Bab61]

C. Babbage. On the Principles and Development of the Calculator.
Dover, 1961. Edited and with an introduction by P. Morrison and E.
Morrison.

[Bab94]

C. Babbage. Passages from the Life of a Philosopher. Rutgers University Press and IEEE-Press, New Brunswick, New Jersey and Piscataway, New Jersey, 1994. Edited with a new introduction by Martin
Campbell-Kelly. Originally published 1864.

[Bal91]V. K. Balakrishnan. Introductory Discrete Mathematics. Dover, 1991.

[Bar84]

H. Barendregt. The Lambda Calculus: Its Syntax and Semantics (2nded.). North-Holland, Amsterdam, 1984.

[Bir98]R. Bird. Introduction to Functional Programming Using Haskell.
Prentice Hall, 1998.

[BM96]

J. Barwise and L. Moss. Vicious Circles: On the Mathematics ofNon-wellfounded Phenomena. CSLI Publications, 1996.

[Bry93]

V. Bryant. Yet another introduction to analysis. Cambridge University
Press, 1993.

[Bur98]

Stanley N. Burris. Logic for Mathematics and Computer Science.
Prentice Hall, 1998.

[CG96]

J.H. Conway and R.K. Guy. The Book of Numbers. Springer, 1996.

[CR78]

R. Courant and H. Robbins. What is Mathematics? An Elementary
Approach to Ideas and Methods. Oxford University Press, Oxford,
1978.
425

426

BIBLIOGRAPHY[CrbIS96]

R. Courant and H. Robbins (revised by I. Stewart). What is Mathematics? An Elementary Approach to Ideas and Methods (Second
Edition). Oxford University Press, Oxford, 1996.

[Doe96]

H.C. Doets. Wijzer in Wiskunde. CWI, Amsterdam, 1996. Lecturenotes in Dutch.

[DP02]

B.A. Davey and H.A. Priestley. Introduction to Lattices and Order(Second Edition). Cambridge University Press, Cambridge, 2002.
First edition: 1990.

[DvDdS78] K. Doets, D. van Dalen, and H. de Swart. Sets: Naive, Axiomatic,and Applied. Pergamon Press, Oxford, 1978.
[Ecc97]

P. J. Eccles. An Introduction to Mathematical Reasoning. Cambridge
University Press, 1997.

[EFT94]

H.-D. Ebbinghaus, J. Flum, and W. Thomas. Mathematical Logic.
Springer-Verlag, Berlin, 1994. Second Edition.

[Euc56]

Euclid. The Thirteen Books of the Elements, with Introduction andCommentary by Sir Thomas L. Heath. Dover, 1956.

[Fok00]W. Fokkink. Introduction to Process Algebra. Springer, 2000.

[GKP89]R.L. Graham, D.E. Knuth, and O. Patashnik. Concrete Mathematics.
Addison Wesley, Reading, Mass, 1989.

[Har87]

D. Harel. Algorithmics: The Spirit of Computing. Addison-Wesley,
1987.

[HFP96]

P. Hudak, J. Fasel, and J. Peterson. A gentle introduction to Haskell.
Technical report, Yale University, 1996. Available from the Haskell
homepage: http://www.haskell.org.

[Hin97]

J. Roger Hindley. Basic Simple Type Theory. Cambridge University
Press, 1997.

[HO00]

C. Hall and J. O’Donnell. Discrete Mathematics Using A Computer.
Springer, 2000.

[HR00]

M. Huth and M. Ryan. Logic in Computer Science: Modelling andReasoning about Systems. Cambridge University Press, 2000.

[HT]

The Haskell Team. The Haskell homepage. http://www.haskell.
org.

BIBLIOGRAPHY

427[Hud00]P. Hudak. The Haskell School of Expression: Learning Functional
Programming Through Multimedia. Cambridge University Press,
2000.

[Jon03]

S. Peyton Jones, editor. Haskell 98 Language and Libraries; The
Revised Report. Cambridge University Press, 2003.

[JR+ ]Mark P. Jones, Alastair Reid, et al. The Hugs98 user manual. http://www.haskell.org/hugs/.

[Kar97]

J. Karczmarczuk. Generating power of lazy semantics. TheoreticalComputer Science, 187, 1997.

[Knu92]

D.E. Knuth. Literate Programming. CSLI Lecture Notes, no. 27.
CSLI, Stanford, 1992.

[Lar34]

D. Lardner. Babbage’s calculating engine. Edinburgh Review, 1834.

[McI99]

M.D. McIlroy. Power series, power serious. Journal of Functional
Programming, 9:323–335, 1999.

[McI00]M.D. McIlroy. The music of streams. Elsevier Preprints, 2000.

[Mil99]R. Milner. Communicating and Mobile Systems: the π Calculus.
Cambridge University Press, 1999.

[Ore88]

O. Ore. Number Theory and its History. Dover, 1988.

[Pol57]

G. Polya. How to Solve It. A New Aspect of Mathematical Method.
Princeton University Press, Princeton, 1957.

[RL99]

F. Rabhi and G. Lapalme. Algorithms: a Functional Programming
Approach. Addison-Wesley, 1999.

[Rus67]

B. Russell. Letter to Frege. In J. van Heijenoord, editor, From Frege
to Gödel, pages 124–125. Harvard University Press, 1967.

[Rut00]J.J.M.M. Rutten. Behavioural differential equations: a coinductive
calculus of streams, automata, and power series. Report SEN-R0023,
CWI, 2000. Available at URL: www.cwi.nl. To appear in TheoreticalComputer Science.

[SKS01]

A. Silberschatz, H.F. Korth, and S. Sudarshan. Database System Concepts (4th edition). McGraw-Hill, 2001.

[Tho99]

S. Thompson. Haskell: the craft of functional programming (second
edition). Addison Wesley, 1999.

428[Vel94]

BIBLIOGRAPHY
D.J. Velleman. How to Prove It. A Structured Approach. Cambridge
University Press, Cambridge, 1994.

Index(a, b), 136:=, 32:≡, 61
A/R, 194
Ac , 131
R−1 , 163
∆A , 163∩, 127∪, 127dom (R), 162∅, 126∧, 29, 31∨, 29, 32⇔, 29, 35¬, 29, 30⊕.n, 247℘(X), 132⇒, 29, 33
ran(R), 162{a}, 125{x ∈ A | P }, 118{x | P }, 118|Pa |, 192nk=1 ak , 54, 241&&, 17, 31, 141||, 32(.), 69(.*), 354
(op x), 21(x op), 21(x1,x2), 139+, 6--, 122->, 9, 142., 69, 222/, 15/=, 5, 124
::, 9
:l, 2
:r, 5
:t, 9
<, 124<+>, 36<=, 124<=, 13<=>, 36=, 5==, 5, 124==>, 33<, 8>=, 8@, 175[a], 18[n..m], 120
Zn , 195\\, 150\, 46_, 141nk , 346⊥, 46, 84
dom (f ), 206≡n , 189
λx.t, 58
(mod n), 189
429

430
ran(f
n ), 206
k , 197>, 46^, 6, 119d|n, 56e, 221n!, 213|, 6, 246∼, 407‘abc’-formula, 59
absReal, 207
add, 267
addElem, 151
adjustWith, 340
Alexander the Great, v
algebraof sets, 127
all, 68
and, 42, 268
antisymmetric relation, 167any, 68
approx, 375
approximate, 319
apprx, 319
Apt, K.R., ix
arbitrary object, 92
assignment statement, 24
asymmetric relation, 167
average, 15
axiomatics, 114
Babbage, C., 334
backsubst, 342
ballgame, 406
bell, 198
Bell numbers, 198
Benthem, J. van, ix
Bergstra, J., ix
bijection, 218
bijective, 219
bijectivePairs, 219

INDEX
binary, 288
binding power of operators, 33
binomial theorem, 346, 347
BinTree, 257
bisimulation, 380
black hole, 276
Bool, 8, 30
Boolean function, 39
brackets, 215
Brouwer, L.E.J, 29
Brunekreef, J., ix
Cantor’s continuum hypothesis, 416
Cantor’s theorem, 414
Cantor, G., 114
Cantor-Bernstein theorem, 416
cardinal number, 420
case, 142cat, 265
Catalan numbers, 254, 394
chain, 169
characteristic function, 182
chr, 228
Church, A., 1, 64
class, 124class, 227
clock, 367
closed form, 212
closures of a relation, 170
co-domainfunction, 209
co-image, 214
co-prime, 291
coprime, 291
Cohen, P.J., 416
coImage, 215
coImagePairs, 216
coinduction, 382
cols, 339
comp, 223, 389
compare, 141

INDEX
comparison property, 169
completeness, 168
complex numbers, 320
complR, 178
composition
function –, 69, 222
confluence, 404
congruence, 236
conjunction, 31
constructor, 12constructor identifier, 12
continue, 335
continuity, 66, 315
continuum problem, 416
contradiction, 48
contraposition, 35
converse, 35
conversion, 15
convolution, 354
Coquand, T., ix
corecursion, 362
corecursive definition, 362
countable, 412
curry, 183
Curry, H.B., 1, 183
curry3, 210
CWI, ix
data, 139, 257
data, 30, 145
database query, 145
decExpand, 307
decForm, 308
decodeFloat, 311
deduction rule, 78
default, 362
delete, 149deleteSet, 156delta, 355
denumerable, 412
deriv, 358

431
deriving, 140
destructive assignment, 24
diagonal, 419
difLists, 334
difs, 332
disjunction, 32
display, 158div, 20
divergence, 122
divides, 5, 183
domain, 162
Double, 17, 311
e2o, 397
echelon, 341
echelon matrix form, 338
elem, 123
elem, 151
elem’, 150
elemIndex, 308
eliminate, 341
else, 230
emptySet, 156
encodeFloat, 312
Enum, 227
enum_2, 107
EQ, 141
Eq, 124, 246
eq, 184
eq1, 298
equalSize, 190
equation guarding, 6equational reasoning, 24
equipollent, 406
equiv2listpart, 199
equiv2part, 199
equivalence, 35
equivalence, 188
equivalence’, 188
Eratosthenes
sieve of —, 105

INDEX

432
error, 8, 230
Erven, T. van, ix
Euclid, 103, 290
Euclid’s GCD algorithm, 290
Euler, L., 104even, 81, 222
evens, 81, 362
evens1, 118
evens2, 119every, 69
exception handling, 230
exclaim, 252
exp, 221
expn, 253
exponent, 312
fac, 213fac’, 213
False, 4, 30
fasterprimes, 107
fct2equiv, 232
fct2list, 207
fct2listpart, 234
Fermat, P. de, 104fib, 254
fib’, 254
Fibonacci numbers, 253, 254, 363
field, 301
filter, 22
fixity declaration, 33
flip, 184
Float, 15, 311
floatDigits, 312
Floating, 311
floatRadix, 311
Fokkink, W., ix
foldl, 269
foldn, 252
foldr, 266
foldr1, 268
foldT, 262

forall, 12Fraenkel, A., 114
fromEnum, 227
fromInt, 15
fromTower, 281
fst, 139
function, 205
domain, 206
function composition, 69, 222
fundamental theoremof algebra, 320of arithmetic, 293
Gödel, K., 416
Gaussian elimination, 337
gcd, 291
genDifs, 335
genMatrix, 339
GIMPS, 108
GNU’s Not Unix, 362
Goldreyer, D., 28
Goris, E., ix
greatest common divisor
definition, 290
Euclid’s algorithm for —, 290
properties, 405
GT, 141
gt1, 299
guard, 7
guarded equation, 6
Haan, R. de, ix
halting problem, 121
hanoi, 276
hanoi’, 280
hanoiCount, 280
Haskell, 1
head, 142
Heman, S., ix
hex, 289
Hoogland, E., ix

INDEXid, 207
idR, 175
Iemhoff, R., ix
if then else, 230
iff, 30
ILLC, ix
image, 214
image, 215
imagePairs, 216
implication, 33
import, 72
in, 15
induction, 239, 240, 400
strong, 401
infinityof primes, 103
infix, 4
infix, 33
infix notation, 21
infixl, 156
infixr, 156
init, 143
injection, 218
injective, 219
injectivePairs, 219
injs, 221
inorder tree traversal, 262
inR, 178
insertSet, 156
inSet, 155
instanceof a type class, 124
instance, 140
Int, 11int, 388
Integer, 9, 11
integers, 294
Integral, 11
integral rational functions, 331
intersect, 150
intransitive relation, 168

433
intToDigit, 288, 289
intuitionism, 29
inverse, 184
invR, 178
irrational numbers, 311
irrationality√of 2, 309
irreflexive relation, 166irreflR, 179
isAlpha, 232
isEmpty, 156
iterate, 313, 336, 363
Jongh, D. de, ix
Just, 231
Kaldewaij, A., ix
labeled transition system, 365
lambda abstraction, 46, 58, 145, 211
lambda term, 58
last, 143lawassociativity, 46, 247, 249
commutativity, 46, 247, 249
contraposition, 45
DeMorgan, 46, 132
distribution, 46
distributivity, 249
dominance, 48
double negation, 45
excluded middle, 48
idempotence, 45
identity, 48
lazy list, 23, 105
lazy pattern, 373LD, 4ldp, 23
ldpf, 23
leaf tree, 263
LeafTree, 263
left triangular matrix form, 338

INDEX434
len, 265
length, 15leq, 250
leq1, 299
lessEq, 184let, 15
lexicographical order, 141
limit, 315
linear relation, 169
list, 16, 139
list comprehension, 42, 54, 118
list2fct, 207
list2set, 156
listpart2equiv, 199
listPartition, 199
listRange, 208
lists, 139
listValues, 208ln, 267ln’, 272
load Haskell file, 2
logBase, 281
LT, 141
Lucas numbers, 394
Main>, 5mantissa, 311
map, 21, 151, 272
mapLT, 264
mapR, 264
mapT, 261
Matrix, 338
matrix, 338
maxBound, 208
Maybe, 231
maybe, 231
mechanic’s rule, 313
mechanics, 313
mechanicsRule, 313
Menaechmus, v
Mersenne, M., 104min, 13
minBound, 208
mkStdGen, 366
mlt, 267
mnmInt, 13
mod, modulo, 190
module, 2
modulo, 190
Modus Ponens, 80
mult, 252
mult1, 298
mySqrt, 319
n-tuples, 139
Napier’s number, 221, 395
natpairs, 418
natstar, 418
Natural, 246
natural logarithm, 221
natural number, 246
naturals, 118, 362
necessary condition, 35
negate, 222
negation, 30
Newman’s Lemma, 404
Newman, B., 28
Newton’s method, 313
Newton, I., 346
newtype, 153
next, 335
nextD, 335
nondeterminism, 365
not, 30
notElem, 151
Nothing, 231
Nualláin, B. Ó, ixnub, 144
null, 143
Num, 124o2e, 396
odd, 222

INDEX
odds, 362
odds1, 119
oddsFrom3, 107of, 142
one-to-one, 218
ones, 125
ones, 362
onto, 218
Oostrom, V. van, ix
open problems, 109, 111
operation, 236
operator precedence, 42
or, 268
Ord, 124
Ord, 141ord, 228
order
partial —, 168
strict partial —, 168
total —, 169
Ordering, 141
otherwise, 7
overloading, 125
p2fct, 343
pair, 418
pairs, 136
paradoxhalting —, 121
Russell —, 120
part2error, 231partial functions, 229
partial order, 168
Pascal’s triangle, 348
Pascal, B., 348
pattern matching, 13, 143, 250
perms, 221
Platonism, 28
plus, 247
plus, 252
plus1, 298

435
po-set reflection, 195
polynomial, 320
polynomials, 331
polynomials and coefficient lists, 344
Ponse, A., ix
postorder tree traversal, 262
power series, 387
powerList, 151
powerSet, 156pre, 253
pre-order, 168
pred, 227
prefix, 4
prefix, 17
prefix notation, 21
Prelude>, 2Prelude, 26preorder tree traversal, 262
primCompAux, 142
prime, 23prime factorization algorithm, 19
prime’’, 165
prime0, 8
primesdefinition, 60
Mersenne —, 105
primes, 106
primes’, 364
primes0, 22
primes1, 23
principle
comprehension —, 114
minimality —, 402
Process, 366
productof sets, 136
product, 213
propositional function, 39
ptd, 369
quasi-order, 168

INDEX436
quotient, 194
quotient, 251
quotRem, 286
raccess, 194
random numbers, 366
random streams, 366
Random.hs, 361
randomInts, 366
randomRs, 366
ranPairs, 208
Rational, 341
rationals, 299
countability of —, 413
rationals, 418
rclosR, 182
reals, 310
uncountability of —, 414
recip, 313
recurrence, 212
recursion, 246
recursive definition, 7, 246
reduce1, 298
reflect, 264
reflexive closure, 171
reflexive relation, 166
reflexive transitive closure, 171
reflR, 178
relationantisymmetric —, 167
asymmetric —, 167
between, 162
domain, 162
from—to, 162
intransitive —, 168
irreflexive —, 166
linear —, 169
range, 162reflexive —, 166
symmetric —, 166transitive —, 167

relatively prime, 291
reload Haskell file, 5rem, 5remainder, 251removeFst, 14
reserved keywords, 12
restrict, 214
restrictPairs, 214
restrictR, 182rev, 269
rev’, 270
rev1, 272
Rodenburg, P., ix
rose tree, 264
rows, 339
royal road to mathematics, v
Russell, B., 57, 120
Rutten, J., ix
sclosR, 182
sections, 21
Set a, 153
Show, 246
showDigits, 289
showSet, 155
sieve, 106, 364
sieve of Eratosthenes, 105sieve’, 364
significand, 312
sin, 221
singleton, 125
small_squares1, 119
small_squares2, 120
Smullyan’s ball game, 405
Smullyan, R., 405
snd, 139
solveQdr, 59
solveSeq, 342
solving quadratic equations, 59
some, 69
soundness, 168

INDEX
space leak, 276
split, 263
splitAt, 308
splitList, 144
sqrt, 59, 221
sqrtM, 313
srtInts, 14
start, 366
stirling, 198
Stirling set numbers, 198
stream, 362
stream bisimulation, 383
strict partial order, 168
String, 16
String, 16
stringCompare, 232
sub domain, 53
substitution principle, 48
subtr, 251
subtr, 253
subtr1, 298
succ, 227
successor, 246
sufficient condition, 35
sum, 241
sum, 15, 54, 241
sumCubes, 245
sumCubes’, 245
sumEvens, 242
sumEvens’, 242
sumInts, 242
sumOdds, 241
sumOdds’, 241
sumSquares, 243
sumSquares’, 243surjection, 218
surjective, 219
surjectivePairs, 219
Swaen, M., ix
symmetric closure, 171
symmetric difference, 132437
symmetric relation, 166
symR, 179
tail, 142
take, 107
takeWhile, 313
tan, 221
tclosR, 182
Terlouw, J., ix
theFibs, 363
then, 230
theNats, 363
theNats1, 363
theOdds, 363
theOnes, 363
toBase, 289
toEnum, 227
total order, 169
totalR, 178
toTower, 281
tower of Hanoi, 273
transClosure’, 187
transition system, 365transitive closure, 171
transitive relation, 167
transR, 179
treeleaf –, 263
rose —, 264
tree traversal, 262
trivially true statements, 34
Tromp, J., ix
True, 4, 30
truncate, 281
truth function, 39
Turing, A., 64
type, 8, 17, 53, 121, 140
type, 145
type conversion, 15
type declaration, 9type judgment, 10

438
type variables, 18
uncountable, 413
uncurry, 183
uncurry3, 210
undefined, 122
undefined, 375
union, 150
Van Benthem, J., ix
variable identifier, 12vending, 367
Venema, Y., ix
Visser, A., ix
Vries, F.J. de, ix
weak confluence, 404
Wehner, S., ix
well-founded, 245, 403
where, 2, 14
Who is Afraid of Red, Yellow andBlue, 28
wild card, 12
wildcard, 141z, 353Zermelo, E., 114
zipWith, 339, 363

INDEX