importance of primitive recursion

importance of primitive recursion

In this entry, we show that the

operation

of

primitive recursion

is an important ingredient in the

definition

of

primitive recursive functions

. Very basic

arithmetic functions

, such as

addition

, depend on primitive recursion in an essential way. One can not go very far without it.

One way to prove this is to come up with a

property

on those

functions

which are

derivable

from the initial functions using

functional

composition

alone, and then find a primitive recursive function that fails this property. Thus, we begin with the following definition:

Definition

. Let

𝒫

⁢

ℛ

0

be the smallest subset of the set

𝒫

⁢

ℛ

of primitive recursive functions such that

1.

(initial functions):

z

,

s

,

p

m

k

∈

𝒫

⁢

ℛ

0

for all

positive integers

m

,

k

with

m

≤

k

.

2.

(

closure

under functional composition): if

f

,

g

1

,

…

,

g

m

∈

𝒫

⁢

ℛ

0

with

f

m

-ary and

g

i

n

-ary, then the

n

-ary function

h

:=

f

⁢

(

g

1

,

…

,

g

m

)

∈

𝒫

⁢

ℛ

0

as well.

From the definition above, we observe that any function

f

⁢

(

x

1

,

…

,

x

n

)

obtained without the use of the successor function

s

must be

bounded

by

max

⁡

{

x

1

,

…

,

x

n

}

. Now, since one application of

s

increments the input by

1

, a finite applications of

s

in the construction of

f

results in

f

being bounded by

max

⁡

{

x

1

,

…

,

x

n

}

plus some fixed

integer

. This is the property we seek, and we prove it formally:

Proposition 1

.

For any

f

∈

P

⁢

R

0

with arity

n

, the following condition is satisfied:

(*) there is an integer

r

≥

0

such that

f

⁢

(

x

1

,

…

,

x

n

)

≤

max

⁡

{

x

1

,

…

,

x

n

}

+

r

.

Proof.

For each integer

n

≥

0

, let

𝒳

n

be the subset of

𝒫

⁢

ℛ

0

that contains all the initial functions, as well as functions derivable from the initial functions by applying functional composition at most

n

times. The following facts are clear:

1.

𝒳

0

is the set of all initial functions,

2.

𝒳

n

⊆

𝒳

n

+

1

,

3.

and

𝒫

⁢

ℛ

0

=

⋃

{

𝒳

n

∣

n

∈

ℕ

}

, where

ℕ

contains

0

in this

context

.

We prove our assertion by

induction

on

n

. In fact, we will prove a stronger assertion:

in the conditon (*) above,

r

=

n

+

1

can be picked for every function in

𝒳

n

.

•

For

n

=

0

, it is clear that all initial functions have the desired property:

r

=

0

for

z

and

p

m

k

, and

r

=

1

for

s

. In either case, we can pick

r

=

1

.

•

Suppose now that we can pick

r

=

i

+

1

for any function in

𝒳

i

.

•

Let

f

be an

k

-ary function in

𝒳

i

+

1

. If

f

∈

𝒳

i

, then

r

=

i

+

1

can be picked by

assumption

. Otherwise,

f

=

h

⁢

(

g

1

,

…

,

g

m

)

, where

h

is

m

-ary, and each

g

j

is

k

-ary. Furthermore,

g

j

∈

𝒳

i

j

, each

i

j

≤

i

, and

h

∈

𝒳

t

, where

t

=

i

-

max

⁡

{

i

1

,

…

,

i

m

}

. Then, by assumption

g

j

⁢

(

x

1

,

…

,

x

k

)

≤

x

+

i

j

+

1

≤

x

+

ℓ

+

1

for all

j

∈

{

1

,

…

,

m

}

, where

x

=

max

⁡

{

x

1

,

…

,

x

k

}

and

ℓ

=

max

⁡

{

i

1

,

…

,

i

m

}

. As a result,

f

⁢

(

x

1

,

…

,

x

k

)

=

h

⁢

(

g

1

⁢

(

x

1

,

…

,

x

k

)

,

…

,

g

m

⁢

(

x

1

,

…

,

x

k

)

)

≤

max

⁡

{

g

j

⁢

(

x

1

,

…

,

x

k

)

∣

j

=

1

,

…

,

m

}

+

t

+

1

≤

(

x

+

ℓ

+

1

)

+

t

+

1

=

x

+

(

i

+

1

)

+

1

.

This proves the stronger assertion.

Now, if we pick any

f

∈

𝒫

⁢

ℛ

0

, then

f

∈

𝒳

m

for some

m

by the third fact above. The proof is now

complete

by setting

r

=

m

+

1

.
∎

Now, we show the following:

Proposition 2

.

𝒫

⁢

ℛ

0

⊂

𝒫

⁢

ℛ

.

Proof.

We prove that

add

⁡

(

x

,

y

)

=

x

+

y

does not have property (*) above. But if it does, then

add

⁡

(

x

,

y

)

≤

max

⁡

{

x

,

y

}

+

r

for some non-negative integer

r

. Then, for an arbitrary non-negative integer

n

,

2

⁢

n

=

add

⁡

(

n

,

n

)

≤

max

⁡

{

n

,

n

}

+

r

=

n

+

r

, so that

n

≤

r

, a

contradiction

(pick

n

=

r

+

1

)! ∎

In fact, the above proof shows the function

f

⁢

(

x

)

=

2

⁢

x

, simpler than addition in the sense that

f

is unary, can not be defined in terms of the initial functions and functional composition alone. Primitive recursion is required at least once to define

f

.

Remark

. In fact, one can obtain an

infinite chain

of subsets of

𝒫

⁢

ℛ

in the following manner:

𝒫

⁢

ℛ

n

is the smallest set of all primitive recursive functions generated by the initial functions, closed under functional composition, and at most

n

applications of primitive recursion. Then

𝒫

⁢

ℛ

n

⊂

𝒫

⁢

ℛ

n

+

1

and

𝒫

⁢

ℛ

=

⋃

i

=

0

∞

𝒫

⁢

ℛ

i

For example, addition is in

𝒫

⁢

ℛ

1

-

𝒫

⁢

ℛ

0

, as was shown earlier. Furthermore, it is not hard to show that

multiplication

is in

𝒫

⁢

ℛ

2

-

𝒫

⁢

ℛ

1

, and exponentiation is in

𝒫

⁢

ℛ

3

-

𝒫

⁢

ℛ

2

. In fact, the

Ackermann function

can be utilized to show that each

𝒫

⁢

ℛ

n

⊂

𝒫

⁢

ℛ

n

+

1

is proper. Using this technique, it is possible to show that the Ackermann function is a

total recursive function

that is not primitive recursive. With respect to the set

ℰ

⁢

ℛ

of

elementary recursive functions

, it can be shown that

ℰ

⁢

ℛ

=

𝒫

⁢

ℛ

3

.

Title

importance of primitive recursion

Canonical

name

ImportanceOfPrimitiveRecursion

Date of creation

2013-03-22 19:05:21

Last modified on

2013-03-22 19:05:21

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

12

Author

CWoo (3771)

Entry type

Feature

Classification

msc 03D20

Generated on Thu Feb 8 20:03:45 2018 by

LaTeXML