nyuuzyou/wikis huggingface dataset:
Principle of Mathematical Induction/Warning/Example 2

== Example of Incorrect Use of Principle of Mathematical Induction ==
We are to prove that:
:$\dfrac 1 {1 \times 2} + \dfrac 1 {2 \times 3} + \dotsb + \dfrac 1 {\paren {n - 1} \times n} = \dfrac 3 2 - \dfrac 1 n$
For $n = 1$ we have:
:$\dfrac 3 2 - \dfrac 1 n = \dfrac 1 2 = \dfrac 1 {1 \times 2}$
Assuming true for $k$, we have:
But clearly this is wrong, because for $n = 6$:
:$\dfrac 1 2 + \dfrac 1 6 + \dfrac 1 {12} + \dfrac 1 {30} = \dfrac 5 6$
on the , but:
:$\dfrac 3 2 - \dfrac 1 6 = \dfrac 4 3$
on the .
== Refutation ==
The supposed sequence of terms on the starts at $n = 2$.
It can be seen that it is meaningless (or has no terms for $n = 1$.
Hence in the first statement:
:$\dfrac 3 2 - \dfrac 1 n = \dfrac 1 2 = \dfrac 1 {1 \times 2}$
the term on the is the term for $n = 2$, making this equation invalid.
The correct result is Sum of Sequence of Products of Consecutive Reciprocals:
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $3$
Category:Principle of Mathematical Induction
Principle of Mathematical Induction/Warning/Example 3

== Example of Incorrect Use of Principle of Mathematical Induction ==
We are to prove that:
:$1 + 3 + 5 + \dotsb + \paren {2 n - 1} = n^2 + 3$
We establish as an induction hypothesis:
:$1 + 3 + 5 + \dotsb + \paren {2 k - 1} = k^2 + 3$
Then:
But clearly this is wrong, because for $n = 2$, say:
on the , but:
:$2^2 + 3 = 7$
on the .
== Refutation ==
The basis for the induction has not been established.
It is in fact not possible to find a value of $n$ for which $1 + 3 + 5 + \dotsb + \paren {2 n - 1} = n^2 + 3$ actually holds.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction
Category:Principle of Mathematical Induction
Principle of Mathematical Induction/Well-Ordered Integral Domain

== Theorem ==
Let $\struct {D, +, \times, \le}$ be a well-ordered integral domain whose zero is $0_D$.
Let the unity of $D$ be $1_D$.
Let $S \subseteq D$ be such that:
:$1_D \in S$
:$a \in S \implies a + 1_D \in S$
Then:
:$D_{> 0_D} \subseteq S$
where $D_{> 0_D}$ denotes all the elements $d \in D$ such that $\map P d$.
That is, $D_{> 0_D}$ is the set of all (strictly) positive elements of $D$.
== Proof ==
Let $\overline S$ be the set of all elements of $D_{> 0_D}$ that are not in $S$:
:$\overline S = D_{> 0_D} \setminus S$
 $\overline S$ is not empty.
Then as $D$ is well-ordered, it follows that $\overline S$ has a minimal element, which we will call $m$.
Then $m - 1 \notin \overline S$.
But $1 < m$ as $1 \in S$, and from One Succeeds Zero in Well-Ordered Integral Domain, $1$ is the minimal positive element of $D$.
So $0 < m - 1$ and so $m - 1$ is strictly positive.
Because $m - 1 \notin \overline S$, it follows that $m - 1 \in S$.
By construction of $S$ it follows that $m \in S$.
So $m \in S$ and $m \in \overline S$, that is:
:$m \notin S$
From this contradiction we deduced that there can therefore be no such $m$.
Hence:
:$D_{\ge 0_D} \setminus S = \O$.
From Set Difference with Superset is Empty Set it follows that $D_{\ge 0_D} \subseteq S$.
== Also see ==
This proof is (trivially) the same as the Principle of Mathematical Induction, but whereas this one takes as the basis the concept of the well-ordered integral domain, the latter proof uses the naturally ordered semigroup.
Both approaches are tantamount to the same thing, as they both lead (ultimately) to the uniqueness of the set of natural numbers $\N$ and also the set of integers $\Z$.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 8$. Well-Order: Theorem $13$
Category:Well-Ordered Integral Domains
Category:Principle of Mathematical Induction
Principle of Mathematical Induction/Well-Ordered Set

== Theorem ==
Let $\struct {S, \preceq}$ be a well-ordered set.
Let $T \subseteq S$ be a subset of $S$ such that:
:$\forall s \in S: \paren {\forall t \in S: t \prec s \implies t \in T} \implies s \in T$
Then $T = S$.
== Proof ==
 that $T \ne S$.
From Set Difference is Subset, $S \setminus T \subset S$.
From Set Difference with Proper Subset, $S \setminus T \ne \O$.
By the definition of a well-ordered set, there exists a smallest element $s$ of $S \setminus T$.
As $s \in S$, it follows from the definition of $T$ that:
:$\forall t \in S: t \prec s \implies t \in T$
But then $s \in T$ by hypothesis, contradicting the definition of $s$.
Hence the result, by Proof by Contradiction.
== Also see ==
* Well-Ordered Induction
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.1$
Category:Well-Orderings
Category:Principle of Mathematical Induction
Principle of Mathematical Induction/Zero-Based

== Theorem ==
Let $\map P n$ be a propositional function depending on $n \in \N$.
Suppose that:
:$(1): \quad \map P 0$ is true
:$(2): \quad \forall k \in \N: k \ge 0 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N$.
== Proof ==
Consider $\N$ defined as a Peano structure.
The result follows from Principle of Mathematical Induction for Peano Structure.
== Also see ==
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
* : Appendix $\text{A}.12$: Induction
* : Appendix $\text{A}$: Set Theory: Induction
* : Appendix $\text{A}.6$
Category:Principle of Mathematical Induction
Principle of Mathematical Induction for Minimally Inductive Set

== Theorem ==
Let $\omega$ be the minimally inductive set.
Let $S \subseteq \omega$.
Suppose that:
:$(1): \quad \O \in S$
:$(2): \quad \forall x: x \in S \implies x^+ \in S$
where $x^+$ is the successor set of $x$.
Then:
:$S = \omega$
</onlyinclude>
== Proof ==
The hypotheses state precisely that $S$ is an inductive set.
Then the minimally inductive set $\omega$ being defined as the intersection of all inductive sets, we conclude that:
:$\omega \subseteq S$
by Intersection is Subset: General Result.
Thus, by definition of set equality:
:$S = \omega$
== Sources ==
* : $\S 12$: The Peano Axioms
* : $\S 7.31$
Category:Minimally Inductive Set
Principle of Mathematical Induction for Natural Numbers in Real Numbers

== Theorem ==
Let $\struct {\R, +, \times, \le}$ be the field of real numbers.
Let $\N$ be the natural numbers in $\R$.
Suppose that $A \subseteq \N$ is an inductive set.
Then $A = \N$.
== Proof ==
By definition of the natural numbers in $\R$:
:$\N = \ds \bigcap \II$
where $\II$ is the set of inductive sets in $\R$.
Since $A$ was supposed to be inductive, it follows that:
:$\N \subseteq A$
from Intersection is Subset: General Result.
Hence by definition of set equality:
:$A = \N$
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
Category:Natural Numbers in Real Numbers
Principle of Non-Contradiction

== Theorem ==
The Principle of Non-Contradiction is a valid argument in types of logic dealing with negation $\neg$ and contradiction $\bot$.
This includes classical propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
=== Sequent Form ===
== Explanation ==
== Also known as ==
== Also see ==
* Definition:Logical Not
* Definition:Contradiction
* Law of Excluded Middle
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.6 \ \text{(b)}$
* 
* 
Category:Principle of Non-Contradiction
Category:Rules of Inference
Category:Contradiction
Principle of Non-Contradiction/Explanation

== Theorem ==
The Principle of Non-Contradiction can be expressed in natural language as follows:
:A statement can not be both true and not true at the same time.
This means: if we have managed to deduce that a statement is both true and false, then the sequence of deductions show that the pool of assumptions upon which the sequent rests contains assumptions which are mutually contradictory.
Thus it provides a means of eliminating a logical not from a sequent.
It is one of the cornerstones of Aristotelian logic, along with the law of the excluded middle.
Category:Principle of Non-Contradiction
Principle of Non-Contradiction/Proof Rule

== Theorem ==
The principle of non-contradiction is a valid argument in types of logic dealing with negation $\neg$ and contradiction $\bot$.
This includes classical propositional logic and predicate logic, and in particular natural deduction.
As a proof rule it is expressed in the form:
:If we can conclude both $\phi$ and $\neg \phi$, we may infer a contradiction.
It can be written:
:$\ds {\phi \quad \neg \phi \over \bot} \neg_e$
=== Tableau Form ===
== Explanation ==
== Also known as ==
== Also see ==
* This is a rule of inference of the following proof systems:
** Definition:Natural Deduction
== Technical Note ==
== Sources ==
* : $\S 1.2.1$: Rules for natural deduction
Category:Rules of Inference
Category:Principle of Non-Contradiction
Principle of Non-Contradiction/Proof Rule/Technical Note

== Technical Note on ==
When invoking the in a tableau proof, use the template:
:{{NonContradiction|line|pool|first|second}}
or:
:{{NonContradiction|line|pool|first|second|comment}}
where:
:line is the number of the line on the tableau proof where the is to be invoked
:pool is the pool of assumptions (comma-separated list)
:first is the first of the two lines of the tableau proof upon which this line directly depends
:second is the second of the two lines of the tableau proof upon which this line directly depends
:comment is the (optional) comment that is to be displayed in the Notes column.
Category:Technical Notes
Principle of Non-Contradiction/Sequent Form/Formulation 1/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
$\begin{array}{|cccc||c|} \hline
p & \land & \neg & p & \bot \\
\hline
F & F & T & F & F \\
T & F & F & T & F \\
\hline
\end{array}$
As can be seen by inspection, the truth value of the main connective, that is $\land$, is $F$ for each boolean interpretation for $p$.
Category:Truth Table Proofs
Category:Principle of Non-Contradiction
Principle of Non-Contradiction/Sequent Form/Formulation 2

== Theorem ==
:$\vdash \neg \paren {p \land \neg p}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S \text{II}.13$: Symbolism of sentential calculus
* : $\S 3.6$: Reference Formulae: $RF \, 1$
Category:Principle of Non-Contradiction
Principle of Non-Contradiction/Sequent Form/Formulation 2/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition $\neg \left({p \land \neg p}\right)$.
As can be seen by inspection, the truth value of the main connective, that is $\neg$, is $T$ for each boolean interpretation for $p$.
$\begin{array}{|ccccc|} \hline
\neg & (p & \land & \neg & p)\\
\hline
T & F & F & T & F \\
T & T & F & F & T \\
\hline
\end{array}$
Category:Principle of Non-Contradiction
Category:Truth Table Proofs
Principle of Open Induction for Real Numbers

== Theorem ==
Let $a < b$ be real numbers.
Let $S$ be an open set of real numbers.
Suppose that, for every $x \in \closedint a b$ such that:
:$\hointr a x \subseteq S$
it also holds that:
:$x \in S$
Then, $\closedint a b \subseteq S$.
== Proof ==
 there exists some $x \in \closedint a b$ such that:
:$x \notin S$
Let:
:$T := \closedint a b \setminus S$
be the set of all such $x$.
By Set Difference is Subset:
:$T \subseteq \closedint a b$
so by Subset of Bounded Below Set is Bounded Below:
:$T$ is bounded below.
Since $T$ is non-empty by assumption, it follows from the Greatest Lower Bound Property that:
:$T$ admits an infimum.
Let $x_0 := \inf T$ be the infimum of $T$.
By definition, $x_0$ is a lower bound of $T$.
Therefore:
:$\forall x \in \closedint a b: x < x_0 \implies x \in S$
or equivalently:
:$\hointr a {x_0} \subseteq S$
Hence, :
:$x_0 \in S$
By definition of open set, there exists some $\epsilon > 0$ such that:
:$\openint {x_0 - \epsilon} {x_0 + \epsilon} \subseteq S$
By Union of Subsets is Subset:
:$\hointr a {x_0} \cup \openint {x_0 - \epsilon} {x_0 + \epsilon} = \hointr a {x_0 + \epsilon} \subseteq S$
Thus, by Set Difference with Subset is Superset of Set Difference:
:$T = \paren {\closedint a b \setminus S} \subseteq \paren {\closedint a b \setminus \hointr a {x_0 + \epsilon}} = \closedint {x_0 + \epsilon} b$
Therefore, $x_0 + \epsilon$ is a lower bound of $T$.
Hence, by definition of infimum, it must hold that:
:$x_0 + \epsilon \le x_0$
But that implies that:
:$\epsilon \le 0$
contradicting the definition of $\epsilon$.
We are forced to conclude that our assumption:
:$\exists x \in \closedint a b \setminus S$
was incorrect.
Therefore, by Proof by Contradiction:
:$\closedint a b \subseteq S$
== Sources ==
* 
Category:Mathematical Induction
Category:Real Analysis
Principle of Parsimony

== Mathematical Principle ==
Let $P$ be a stochastic process which is being modelled by means of a stochastic model $M$.
$M$ will necessarily use a number of constants and parameters whose values are to be determined by estimation from the data.
The principle of parsimony dictates that $M$ employs the smallest possible number of parameters such that $M$ will adequately represent the behaviour of $P$.
== Examples ==
== Sources ==
* 
* :
::$1$: Introduction:
:::$1.3$ Basic Ideas in Model Building:
::::$1.3.1$ Parsimony
Category:Mathematical Models
Category:Principle of Parsimony
Category:Named Theorems
Principle of Parsimony/Examples/Arbitrary Dynamic Model

== Examples of Use of Principle of Parsimony ==
Let $M$ be a dynamic model of the form:
:$(1): \quad Y_t = \paren {\omega_0 - \omega_1 B - \omega_2 B^2 - \dotsb - \omega_s B^2} X_t$
when dealing with a system that could be adequately presented by:
:$(2): \quad \paren {1 - \delta B}^{-1} \omega_0 X_t = \omega_0 \paren {1 + \delta B + \delta^2 B^2 + \dotsb} X_t$
where $\size \delta < 1$.
Because of experimental error, it is quite possible to miss noticing the relationship between the coefficients in the fitted equation.
Hence we may end up fitting a relationship like $(1)$ which contains $s + 1$ parameters, where the simpler form $(2)$, which has only $2$ parameters, would have been adequate.
Hence the estimation of the output $Y_t$ may be unnecessarily poor for the given values of $X_t, X_{t - 1}, \dotsb$.
== Sources ==
* :
::$1$: Introduction:
:::$1.3$ Basic Ideas in Model Building:
::::$1.3.1$ Parsimony
Category:Principle of Parsimony
Principle of Recursive Definition

== Theorem ==
Let $\N$ be the natural numbers.
Let $T$ be a class (which may be a set).
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:$\forall x \in \N: \map f x = \begin{cases}
a & : x = 0 \\
\map g {\map f n} & : x = n + 1
\end{cases}$
=== General Result ===
=== Strong Version ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also presented as ==
== Fallacious Proof ==
The proofs given (hidden behind the links) are necessarily long, precise and detailed.
There is a temptation to take short cuts and gloss over the important details.
The following argument, for example, though considerably shorter, is incorrect.
== Also known as ==
This result is often referred to as the Recursion Theorem.
Some sources only cover the general result.
== Also see ==
* Second Principle of Recursive Definition
* Transfinite Recursion Theorem
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion: Theorem $8.1$
Category:Named Theorems
Category:Mapping Theory
Category:Natural Numbers
Category:Recursive Definitions
Category:Principle of Recursive Definition
Principle of Recursive Definition/Also presented as

== Theorem ==
The Principle of Recursive Definition can also be presented as:
:For any mapping $f: T \to T$ and any $a \in T$, there exists an infinite sequence $a_0, a_1, \ldots, a_n, a_{n + 1}, \ldots$ such that:
::$a_0 = a$
::$a_{n + 1} = \map g {a_n}$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion
Category:Principle of Recursive Definition
Principle of Recursive Definition/Fallacious Proof

== Theorem ==
== Proof ==
Consider $\N$, defined as a naturally ordered semigroup $\struct {S, \circ, \preceq}$.
Let the mapping $f$ be defined as:
:$\map f x = \begin{cases}
a & : x = 0 \\
\map s {\map f n} & : x = n \circ 1 \end{cases}$
if $\map f n$ is defined.
Let $S' = \set {n \in S: \map f n \text{ is defined} }$.
Then:
:$0 \in S'$
and:
:$n + 1 \in S'$
so by Principle of Mathematical Induction for Naturally Ordered Semigroup:
:$S' = S$
Thus the domain of $f$ is $S$.
Consequently, $f$ is a mapping from $S$ into $T$ which satisfies:
:$\map f 0 = a$
and:
:$\map f {n \circ 1} = \map s {\map f n}$
for all $n \in S$.
=== Objections ===
$(1):$
In the above argument, $S'$ is not precisely defined.
In order for a set to be defined, that definition should be able to be expressed solely in terms of logic and definitions from set theory.
In this case, the expression "is defined" does not meet that criterion.
$(2):$
The mapping $f$ is not defined properly.
A mapping needs to be defined as a set of ordered pairs which needs to satisfy a condition.
In the above, it is indeed specified that
:$\tuple {0, a} \in f$
and:
:$\tuple {n \circ 1, \map s x} \in f$ whenever $\tuple {n, x} \in f$
Thus it appears either that:
:$f$ itself is used to define $f$
or else that:
:$f$ itself changes during the process in which it is being defined.
Neither of these possibilities can be accepted.
$(3):$
The only property of $\struct {S, \circ, \preceq}$ used in the argument is that it satisfies Principle of Mathematical Induction for Naturally Ordered Semigroup.
However, consider the commutative semigroup $\struct {D, +}$ which has elements $0$ and $1$, such that:
:$D$ is the only subset of $D$:
::containing $0$
:and
::containing $x + 1$ whenever it contains $x$.
Then:
:for any set $T$
:any mapping $g: T \to T$
:any element $a \in T$
there exists a mapping $f: D \to T$ such that:
:$\map f y = \begin{cases}
a & : y = 0 \\
\map s {\map f x} & : y = x + 1 \end{cases}$
But consider the additive group of integers modulo $2$:
:$\struct {\Z_2, +_2}$
which is indeed a commutative semigroup containing $0$ and $1$ which satisfies the hypothesis.
But if $g: \N \to \N$ is the mapping defined as:
:$\map g n = n + 1$
there is no mapping $f: \Z_2 \to \N$ which satisfies:
:$\map f y = \begin{cases}
0 & : y = 0 \\
\map s {\map f x} & : y = x +_2 1 \end{cases}$
for all $x \in \Z_2$.
Hence the argument is invalid.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
* : Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion
Category:Principle of Recursive Definition
Principle of Recursive Definition/Proof 4

== Theorem ==
== Proof ==
From the Principle of Recursive Definition: Strong Version:
Let $h: A \to A$ be defined as:
:$\forall x \in A: \map h x := \map g {a, x}$ for arbitrary $a \in \omega$
That is:
:$\forall y \in \omega: \map g {y, x} = \map h x$
Then there exists exactly one mapping $f: \omega \to A$ such that:
:$\forall x \in \omega: \map f x = \begin {cases}
c & : x = \O \\
\map h {\map f n} & : x = n^+
\end {cases}$
The result follows on identifying $\omega$ with $\N$, $c$ with $a$, $A$ with $T$, $\O$ with $0$ and $n^+$ with $n + 1$.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion: Exercise $8.3$
Category:Principle of Recursive Definition
Principle of Recursive Definition for Minimally Inductive Set

== Theorem ==
Let $\omega$ be the minimally inductive set.
Let $T$ be a set.
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \omega \to T$ such that:
:$\forall x \in \omega: \map f x = \begin {cases}
a & : x = \O \\
\map g {\map f n} & : x = n^+
\end {cases}$
where $n^+$ is the successor set of $n$.
== Proof ==
Take the function $F$ generated in Second Principle of Transfinite Recursion.
Set $f = F {\restriction_\omega}$.
Therefore, such a function exists.
Now, suppose there are two functions $f$ and $f'$ that satisfy this:
:$\map f \O = \map {f'} \O$
Then:
This completes the proof.
</onlyinclude>
== Sources ==
* : $\S 12$: The Peano Axioms
Category:Mapping Theory
Category:Minimally Inductive Set
Category:Principle of Recursive Definition
Principle of Recursive Definition for Peano Structure

== Theorem ==
Let $\struct {P, 0, s}$ be a Peano structure.
Let $T$ be a set.
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: P \to T$ such that:
:$\forall x \in P: \map f x = \begin{cases}
a & : x = 0 \\
\map g {\map f n} & : x = \map s n
\end{cases}$
== Proof ==
For each $n \in P$, define $\map A n$ as:
:$\map A n = \set {h: P \to T \mid \map h 0 = a \land \forall m < n: \map h {\map s n} = \map g {\map h m} }$
First, we prove for all $n \in P$ that $\map A n$ is not empty.
More formally, we prove that $A = \set {n \in P: \map A n \ne \O} = P$.
For $n = 0$, there are no $m < n$, so any mapping $h: P \to T$ with $\map h 0 = a$ is an element of $\map A 0$.
Since the constant mapping $\map h n = a$ satisfies this condition, it follows that $0 \in A$.
Now suppose that $n \in A$, and let $h \in \map A n$.
Define $h': P \to T$ by, for all $n' \in P$:
:$\map {h'} {n'} = \begin {cases}
\map h {n'} & : n' \le n \\
\map g {\map h n} & : n' > n
\end {cases}$
To check that $h' \in \map A {\map s n}$, we have to verify that:
:$\forall m < \map s n: \map {h'} {\map s m} = \map g {\map {h'} m}$
Since $h \in \map A n$, only the case $m = n$ needs to be verified:
Therefore $h' \in \map A {\map s n}$, and so $\map A {\map s n} \ne \O$.
That means $\map s n \in A$, and by Axiom $(\text P 5)$, we conclude $A = P$.
Now let $A' = \set {n \in P: \forall h, h' \in \map A n: \map h n = \map {h'} n}$.
Then by definition of $\map A )$, it follows that $0 \in A'$.
Suppose now that $n \in A'$, and let $h, h' \in \map A {\map s n}$.
Then:
Hence $\map s n \in A'$, and by Axiom $(\text P 5)$, we conclude $A' = P$.
Because any $f: \N \to T$ as in the theorem statement needs to be in all $\map A n$, it follows that such an $f$ is necessarily unique.
Finally, we can define $f: P \to T$ by:
:$\map f n = \map {h_n} n$
where $h_n \in \map A n$.
It is immediate from the definition of the $\map A n$ that:
:$\forall m, n, m < n: \map A n \subseteq \map A m$
Hence, for every $m, n$ such that $m < n$:
:$\map {h_m} m = \map {h_n} m$
Thus, for all $m < n$:
:$\map f m = \map {h_n} m$
Since $h_n \in \map A n$, it follows that also:
:$f \in \map A n$
Thus, since $n$ was arbitrary, it follows that for all $n \in P$:
:$\map f {\map s n} = \map g {\map f n}$
as desired.
Category:Principle of Recursive Definition
Category:Mapping Theory
Category:Peano's Axioms
Principle of Recursive Definition for Well-Ordered Sets

== Theorem ==
Let $J$ be a well-ordered set.
Let $C$ be any set.
Let $\FF$ be the set of all functions that map initial segments $S_a$ of $J$ into $C$.
Then for any function of the form:
:$\rho: \FF \to C$
there exists a unique function:
:$h: J \to C$
satisfying:
:$\forall \alpha \in J: \map h \alpha = \map \rho {h \restriction_{S_\alpha} }$
where $\restriction$ denotes the restriction of a mapping.
== Proof ==
== Also see ==
* Principle of Recursive Definition
== Sources ==
* : Supplementary Exercise $1.1$
Category:Well-Orderings
Principle of Stationary Action

== Physical Law ==
The principle of stationary actions states that the equations of motion of a physical system can be acquired by finding a stationary point of the action.
In other words, the first variation of the action has to vanish.
== Also known as ==
The principle of stationary action is also known as the principle of least action.
== Notes ==
The principle of least action attained its name due to classical problems of minimization.
However, if broken trajectories are allowed, the action can sometimes acquire lower values than for any allowed smooth trajectory.
Since smooth trajectories are more realistic, leastness has been weakened to stationarity.
Category:Mechanics
Category:Physics
Category:Applied Mathematics
Category:Lagrangian Mechanics