# Principle of Recursive Definition/Strong Version

## Theorem

Let $\omega$ denote the natural numbers as defined by the von Neumann construction.

Let $A$ be a class.

Let $c \in A$.

Let $g: \omega \times A \to A$ be a mapping.

Then there exists exactly one mapping $f: \omega \to A$ such that:

$\forall x \in \omega: \map f x = \begin{cases} c & : x = \O \\ \map g {n, \map f n} & : x = n^+ \end{cases}$

## Proof

From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is a minimally inductive class under the successor mapping.

First an admissible mapping is defined.

Let $n \in \omega$.

The mapping $h: n \to A$ is defined as an admissible mapping for $n$ if and only if:

$\forall r \subseteq n: \map h r = \begin{cases} c & : r = 0 \\ \map g {y, \map h y} & : r = y^+ \end{cases}$

Define:

$S = \set {x \in \omega: x \text { has exactly one admissible mapping} }$

We now use the Principle of Finite Induction for Peano Structure to prove that $S = \omega$.

### Basis for the Induction

Consider the empty set $\O$.

Let the mapping $h_0: \O \to A$ be defined as:

$\map {h_0} \O = c$

By definition, $h_0$ is the unique admissible mapping for $\O$.

Thus $\O \in S$.

This is the basis for the induction.

### Induction Hypothesis

This is the induction hypothesis.

Let $n \in S$.

That is, there exists a unique admissible mapping for $n$.

### Induction Step

It is now to be shown that:

$n^+ \in S$

From the induction hypothesis:

There exists a unique admissible mapping $h_n$ for $n$.

Let us define a mapping $\phi: n^+ \to A$ by:

$\forall r \subseteq n^+: \map \phi r = \begin{cases} \map {h_n} r & : r \subseteq n \\ \map g {n, \map {h_n} n} & : r = n^+ \end{cases}$

By definition, $\phi$ is an admissible mapping for $n^+$.

Let $\phi'$ be an admissible mapping for $n^+$.

Uniqueness of $\phi$ is then proved by showing that $\phi' = \phi$, as follows.

The restriction of $\phi'$ to $n$ is an admissible mapping for $n$

Thus by the induction hypothesis:

$\phi' {\restriction_n} = \phi {\restriction_n}$

It follows that:

 $\ds \map {\phi'} {n^+}$ $=$ $\ds \map g {n, \map {\phi'} n}$ $\ds$ $=$ $\ds \map g {n, \map h n}$ $\ds$ $=$ $\ds \map \phi {n^+}$

Thus as:

$\phi' {\restriction_n} = \phi {\restriction_n}$

and:

$\map {\phi'} {n^+} = \map \phi {n^+}$

it follows by Equality of Mappings that:

$\phi' = \phi$

Thus:

$n^+ \in S$

Thus by the Principle of Finite Induction for Peano Structure it follows that:

$S = \omega$

$\Box$

For all $n \in \omega$, let $h_n$ be the unique admissible mapping for $n$.

Let $n \in \omega$.

The restriction of $h_{n^+}$ to $n$ is an admissible mapping for $n$.

Therefore:

$h_{n^+} {\restriction_n} = h_n$

So:

$\forall r \subseteq n: \map {h_{n^+} } r = \map {h_n} r$

and in particular:

$\map {h_{n^+} } n = \map {h_n} n$

Let $f: \omega \to T$ be the mapping defined as:

$\forall n \in \omega: \map f n = \map {h_n} n$

Then:

$\map f \O = \map {h_0} \O = c$

and, for all $n \in \omega$:

 $\ds \map f {n^+}$ $=$ $\ds \map {h_{n^+} } {n^+}$ $\ds$ $=$ $\ds \map g {n, \map {h_{n^+} } n}$ $\ds$ $=$ $\ds \map g {n, \map {h_n} n}$ $\ds$ $=$ $\ds \map g {n, \map f n}$

Thus $f: \omega \to A$ is a mapping with the desired properties.

$\Box$

Let $f'$ be any mapping satisfying the desired properties.

It remains to be proved that:

$f' = f$

that is, that $f$ is the only mapping with the desired properties.

By the foregoing, for all $n \in \omega$, the restriction of $f'$ to $n$ is an admissible mapping for $n$.

Therefore:

$f' = h_n$

So:

$\forall n \in \omega: \map {f'} n = \map {h_n} n = \map f n$

and the proof is complete.

$\blacksquare$