Principle of Recursive Definition/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\N$ be the natural numbers.

Let $T$ 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:

$\forall x \in \N: \map f x = \begin{cases} a & : x = 0 \\ \map g {\map f n} & : x = n + 1 \end{cases}$


Proof

Consider $\N$ defined as the von Neumann construction of the natural numbers.

The result follows from Principle of Recursive Definition for Minimal Infinite Successor Set.

$\blacksquare$