Principle of Recursive Definition/Proof 4

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 $n + 1$.