Principle of Recursive Definition/Proof 2

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: f \left({x}\right) = \begin{cases}

a & : x = 0 \\ g \left({f \left({n}\right)}\right) & : x = n + 1 \end{cases}$

Proof
Consider $\N$ defined as elements of the minimal infinite successor set $\omega$.

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