Transfinite Recursion Theorem/Formulation 5

Theorem
Let $\On$ denote the class of all ordinals.

Let $h$ be a mapping.

Then there exists a unique mapping $F$ such that:


 * $\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$

Proof
This is a special case of the Transfinite Recursion Theorem: Formulation $2$.

Recall:

Consider $\map g x$.

If $x$ is a mapping, take $\map g x = \map h {\Img x}$.

If $x$ is not a mapping, take $\map g x = x$.

The result follows.