Transfinite Recursion Theorem/Theorem 2

Theorem
Let $\operatorname{Dom} (x)$ denote the domain of $x$. Let $\operatorname{Im} (x)$ denote the image of the mapping $x$.

Let $G$ be a collection of ordered pairs $(x,y)$ satisfying at least one of the following conditions:


 * $x = \varnothing$ and $y = a$
 * $\exists \beta: \operatorname{Dom} (x) = \beta^+$ and $y = H ( x ( \bigcup \operatorname{Dom} (x) ) )$
 * $\operatorname{Dom} (x)$ is a limit ordinal and $y = \bigcup \operatorname{Ran} (x)$.

Let $F$ be a mapping and let the domain of $F$ be the ordinals.

Let $F(\alpha) = G(F \restriction \alpha)$ for all ordinals $\alpha$. Then,


 * $F(\varnothing) = a$
 * $F(\beta^+) = H(F(\beta))$
 * For limit ordinals $\beta$, $F(\beta) = \bigcup_{\gamma \in \beta} F(\gamma)$
 * $F$ is unique. That is, if there is another function $A$ satisfying the above three properties, then $A = F$.

Proof
This proves the first part.

This proves the second part.

This proves the third part.

We can proceed in the fourth part by Transfinite Induction.

This proves the base case.

This proves the inductive step.

This proves the limit case. This completes the proof.