Transfinite Recursion Theorem/Theorem 2

Theorem
Let $\operatorname{Dom} \left({x}\right)$ denote the domain of $x$.

Let $\operatorname{Im} \left({x}\right)$ denote the image of the mapping $x$.

Let $G$ be a class of ordered pairs $\left({x, y}\right)$ satisfying at least one of the following conditions:
 * $(1): \quad x = \varnothing$ and $y = a$


 * $(2): \quad \exists \beta: \operatorname{Dom} \left({x}\right) = \beta^+$ and $y = H \left({x \left({\bigcup \operatorname{Dom} \left({x}\right)}\right)}\right)$


 * $(3): \quad \operatorname{Dom} \left({x}\right)$ is a limit ordinal and $y = \bigcup \operatorname{Rng} \left({x}\right)$.

Let $F \left({\alpha}\right) = G \left({F \restriction \alpha}\right)$ for all ordinals $\alpha$.

Then:


 * $F$ is a mapping and the domain of $F$ is the ordinals, $\operatorname{On}$.
 * $F \left({\varnothing}\right) = a$
 * $F \left({\beta^+}\right) = H \left({F \left({\beta}\right)}\right)$
 * For limit ordinals $\beta$, $\displaystyle F \left({\beta}\right) = \bigcup_{\gamma \mathop \in \beta} F \left({\gamma}\right)$
 * $F$ is unique.
 * That is, if there is another function $A$ satisfying the above three properties, then $A = F$.

Proof
We can proceed in the fourth part by Transfinite Induction.

Basis for the Induction
This proves the basis for the induction.

Induction Step
This proves the induction step.

Limit Case
This proves the limit case.