Transfinite Recursion Theorem/Formulation 2/Lemma

Lemma for Transfinite Recursion Theorem: Formulation $2$
Let $\On$ denote the class of all ordinals.

Let $S$ denote the class of all ordinal sequences.

Let $g$ be a mapping such that $S \subseteq \Dom g$.

There exists an extending operation $E$ such that for every mapping $F$ on $\On$:


 * $\forall \alpha \in \On: \map {\paren {\map E {F \restriction \alpha} } } \alpha = \map g {F \restriction \alpha}$

Proof
Let $\theta_\alpha$ be an arbitrary ordinal sequence of length $\alpha$.

Let $\map E {\theta_\alpha}$ be the extending operation defined as:
 * $\map E {\theta_\alpha} := \theta_\alpha \cup \set {\tuple {\alpha, \map g {\theta_\alpha} } }$

Thus $\map E {\theta_\alpha}$ is an ordinal sequence of length $\alpha^+$.

Also, for $\beta \in \On$ such that $\beta < \alpha$:
 * $\map {\paren {\map E {\theta_\alpha} } } \beta = \map {\theta_\alpha} \beta$

but:
 * $\map {\paren {\map E {\theta_\alpha} } } \alpha = \map g {\theta_\alpha}$

Then for any mapping $F$ on $\On$:
 * $F \restriction \alpha$ is an ordinal sequence of length $\alpha$.

Hence:
 * $\map {\paren {\map E {F \restriction \alpha} } } \alpha = \map g {F \restriction \alpha}$