=== $K$ is a chain === First, it is necessary to must show that $K$ is a chain of mappings under the subset relation. Take any $f, g \in K$. Set $B$ equal to the domain of $f$ and set $C$ equal to the domain of $g$. From Relation between Two Ordinals: :$B \subseteq C \lor C \subseteq B$ From Lemma 1: : $\forall x \in B \cap C: f \left({x}\right) = g \left({x}\right)$ So depending on whether $B \subseteq C$ or $C \subseteq B$: : $f \subseteq g \lor g \subseteq f$ From Subset Relation is Ordering, it follows that $\subseteq$ is a total ordering on $K$. === Domain of $F$ is an ordinal === $K$ is a chain of mappings. Since the union of a chain of mappings is a mapping, $F$ is a mapping. Moreover, $\forall x \in B: f \left({x}\right) = F \left({x}\right)$, because $F$ is an extension of the mapping $f$. The domain of $F$ is a subset of the ordinals because it is the union of a collection of mappings whose domains are themselves subsets of ordinals. {{begin-eqn}} {{eqn | l = a \in \operatorname{Dom} \left({F}\right) \land x \in a       | o = \implies       | r = \exists f: \exists \beta \in \operatorname{On}: \left({f: \beta \to \operatorname{Im} \left({f}\right) \land a \in \beta }\right)       | c = Definition of $F$ (above) }} {{eqn | o = \implies       | r = a \in \operatorname{On} \land x \in \beta       | c = Transitivity of $\beta$ and $\operatorname{On}$ }} {{eqn | o = \implies       | r = x \in \operatorname{Dom} \left({F}\right)       | c =  }} {{end-eqn}} From this it follows that $\operatorname{Dom} \left({F}\right)$ is an ordinal, since it is a transitive subset of $\operatorname{On}$. === Domain of $F$ is $\operatorname{On}$ === Assume the domain of $F$ is a set. Since $F$ is a mapping, the range of $F$ is a set and $F$ is a set. Then $\operatorname{Dom} F = \gamma$ for some ordinal $\gamma$.  Define $C$ as follows: :$C = F \cup \left\{{\left({\gamma, G \left({F {\restriction_\gamma}}\right)}\right)}\right\}$ Then $C$ is a mapping with domain $\gamma^+$ and satisfies: : $\forall x < \gamma^+: C \left({x}\right) = G \left({F {\restriction_\gamma} }\right)$ Therefore $C$ is a member of $K$ (note that $F {\restriction_\gamma} = C {\restriction_\gamma}$). :$\displaystyle C \in K \implies C \subseteq \bigcup K$ So $C \subseteq F$. But if $F$ is a set, $F \subsetneq C$, a contradiction. Therefore, $\operatorname{Dom} \left({F}\right)$ cannot be a set and must therefore be the set of all ordinals $\operatorname{On}$. {{qed|lemma}} === Value of $F$ is $G \left({F {\restriction_\alpha} }\right)$ === :$\forall x \in \operatorname{On}: \exists f \in K: f \left({x}\right) = F \left({x}\right)$ Since $f \in K$: : $f \left({x}\right) = G \left({f {\restriction_x} }\right)$ But $\forall y < x: f \left({y}\right) = F \left({y}\right)$, so by Equality of Restrictions: : $f {\restriction_x} = F {\restriction_x}$ Therefore,: :$F \left({\alpha}\right) = G \left({F {\restriction_\alpha} }\right)$ {{qed|lemma}} === $F$ is Unique === Finally, assume there is another mapping $H$ that satisfies the first two properties. Then $H$ is equal to $F$ by transfinite induction: {{begin-eqn}} {{eqn | l = \forall x \in y: H \left({x}\right) = F \left({x}\right)       | o = \implies       | r = \left({H \restriction y}\right) = \left({F \restriction y}\right)       | c = Equality of Restrictions }} {{eqn | o = \implies       | r = G \left({H \restriction y}\right) = G \left({F \restriction y}\right)       | c = by hypothesis }} {{eqn | o = \implies       | r = H \left({y}\right) = F \left({y}\right)       | c = by hypothesis }} {{end-eqn}} So for all ordinals: : $H \left({y}\right) = F \left({y}\right)$ Thus $H = F$. {{qed}}