{{NotZFC}} === Lemma === {{:Order Isomorphism between Ordinals and Proper Class/Lemma}}{{qed|lemma}} Assume that: :$\exists x: \left({A \setminus \operatorname{Im} \left({x}\right) = \varnothing}\right)$ Then: :$A \subseteq \operatorname{Im} \left({x}\right)$ Therefore, $A$ is a set. This contradicts the fact that $A$ is a proper class. Therefore by Axiom of Subsets Equivalents and De Morgan's Laws (Predicate Logic): :$\forall x: \left({A \setminus \operatorname{Im} \left({x}\right) \ne \varnothing}\right)$ Then: {{begin-eqn}} {{eqn | r = \forall x: \left({\left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing}\right)       | o =  }} {{eqn | n = 1       | o = \implies       | r = F \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \land \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \cap A_{F \left({x}\right)} \ne \varnothing       | c = Lemma }} {{eqn | o = \implies       | r = f: \operatorname{On} \rightarrowtail A       | c = Condition for Injective Mapping on Ordinals }} {{end-eqn}} where $\rightarrowtail$ denotes an injection. Now to prove that $F$ is surjective: {{begin-eqn}} {{eqn | l = y \in \operatorname{Im} \left({F}\right)       | o = \implies       | r = \exists x: y = F \left({x}\right)       | c = Definition of Image }} {{eqn | o = \implies       | r = \exists x: \left({\left({A \setminus \operatorname{Im} \left({x}\right)}\right) \cap A_y = \varnothing \land y = F \left({x}\right)}\right)       | c = from $(1)$ }} {{eqn | o = \implies       | r = \exists x: \forall z: \left({\left({z \prec y \land x \in A}\right) \implies z \in \operatorname{Im} \left({x}\right)}\right)       |c = }} {{eqn | o = \implies       | r = \forall z: \left({\left({z \prec y \land x \in A}\right) \implies z \in \operatorname{Im} \left({F}\right)}\right)       | c = }} {{eqn | o = \implies       | r = \operatorname{Im} \left({F}\right) = A \lor \exists x \in A: \operatorname{Im} \left({F}\right) = \left({A \cap A_x}\right)       | c = Well-Ordered Transitive Subset is Equal or Equal to Initial Segment }} {{end-eqn}} But if $\operatorname{Im} \left({F}\right) = \left({A \cap A_x}\right)$, then it is equal to some initial segment of $A$. This would imply that $\operatorname{Im} \left({F}\right)$ is a set, which is a contradiction. Therefore $A = \operatorname{Im} \left({F}\right)$ and the function is bijective. {{begin-eqn}} {{eqn | l = x \in y       | o = \implies       | r = \operatorname{Im} \left({x}\right) \subseteq \operatorname{Im} \left({y}\right)       | c = Transitive Set is Proper Subset of Ordinal iff Element of Ordinal and the fact that the image preserves subsets }} {{eqn | o = \implies       | r = \left({A \setminus \operatorname{Im} \left({y}\right)}\right) \subseteq \left({A \setminus \operatorname{Im} \left({x}\right)}\right)       | c = }} {{eqn | o = \implies       | r = F \left({y}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)       | c = from $(1)$ }} {{eqn | o = \implies       | r = F \left({y}\right) \notin A_{F \left({x}\right)}       | c = from $(1)$ }} {{eqn | o = \implies       | r = F \left({x}\right) \prec F \left({y}\right) \lor F \left({x}\right) = F \left({y}\right)       | c = $\prec$ is a Strict Well-Ordering }} {{eqn | o = \implies       | r = F \left({x}\right) \prec F \left({y}\right)       | c = $F$ is injective and $x \ne y$ }} {{end-eqn}} Conversely, assume $F \left({x}\right) \prec F \left({y}\right)$. Then $y \in x$ and $x = y$ lead to contradictory conclusions. By Ordinal Membership is Trichotomy, we may conclude that $x \in y$. {{qed}}