By transfinite induction on $y$. === Base Case === {{begin-eqn}} {{eqn | l = x \le \O       | o = \leadsto       | r = x = \O       | c = Subset of Empty Set }} {{eqn | o = \leadsto       | r = \paren {x + z} = z       | c = {{Defof|Ordinal Addition}} }} {{eqn | o = \leadsto       | r = \paren {\paren {x + z} = \O \iff z = \O}       | c = Equality is Equivalence Relation }} {{end-eqn}} === Inductive Case === {{begin-eqn}} {{eqn | l = x \le y^+       | o = \leadsto       | r = \paren {x \le y \lor x = y^+}       | c = Ordinal is Subset of Successor }} {{eqn | l = x \le y       | o = \leadsto       | r = \exists ! z \in \On: \paren {x + z} = y       | c = By Hypothesis }} {{eqn | o = \leadsto       | r = \exists ! z \in \On: \paren {x + z}^+ = y^+       | c = Equality of Successors }} {{eqn | o = \leadsto       | r = \exists ! w \in \On: \paren {x + w} = y^+       | c = Equality of Successors }} {{end-eqn}} The last step is justified because of: :$\paren {x + z^+} = y^+$ which guarantees existence, and by Ordinal Addition is Left Cancellable: :$\paren {x + w} = \paren {x + z^+} \implies w = z^+$ which guarantees uniqueness. === Limit Case === {{begin-eqn}} {{eqn | l = \forall w \in y: x \le w       | o = \leadsto       | r = \exists ! z: \paren {x + z} = w       | c = By Hypothesis }} {{eqn | l = x \le y       | o = \leadsto       | r = x = y \lor x \lt y       | c = }} {{eqn | l = x = y       | o = \leadsto       | r = x + \O = y       | c = {{Defof|Ordinal Addition}} }} {{eqn | o = \leadsto       | r = x + z = y \iff x + \O = x + z       | c = Equality is Equivalence Relation }} {{eqn | o = \leadsto       | r = x + z = y \iff z = \O       | c = Ordinal Addition is Left Cancellable }} {{eqn | o = \leadsto       | r = \exists ! z: x + z = y       | c = $z$ must be equal to $\O$ }} {{eqn | l = x < y       | o = \leadsto       | r = \exists w < y: x < w       | c = Union of Limit Ordinal }} {{end-eqn}} Set $A = \set {w : x < w \land w < y}$. The above statement shows that $A$ is nonempty. Then: :$\forall w \in A: \exists ! z: \paren {x + z} = w$ Create a mapping $F$, that sends each $w \in A$ to the unique $z$ that satisfies $\paren {x + z} = w$. {{explain|Specify domain and range of $F$.}} {{begin-eqn}} {{eqn | l = x < w \land w < y       | o = \leadsto       | r = x + \map F w = w       | c = Definition of $F$ above }} {{eqn | o = \leadsto       | r = \bigcup_{w \mathop \in A} \paren {x + \map F w} = \bigcup_{w \mathop \in A} w       | c = Indexed Union Equality }} {{eqn | o = \leadsto       | r = \bigcup_{w \mathop \in A} \paren {x + \map F w} = y       | c = Union of Limit Ordinal }} {{end-eqn}} Finally, we must prove that: :$\displaystyle \bigcup_{w \mathop \in A} \paren {x + \map F w} = \paren {x + \bigcup_{w \in A} F \map F w}$ It suffices to prove that $\displaystyle \bigcup_{w \mathop \in A} \map F w$ is a limit ordinal. Let $w = x^+$. Then $\map F w = 1$, and Union of Ordinals is Least Upper Bound. Thus: :$\displaystyle \bigcup_{w \mathop \in A} \map F w \ne \O$ {{begin-eqn}} {{eqn | l = z \in \bigcup_{w \mathop \in A} \map F w       | o = \leadsto       | r = \exists w \in A: z < \map F w       | c = {{Defof|Union of Family}} }} {{eqn | o = \leadsto       | r = \exists w \in A: z^+ < \map F {w^+}       | c = }} {{eqn | o = \leadsto       | r = z^+ \in \bigcup_{w \mathop \in A} \map F w       | c = }} {{end-eqn}} Thus: :$\displaystyle \bigcup_{w \mathop \in A} F \map F w \ne z^+$ Therefore $\bigcup_{w \mathop \in A} \map F w$ must be a limit ordinal. To prove uniqueness, assume $x + z = y$. Then by Ordinal Addition is Left Cancellable: :$\displaystyle x + z = x + \bigcup_{w \mathop \in A} \map F w \implies z = \bigcup_{w \mathop \in A} \map F w$ {{qed}}