The proof shall proceed by Transfinite Induction, as follows: === Basis for the Induction === Let $0$ denote the ordinal zero. {{begin-eqn}} {{eqn|l = x \times \left({ y + 0 }\right)      |r = x \times y      |c = Definition of Ordinal Addition }} {{eqn|r = \left({ x \times y }\right) + 0      |c = Definition of Ordinal Addition }} {{eqn|r = \left({ x \times y }\right) + \left({ x \times 0 }\right)      |c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the basis for the induction. === Induction Step === {{begin-eqn}} {{eqn|l = x \times \left({ y + z }\right)       |r = \left({ x \times y }\right) + \left({ x \times z }\right)      |c = Inductive Hypothesis }} {{eqn|ll = \implies      |l = x \times \left({ y + z^+ }\right)      |r = x \times \left({ y + z }\right)^+      |c = Definition of Ordinal Addition }} {{eqn|r = \left({ x \times \left({ y + z }\right) }\right) + x      |c = Definition of Ordinal Multiplication }} {{eqn|r = \left({ \left({ x \times y }\right) + \left({ x \times z }\right) }\right) + x      |c = Inductive Hypothesis }} {{eqn|r = \left({ x \times y }\right) + \left({ \left({ x \times z }\right) + x }\right)      |c = Ordinal Addition is Associative }} {{eqn|r = \left({ x \times y }\right) + \left({ x \times z^+ }\right)      |c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the induction step. === Limit Case === The inductive hypothesis for the limit case states that: :$x \times \left({ y + w }\right) = \left({ x \times y }\right) + \left({ x \times w }\right)$ for all $w \in z$ and $z$ is a limit ordinal. The proof shall proceed by cases: === Case 1 === Suppose $x = 0$. {{begin-eqn}} {{eqn|l = x \times \left({ y + z }\right)      |r = 0      |c = Ordinal Multiplication by Zero }} {{eqn|r = 0 + 0      |c = Definition of Ordinal Addition }} {{eqn|r = \left({ x \times y }\right) + \left({ x \times z }\right)      |c = Ordinal Multiplication by Zero }} {{end-eqn}} === Case 2 === Suppose that $x \ne 0$. Since $w$ is a limit ordinal, $y + w$ and $x \times w$ are limit ordinals by Limit Ordinals Preserved Under Ordinal Addition and Limit Ordinals Preserved Under Ordinal Multiplication. {{begin-eqn}} {{eqn|l = x \times \left({ y + z }\right)      |r = \bigcup_{w \mathop \in \left({ y + z }\right)} \left({ x \times w }\right)      |c = Definition of Ordinal Multiplication }} {{eqn|l = \left({ x \times y }\right) + \left({ x \times z }\right)      |r = \bigcup_{v \mathop \in \left({ x \times z }\right)} \left({ x \times y }\right) + v      |c = Definition of Ordinal Addition }} {{end-eqn}} Take any $w \in \left({ y + z }\right)$. It follows that $w \in y \lor \left({ y \subseteq w \land w \in \left({ y + z }\right) }\right)$ by Relation between Two Ordinals and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. Thus, $\left({ w \in y \lor w = \left({ y + u }\right) }\right)$ for some $u \in z$ by Ordinal Subtraction when Possible is Unique. If $w < y$, then: {{begin-eqn}} {{eqn|l = \left({ x \times w }\right)      |o = \in      |r = \left({ x \times y }\right)      |c = Membership is Left Compatible with Ordinal Multiplication }} {{eqn|o = \subseteq      |r = \left({ x \times y }\right) + v      |c = Ordinal is Less than Sum }} {{end-eqn}} If $w = \left({ y + u }\right)$, then: {{begin-eqn}} {{eqn|l = x \times w      |r = x \times \left({ y + u }\right)      |c = definition of $w$ }} {{eqn|r = \left({ x \times y }\right) + \left({ x \times u }\right)      |c = Inductive Hypothesis }} {{eqn|l = \left({ x \times u }\right)      |o = \in      |r = \left({ x \times z }\right)      |c = Membership is Left Compatible with Ordinal Multiplication }} {{eqn|r = \left({ x \times y }\right) + v      |c = setting $v$ to $x \times u$ }} {{end-eqn}} By Supremum Inequality for Ordinals: : $x \times \left({ y + z }\right) \subseteq \left({ x \times y }\right) + \left({ x \times z }\right)$ Conversely, if $v \in \left({ x \times z }\right)$, then: {{begin-eqn}} {{eqn|ll = \exists w \in z:      |l = v      |o = \in      |r = \left({ x \times w }\right)      |c = Ordinal is Less than Ordinal times Limit }} {{eqn|ll = \implies      |l = \left({ x \times y }\right) + v      |r = \left({ x \times y }\right) + \left({ x \times w }\right)      |c = Substitutivity of Class Equality }} {{eqn|r = x \times \left({ y + w }\right)      |c = Inductive Hypothesis }} {{end-eqn}} By Supremum Inequality for Ordinals: : $\left({ x \times y }\right) + \left({ x \times z }\right) \subseteq x \times \left({ y + z }\right)$ By definition of set equality: : $x \times \left({ y + z }\right) = \left({ x \times y }\right) + \left({ x \times z }\right)$ This proves the limit case. {{qed}}