Ordinal Multiplication is Left Distributive

Theorem
Let $x$, $y$, and $z$ be ordinals.

Let $\times$ denote ordinal multiplication.

Let $+$ denote ordinal addition.

Then:
 * $x \times \left({ y + z }\right) = \left({ x \times y }\right) + \left({ x \times z }\right)$

Proof
The proof shall proceed by Transfinite Induction, as follows:

Basis for the Induction
Let $0$ denote the ordinal zero.

This proves the basis for the induction.

Induction Step
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 $x = 0$.

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.

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:

If $w = \left({ y + u }\right)$, then:

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:

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.