Ordinal Multiplication is Left Distributive

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

Let $\times$ denote ordinal multiplication.

Let $+$ denote ordinal addition.

Then:
 * $x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$

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:


 * $x \times \paren {y + w} = \paren {x \times y} + \paren {x \times w}$ 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 $w \in y + z$.

It follows that $w \in y \lor \paren {y \subseteq w \land w \in y + z}$ by Relation between Two Ordinals and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.

Thus, $w \in y \lor w = y + u$ for some $u \in z$ by Ordinal Subtraction when Possible is Unique.

If $w < y$, then:

If $w = y + u$, then:

By Supremum Inequality for Ordinals:
 * $x \times \paren {y + z} \subseteq \paren {x \times y} + \paren {x \times z}$

Conversely, if $v \in x \times z$, then:

By Supremum Inequality for Ordinals:
 * $\paren {x \times y} + \paren {x \times z} \subseteq x \times \paren {y + z}$

By definition of set equality:
 * $x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$

This proves the limit case.