Ordinal Multiplication is Associative

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

Let $\times$ denote ordinal multiplication.

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

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

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

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:


 * $\forall w \in Z: x \times \left({y \times w}\right) = \left({x \times y}\right) \times w$

where $z$ is a limit ordinal.

The proof shall proceed by cases:

Case 1
If $y = 0$, then:

Case 2
If $y \ne 0$, then $y \times z$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication.

It follows that:

If $u < \left({y \times z}\right)$, then $u < \left({y \times w}\right)$ for some $w \in z$ by Ordinal is Less than Ordinal times Limit.

Generalizing, the result follows for all $u \in \left({y \times z}\right)$, so by Supremum Inequality for Ordinals:


 * $x \times \left({y \times z}\right) \le \left({x \times y}\right) \times z$

Conversely, take any $w < z$.

It follows by Supremum Inequality for Ordinals that:


 * $\left({x \times y}\right) \times z \le x \times \left({y \times z}\right)$

By definition of set equality:
 * $x \times \left({y \times z}\right) = \left({x \times y}\right) \times z$

This proves the limit case.