Subset is Right Compatible with Ordinal Multiplication

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


 * $\displaystyle x \le y \implies \left({ x \cdot z }\right) \le \left({ y \cdot z }\right)$

Proof
The proof shall proceed by Transfinite Induction on $z$.

Basis for the Induction

 * $\left({ x \cdot 0 }\right) = 0$ by Definition:Ordinal Multiplication, so the statement simply reduces to:


 * $0 \le 0$

This proves the basis for the induction.

Induction Step
This proves the induction step.

Limit Case
This proves the limit case.

Also see

 * Subset Right Compatible with Ordinal Addition