Membership is Left Compatible with Ordinal Multiplication

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


 * $\displaystyle \left({ x < y \land z > 0 }\right) \iff \left({ z \cdot x }\right) < \left({ z \cdot y }\right)$

Sufficient Condition
The proof of the sufficient condition shall proceed by Transfinite Induction on $y$.

Basis for the Induction
Both $x < 0$ and $\left({ x \cdot z }\right) < \left({ 0 \cdot z }\right)$ are contradictory, so the iff statement holds for the condition that $y = 0$.

This proves the basis for the induction.

Induction Step
Suppose the biconditional statement holds for $y$. Then:

In either case, $\left({ z \cdot x }\right) < \left({ z \cdot y^+ }\right)$

This proves the induction step.

Limit Case
Suppose $y$ is a limit ordinal:

This proves the limit case.

Necessary Condition
Conversely, suppose $\left({ z \cdot x }\right) < \left({ z \cdot y^+ }\right)$.

Then, $z \ne 0$ because if it were equal, both sides of the inequality would be $0$.

So $z > 0$.

Furthermore:


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


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

So if $\left({ z \cdot x }\right) < \left({ z \cdot y }\right)$, then $y \ne x$ and $y \not < x$, so $x < y$ by Ordinal Membership Trichotomy.