Membership is Left Compatible with Ordinal Multiplication

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

Then:
 * $\paren {x < y \land z > 0} \iff \paren {z \cdot x} < \paren {z \cdot y}$

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

Basis for the Induction
Both $x < 0$ and $\paren {x \cdot z} < \paren {0 \cdot z}$ are contradictory, so the 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:
 * $\paren {z \cdot x} < \paren {z \cdot y^+}$

This proves the induction step.

Limit Case
Suppose $y$ is a limit ordinal:

This proves the limit case.

Necessary Condition
Conversely, suppose $\paren {z \cdot x} < \paren {z \cdot y^+}$.

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

So $z > 0$.

Furthermore:


 * $y < x \implies \paren {z \cdot y} < \paren {z \cdot x}$


 * $y = x \implies \paren {z \cdot y} = \paren {z \cdot x}$

So if $\paren {z \cdot x} < \paren {z \cdot y}$, then $y \ne x$ and $y \not < x$, so $x < y$ by Ordinal Membership is Trichotomy.