# Subset is Right Compatible with Ordinal Multiplication

## Theorem

Let $x, y, z$ be ordinals.

Then:

$x \le y \implies \paren {x \cdot z} \le \paren {y \cdot z}$

## Proof

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

### Basis for the Induction

By definition of ordinal multiplication:

$x \cdot 0 = 0$

so the statement simply reduces to:

$0 \le 0$

This proves the basis for the induction.

### Induction Step

 $\ds x$ $\le$ $\ds y$ $\ds \leadsto \ \$ $\ds x \cdot z$ $\le$ $\ds y \cdot z$ Inductive Hypothesis $\ds \leadsto \ \$ $\ds \paren {x \cdot z} + x$ $\le$ $\ds \paren {y \cdot z} + y$ Membership is Left Compatible with Ordinal Addition and Subset is Right Compatible with Ordinal Addition $\ds \leadsto \ \$ $\ds x \cdot z^+$ $\le$ $\ds y \cdot z^+$ Definition of Ordinal Multiplication

This proves the induction step.

### Limit Case

 $\ds x$ $\le$ $\ds y$ $\ds \leadsto \ \$ $\ds \forall w \in z: \,$ $\ds x \cdot w$ $\le$ $\ds y \cdot w$ Inductive Hypothesis $\ds \leadsto \ \$ $\ds \bigcup_{w \mathop \in z} \paren {x \cdot w}$ $\le$ $\ds \bigcup_{w \mathop \in z} \paren {y \cdot w}$ Indexed Union Subset $\ds \leadsto \ \$ $\ds x \cdot z$ $\le$ $\ds y \cdot z$ Definition of Ordinal Multiplication

This proves the limit case.

$\blacksquare$