Ordinal Multiplication is Left Cancellable

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $x, y, z$ be ordinals.

Let $z \ne 0$.


Then:

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

That is, ordinal multiplication is left cancellable.


Proof

For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably.

This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.


Since $z \ne 0$ and $z \not < 0$, $0 < z$ by Ordinal Membership is Trichotomy.

Note that:

\(\ds x < y \land z > 0\) \(\implies\) \(\ds \paren {z \cdot x} < \paren {z \cdot y}\) Membership is Left Compatible with Ordinal Multiplication
\(\ds y < x \land z > 0\) \(\implies\) \(\ds \paren {z \cdot y} < \paren {z \cdot x}\) Membership is Left Compatible with Ordinal Multiplication

In addition:

\(\ds \paren {z \cdot x} = \paren {z \cdot y}\) \(\implies\) \(\ds \paren {z \cdot x} \not < \paren {z \cdot y} \land \paren {z \cdot y} \not < \paren {z \cdot x}\) No Membership Loops

This contradicts the consequents of the first two equations, so:

\(\ds \) \(\leadsto\) \(\ds x \not < y \land y \not < x\) Rule of Transposition, Praeclarum Theorema
\(\ds \) \(\leadsto\) \(\ds x = y\) Ordinal Membership is Trichotomy

$\blacksquare$


Also see


Sources