Ordinal Multiplication is Left Cancellable
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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
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.20$