Ordinal Addition is Left Cancellable

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $x, y, z$ be ordinals.


Then:

$\paren {z + x} = \paren {z + y} \implies x = y$

That is, ordinal addition 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.


Note that:

\(\ds x < y\) \(\implies\) \(\ds \paren {z + x} < \paren {z + y}\) Membership is Left Compatible with Ordinal Addition
\(\ds y < x\) \(\implies\) \(\ds \paren {z + y} < \paren {z + x}\) Membership is Left Compatible with Ordinal Addition


However:

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


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

\(\ds \) \(\implies\) \(\ds x \not < y \land y \not < x\) Rule of Transposition
\(\ds \) \(\implies\) \(\ds x = y\) Ordinal Membership is Trichotomy

$\blacksquare$


Warning


This page or section has statements made on it that ought to be extracted and proved in a Theorem page.
In particular: Extract this into its own page and link to it in also see
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by creating any appropriate Theorem pages that may be needed.
To discuss this page in more detail, feel free to use the talk page.


Ordinal addition is not right cancellable.

By Finite Ordinal Plus Transfinite Ordinal

$\paren {1 + \omega} = \paren {0 + \omega}$

but $1 \ne 0$.


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Ordinal_Addition_is_Left_Cancellable&oldid=567898"