Ordinal is Less than Sum

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

Then:
 * $x \le \left({x + y}\right)$
 * $x \le \left({y + x}\right)$

Proof
By Ordinal Addition by Zero:


 * $x = \left({x + \varnothing}\right) = \left({\varnothing + x}\right)$

By Membership is Left Compatible with Ordinal Addition:


 * $\varnothing < y \implies x < \left({x + y}\right)$

But if $y = \varnothing$, then it is clear the inequality holds as well.

One of these cases must hold by Empty Set is Subset of All Sets.

So in either case:


 * $x \le \left({x + y}\right)$

Similarly, by Subset Right Compatible with Ordinal Addition, we have:


 * $\varnothing \le y \implies x \le \left({y + x}\right)$

The fact that $\varnothing \le y$ is clear from Empty Set is Subset of All Sets, so:


 * $x \le \left({y + x}\right)$