Ordinal is Less than Sum

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


 * $\displaystyle x \le ( x + y )$
 * $\displaystyle x \le ( y + x )$

Proof
By Ordinal Addition by Zero:


 * $\displaystyle x = ( x + \varnothing ) = ( \varnothing + x )$

By Membership is Left Compatible with Ordinal Addition:


 * $\displaystyle \varnothing < y \implies x \lt ( x + y )$

But if $y = \varnothing$, then it is clear the inequality holds as well. And one of these cases must hold by Empty Set Subset of All. So in either case:


 * $\displaystyle x \le ( x + y )$

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


 * $\displaystyle \varnothing \le y \implies x \le ( y + x )$

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


 * $\displaystyle x \le ( y + x )$