Membership is Left Compatible with Ordinal Addition

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

Let $<$ denote membership $\in$, since $\in$ is a strict well-ordering on the ordinals.

Then:


 * $x < y \implies \paren {z + x} < \paren {z + y}$

Proof
The proof proceeds by transfinite induction on $y$.

In the proof, we shall use $\in$, $\subsetneq$, and $<$ interchangeably.

We are justified in this by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.

Base Case
The conclusion:
 * $x < \O \implies \paren {z + x} < \paren {z + \O}$

follows from propositional logic.