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:


 * $\displaystyle x < y \implies ( z + x ) < ( z + y )$

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 $x < \varnothing \implies ( z + x ) < ( z + \varnothing )$ follows from propositional logic.