Condition for Membership is Right Compatible with Ordinal Exponentiation

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

Let $z$ be the successor of some ordinal $w$.

Then:


 * $x < y \iff x^z < y^z$

Sufficient Condition
By Subset Right Compatible with Ordinal Exponentiation, it follows that $x^w \le y^w$

Thus, the sufficient condition is satisfied.

Necessary Condition
Suppose $x^z < y^z$.


 * $y \le x \implies y^z \le x^z$

This contradicts the fact that $x^z < y^z$ so $y \not \le x$.

Therefore, $x < y$ by Ordinal Membership Trichotomy.