Membership is Left Compatible with Ordinal Exponentiation

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

Suppose $1 < z$.

Then:


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

Sufficient Condition
The proof shall proceed by Transfinite Induction on $y$.

Basis for the Induction
It is a contradiction that $x < 0$ by the definition of empty set.

This proves the basis for the induction.

Induction Step
The inductive step states that $x < y \implies z^x < z^y$.

Then, if $x < y^+$, then $x = y \lor x < y$.

Lemma
$z^y < z^y \times y$ by Membership is Left Compatible with Ordinal Multiplication.

So $z^y < z^{y^+}$ by the definition of ordinal exponentiation.

Suppose $x = y$. Then $z^x < z^{y^+}$.

Suppose $x < y$. Then:

In either case, the inductive step holds.

This proves the induction step.

Limit Case
The inductive hypothesis states that:


 * $\forall w \in y: z^x < z^w$ for $x < w$

Take any $x < y$.

By Limit Ordinal Equals its Union, it follows that $x < w$ for some $w \in y$.

By the inductive hypothesis, $z^x < z^w$.

Since $w \in y$, it follows that:

This proves the limit case.

Necessary Condition
Note that the last part proves that:


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

Moreover:


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

Therefore, by the fact that $\in$ is a strong ordering:


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

By the Rule of Transposition:


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