Ordinal Sum of Powers

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

Then:


 * $x^y \times x^z = x^{y + z}$

Proof
The proof shall proceed by Transfinite Induction on $z$.

Basis for the Induction

 * $x^0 = 1$ for all $x$

This proves the basis for the induction.

Induction Step
Suppose that $x^y \times x^z = x^{y + z}$.

Then:

This proves the induction step.

Limit Case
Suppose that $\forall w \in z: x^y \times x^w = x^{y + w}$ for limit ordinal $z$.

Conversely:

But this means that $u$ is bounded above by $x^v$ for some $v \in z$.

Thus there exists a $v \in z$ such that:


 * $w \le \left({ x^y \times x^v }\right)$

By Supremum Inequality for Ordinals, it follows that:


 * $\left({ x^y \times x^z }\right) \le \bigcup_{w \in z} \left({ x^y \times x^w }\right)$

Conversely:

Thus, $u$ is bounded above by $\left({ y + v }\right)$ for some $v \in z$.

Therefore:


 * $x^u \le x^{y + v}$

By Supremum Inequality for Ordinals, it follows that:


 * $x^{y + z} \le \bigcup_{w \in z} x^{y + w}$

Thus, by definition of set equality:


 * $x^{y + z} = \bigcup_{w \in z} x^{y + w}$

Combining the results of the first and second lemmas for the limit case:


 * $x^{y+z} = x^y \times x^z$

This proves the limit case.