# Ordinal Exponentiation via Cantor Normal Form/Limit Exponents

## Theorem

Let $x$ and $y$ be ordinals.

Let $x$ and $y$ be limit ordinals.

Let $\left\langle{a_i}\right\rangle$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.

Let $\left\langle{b_i}\right\rangle$ be a sequence of natural numbers.

Then:

$\displaystyle \left({\sum_{i \mathop = 1}^n x^{a_i} \times b_i}\right)^y = x^{a_1 \mathop \times y}$

## Proof

$\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_i} \times b_i }\right) \le x^{a_1} \times \left({b_1 + 1}\right)$

Furthermore:

$\displaystyle x^{a_1} \le \sum_{i \mathop = 1}^n \left({ x^{a_i} \times b_i }\right)$

It follows that:

 $\displaystyle \left({ x^{a_1} }\right)^y$ $\le$ $\displaystyle \left({\sum_{i \mathop = 1}^n x^{a_i} \times b_i}\right)^y$ Subset is Right Compatible with Ordinal Exponentiation $\displaystyle$ $\le$ $\displaystyle \left({x^{a_1} \times \left({b_1 + 1}\right)}\right)^y$ Subset is Right Compatible with Ordinal Exponentiation $\displaystyle$ $=$ $\displaystyle x^{a_1 \times y}$ Ordinal Exponentiation of Terms

It follows that:

 $\displaystyle x^{a_1 \mathop + y}$ $\le$ $\displaystyle \left({ \sum_{i \mathop = 1}^n x^{a_i} b_i }\right)^y$ by Ordinal Power of Power $\displaystyle$ $\le$ $\displaystyle x^{a_1 \mathop + y}$ Proven above $\displaystyle \implies \ \$ $\displaystyle \left({\sum_{i \mathop = 1}^n x^{a_i} b_i}\right)^y$ $=$ $\displaystyle x^{a_1 \mathop + y}$ Definition of Set Equality

$\blacksquare$