Ordinal Multiplication via Cantor Normal Form/Limit Base

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

Let $x$ be a limit ordinal and let $y > 0$.

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

Let $\langle b_i \rangle$ be a sequence of natural numbers.

Then:


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

Proof
The proof shall proceed by finite induction on $n$:

For all $n \in \N_{\le 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_i} b_i }\right) \times x^y = x^{a_1 \mathop + y}$

Since $x$ is a limit ordinal, it follows that $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiation.

Basis for the Induction

 * $P\left({ 1 }\right)$ says that:


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

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{i \mathop = 1}^k \left({ x^{a_i} b_i }\right) \times x^y = x^{a_1 + y}$

Then we need to show:
 * $\displaystyle \sum_{i \mathop = 1}^{k \mathop + 1} \left({ x^{a_i} b_i }\right) \times x^y = x^{a_1 \mathop + y}$

Induction Step
This is our induction step:

By Upper Bound of Ordinal Sum, it follows that:


 * $\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_{i\mathop + 1} } b_{i \mathop + 1} }\right) < x^{a_1}$

By Membership is Left Compatible with Ordinal Multiplication:

But:

Therefore:

Also see

 * Definition:Cantor Normal Form