Ordinal Multiplication via Cantor Normal Form/Limit Base

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

Let $x$ be a limit ordinal.

Let $y > 0$.

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

Let $\sequence {b_i}$ be a sequence of finite ordinals.

Then:


 * $\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$

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

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

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
$\map P 1$ is the statement:


 * $\ds \sum_{i \mathop = 1}^1 \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\ds \sum_{i \mathop = 1}^k \paren {x^{a_i} b_i} \times x^y = x^{a_1 + y}$

Then we need to show:
 * $\ds \sum_{i \mathop = 1}^{k \mathop + 1} \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$

Induction Step
This is our induction step:

By Upper Bound of Ordinal Sum, it follows that:


 * $\ds \sum_{i \mathop = 1}^n \paren {x^{a_{i \mathop + 1} } b_{i \mathop + 1} } < x^{a_1}$

By Membership is Left Compatible with Ordinal Multiplication:

But:

Therefore:

Also see

 * Definition:Cantor Normal Form