Upper Bound of Ordinal Sum

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

Suppose $x > 1$.

Let $\langle a_n \rangle$ be a finite sequence of ordinals such that:


 * $a_n < x$ for all $n$

Let $\left\langle{b_n}\right\rangle$ be a strictly decreasing finite sequence of ordinals such that:


 * $b_n < y$ for all $n$

Then:


 * $\displaystyle \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$

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:
 * $\displaystyle \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$

Basis for the Induction

 * $P \left({0}\right)$ says that:


 * $\displaystyle \sum_{i \mathop = 1}^0 x^{b_i} a_i < x^y$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{i \mathop = 1}^k x^{b_i} a_i < x^y$

Then we need to show:
 * $\displaystyle \sum_{i \mathop = 1}^{k \mathop + 1} x^{b_i} a_i < x^y$

Induction Step
This is our induction step:

By the inductive hypothesis:


 * $\displaystyle \sum_{i \mathop = 1}^k x^{b_{i + 1}} a_{i + 1} < x^{b_1}$ since $b_1$ is greater than all $b_i$ for $i > 1$.

Then:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \sum_{i \mathop = 1}^n a_i x^{b_i} < x^y$