Upper Bound of Ordinal Sum

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

Suppose $x > 1$.

Let $\sequence {a_n}$ be a finite sequence of ordinals such that:


 * $a_n < x$ for all $n$

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


 * $b_n < y$ for all $n$

Then:


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

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

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

Basis for the Induction

 * $\map P 0$ says that:


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

This is our basis for the induction.

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

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

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

Induction Step
This is our induction step:

By the inductive hypothesis:


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

Then:

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

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