Summation of General Logarithms

Theorem
Let $R: \Z \to \left\{ {\mathrm T, \mathrm F}\right\}$ be a propositional function on the set of integers.

Let $\displaystyle \prod_{R \left({i}\right)} a_i$ denote a product over $R$.

Let the fiber of truth of $R$ be finite.

Then:
 * $\displaystyle \log_b \left({\prod_{R \left({i}\right)} a_i}\right) = \sum_{R \left({i}\right)} \log_b a_i$

Proof
The proof proceeds by induction.

First let:
 * $S := \left\{ {a_i: R \left({i}\right)}\right\}$

We have that $S$ is finite.

Hence the contents of $S$ can be well-ordered, by Finite Totally Ordered Set is Well-Ordered.

Let $S$ have $m$ elements, identified as:
 * $S = \left\{ {s_1, s_2, \ldots, s_m}\right\}$

For all $n \in \Z_{\ge 0}$ such that $n \le m$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \log_b \left({\prod_{i \mathop = 1}^n s_i}\right) = \sum_{i \mathop = 1}^n \log_b s_i$

$P \left({0}\right)$ is the case:

Basis for the Induction
$P \left({1}\right)$ is the case:

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\displaystyle \log_b \left({\prod_{i \mathop = 1}^k s_i}\right) = \sum_{i \mathop = 1}^k \log_b s_i$

from which it is to be shown that:
 * $\displaystyle \log_b \left({\prod_{i \mathop = 1}^{k + 1} s_i}\right) = \sum_{i \mathop = 1}^{k + 1} \log_b s_i$

Induction Step
This is the induction step:

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

Therefore:
 * $\displaystyle \log_b \left({\prod_{i \mathop = 1}^n s_i}\right) = \sum_{i \mathop = 1}^n \log_b s_i$

Hence, by definition of $S$:


 * $\displaystyle \log_b \left({\prod_{R \left({i}\right)} a_i}\right) = \sum_{R \left({i}\right)} \log_b a_i$