Summation of General Logarithms

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

Let $\ds \prod_{\map R i} a_i$ denote a product over $R$.

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

Then:
 * $\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$

Proof
The proof proceeds by induction.

First let:
 * $S := \set {a_i: \map R i}$

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:
 * $S = \set {s_1, s_2, \ldots, s_m}$

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

$\map P 0$ is the case:

Basis for the Induction
$\map P 1$ is the case:

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\ds \map {\log_b} {\prod_{i \mathop = 1}^k s_i} = \sum_{i \mathop = 1}^k \log_b s_i$

from which it is to be shown that:
 * $\ds \map {\log_b} {\prod_{i \mathop = 1}^{k + 1} s_i} = \sum_{i \mathop = 1}^{k + 1} \log_b s_i$

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \map {\log_b} {\prod_{i \mathop = 1}^n s_i} = \sum_{i \mathop = 1}^n \log_b s_i$

Hence, by definition of $S$:


 * $\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$