Count of a's and b's in Fibonacci String

Theorem
Let $S_n$ denote the $n$th Fibonacci string.

Then for $n \ge 3$, $S_n$ has:
 * $F_{n - 2}$ instances of $\text a$
 * $F_{n - 1}$ instances of $\text b$.

Proof
The proof proceeds by strong induction.

For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition:
 * $S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$.

Basis for the Induction
$P \left({3}\right)$ is the case:
 * $S_n = \text {ba}$

It can be seen that $S_n$ has $F_1 = 1$ instance of $\text a$ and $F_2 = 1$ instance of $\text b$.

Thus $P \left({3}\right)$ is seen to hold.

$P \left({4}\right)$ is the case:
 * $S_n = \text {bab}$

It can be seen that $S_n$ has $F_2 = 1$ instance of $\text a$ and $F_3 = 2$ instances of $\text b$.

Thus $P \left({4}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $4 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true.

This is the induction hypothesis:
 * $S_k$ has $F_{k - 2}$ instances of $\text a$ and $F_{k - 1}$ instances of $\text b$

and:
 * $S_{k - 1}$ has $F_{k - 3}$ instances of $\text a$ and $F_{k - 2}$ instances of $\text b$

from which it is to be shown that:
 * $S_{k + 1}$ has $F_{k - 1}$ instances of $\text a$ and $F_k$ instances of $\text b$.

Induction Step
This is the induction step:

By definition of Fibonacci string:
 * $S_{k + 1} = S_k S_{k - 1}$

concatenated.

By the induction hypothesis:
 * $S_k$ has $F_{k - 2}$ instances of $\text a$ and $F_{k - 1}$ instances of $\text b$

and:
 * $S_{k - 1}$ has $F_{k - 3}$ instances of $\text a$ and $F_{k - 2}$ instances of $\text b$

So:
 * $S_{k + 1}$ has $F_{k - 2} + F_{k - 3}$ instances of $\text a$

and so by definition of Fibonacci numbers:
 * $S_{k + 1}$ has $F_{k - 1}$ instances of $\text a$

and:
 * $S_{k + 1}$ has $F_{k - 1} + F_{k - 2}$ instances of $\text b$

and so by definition of Fibonacci numbers:
 * $S_{k + 1}$ has $F_k$ instances of $\text b$.

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

Therefore:
 * for all $n \in \Z$ such that $n \ge 3$: $S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$.