Fibonacci String Begins with ba

Theorem
Let $S_n$ be a Fibonacci string of length $n$.

Then for $n \ge 3$, $S_n$ begins with $\text {ba}$.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
 * $S_n$ begins with $\text {ba}$

We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these begin with $\text{ba}$.

Basis for the Induction
$\map P 3$ is the case:
 * $S_3 = \text {ba}$

Thus $\map P 3$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $S_k$ begins with $\text {ba}$

from which it is to be shown that:
 * $S_{k + 1}$ begins with $\text {ba}$

Induction Step
This is the induction step:

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

concatenated.

So $S_{k + 1}$ begins with $S_k$.

By the induction hypothesis, $S_k$ begins with $\text {ba}$.

Thus $S_{k + 1}$ likewise begins with $\text {ba}$.

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

Therefore:
 * for all $n \in \Z$ such that $n \ge 3$, $S_n$ begins with $\text {ba}$.