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 $P \left({n}\right)$ 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
$P \left({3}\right)$ is the case:
 * $S_3 = \text{ba}$

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

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ 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 $P \left({k}\right) \implies P \left({k + 1}\right)$ 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 $\text{ba}$.