Prefix of Fibonacci String

Theorem
Let $n \in \Z_{>1}$.

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

Let $m \in \Z$ such that $1 < m \le n$.

Let $F_m$ denote the $m$th Fibonacci number.

The initial part of $S_n$ of length $F_m$ is the Fibonacci string $S_m$.

Proof
The proof proceeds by strong induction.

For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition:
 * for all $m \in \Z$ such that $1 < m \le n$, the initial part of $S_n$ of length $F_m$ is $S_m$.

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

For $m = 2$, $S_3$ starts with $S_2 = \text{b}$, which is the initial part of $S_3$ of length $F_2 = 1$.

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

$P \left({4}\right)$ is the case:
 * $S_4 = \text{bab}$

For $m = 2$, $S_4$ starts with $S_2 = \text{b}$, which is the initial part of $S_4$ of length $F_2 = 1$.

For $m = 3$, $S_4$ starts with $S_3 = \text{ba}$, which is the initial part of $S_4$ of length $F_3 = 2$.

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:
 * for all $m \in \Z$ such that $1 < m \le k$, the initial part of $S_k$ of length $F_m$ is $S_m$.

from which it is to be shown that:
 * for all $m \in \Z$ such that $1 < m \le k + 1$, the initial part of $S_{k + 1}$ of length $F_m$ is $S_m$.

Induction Step
This is the induction step:

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

concatenated.

Thus the initial part of $S_{k + 1}$ of length $F_k$ is $S_k$.

By the induction hypothesis, for all $m \in \Z$ such that $1 < m < k - 1$, the initial part of $S_k$ of length $F_m$ is $S_m$.

But we also have that the initial part of $S_{k + 1}$ of length $F_k$ is $S_k$.

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 $m \in \Z$ such that $1 < m \le n$, the initial part of $S_n$ of length $F_m$ is $S_m$.