Equivalence of Definitions of Balanced String

Proof
The proof proceeds by strong induction on the length of a string.

For all $n \in \N$, let $\map P n$ be the proposition:
 * Every balanced string $S_n$ of length $n$ is a balanced string by definition $1$ it been generated by definition $2$.

Basis for the Induction
$\map P 0$ is the case:
 * Every balanced string $S_0$ of length $0$ is a balanced string by definition $1$ it been generated by definition $2$.

By definition $2$ of a balanced string:
 * $(1): \quad$ The null string $\epsilon$ is a balanced string.

This is the only string of length $0$.

$\epsilon$ is seen vacuously to be a balanced string by definition $1$.

Thus $\map P 0$ is seen to hold.

$\map P 1$ is the case:
 * Every balanced string $S_1$ of length $n$ is a balanced string by definition $1$ it been generated by definition $2$.

Let $a$ be an arbitrary symbol.

We use $\map P 0$ and $\map P 1$ together to form the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$ and $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

This is the induction hypothesis:
 * Every balanced string $S_j$ of length $j$ such that $j \le k$ is a balanced string by definition $1$ it been generated by definition $2$.

from which it is to be shown that:
 * Every balanced string $S_{k + 1}$ of length $k + 1$ is a balanced string by definition $1$ it been generated by definition $2$.

Induction Step
This is the induction step:

Let $x$ be a balanced string of length $k - 1$ according to either definition $1$ or definition $2$.

By the induction hypothesis, $x$ is therefore a balanced string by both definition $1$ and definition $2$.

Thus we have shown that for all $k \in \N$, if $\sqbrk y$ reads the same backwards as it does forwards, then it has been created by one of the rules $(1)$, $(2)$ or $(3)$.

So $\paren {\forall j: 0 \le j \le k: \map P j} \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:
 * For all $n \in \N$: every balanced string $S_n$ of length $n$ is a balanced string by definition $1$ it been generated by definition $2$.