WFF of PropLog is Balanced

Theorem
Every WFF of propositional calculus has the same number of left brackets as right brackets.

A string with the same number of left as right brackets is (in this context) referred to as balanced.

Proof
Let:
 * $l \left({\mathbf Q}\right)$ denote the number of left brackets in a string $\mathbf Q$
 * $r \left({\mathbf Q}\right)$ denote the number of right brackets in a string $\mathbf Q$.

The only WFFs of PropCalc of Length 1‎ are elements of the alphabet and the symbols $\bot$ and $\top$.

None of these consist of brackets.

So every WFF $\mathbf D$ of length $1$ have $l \left({\mathbf D}\right) = r \left({\mathbf D}\right) = 0$.

So every WFF of length $1$ is balanced.

Now, suppose all strings of length $k$ and less are balanced.

Let $\mathbf A$ be a WFF of length $k+1$.

There are two cases:


 * $\mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$ and hence balanced.

So $\mathbf A$ contains the same brackets as $\mathbf B$ and so must itself be balanced.


 * $\mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ where $\circ$ is one of the binary connectives.

Then:
 * $l \left({\mathbf A}\right) = l \left({\mathbf B}\right) + l \left({\mathbf C}\right) + 1$
 * $r \left({\mathbf A}\right) = r \left({\mathbf B}\right) + r \left({\mathbf C}\right) + 1$.

Now, $\mathbf B$ and $\mathbf C$ are both shorter than $k+1$ and are therefore balanced.

So $l \left({\mathbf B}\right) = r \left({\mathbf B}\right)$ and $l \left({\mathbf C}\right) = r \left({\mathbf C}\right)$

It follows that $l \left({\mathbf A}\right) = r \left({\mathbf A}\right)$ and so $\mathbf A$ is balanced.

We assumed that all WFFs of length $k$ and less are balanced, and demonstrated that as a consequence all WFFs of length $k+1$ are balanced.

The result follows by strong induction.