WFF of PropLog is Balanced

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Theorem

Let $\mathbf A$ be a WFF of propositional logic.


Then $\mathbf A$ is a balanced string.


Proof

We will prove by strong induction on $n$ that:

All WFFs of length $n$ are balanced.


Let $\map l {\mathbf A}$ denote the number of left brackets in a string $\mathbf A$.

Let $\map r {\mathbf A}$ denote the number of right brackets in a string $\mathbf A$.


Induction Basis

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

By definition, none of these consist of brackets.


So every WFF $\mathbf A$ of length $1$ has $\map l {\mathbf A} = \map r {\mathbf A} = 0$.

So every WFF of length $1$ is balanced.

This provides the induction basis.


Induction Hypothesis

Now, suppose all WFFs of propositional logic of length at most $n$ are balanced.

This provides the induction hypothesis.


Induction Step

Let $\mathbf A$ be a WFF of propositional logic of length $n + 1$.


The following cases are to be considered (with $\mathbf B$ and $\mathbf C$ WFFs):

$\mathbf A = \neg \mathbf B$
$\mathbf A = \paren {\mathbf B \circ \mathbf C}$, where $\circ$ is one of the binary connectives


Suppose $\mathbf A = \neg \mathbf B$.

Then $\mathbf B$ is a WFF of length $n$.

By the induction hypothesis, it is hence balanced.

Since $\mathbf A$ contains the same brackets as $\mathbf B$, it must also be balanced.


Suppose now $\mathbf A = \paren {\mathbf B \circ \mathbf C}$, where $\circ$ is one of the binary connectives.

Then:

$\map l {\mathbf A} = \map l {\mathbf B} + \map l {\mathbf C} + 1$
$\map r {\mathbf A} = \map r {\mathbf B} + \map r {\mathbf C} + 1$.

Now, $\mathbf B$ and $\mathbf C$ are both of length strictly less than $n + 1$, and thus balanced, by the induction hypothesis.

That is:

$\map l {\mathbf B} = \map r {\mathbf B}$
$\map l {\mathbf C} = \map r {\mathbf C}$

It follows that also:

$\map l {\mathbf A} = \map r {\mathbf A}$

We also note that the first symbol of $\paren {\mathbf B \circ \mathbf C}$ is $($.

Hence as $\mathbf B$ and $\mathbf C$ are both balanced, every prefix of $\paren {\mathbf B \circ \mathbf C}$ must have strictly more left brackets than right brackets.

This precisely states that $\mathbf A$ is balanced.


We assumed that all WFFs of length $n$ and less are balanced.

We have proved as a consequence all WFFs of length $n + 1$ are balanced.

$\Box$


The result now follows by strong induction.

$\blacksquare$


Sources

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