Prefix of WFF of PropLog is not WFF

Theorem
Let $\mathbf A$ be a propositional WFF.

Let $\mathbf S$ be an initial part of $\mathbf A$.

Then $\mathbf S$ is not a propositional WFF.

Proof
Let $l \left({\mathbf Q}\right)$ denote the length of a string $\mathbf Q$.

By definition, $\mathbf S$ is an initial part of $\mathbf A$ iff $\mathbf A = \mathbf{ST}$ for some non-null string $\mathbf T$.

Thus we note that $l \left({\mathbf S}\right) < l \left({\mathbf A}\right)$.

Let $\mathbf A$ be a WFF such that $l \left({\mathbf A}\right) = 1$.

Then for an initial part $\mathbf S$, $l \left({\mathbf S}\right) < 1 = 0$.

That is, $\mathbf S$ must be the null string, which is not a WFF.

So the result holds for WFFs of length $1$.

Now, we assume an induction hypothesis: that the result holds for all WFFs of length $k$ or less.

Let $\mathbf A$ be a WFF such that $l \left({\mathbf A}\right) = k+1$.

Suppose $\mathbf D$ is an initial part of $\mathbf A$ which happens to be a WFF.

That is, $\mathbf A = \mathbf{DT}$ where $\mathbf T$ is non-null.

There are two cases:


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

$\mathbf D$ is a WFF starting with $\neg$, so $\mathbf D = \neg \mathbf E$ where $\mathbf E$ is also a WFF.

We remove the initial $\neg$ from $\mathbf A = \mathbf{DT}$ to get $\mathbf B = \mathbf{ET}$.

But then $\mathbf B$ is a WFF of length $k$ which has $\mathbf E$ as an initial part which is itself a WFF.

This contradicts the induction hypothesis.

Therefore no initial part of $\mathbf A = \neg \mathbf B$ can be a WFF.


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

In this case, $\mathbf D$ is a WFF starting with $($, so $\mathbf D = \left({\mathbf E * \mathbf F}\right)$ for some binary connective $*$ and some WFFs $\mathbf E$ and $\mathbf F$.

Thus $\mathbf B \circ \mathbf C) = \mathbf E * \mathbf F) \mathbf T$.

Both $\mathbf B$ and $\mathbf E$ are WFFs of length less than $k+1$.

By the inductive hypothesis, then, neither $\mathbf B$ nor $\mathbf E$ can be an initial part of the other.

But since both $\mathbf B$ and $\mathbf E$ start at the same place in $\mathbf A$, they must be the same: $\mathbf B = \mathbf E$.

Therefore $\mathbf B \circ \mathbf C) = \mathbf B * \mathbf F) \mathbf T$.

So $\circ = *$ and $\mathbf C) = \mathbf F) \mathbf T$.

But then the WFF $\mathbf F$ is an initial part of the WFF $\mathbf C$ of length less than $k+1$.

This contradicts our inductive hypothesis.

Therefore no initial part of $\mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ can be a WFF.

So no initial part of any WFF of length $k+1$ can be a WFF.

The result follows by strong induction.