Prefix of WFF of PropLog is not WFF

Theorem
Let $$\mathbf{A}$$ be a WFF of propositional calculus.

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

Then $$\mathbf{S}$$ is not a WFF of propositional calculus.

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

By definition, $$\mathbf{S}$$ is an initial part of $$\mathbf{A}$$ if $$\mathbf{A} = \mathbf{ST}$$ 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.