Prefix of WFF of Predicate Logic is not WFF

Theorem
Let $\mathbf A$ be a WFF of predicate calculus.

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

Then $\mathbf S$ is not a WFF of predicate 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 $\mathbf A$ iff $\mathbf A = \mathbf{ST}$ for some non-null string $\mathbf T$.

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

The proof proceeds by induction on $l \left({\mathbf A}\right)$.

Basis for the Induction
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 $1$.

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge n_0$.

Assume the result holds for all WFFs of length $k$ or less.

This is our induction hypothesis.

Induction Step
This is our induction step:

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.

We need to investigate the following cases:
 * $(1): \quad \mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$.
 * $(2): \quad \mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ where $\circ$ is one of the binary connectives.
 * $(3): \quad \mathbf A = p \left({u_1, u_2, \ldots, u_n}\right)$, where $u_1, u_2, \ldots, u_n$ are individual symbols, and $p \in \mathcal{P}_n$.
 * $(4): \quad \mathbf A = ( Q x: \mathbf B )$, where $\mathbf B$ is a WFF of length $k-5$, $Q$ is a quantifier ($\forall$ or $\exists$) and $x$ is a variable.

We deal with these one by one.

Cases $(1)$ and $(2)$ are covered by the argument in No Initial Part of a WFF of PropCalc is a WFF.


 * $(3): \quad$ The atomic WFF $\mathbf A = p \left({u_1, u_2, \ldots, u_n}\right)$:

Here we have that $\mathbf D$ would be a string of the form:
 * $p$ where $p$ is an $n$-ary predicate symbol. This can not be a WFF.
 * $p ($ which is also not a WFF.
 * $p (u_1, u_2, \ldots, u_k$ which can not be a WFF.
 * $p (u_1, u_2, \ldots, u_k,$ which can not be a WFF.


 * $(4): \quad \mathbf A = ( Q x: \mathbf B )$:

$\mathbf D$ can not be $($, $( Q$, $( Q x$ or $( Q x:$ as none of these are WFFs.

Thus $\mathbf D$ is a WFF, starting with $( Q x: $; hence $\mathbf D = ( Q x: \mathbf E )$, where $\mathbf E$ is also a WFF.

We remove the initial $( Q x: $ and the last closing parenthesis $)$ from $\mathbf A = \mathbf{DT}$ to get $\mathbf B = \mathbf{ET}'$ (where $\mathbf T'$ results from $\mathbf T$ by removing the last closing parenthesis).

Recall that $\mathbf B$ is a WFF of length $k-5$.

It has $\mathbf E$ as an initial part, which is itself a WFF.

This contradicts the induction hypothesis.

Therefore no initial part of $\mathbf A = ( Q x: \mathbf{B} )$ can be a WFF.

Thus all four cases have been investigated, and we have found that no initial part of any WFF of length $k+1$ can be a WFF.

The result follows by the Second Principle of Mathematical Induction.