The proof proceeds by strong induction on the length of a WFF of propositional logic. Let $l \left({\mathbf Q}\right)$ denote the length of a string $\mathbf Q$. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :The initial part of $\mathbf A$ such that $l \left({\mathbf A}\right) = n$ is not a WFF of propositional logic. By definition, $\mathbf S$ is an initial part of $\mathbf A$ {{iff}} $\mathbf A = \mathbf {S T}$ for some non-null string $\mathbf T$. Thus we note that $l \left({\mathbf S}\right) < l \left({\mathbf A}\right)$. === 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 $1$. That is, $P \left({1}\right)$ is true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that for $k \ge 1$, if $P \left({j}\right)$ is true for all $j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :The initial part of $\mathbf A$ such that $l \left({\mathbf A}\right) = k$ is not a WFF of propositional logic. Then we need to show: :The initial part of $\mathbf A$ such that $l \left({\mathbf A}\right) = k + 1$ is not a WFF of propositional logic. === 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, $\mathbf A = \mathbf{D T}$ where: : $\mathbf D$ is a WFF : $\mathbf T$ is non-null. There are two cases: $(1): \quad \mathbf A = \neg \mathbf B$ where $\mathbf B$ is a WFF of length $k$. Thus $\mathbf D$ 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 $\mathbf A = \mathbf{D T}$ to get: :$\mathbf B = \mathbf{E T}$ 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. {{qed|lemma}} $(2): \quad \mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ 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 connectives $*$ 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 $k + 1$. Therefore, by the induction hypothesis, 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 induction hypothesis. Therefore no initial part of $\mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ can be a WFF. {{qed|lemma}} As $\mathbf A$ is arbitrary, it follows that no initial part of any WFF of length $k + 1$ can be a WFF. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by strong induction. Therefore, for all $n \in \N_{> 0}$, the initial part of $\mathbf A$ such that $l \left({\mathbf A}\right) = n$ is not a WFF of propositional logic. Hence the result. {{qed}}