I assume that a wff is either
- a sentence symbol of length 1
- of the form $(\neg \alpha)$ of length $L+3$ where $\alpha$ is a wff of length $L$
- of the form $(\alpha\circ\beta)$ of length $L_1+L_2+3$, where $\alpha,\beta$ are wff of lengths $L_a,L_2$ and $\circ\in\{\land\lor,\to,\leftrightarrow\}$
Then use structiral induction to show
Proposition. If $\alpha$ is a wff then its length $L(\alpha)$ is in $\mathbb N\setminus\{2,3,6\}$
Proof: We assume that $L(\alpha)\in\mathbb N$ is already known, so it suffices to show $L(\alpha)\notin\{2,3,6\}$.
- If $\alpha$ is a sentence symbol then $L(\alpha)=1\in\mathbb N\setminus\{2,3,6\}$
- If $\alpha$ is of the form $(\neg\beta)$, then $L(\alpha)=L(\beta)+3$. If we assume that $L(\alpha)\in\{2,3,6\}$ we conclude $L(\beta)\in \{-1,0,3\}$, contradicting the induction hypothesis $L(\beta)\in\mathbb N\setminus\{2,3,6\}$. Therefore $L(\alpha)\in\mathbb N\setminus\{2,3,6\}$ also in this case
- If $\alpha$ is of the form $(\beta\circ\gamma)$, we have $L(\alpha)=L(\beta)+L(\gamma)+3$, especially $L(\alpha)\ge 5$. We need only exclude $L(\alpha)=6$, which would require that of of the sub-lengths is $1$ and the other os $2$, but that is not possible.
Proposition. If $n\in\mathbb N\setminus\{2,3,6\}$, then there exists a wff $\alpha$ with $L(\alpha)=n$.
Proof: $A$, $(\neg A)$, $(A\land A)$ are examples of lengths $1,4,5$.
$((A\land A)\land A)$ is of length $9$
Since negating increases the length by $3$, we can obtain any length $n=4+3k$ and any length $n=5+3k$ and any lenbgth $n=9+3k$ with $k\ge 0$.
The second step is trivial - $\alpha$ has no negations that weren't in $\alpha_1$ and $\alpha_2$. So any negation in $\alpha$ is in either $\alpha_1$ or $\alpha_2$; and any negation in $\alpha_1$ or $\alpha_2$ is applied only to sentence symbols.
In the first case, there are several sub-cases. Suppose $\beta$ is just a sentence symbol - then we're done. Suppose instead that $\beta$ is of the form $\gamma \wedge \delta$. Then $\neg\beta = \neg(\gamma \wedge \delta) \equiv \neg\gamma \vee \neg\delta$. Since $\gamma$ and $\delta$ are of strictly lower complexity than $\beta$, by induction $\neg\gamma$ and $\neg\delta$ have the desired property, so $\neg\gamma\vee\neg\delta$ does as well.
I'll let you work out the other sub-cases for yourself - they'll be similar.
EDIT: In more detail:
A wff has complexity zero if it is a sentence symbol. If $\alpha$ has complexity $n$, then $\neg\alpha$ has complexity $n+1$. The complexity of $\beta\square\gamma$ for $\square$ a connective is $1+n$, where $n$ is the maximum of the complexities of $\beta$ and $\gamma$.
We prove by induction on complexity of formula that every wff is equivalent to a wff in which negations apply only to sentence symbols.
The case of complexity zero is trivial: if $\alpha$ has complexity zero, $\alpha$ is a sentence symbol, so $\neg\alpha$ is a wff in which negations apply only to sentence symbols.
Suppose now that the claim is true for wffs of complexity at most $n$, and let $\alpha$ be a wff of complexity $n+1$. Then $\alpha$ takes one of the forms $\neg\beta$, $\gamma\wedge\delta$, $\gamma\vee\delta$, etc.
Suppose $\alpha$ is $\gamma\square\delta$ for some connective $\square$. $\gamma$ and $\delta$ both have complexity at most $n$, since $\alpha$ has complexity $n+1$; so by the induction hypothesis, $\gamma$ and $\delta$ are equivalent to wffs $\gamma'$ and $\delta'$ respectively so that in each negations only apply to sentence symbols. Then $\gamma'\square\delta'$ is equivalent to $\alpha$ and has the desired property.
Suppose instead that $\alpha$ is $\neg\beta$. Then $\beta$ has complexity $n$. If $\beta$ is of the form $\neg\gamma$, then $\gamma$ has complexity less than $n$, so has an equivalent formula $\gamma'$ with the desired property; but $\gamma'$ is equivalent to $\alpha$, so we're done.
On the other hand, if $\beta$ takes the form (for example) $\gamma\wedge\delta$, then $\gamma$ and $\delta$ each have complexity less than $n$. $\neg(\gamma\wedge\delta) \equiv \neg\gamma\vee\neg\delta$, by DeMorgan's laws. $\neg\gamma$ has complexity one more than that of $\gamma$; but $\gamma$ had complexity strictly less than $n$, so $\neg\gamma$ has complexity at most $n$. Therefore, by the induction hypothesis, $\neg\gamma$ is equivalent to a wff $\gamma'$ in which negation applies only to sentence symbols. Likewise, $\neg\delta$ has an equivalent wff $\delta'$ of the desired sort. Then $\gamma'\vee\delta'$ is equivalent to $\alpha$, and has the desired property.
It remains to show the induction for the connectives $\vee$, $\to$, and $\leftrightarrow$.
Best Answer
You certainly have the right idea ... but there is just one small problem with the case of the negation. You say to consider the 'two predecessors' of $\beta$ ... but what if $\beta$ is a negation itself?
Fortunately, the proof is easily fixed ... instead of performing induction on the number of sentence symbols, simply perform induction on the length of the statements, i.e. the number of sentence symbols plus the number of logical operators that are in the expression.