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 $k\ge 0$.
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.
Best Answer
HINT: I assume that these are the well-formed formulas of propositional logic. These are defined by structural recursion:
To prove that the family of WFFs has some property $P$, one uses structural induction; this is a kind of induction argument that parallels the recursive definition of the family of WFFs. The steps are:
If you can do all this, you can conclude by (structural) induction that all WFFs have $P$.
In your particular case the first (or basis) step is easy: each atom has $P$ vacuously, because no atom contains two atoms at all. Now see if you can carry out the two parts of the induction step: both are very straightforward.