I have the three steps to prove the statement using structural induction:
1 - every propositional variable has property Q
2 - if A is -B, and B has Q, then A has Q
3 - if A is A v B, etc, and B and C have Q, then A has Q.
I would say that Q: if the formula has 'v' then the formula has at least two propositional variables. But statement one only include variables, so how can a variable contain a connective?
Thank you!
We can apply a quasi-inductive reasoning, based on the number of occurrences of the connectives, provided that the formula has one occurrence of the connective $\lor$.
(i) base : if $\varphi$ has only one occurrence of connectives, this must be $\lor$; then it must be $\phi_1 \lor \phi_2$ and thus the result holds, because the two sub-formulae cannot be empty.
Now, trivially, adding more connectives the number of propositional variables cannot decrease, and thus if it holds for $n$ occurrences, it holds also for $n+1$.
Note. We have to specify that there must be one occurrence of $\lor$; otherwise, we may have a formula like $\lnot ( \lnot ( \lnot ( \lnot p_i)))$ and the result will not hold.