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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!

1 Answers 1

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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.

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    Thank you. I was headed in that direction and this just clarifies it for me. :) – 2017-02-15