There is this question I am currently facing: Find the value of $f(n)$, which is the maximum number of negation sign one can have in a wff with $n$ atoms, without the subformula of the form $(¬(¬ \alpha))$, in terms of $n$.
This is my approach: consider $n=1$. Then I can have only 1 negation sign, by having $¬(A_1)$. Then by using recurrence, let $\omega$ be a wff with $n$ atoms and maximum number of negation sign, say $x_n$. Then for wff with $n+1$ atoms, I can take $¬(\omega ∧ ¬(A_{n+1}))$, so $x_{n+1}=x_n+2$. Solving it yields $f(n)=2n-1$. I did not consider $¬(¬(\omega) ∧ ¬(A_{n+1}))$ because $\omega$ will be starting with a ¬, so $¬(¬(\omega) ∧ ¬(A_{n+1}))$ will have the form of $(¬(¬ \alpha))$.
Is this correct?
Edit: $n$ is a natural number. And no limits of signs.