If $\mathcal{L}$ is any first-order language then I want to prove that no wff have $\neg$ and $)$ next to each other (in any order) in the string of symbols that make up the wff. As a hint it has been given that I should start by proving that no wff ends with the symbol $\neg$.
Intuitively, this makes sense as I think $\neg$ should be placed in front of something in order to negate it. However, I struggle to do this more formally.
My idea is the following:
By Enderton (p.74) I have that "The set of well-formed formulas (wffs, or just formulas) is the set of expressions that can be built up from the atomic formulas by applying (zero or more times) the operations $E_{\neg}$, $E_\to$,and $Q_i$ (i=1,2,…)" where
$E_{\neg}(\gamma)=(\neg \gamma)$
$E_\to(\gamma,\delta)=(\gamma \to \delta)$
$Q_i(\gamma)= \forall v_i \gamma$
I think it is rather clear that I cannot form $\gamma \neg$ from either $E_{\neg}$, $E_\to$ or $Q_i$ if $\gamma$ is an atomic formula. But I think there should be something else to the solution. My questions are therefore:
1) Is there another way to get started with this problem as my current approach only provides me with a more intuitive feeling?
2) $\neg$ is a unary operation according to Enderton but does this means that $\neg$ "needs" some terms on the right side of it in order to be an atomic formula?
I am quite new to mathematical logic hence the probably basic questions.
Best Answer
From the substitution rules, we see that in any wff
This immediately shows that $\neg$ can never appear beside $)$ in a wff.
$\neg$ is not an atomic formula because $\neg$ is a logical connective that just happens to connect to only one argument. It does not constitute a predicate.