In no wff are the symbols $\neg$ and $)$ next to each other

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.

• The most common way of proving this type of problem is by induction on formulas, most mathematical logic textbooks include a section on the topic.
– Sam
Apr 22, 2020 at 11:17

• $$\neg$$ can only be preceded by $$($$ and be followed by the start of a wff
• the start of a wff is either an atom or $$($$
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.