$$\{x\in X \mid \exists V{\subseteq} E\;\; V{\in} \mathscr N(x)\} \tag{1}$$ $$\{x\in E \mid \exists V{\subseteq} E\;\; V{\in} \mathscr N(x)\} \tag{2}$$ These two sets are equal to $E°.$
I know that the correct negation of $(1)$ and $(2)$ are $$ \{x\in X \mid \forall V, V\cap E^c \neq \emptyset, V{\in} \mathscr N(x)\} \tag{3},$$ which is is correct, and $$ \{x\in E \mid \forall V, V\cap E^c \neq \emptyset, V{\in} \mathscr N(x)\} \tag{4},$$ which isn't correct (the complement of $E°$ cannot be made elements of $E$).
What is the coherent rule to apply when taking $x$ not belonging to these sets? What is the systematic rule to apply to $(1)$ to obtain $(3)?$ I know that ∃ becomes ∀, etc, but what about x∈X? It that seems we have not changed this part.
Let $P(x)$ be a propositional function of $x,$ for example, $\exists k{\in}\mathbb Z\:\,x=7k.$
Then the complement of the set $$\left\{x\in A\mid P(x)\right\}$$ is $$\left\{x\in A \mid P(x)\right\}^\complement\\ =\left\{x\mid x\in A\quad\textit{and}\quad P(x)\right\}^\complement\\ =\left\{x\mid \textit{not}\;\big(x\in A\quad\textit{and}\quad P(x)\big)\right\}\\ =\left\{x\mid \big(x\not\in A\quad\textit{or}\quad \textit{not}\;P(x)\big)\right\}\\ =A^\complement\cup\left\{x\mid \textit{not}\;P(x) \right\}. $$
P.S. We negate propositions and propositional functions, and we take complement of sets.
P.P.S. The negation of a tautology is a contradiction; on the other hand, the complement of the set of tautologies contains contingent sentences as well as contradictions.