To negate a statement, why do you have to invert each quantifier and negate each sub-expression

Question

To negate a statement, you have to invert each quantifier and negate each sub-expression. Example:
\begin{align}
\DeclareMathOperator{\epsilon}{\varepsilon}
f \text{ is continuous at $a$} &\iff \forall\epsilon>0\exists\delta>0\forall x\bigl(|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon\bigr)\\[4pt]
f \text{ is discontinuous at $a$} &\iff \exists\epsilon>0\forall\delta>0\exists x\bigl(|x-a|<\delta\implies |f(x)-f(a)|\ge\varepsilon\bigr)
\end{align}
I'm trying to justify why this is the case. It seems intuitively clear that $\neg\forall xP(x)\iff \exists x\bigl(\neg P(x)\bigr)$: if a proposition is not true for all $x$, then there just needs to be one counterexample. Moreover, $\neg \exists xP(x)\iff \forall x\bigl(\neg P(x)\bigr)$ because both statements say that there aren't any examples of a proposition being true.

My idea is to use these two "rules" recursively:
\begin{align}
f \text{ is discontinuous at $a$} &\iff \neg\Bigl(\forall\epsilon>0\exists\delta>0\forall x\bigl(|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon\bigr)\Bigr)\\[4pt]
&\iff \exists\epsilon>0\neg\Bigl(\exists\delta>0\forall x\bigl(|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon\bigr)\Bigr)\\[4pt]
&\iff \exists\epsilon>0\forall\delta>0\neg\Bigl(\forall x\bigl(|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon\bigr)\Bigr)\\[4pt]
&\iff \exists\epsilon>0\forall\delta>0\exists x\neg\bigl(|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon\bigr)\\[4pt]
&\iff \exists\epsilon>0\forall\delta>0\exists x\Bigl(|x-a|<\delta\implies\neg\bigl(|f(x)-f(a)|<\varepsilon\bigr)\Bigr)\\[4pt]
&\iff \exists\epsilon>0\forall\delta>0\exists x\bigl(|x-a|<\delta\implies|f(x)-f(a)|\ge\varepsilon\bigr)\\[4pt]
\end{align}
Does this line of reasoning make sense? Moreover, can the "invert each quantifier and negate each sub-expression" rule be applied in any situation?

Answer 1

To negate a statement, we have to invert the quantifier and negate the resulting sub-expression.

Correct, because $¬∀x$ is the sane as $∃x¬$ and similar for the case with the negated existential quantifier.

The intuition is quite simple: "it is not true that all cats are black" means that "there are some cat that is not black".

The "procedure" must be iterated starting from the outermost quantifier:

$\lnot \forall \varepsilon \ \exists \delta \ \forall x \ \mathcal A \ \ $ will be : $\ \ \exists \varepsilon \ \forall \delta \ \exists x \ \lnot \mathcal A$.