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

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?

• Yes, because $\lnot \forall$ is the sane as $\exists \lnot$ and similar with the negated existential quantifier. Intuition: "it is not true that all cats are black" means that "there are some cat that is not black" Jun 29 at 11:57
• Could the downvoted please justify it?
– Joe
Jun 29 at 12:01
• Negation of continuity is wrong: You should take negation of last sentence too. Jun 29 at 12:36
• @zkutch: Is that because the negation of $p\rightarrow q$ is $p \wedge \neg q$? The negation of $p \rightarrow q$ is not $p \rightarrow \neg q$.
– Joe
Jun 29 at 12:57
• @MauroALLEGRANZA: Thanks for clarifying. Is it possible if post your comment as an answer so that we can remove this question from the "unanswered questions" list?
– Joe
Jun 30 at 12:30

Correct, because $$¬∀x$$ is the sane as $$∃x¬$$ and similar for the case with the negated existential quantifier.
$$\lnot \forall \varepsilon \ \exists \delta \ \forall x \ \mathcal A \ \$$ will be : $$\ \ \exists \varepsilon \ \forall \delta \ \exists x \ \lnot \mathcal A$$.