Trying to understand negation of quantifiers

Question

Trying to understand the negation of the following:

For this: ∀x~P(x) I have this as negation: ~∃xP(x)

For this: ~∃x(∀yP(y) Λ Q(x)) I have this: ∀x(~∃yP(y) V ~Q(x))

Are these correct? If not please provide the right negation

Moving the negation signs:

∃x~P(x) ☰ ~∃xP(x)?

∀x(~∃yP(y) V ~Q(x)) ☰ ∀x~(∀yP(y) Λ Q(x)) ☰ ~∃x(∀yP(y) Λ Q(x))??

Are the above still equivalent?

Answer 1

$\forall x.\neg P(x)$ and $\neg \exists x.P(x)$ are not each other's negations -- on the contrary they are equivalent.

If you negate $\forall x.\neg P(x)$ you get either $\neg\forall x.\neg P(x)$ which is equivalent to $\exists x.P(x)$.

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$\exists x.\neg P(x)$ is not equivalent to $\neg\exists x.P(x)$.

$\exists x.\neg P(x)$ is equivalent to $\neg\forall x.P(x)$. When you move a negation through a quantifier, the quantifier changes from $\exists$ to $\forall$ or vice versa.

Answer 2

Informally, $\forall x(\lnot P(x))$ precisely if $\exists x P(x)$. So $\exists xP(x)$ is a correct negation of $\forall x(\lnot P(x))$

More formally, $\lnot\forall x\varphi$ if and only if $\exists x(\lnot\varphi)$. Using $\varphi=\lnot P(x)$, we get $\lnot\lnot P(x)$, or equivalently $P(x)$.

The second is simpler. If you want to negate $\lnot\exists x(\forall yP(y)\land Q(x))$, you just remove the $\lnot$ in front.

Remark: It is not clear what you are asking for the second problem, for negating $\lnot\exists x(\forall yP(y)\land Q(x))$ is perhaps too simple to be asked as a question. If you want to negate $\exists x(\forall yP(y)\land Q(x))$, the first step is to write $\forall x(\lnot (\forall yP(y)\land Q(x)) )$. Then if we feel like it we can replace $\lnot (\forall yP(y)\land Q(x))$ by $(\lnot \forall yQ(y))\lor (\lnot Q(x))$, and then if we wish replace $\lnot\forall yQ(y)$ by $\exists y\lnot Q(y)$.