Trying to understand negation of quantifiers 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?
 A: 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 $\exists y\lnot Q(y)$.    
A: $\forall x.\neg P(x)$ 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 $\exists x.P(x)$.

$\exists x.\neg P(x)$ 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.
