What is the negation of ∀x∃y¬P(x,y) without using ¬?

Question

Found it to be ∃x∀yP(x,y).

Is this right?

Answer 1

Here's one possible way of reasoning, passing the $\lnot$ through the quantors:

$\lnot\forall x: \exists y:\lnot P(x,y) ~\Leftrightarrow~ \exists x: \lnot\exists y: \lnot P(x, y) ~\Leftrightarrow~ \exists x : \forall y : \lnot\lnot P(x, y)~\Leftrightarrow~ \exists x : \forall y : P(x,y)$

Answer 2

Yes, the answer you gave is right.