Is it right to render the quantifier of existence through a disjunction during skolemization

I am to solve a problem:

$$\forall x \exists y \neg P(x,y) \rightarrow \exists x \forall y Q(x,y)$$

I'm getting rid of implication:

$$\exists x \forall y P(x,y) \lor \exists x \forall y Q(x,y)$$

And now I'm using this formula: $$\exists x A(x) \lor \exists B(x)$$ = $$\exists x( A(x) \lor B(x))$$

$$\exists x (\forall y P(x,y) \lor \forall y Q(x,y))$$

Then I resolve variable conflicts: y->t

$$\exists x (\forall y P(x,y) \lor \forall t Q(x,t))$$

And moving all quantifiers to the very begining

$$\exists x \forall y \forall t (P(x,y) \lor Q(x,t)))$$

The existence quantifier should be changed to the constanta: x->c

$$\forall y \forall t (P(c,y) \lor Q(c,t)))$$

And then eliminating the $$\forall$$ quantifiers

$$(P(c,y) \lor Q(c,t))$$

Am I right at my solution? Can anybody please check the answer?