For the negation of statement

Question

∃y ∈ Z such that ∀x ∈ Z, R(x + y) ∀x ∈ Z, ∃y ∈ Z such that R(x + y)

Base on this two rule "For all x, A(x)" negation:"There exist x such that not A(x)" "There exists x such that A(x)" negation:"For every x, not A(x)"

what I did is "∀y ∈ Z, not ∀x ∈ Z, R(x + y)" "∃x ∈ Z, ∀y ∈ Z such that R(x + y) " I wish anyone could help me to check whether this is right or not. Thank you!

Answer 1

$\forall y\in Z \text{ s.t. } \exists x\in Z, \text{ not } R(x+y)$

$\exists x\in Z, \forall y\in Z, \text{ s.t. not } R(x+y)$

Answer 2

The negation of $\exists y\in Z\forall x\in Z\;P(x,y)$ is:$$\forall y\in Z\exists x\in Z\neg P(x,y)$$

Further this is not the same statement as $\forall x\in Z\exists y\in Z\neg P(x,y)$