Negating ∃x∀z∃y(S(x,y) ∧ C(y,z))

Question

Negating ∃x∀z∃y(S(x,y) ∧ C(y,z)) would be logically equivalent to ¬∃x∀z∃y(S(x,y) ∧ C(y,z))

however would ∀x∃z∀y¬(S(x,y) ∧ C(y,z)) also be logically equivalent?

Answer 1

The final answer will use De Morgan's law to get

$$(\forall x )\;\;(\exists z) \;\; : $$ $$\;\; (\forall y ) \;\;\lnot S(x,y) \; \vee \lnot C(y,z).$$

Answer 2

Yes. You can keep pushing the negation in:

$\neg \exists x \forall z \exists y (S(x,y) ∧ C(y,z)) \Leftrightarrow$

$\forall x \neg \forall z \exists y (S(x,y) ∧ C(y,z)) \Leftrightarrow$

$\forall x \exists z \neg \exists y (S(x,y) ∧ C(y,z)) \Leftrightarrow$

$\forall x \exists z \forall y \neg (S(x,y) ∧ C(y,z)) $