# How can I deny the formula $(\exists x)(p(x)\vee(\forall y)h(y)) \leftrightarrow q$

Can anyone explain me how can I deny this propositional formula?

$$(\exists x)(p(x)\vee(\forall y)h(y)) \;\leftrightarrow\; q$$

According to my textbook, the answer would be: $$(\forall x)(\sim p(x)\wedge(\exists y)\sim h(y)) \;\leftrightarrow\; \sim q$$

The negation of $$A \leftrightarrow B$$ will be : $$\lnot A \leftrightarrow B$$ (or, equivalently : $$A \leftrightarrow \lnot B$$); you can check it with truth table.

Thus, the negation of the original formula will be :

$$(∃x)(p(x) \lor (∀y)h(y)) ↔ \lnot q$$

or, equivalently :

$$\lnot (∃x) (p(x) \lor (∀y)h(y)) ↔ q$$

This in turn is equivalent to :

$$(∀x) \lnot (p(x) \lor (∀y)h(y)) ↔ q$$

and thus, using De Morgan's laws, to :

$$(∀x)(\lnot p(x) \land (∃y)\lnot h(y)) ↔ q$$.