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: $$(\forall x)(\sim p(x)\wedge(\exists y)\sim h(y)) \;\leftrightarrow\; \sim q $$

Thanks in advance.


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 :

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


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