The negation of $\exists x \forall y \neg \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$

Question

I am trying to understand the negation of $\exists x \forall y \neg \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$

As a side example if I have a statement $\neg \forall x P(x)$, then this is equivalent to $\exists x \neg P(x)$ where the negation is $\forall x P(x)$, motivating what I think is the answer to my overall question.

Then I believe the answer should be: $\forall x \exists y \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$ (i.e simply remove the $\neg$ from the $\forall z$)

but I am a bit paranoid that the answer could be: $\forall x \exists y \forall z(P(x,y) \iff \neg Q(x,y) \lor \neg R(x,y,z))$, passing the negation through to the inner bracket.

Any insights which is correct appreciated.

Please replace one quantifier's $x$ with $z$ in both the original statement and each attempt you make to negate it. — J.G., Jan 25 '21 at 18:14

J.G. · Accepted Answer · 2021-01-25T18:29:08.103

1

Your initial feeling is right:$$\neg\exists x\forall y\neg\forall z\phi(x,\,y,\,z)\iff\forall x\neg\forall y\neg\forall z\phi(x,\,y,\,z)\iff\forall x\exists y\forall z\phi(x,\,y,\,z).$$Alternatively,$$\neg\exists x\forall y\neg\forall z\phi(x,\,y,\,z)\iff\neg\exists x\neg\exists y\exists z\phi(x,\,y,\,z)\iff\forall x\exists y\forall z\phi(x,\,y,\,z).$$

edited Jan 25 '21 at 18:29

answered Jan 25 '21 at 18:21

J.G.

115,835

score 1 · Answer 2 · answered Jan 25 '21 at 18:25

1

Your initial answer is right

The tool in the link is the most helpful thing I have found for such problems.

answered Jan 25 '21 at 18:25

SagarM

1,789

The negation of $\exists x \forall y \neg \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$

2 Answers2