(A) Consider the statement $\forall x \exists y \lnot P(x, y)$. Write down a negation of the statement that does not use the symbol $\lnot$.

I just said that its $\forall x \exists y P(x, y)$ but I'm not entirely sure. Can someone clarify?

(B) Under the interpretation where $x$ and $y$ are in $\mathbb{R} \smallsetminus \{0\}$ and $P(x, y)$ is "$xy ≥ 0$", is the original statement in (B) true or is its negation true?

For this part, I said when $x$ and $y$ are BOTH positive or negative integers, then (B) is true, but when one is positive and the other is negative, then the negation is true? Am I on the right track?

  • $\begingroup$ Concerning B you must wonder: "is $\forall x\exists y\neg P(x,y)$ a true statement in this context?" If so then you choose for "original" as answer. If not then you choose for "negation of original" as answer. $\endgroup$
    – drhab
    Commented Jun 16, 2018 at 7:20

2 Answers 2




Step by step we find that the following are negations of $\forall x\exists y\neg P(x,y)$

  • $\neg\forall x\exists y\neg P(x,y)$
  • $\exists x\neg\exists y\neg P(x,y)$
  • $\exists x\forall y\neg\neg P(x,y)$
  • $\exists x\forall yP(x,y)$

In the last one $\neg$ is not present.


The original statement is true.

For every $x\in\mathbb R-\{0\}$ we can find an $y\in\mathbb R-\{0\}$ such that $xy<0$. We can take $y=-x$.

  • $\begingroup$ Is (b) correct or also wrong wrong because (a) was wrong? $\endgroup$
    – user557403
    Commented Jun 16, 2018 at 7:08
  • $\begingroup$ @Okie Your answer to (b) is not even wrong. The answer required should be just "the statement in A is true" or "the negation of the statement in A is true". $\endgroup$ Commented Jun 16, 2018 at 7:11

(A) The negation of sentences is $\exists x\forall y $ $P(x,y)$

(B) when you interpret the sentence in $\mathbb{R}\backslash\{0\}$ you have that $\forall x\exists y $ $P(x,y)$ because for all $x\in \mathbb{R}\backslash\{0\}$ exists $y=-x$ such that $\neg (xy>0)$


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