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$\neg \forall x \exists y \neg P(x,y)$ is equal to $\exists x \exists y \neg P(x,y)$

I had to make sure, because I wasn't sure at all.

amWhy
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PopularScience
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3 Answers3

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Is $\;\;\neg \forall x \exists y \neg P(x,y)\;\;$ equal to $\;\;\exists x \exists y \neg P(x,y)\;\;$?

No. See the following chain of equivalences to see why not:

$$\neg \forall x \exists y \neg P(x,y) \;\;\equiv \;\;\exists x \neg \exists y \neg P(x, y) \;\; \equiv \;\;\exists x \forall y \neg\neg P(x, y) \;\; \equiv \;\; \exists x\forall y P(x, y)$$

We used the facts that $\lnot \forall \varphi(x) \equiv \exists x \lnot \varphi(x)\;$ and $\;\lnot \exists x \varphi(x) \equiv \forall x \lnot \varphi(x)$.

Alternatively, again using the equivalences $\lnot \forall \varphi(x) \equiv \exists x \lnot \varphi(x)\;$ and $\;\lnot \exists x \varphi(x) \equiv \forall x \lnot \varphi(x)$:

$$\neg \forall x \exists y \neg P(x,y) \;\;\equiv\;\; \neg \forall x \neg \forall y P(x,y) \;\;\equiv\;\; \neg\neg \exists x \forall y P(x,y)\;\;\equiv \;\;\exists x \forall y P(x,y)$$

amWhy
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  • so $\neg (\forall x \exists y) \neg P(x,y) = \neg \forall x \neg \exists y \neg P(x,y)$ ? – PopularScience Dec 07 '12 at 01:12
  • why does the negation sign go from left to right? – PopularScience Dec 07 '12 at 01:16
  • It can be brought from inside out, too, you get the same result: moving a negation sign in, or out, of quantified statements reverses the quantifier. – amWhy Dec 07 '12 at 01:17
  • to deny that a statement is true for all x, you need only assert the existence of an x such that the statement is false. To deny the existence of an x for which something is true, then you assert that for all x, it is not true that that something is true. – amWhy Dec 07 '12 at 01:39
  • @amWhy Don't you have to obtain a contradiction to prove the statement in question false? I don't see how. – Dan Christensen Dec 07 '12 at 05:25
  • @DanChristensen You need only find a counterexample (an interpretation in which the lhs is true, but not the right hand side, or vice versa) to conclude that the statement in question is false. Even if the left hand side of the OPs proposed equivalence "implies" the consequence, that does not prove equivalence. Take P(x, y) to mean "x likes y". See that the LHS $\not\equiv$ RHS of OPs proposed equivalence. – amWhy Dec 07 '12 at 15:28
  • @amWhy Yes. See, for example, my answer at http://math.stackexchange.com/questions/252731/are-these-two-statement-equivalent There, I formally construct a simple counter-example of $\forall x\exists y P(x,y)\rightarrow \exists y\forall x P(x,y)$ – Dan Christensen Dec 07 '12 at 19:30
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To get an intuitive grip on it, simplify matters by letting $Q(x,y)$ mean $\lnot P(x,y)$, so that you can rewrite the statements as $\lnot\forall x\exists y~Q(x,y)$ and $\exists x\exists y~Q(x,y)$, respectively. The second statement is easy to understand: it just says that there’s at least one pair of objects $a$ and $b$ for which $Q(a,b)$ is true.

To see in more intuitive terms what the first statement says, look at its negation, $\forall x\exists y~Q(x,y)$: this just says that for each $x$ there is a $y$ making $Q(x,y)$ true. Its negation (i.e., the original statement $\lnot\forall x\exists y~Q(x,y)$) must therefore say that there is at least one object $a$ such that no $y$ makes $Q(a,y)$ a true statement. What if this $a$ is the only object in the domain of discourse? Then $\lnot\forall x\exists y~Q(x,y)$ is true but $\exists x\exists y~Q(x,y)$ is false.

However, you should also be able to manipulate the expressions formally as in amWhy’s answer, using the facts that $\lnot\exists x~\varphi(x)\equiv\forall x~\lnot\varphi(x)$ and $\lnot\forall x~\varphi(x)\equiv\exists x~\lnot\varphi(x)$. In words, you can move a negation through a quantifier if you change the quantifier from $\forall$ to $\exists$ or from $\exists$ to $\forall$, whichever is relevant.

Brian M. Scott
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Here is a counter-example: Let the domain of quantification be $U=\{ x, y \}$ with distinct $x$ and $y$. And let $P$ be the "is equal to" relation on $U$. Then the LHS is false while the RHS is true.

Dan Christensen
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