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Express in logical form: There is exactly one person John likes.

Doing this out we can write it using the "There is exactly one" quantifier $\exists!$. Letting $L(x,y)$: $x$ likes $y$. This can be written as $\exists!xL(J,x)$. After seeing this I was wondering how we can negate it. If we were to not use the "There is exactly one" quantifiers it would become $\exists x\forall y(L(J,x)\land\lnot (L(J,y)\land y\neq x))$ Then if I were to negate it all we get $\forall x\exists y(\lnot L(J,x)\lor (L(J,y)\land y\neq x))$ Which to me reads as, John does not like everyone or John likes at least one person, and that person is not $x$. The negation that I have seems to lose the fact that John likes more than one person, regardless of if that person is $x$ or not. So how would I go about negating this statement? Is there a quantifier that is the opposite of $\exists!$?

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    $\begingroup$ Well, you have two cases. Either there's nobody John likes or there exist two people whom John likes. $\endgroup$ – lulu Sep 11 '17 at 16:44
  • $\begingroup$ Note that $\exists!x\phi(x)$ is just a short form of $\exists x(\phi(x)\wedge\forall y(\phi(y)\to x=y))$. I think you can negate this. Logically I would say the opposite of "exactly one" is "any other amount than one, e.g. none, or more than one". $\endgroup$ – M. Winter Sep 11 '17 at 16:45
  • $\begingroup$ You are confusing a generic x with a specific specific x. You need notation for "John likes at least two people. "John likes exactly one person" means "It is not the case that John likes nobody and it is not the case that John likes at least two people". The negation is either John likes nobody or John likes at least two people. $\endgroup$ – fleablood Sep 11 '17 at 16:45
  • $\begingroup$ M. Winter gives a good hint. So Joh likes at least two people can be written $\exists x,y| x \ne y \land (L(J,x) \land L(J,y))$ so $!\exists x|L(Jx)$ would be $[\forall x|\lnot(J,x)]\lor[\exists x,y| x \ne y \land (L(J,x) \land L(J,y))]$. $\endgroup$ – fleablood Sep 11 '17 at 16:52
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The negation is correct, though the way you wrote it obfuscates the meaning. I believe the statements are equivalent, but the way you wrote it is less clear. For $\exists!xL(J,x)$, I'd write it as $\exists x(L(J,x)\land \forall y(L(J,y)\implies x=y))$. This gives a negation of $\forall x(\neg L(J,x)\lor \exists y(L(J,y)\land x\neq y))$.

You can then read this as, "For every person, either John doesn't like them, or John likes them and somebody else. Hence, if the second case is ever achieved (John likes that person and another person), then there is no unique person than John likes. If the second case is never achieved, then it is always the first case (John doesn't like that person), so John likes nobody, and so there exists no person that John likes, much less a unique one.

EDIT: fleablood's comment explains your question about your own sentence. If John likes no people, then he certain doesn't like a unique person. If there is somebody he likes, say that person is $z$, then $z$ must satisfy $\exists y(\neg L(J,z)\lor (L(J,y)\land y\neq z))$. Since John likes $z$, the first part of the "or" statement cannot be satisfied, so the second must be, meaning John likes another person who is not $z$. Either way of writing the statement and its negation is accurate.

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    $\begingroup$ Or you can read it as "For every person, either John doesn't like them, or John likes someone else". By logic one can conclude that if John does like someone else then the statement applies to the someone else as well and John must therefore like another someone else as well. So either John dislikes every person or John likes at least two people. $\endgroup$ – fleablood Sep 11 '17 at 17:00
  • $\begingroup$ @fleablood That's true; then we'd write it: $\forall x(\neg L(J,x)\lor \exists y(L(J,y)\land y\neq x)$. Then it has negation $\exists x(L(J,x)\land \forall y(\neg L(J,y)\lor x=y))$, which reflects the fact that $P\implies Q$ is equivalent to $\neg P\lor Q$. $\endgroup$ – Kevin Long Sep 11 '17 at 18:11
  • $\begingroup$ I wasn't implying the op's negation was incorrect. Quite the opposite. It was completely correct. I'm pointing out that although it doesn't literally claim that john either doesn't like a person or likes them and another, it can be inferred. $\endgroup$ – fleablood Sep 11 '17 at 18:32
  • $\begingroup$ @fleablood Maybe I wrote that poorly, because what I was saying was that your comment explains why OP's negation is correct. $\endgroup$ – Kevin Long Sep 11 '17 at 19:01

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