If $L(x,y)$: x loves y.

Use quantifiers to express:

"There is someone who loves no one besides himself or herself"

The answer given by textbook is $∃x∀y(L(x,y) ↔ x=y)$

But I think $∃x∀y(L(x,y) → x=y)$ is also correct, right?

  • 2
    $\begingroup$ no, thats wrong as your expression does not say that $x$ loves himself, even though the difference is subtle I have to admit. $\endgroup$ – noctusraid Oct 23 '16 at 10:56
  • $\begingroup$ so,how about this ∃x∀y(L(x,y) ∧ x=y)? $\endgroup$ – Ned Stack Oct 23 '16 at 11:32
  • $\begingroup$ On a possibly amusing side note, the logician Quine used to put this question on exams for introductory courses: "As the song goes, 'Everybody loves my baby, but my baby don't love nobody but me.' Prove that I am my baby." $\endgroup$ – lulu Oct 23 '16 at 11:52
  • $\begingroup$ Related: math.stackexchange.com/q/1488516/11994 $\endgroup$ – MarnixKlooster ReinstateMonica Oct 25 '16 at 20:05

It depends on how you interpret the English sentence "There is someone who loves no one besides himself or herself". Does the sentence imply that this someone loves himself/herself? If no, your answer is correct; if yes, the textbook's answer is correct.

  • $\begingroup$ how about this ∃x∀y(L(x,y) ∧ x=y), assume that he loves himself. $\endgroup$ – Ned Stack Oct 23 '16 at 11:32
  • $\begingroup$ @Ned: That sentence says in effect that there is exactly one person, and that person loves himself. $\endgroup$ – Brian M. Scott Oct 23 '16 at 12:37
  • $\begingroup$ I don't understand the meaning of ↔ $\endgroup$ – Ned Stack Oct 23 '16 at 13:27
  • $\begingroup$ @NedStack: Then you've identified your problem. Understand exactly what the primitive boolean operations mean, via their truth tables. Only after that then try to figure out how to translate English sentences to logical language. $\endgroup$ – user21820 Oct 24 '16 at 11:40

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