Express the following in predicate logic: "Nobody loves anybody."

$$P(x): \text{x is a person.}$$ $$L(x,y): \text{x loves y.}$$

My attempt was:

$$\neg[\exists x(P(x) \land \forall y P(y) \longrightarrow L(x,y))]$$

Although my instructor wrote it as:

$$\neg[\exists x(P(x) \land ( \forall y P(y) \longrightarrow L(x,y)))]$$

I do not get why he nested the implies in separate parentheses.

More so I am not able to semantically tell the difference between the two formulae.


  • 1
    $\begingroup$ It's about operator precedence. $p\wedge q \to r$ is usually interpreted as $(p\wedge q)\to r$ rather than $p\wedge (q\to r)$ though not always. Personally I'd have wrapped the quantifier scope; $$\neg\exists x~\Big(P(x)\wedge \forall y~\big(P(y)\to L(x,y)\big)\Big)$$ which is "nobody loves everybody," by the way. $\endgroup$ – Graham Kemp Feb 17 '16 at 5:22
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    $\begingroup$ Nobody loves anybody does not have the same meaning as nobody loves any body. $\endgroup$ – André Nicolas Feb 17 '16 at 5:27
  • $\begingroup$ @Sharma: Sorry, miscounted. $\endgroup$ – André Nicolas Feb 17 '16 at 5:31
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    $\begingroup$ @AndréNicolas I did too. The (pointless) outer brackets don't help. $\endgroup$ – BrianO Feb 17 '16 at 5:36

Actually both of the formulas you have are wrong in some way. In both, the $\forall y$ only binds the $y$ in $P(y)$, leaving the $y$ in $L(x,y)$ free. In your professor's sentence, the parenthesis before $\forall y$ should come right after it. Finally, there's no need for the outer brackets. Thus: $$ \neg\exists x~\Big(P(x) \land \forall y \big(P(y) \to L(x,y)\big)\Big) $$ However, this isn't right: it's equivalent to $$\begin{align} \forall x~\Big(P(x) \to \neg\forall y~\big(P(y) \to L(x,y)\big)\Big) &\iff \forall x~\Big(P(x) \to \exists y~\big(P(y) \land \neg L(x,y)\big)\Big) \end{align}$$ which means "everybody doesn't love someone", or equivalently, nobody loves everybody. Everybody does not mean anybody. I am assuming that "nobody loves anybody" does not mean the same thing as "nobody loves just anybody" (i.e. everybody).

I take "nobody loves anybody" to mean for all people $x$ and $y$, $x$ does not love $y$: $$ \forall x~\forall y~\Big(\big(P(x)\land P(y)\big)\to \neg L(x,y)\Big) $$

  • $\begingroup$ Your interpretation is the same to say that love doesn't exist between people? Sad :( $\endgroup$ – Masacroso Feb 17 '16 at 5:30
  • $\begingroup$ @Masacroso Basically yes. The question isn't whether that's true ;) $\endgroup$ – BrianO Feb 17 '16 at 5:31
  • $\begingroup$ I think your interpretation is not good after all (Im not english speaker anyway) because I feel is different to say "nodoby loves someone". It is hard to see these things for not a english speaker. I love you a bit anyway, so it cant be true :) $\endgroup$ – Masacroso Feb 17 '16 at 5:34
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    $\begingroup$ I am a fluent native English speaker. One would never say "Nobody loves someone." $\endgroup$ – BrianO Feb 17 '16 at 5:34
  • $\begingroup$ @GrahamKemp OK I'll go for that, thanks. $\endgroup$ – BrianO Feb 17 '16 at 5:38

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