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The question I am trying to reason about is as follows:

let L(x,y) be the statement "x loves y"

There is someone who loves no one besides himself or herself

I tried to write this as:

∃x ∀y , (L(x,x) ^ ~L(x,y))

But the correct answer is:

∃x ∀y , (L(x,y) <-> x=y)

I can reason about the correct answer when I see it, but I don't think I would come to that conclusion myself. Can someone point out what is wrong with my original answer?

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Your statement says that, for someone $x$, $x$ loves him/herself and $x$ loves nobody. To see this, you can move the $\forall y$ quantifier over the first conjunct because $y$ is not free in $L(x,x)$, so an equivalent way to write it is: $$\exists x\,(L(x,x) \wedge \forall y\, \neg L(x,y)) $$ So for such a person $x$, $\forall y\, \neg L(x,y)$ — $x$ loves nobody. In particular that implies that $x$ does not love himself/herself: take $y = x$. But also, $L(x,x)$, so problems ensue — $0 = 1$ and so on. So there can't be any such $x$, and your statement is just false (its negation is provable, as we just showed).

Finally, although at this stage of learning logic you might not come up with the biconditional to represent the English phrase, the following is equivalent to it and is something that might occur to you: $$ \exists x\,(L(x,x) \wedge \forall y\, (L(x,y) \to y = x)) $$

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  • $\begingroup$ Thanks Brian! So is it basically just a good idea to watch out for logical statements that include similar statements to no one besides? Seems like an easy trick to fall for. $\endgroup$ – Adam Thompson Oct 20 '15 at 3:42
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    $\begingroup$ @AdamThompson Good to be on the alert for those, yes :) See the last paragraph I just added, which shows another way to write the statement. The formula $\forall y\, (L(x,y) \to y = x)$ says "if x loves anyone, it's him/herself." $\endgroup$ – BrianO Oct 20 '15 at 3:50

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