I'm doing some selfstudying and I'm lacking the anwsers to check if I'm correct. So this is why I'm on here so frequently, as I really want to understand the matter.

So here is the sentence I'm trying to convert to predicate logic:

Every boy who loves a girl is also loved by some girl.

DoD = Humans Here's what I got:

$\forall x(Bx \wedge \exists y(Gy\wedge Lxy))\rightarrow \exists z(Gz \wedge Lzx)$

With B being a boy, G being a Girl, and L x Loving y.

Thanks in advance, Rope.

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    $\begingroup$ It should be: $\forall x\left[\left[Bx\wedge\exists y\left[Gy\wedge Lxy\right]\right]\Rightarrow\exists y\left[Gy\wedge Lyx\right]\right]$ $\endgroup$ – drhab Jun 2 '14 at 15:31
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    $\begingroup$ Your translation is correct, apart from a parenthesis problem. $\endgroup$ – André Nicolas Jun 2 '14 at 15:32
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    $\begingroup$ In short you have $A\rightarrow E\left(x\right)$. In $A$ variable $x$ is bounded and in $E\left(x\right)$ it is not. $\endgroup$ – drhab Jun 2 '14 at 15:38
  • $\begingroup$ Ok so by adding the parenthesis I would include x into the last part of the formula? $\endgroup$ – Byebye Jun 2 '14 at 15:45
  • $\begingroup$ Yes, adding parentheses solves it. $\endgroup$ – drhab Jun 2 '14 at 16:00

You are saying here:

If every boy loves a girl then there is a girl that loves some particular boy yet to be specified.

See also my comment.

  • $\begingroup$ Symbol $x$ is not a constant, so it's not "a specific boy $x$", more like "some particular boy yet to be specified". Or perhaps "If every boy loves a girl, then there is a girl that loves <insert name here>." $\endgroup$ – dtldarek Jun 2 '14 at 15:45
  • $\begingroup$ @dtldarek Thanks, repaired. $\endgroup$ – drhab Jun 2 '14 at 16:02

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