# Predicate logic, Every boy who loves a girl is also loved by a girl

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.

• It should be: $\forall x\left[\left[Bx\wedge\exists y\left[Gy\wedge Lxy\right]\right]\Rightarrow\exists y\left[Gy\wedge Lyx\right]\right]$ – drhab Jun 2 '14 at 15:31
• Your translation is correct, apart from a parenthesis problem. – André Nicolas Jun 2 '14 at 15:32
• 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. – drhab Jun 2 '14 at 15:38
• Ok so by adding the parenthesis I would include x into the last part of the formula? – Byebye Jun 2 '14 at 15:45
• Yes, adding parentheses solves it. – drhab Jun 2 '14 at 16:00

• 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>." – dtldarek Jun 2 '14 at 15:45