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

Question

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.

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 02 '14 at 15:31
Your translation is correct, apart from a parenthesis problem. — André Nicolas, Jun 02 '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 02 '14 at 15:38
Ok so by adding the parenthesis I would include x into the last part of the formula? — Byebye, Jun 02 '14 at 15:45
I've never filled a tank with blankets, so obviously this is not a thing that is done. — Matt Samuel, Apr 13 '20 at 01:25

drhab · Accepted Answer · 2014-06-02T16:01:48.367

1

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.

edited Jun 02 '14 at 16:01

answered Jun 02 '14 at 15:41

drhab

151,093

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 ." – dtldarek Jun 02 '14 at 15:45
@dtldarek Thanks, repaired. – drhab Jun 02 '14 at 16:02

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

1 Answers1