Using variables and brackets in Predicate Logic

So, hopefully this is a fitting place to ask.

I have the sentence "If anyone likes Christina, then Didrik does", and the key: Lxy: x likes y d: Didrik c: Christian

I proposed it should be written:

• (∃xLxc) --> (Ldc)

but I friend had chosen to move the brackets, like this:

• ∃x(Lxc --> Ldc)

Unfortunately, we have both only have 1 lecture on Predicate Logic, so our discussion wasn't really that helpful, for none of us were able to really justify our symolization ...

Is anyone here capable explaining if there is any difference, if so, what it is, and perhaps also which is wrong.

Regarding variables: For the sentnece "If someone is attracted to Bjorn, the everyone hates or loves Didrik", with the key: b: Bjorn d: Didrik Axy: x is attracted to y Hxy: x hates y Lxy: x loved y

I proposed:

• ∃xAxb --> ∀x(Hxd v. Lxd)

for I was thinking that if it is the case that at least one person is attracted to Ben, then it is the case the everyone hates or loves Didrik .... (an dhopefully that is what I've written).

At the same time, someone proposed to have different variables for the two qantifiers, like this:

• ∃xAxb --> ∀y(Hyd v. Lyd)

Which one is correct (if any of them are)? Is is correct to say it doesn't matter?

Thank you!

• Welcome to Math.SE! Try to make use of Mathjax here (e.g. $\theta$ for $\theta$). – Shaun Nov 20 '13 at 15:19
• @Shaun Lighten up on your auto prompt to format posts, especially for new users! – Namaste Nov 20 '13 at 15:23
• @amWhy: Okie doke :) – Shaun Nov 20 '13 at 15:25
• @Shaun Your efforts and contributions are appreciated, particularly for those who aren't new! I just, personally, cut a little slack for those who are new or newish! ;-) – Namaste Nov 20 '13 at 15:31

Suppose that $\phi$ and $\psi$ are formulas, and $x$ does not appear as a free variable in $\psi$. Then, in general, we have $$(\exists x)\phi \to \psi$$ is equivalent to $$(\forall x)(\phi \to \psi)$$ and not always equivalent to $$(\exists x)(\phi \to \psi)$$ Here I am using the usual convention that quantifiers bind "tighter" than the $\to$ operator, so $(\exists x)\phi \to \psi$ means $((\exists x)\phi) \to \psi$.
To see why the latter is not equivalent, suppose that $\psi$ is always false, and that $\phi(y)$ holds and $\phi(n)$ does not. Then we have $\phi(n) \to \psi$ holds, so $(\exists x)(\phi(x) \to \psi)$ holds; but $\phi(y) \to \psi$ is false, so $(\exists x)\phi(x) \to \psi$ is false. This is related to the concept of "prenex normal form".
Here is what this means in the context of the first part of the question. Suppose that Yolanda likes Christian, Nancy does not like Christian, and Diderick does not like Christian. Then there is a person such that, if that person likes Christian then so does Diderick: namely, Nancy. So $(\exists x)(Lxc\to Ldc)$ is true. But it is not true that if there exists a person who likes Christian, then Diderick does, because Yolanda likes Christian and Diderick does not. So $(\exists x)Lxc \to Ldc$ is false.