Correct way to use universal quantifiers when specifying a set I have a set:
$X = \{p | p \in P \wedge \forall a(a \in A \wedge p \in F(a) \wedge p \notin F'(a))\}$
This reads (to me): the set containing all p, where p is an element of P, and for all a, a is an element of A and p is in F(a) and not in F'(a).
1) Does this read that for p to be in the set X, $p \in F(a) \wedge p \notin F'(a)$ for every occurrence of $a \in A$?
2) Would it be better as $X = \{p | p \in P \wedge \forall a(a \in A \rightarrow p \in F(a) \wedge p \notin F'(a))\}$ or does this mean that if $a \notin A$ then $p \in X$ because 0 -> anything = 1?
I want it to be the set (informally) of every element where the element is in F(a) and not in F'(a) for all a in A.
Hope that made sense. Thank you.
 A: 
2) Would it be better as $X = \{p | p \in P \wedge \forall a(a \in A \rightarrow p \in F(a) \wedge p \notin F'(a))\}$ 

Yes, you definitely want that formulation.
In standard set theory a formula of the shape
$$ \forall a(a\in A\land \cdots) $$
is always false, because it can only be true if $A$ is such that anything in the world you can think of calling $a$ will be an element of it. And there is no such $A$ in standard set theory. So there will always be at least one $a$ that makes the $a\in A$ part false, and that itself dooms the entire $\forall$ -- it doesn't even matter what the other conjunct is.

or does this mean that if $a \notin A$ then $p \in X$ because 0 -> anything = 1?

No -- the way the $\forall$ quantifier work is that your
$$ \forall a(a \in A \rightarrow p \in F(a) \wedge p \notin F'(a))$$
is true if and only if every $a$ somehow makes the formula in parentheses true. Thus for an $a$ that is not in $A$ you want that formula to be true -- this is how you ensure that those $a$ don't really contribute to the truth value of the $\forall$. When you have a $\forall$, a choice that makes the inner formula true is a "don't care" choice -- what you're interesting is whether there are any counterexamples or not.

As a matter of notation, you would usually write
$$X = \{p \in P \mid \forall a \in A:\,(p \in F(a) \wedge p \notin F'(a))\}$$
which expands to your formulation 2. In particular the bounded quantifier $\forall a\in A$ conventionally abbreviates the $\to$ form.
On the other hand $\exists a\in A:\;(\cdots)$ would abbreviate $\exists a(a\in A\land \cdots)$ because for an $\exists$ it is false that is the "don't care" value, and therefore $\land$ that encodes "ignore $a$s that don't satisfy this".
A: Going off of what you want to create, have you considered the following?
$$P\cap\bigcap_{a\in A}F(a)\setminus F'(a)$$
This equates to your number 2.
