Proof of existence of intersection of non-empty family from the scheme axiom of comprehension I am reading Cori's and Lascar's second volume of "Mathematical Logic". They give the scheme axiom of comprehension : 

saying then we can define subsets $v_{n+2}$ of $v_{n+1}$ by formulas $F(v_0nv_1,\ldots,v_n)$ with variable $v_0$ are parameters $v_1,\ldots,v_n$. They use it then to prove the existence of the intersection of a non empty family of sets :

There are two things I don't understand. The first is the equivalence of the two formulas 

and

The second thing I don't understand is where my error is when I try a direct proof without using their set $c$ (that is, without using that $a$ is non empty): take $n = 2$, and $P(v_0,v_1,v_2) = "\forall v_2, \left( v_2 \in v_1 \Rightarrow v_0 \in v_2 \right)"$ and apply the scheme axiom of comprehension with $v_1 = a$ ensuring that : \begin{equation}
\exists v_3, \forall v_0, \left( v_0 \in v_3 \right) \Leftrightarrow \left( \forall v_2, \left( v_2 \in v_1 \Rightarrow v_0 \in v_2 \right) \right)
\end{equation} is true. Then $v_3$ is the set we are looking for, and is unique by the axiom of extensionality (two sets are equal if and only if they have the same elements.) My proof is obviously false as if $a = v_1 = \varnothing$ we see that the previous statement implies that $v_3$ contains all sets which is a paradox in $\textrm{Z}^{-}$ (the theory with axiom of extensionality, pairs, unions, power set, comprehension).
 A: Intuitively, we want to define the set containing all sets with the property $F(v_0)=\forall v_3(v_3\in a\to v_0\in v_3)$ (which we can abbreviate using the standard $\forall v_3\in a(v_0\in v_3)$), but the problem is that comprehension does not actually allow us to do this, since it only allows us to define subsets of a given set (for good reason!). So we need to change this unbounded comprehension into a bounded comprehension. This is what the set $c\in a$ is for: $\cap a$ will be a subset of any given set in $a,$ so if $c\in a,$ we have $$ \{v_0:F(v_0)\}=\{v_0\in c:F(v_0)\}$$ and the latter is something that is guaranteed to be a set by the comprehension axiom.
So the reason why $\forall v_3\in a(v_0\in v_3)$ is equivalent to $v_0\in c\land \forall v_3\in a(v_0\in v_3)$ is just that: if a set is in the intersection of $a$ then the fact that it is in some given set $c\in a$ is just extraneous information. More formally, if $\forall v_3\in a(v_0\in v_3)$ holds then since $c\in a,$ we can instantiate this to $v_0\in c,$ and thus we have $v_0\in c\land \forall v_3\in a(v_0\in v_3).$ The other direction of the equivalence is trivial.
Your error is that you have ignored the part of the comprehension axiom where we bound the comprehension by a set (as I said in more detail in the comments). As such, what you wrote down isn't an instance of comprehension, rather an instance of unrestricted comprehension, so it should come as no surprise that it results in a contradiction. Yes, naively the intersection of the empty set should contain all sets (since every set satisfies $F(v_0)$ when $a=\emptyset$) but as you say, a set of all sets can't exist. But if you can only use the (bounded) comprehension axiom you can't define the intersection operation on the empty set so you don't wind up with a contradiction. 
(Or rather you can't define it in a way that comports with the fact that morally, the intersection of the empty set should be "everything". For the sake of having a total operation, I've seen some authors define $\cap \emptyset =\emptyset$, which is gross, but you have to handle this issue somehow.)
