Proving that $V=R$ in $ZFC$ Kunen's "Set Theory" on page 21, defines $V:\{x: x=x\}$ and the Russell set $R:\{x:x\notin x\}$.
In discussing $V=R$: Kunen presents this in the context of Lemma I.4.9 that "$V$ doesn't exist, and neither does $R$." He shows using Comprehension if we had a $V$ with $\forall x[x\in V]$, we could form $R$ as $\{x\in V: x\notin x\}$. 
Does this actually prove that $V=R$? 
Thanks 
 A: No, the fact that $V$ and $R$ “don’t exist” has nothing to do with whether or not $V=R.$ Instead of “they don’t exist,” we should probably say “they are not sets” instead for clarity, although it does literally mean they are not in our formal domain of discourse (at least if we are working in ZFC and not a class-set theory).
Quick review of what things mean. Given a property $\phi(x)$ (i.e. a formula in the first order language of set theory), we can (informally) consider the class of sets with this property. Here, the properties are $x=x,$ corresponding to $V,$ and $x\notin x,$ corresponding to $R.$
Sometimes for a given property, there is a set whose elements are exactly the sets with the given property. (For instance, the property $x\neq x$ is never true, so the empty set has exactly the elements with this property.) In that case we say “the class is a set”, or “it exists”, or whatever. But sometimes, we this is not the case as was shown for your two properties/classes.
Two classes being equal means one’s defining property holds if and only if the other’s does. For $V,$ the defining property is $x=x,$ which always holds (thus $V$ is the class of all sets) and for $R$ it is the substantially less trivial $x\notin x,$ which makes $R$ the class of all sets that don’t contain themselves. So $V=R$ means that every set doesn’t contain itself. Expressed in the language of set theory, it is the sentence: $\forall x(x\notin x).$
As I said before, whether this is true or not has nothing to do with the Russel’s paradox arguments for why $V$ and $R$ are not sets. If we have the axiom of Foundation, as we do in ZF, then it is true since Foundation implies a set cannot contain itself. If we don’t have the axiom of foundation, it is undecidable. There are consistent (relative to ZF minus foundation) “anti-foundation” axioms one can assume in place of foundation that imply $V\neq R.$ (Including simply taking $\exists x (x\in x)$ as an axiom, though there are more interesting options.)
