If $V$ is the von Neumann constructible universe, is $\langle V,\in \rangle\models V=L$ true? My question is essentially in the title. I'm reading some notes about the proof of consistency of ZFC assuming the consistency of ZF. The author first assumes theres is a model for ZF-Foundation and proves that the von Neumann universe is then a model for ZF. 
Subsequently the author constructs the constructible universe L and proves that it is a model for the axiom of constructibility V=L and subsequently for AC and GCH.
As far as I understand why L believes in V=L, it's not clear to me whether V is a model of V=L or not. 
To be a model for a sentence means that the sentence is true or false under an interpretation, so it seems to me that there should be a definite answer to that, i.e. either V is a model of V=L or it is not. 
I was unable to find any reference for that in the literature. Would anyone care to help me with that? 
(In particular if V believed in V=L then it would already be a model for ZFC. Or is the answer that we simply cannot deduce $\langle V,\in \rangle\models V=L$ nor V doesn't model V=L? But if so, is it a theorem than we cannot prove any of the above (from ZFC) or we simply don't know?)
 A: Every model of $\sf ZF$ is a von Neumann universe, namely it has a decomposition into a hierarchy of $V_\alpha$'s, such that $V_0=\varnothing$, $V_{\alpha+1}$ is the power set (in $V$) of $V_\alpha$, and at limit ordinals we take limits.
Now, in order to show that $V=L$ is not a theorem of $\sf ZF$ you need to exhibit models where $V\neq L$ holds. For this you need to know two things:


*

*If $M$ and $N$ are two models of $\sf ZFC$ with the same ordinals, then $L$ as constructed in $M$ and in $N$ is the same model. Namely, $L$ depends only on the ordinals of the universe.

*If $\sf ZFC$ is consistent then $\sf ZFC+\lnot CH$ is consistent. Since $V=L$ implies $\sf CH$, this means that $V=L$ is not provable from $\sf ZFC$. The same can be deduced by the theorem that $\sf ZF+\lnot AC$ is consistent if and only if $\sf ZFC$ is consistent, since $L$ satisfies $\sf AC$. For this you need to know what forcing is, which is generally something much more advanced than just the construction of $L$.
To sum up, every model of $\sf ZF$ is a von Neumann universe, but not every model of $\sf ZF$ satisfies $V=L$.
