Where in the analytic hierarchy is the theory of true set theory? Where in the analytic hierarchy is the theory of all true sentences in ZFC? In higher-order ZFC?  In ZFC plus large cardinal axioms?
Edit:  It seems that this is ill-defined.  Why is this ill-defined for ZFC, but true for weaker theories like Peano arithmetic and higher-order arithmetic?
 A: It depends what you mean by "true sentences of ZFC."
If you mean the set of true sentences in the language of set theory - that is, the theory of the ambient model of set theory $V$ - then this isn't in the analytic hierarchy at all. This is because in ZFC we can define the true theory of $V_{\omega+1}$, which consists essentially of the true analytic sentences. So the theory of $V$ is strictly more complicated than anything in the analytic hierarchy. Indeed, any "reasonable" hierarchy will fail to reach the complexity of the theory of $V$, for a much more fundamental reason: the theory of $V$ can't be definable in $V$, by Tarski's theorem! So any complexity hierarchy, all of whose levels are definable, can't capture $Th(V)$. For example, $Th(V)$ is not


*

*arithmetic,

*analytic,

*$\Pi^m_n$ for any $m, n$ (note that already $\Pi^2_1$ exhausts the arithmetic, analytic, and much more)

*or computable from $Th(D, \in)$ for any definable set $D$. Note that we can take $D$ to be something like "$V_\kappa$ for the first inaccessible $\kappa$," or similarly with "inaccessible" replaced with any other definable large cardinal notion. So even $Th(V_\kappa, \in)$ for "big" $\kappa$ is much, much less complicated than $Th(V)$.
Meanwhile, exactly how complicated $Th(V)$ is depends on $V$ - see Mitchell's answer.
If, on the other hand, you mean the set of consequences of ZFC, then this is just at the level of $\Sigma^0_1$ (or $0'$) - not even into the analytic hierarchy, just the first nontrivial level of the arithmetic hierarchy.
A: Let $\Gamma$ be the set of sentences of ZFC that happen to be true.  I doubt that you can determine much about this set, but you can see that it's consistent with ZFC that $\Gamma$ is as complicated as you want.
For instance, using Easton forcing, you can find a model $M$ of ZFC in which $2^\kappa$ has any values you want for regular $\kappa,$ subject only to the limitations that it's a non-decreasing function of $\kappa$ and $\kappa\lt \operatorname{cf}(2^\kappa).$  So for any set $A\subseteq\omega$ (for instance, pick $A$ not to be in the analytical hierarchy at all), you can code $A$ in the values of $2^{\aleph_n},$ and ensure in that way that $A$ is Turing-reducible to $\Gamma$ in $M.$
Here's one way the coding could work: Define $f(n)=n+1+\operatorname{card}(A\cap n),$ and then arrange things so that $2^{\aleph_n}=\aleph_{f(n)}$ for all $n\lt\omega.$  (To be completely accurate here, we need to make sure that we're not changing the analytical hierarchy by the forcing argument, so start with a model of $V=L,$ and only change  $2^{\aleph_n}$ for $n$ large enough that we don't add any new reals or sets of reals.)
As for large cardinals, the same construction works since you can typically change $2^{\aleph_n}$ for $n\lt\omega$ without losing any large cardinal properties.  (And starting with a model of GCH was just a convenience; it's not essential.)
A: Truth has to be defined relative to an intended interpretation. Unfortunately, ZFC doesn't have one; in particular, the usual story that ZFC is intended as a theory of the von Neumann universe fails to tell us which large cardinals exist. So in some sense, we don't really know what ZFC is supposed to "mean".
But suppose you decide that you're interested in the theory of $V_\kappa$, where $\kappa$ is the least inaccessible cardinal. So define $$T:=\mathrm{ZFC}+\mbox{(there exists an inaccessible)}.$$ Then $T$ ensures that there's a true theory of $V_\kappa$. You won't be able to prove much about it though; for example, you won't be able to decide whether or not GCH is an element of this theory. That's because in some models of $T$, the set $V_\kappa$ does satisfy GCH, and in other models of $T$, the set $V_\kappa$ doesn't satisfy GCH.
