What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory? Say that "U" is the axiom that "For each set x, there exists a Grothendieck universe U such that x $\in$ U", where Grothendieck universes are defined in the usual way (or, if that's unclear, in this way). Also, say that "Ca" is the axiom that "For each cardinal κ, there is a strongly inaccessible cardinal $\lambda$ which is strictly larger than κ."
It's known that ZFC+U and ZFC+Ca are completely equivalent and prove the same sentences. A sentence is a theorem of ZFC+U iff it's a theorem of ZFC+Ca.
In addition to the above, there's also Tarski-Grothendieck set theory, which can be found here. The axioms of TG are

*

*The axiom stating everything is a set

*The axiom of extensionality from ZFC

*The axiom of regularity from ZFC

*The axiom of pairing from ZFC

*The axiom of union from ZFC

*The axiom schema of replacement from ZFC

*Tarski's axiom A

Tarski's axiom A states that for any set $x$, there exists a set $y$ containing, $x$ itself, every subset of every member of $y$, the power set of every member of $y$, and every subset of $y$ of cardinality less than $y$.
These three axioms from ZFC are then implied as theorems of TG:

*

*The axiom of infinity

*The axiom of power set

*The axiom schema of specification

*The axiom of choice

My question is as follows: is TG also completely equivalent to ZFC+U and ZFC+Ca, equivalent in the same sense that something is a theorem of TG iff it's a theorem of the other two? Is TG just an axiomatization of ZFC+U/ZFC+Ca which removes redundant axioms and allows them to just be theorems? Or is there some subtle difference between TG and ZFC+U/ZFC+Ca, in that there's some sentence which TG proves that's undecidable in ZFC+U/ZFC+Ca or vice versa?
In other words, instead of typing ZFC+Grothendieck, can I just type TG and be referencing a different axiomatization of the exact same thing?
 A: Yes.  Assume ZFC.  If there is a proper class of inaccessible cardinals, then Tarski's Axiom A holds because whenever $\kappa$ is inaccessible, the rank initial segment $V_\kappa$ of $V$ is a Tarski set.  Conversely, if Tarski's Axiom A holds then for every set $x$ there is a Tarski set $y$ with $x \in y$.  We will show that $|y|$ is an inaccessible cardinal greater than $|x|$, proving the existence of a proper class of inaccessible cardinals.
To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, given $\zeta < \kappa$ we take a subset $z$ of $y$ of size $\zeta$.  We have $z \in y$ because $y$ contains its small subsets.  Then we have $\mathcal{P}(z) \in y$ because $y$ is closed under the power set operation.  Finally $\mathcal{P}(\mathcal{P}(z)) \subset y$ because $y$ contains all subsets of its elements.  This shows that $2^{2^{\zeta}} \le \kappa$ and therefore that $2^{\zeta} < \kappa$.
To show that the cardinality $\kappa$ of $y$ is regular, notice that if $\kappa$ is singular then by the closure of $y$ under small subsets we can get a family of $\kappa^{cof(\kappa)}$ many distinct sets in $y$, contradicting the fact that $\kappa^{cof( \kappa)} > \kappa$ (which is an instance of Koenig's Theorem.)
