Lets add a constant symbol $V$ to the signature of the language of set theory. So working in first order logic with equality, add the following axioms about $\in $ and $V$.

**Extensionality:** $\forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \to x=y)$

**Set construction (reflection):** if $\phi$ is a formula in which all and only symbols $y,x_1,..,x_n$ occur free, and non of them occur bound, in which the symbol $V$ doesn't occur, then: $$\forall x_1 \in V,...,\forall x_n \in V \\ \forall y (\phi \to y \in V) \to \exists x \in V \forall y (y \in x \leftrightarrow \phi)$$; is an axiom.

**Set-hood:** $\forall x \ (x \subsetneq V \leftrightarrow x \in V)$

Now this theory is formulated in the same langauge of Ackermann's, it share with it the first two axioms and the right to left implication of the third axiom. However the left to right implication of the third axiom is in some sense daunting! The idea here is that this theory doesn't have comprehension axioms about proper classes, clearly by the third axiom all classes the second axiom constructs are sets! If we just add the class comprehension schema of Ackermann, then we immediately get a contradiction, since the Russell class would be a set. Now this theory easily prove all axioms of Zermelo set theory, and I'd think that (if consistent) it might even be equi-interpretable with the full Ackermann's set theory itself. Its also to be noted that if instead of adding the symbol $V$ as a constant, we added it as a one place predicate symbol (and so every formula $x \in V$ would be turned to $V(x)$) [as it is the case in the original formulation of Ackermann's where he actually used the symbol $\mathcal M$ for that], then it appears that Infinity would not be provable, which proves that the way how $V$ is added as a primitive does matter as regards the consistency strength of extensions of fragments of Ackermann's set theory.

Question: What's the exact consistency strength of this theory?

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