# Can this extension of a fragment of Ackermann set theory survive inconsistency?

Lets work in the language of Ackermann set theory., which is first order logic with equality $$="$$, class membership $$\in"$$, and sub-world $$V"$$, where the last is a constant symbol informally standing for the class of all sets.

Now add to it identity axioms in the usual manner. The rest of axioms are:

1. Extensionality: $$\forall x (x \in a \leftrightarrow x \in b) \to a=b$$

2. Reflection axiom schema: if $$\phi$$ is a formula that doesn't use the symbol $$V$$, in which $$y$$ is free, and $$x$$ not free, having its parameters among $$\vec{p}$$, then all closures of:

$$\vec{p} \in V \wedge \forall y (\phi \to y \in V) \to \exists x \in V \ \forall y (y \in x \leftrightarrow \phi)$$; are axioms.

3. Set-hood: $$\forall x [x \neq V \to (x \subseteq V \leftrightarrow x \in V)]$$

Questions:

1. Is there an obvious inconsistency with this theory?

2. If not, then is this theory equivalent to ZF in the same sense Ackermann is equivalent to ZF?

I mean this theory clearly interpret both of Ackermann's completeness axioms for $$V$$, and also foundation, so it can interpret Ackermann minus class comprehension schema. It clearly can interpret all axioms of Empty, Pairing, Set union, Power, Separation, Infinity, and Replacement for formulas in the pure set langauge that are closed on the set world, and I conjecture that this is all of what's needed to interpret ZF. Now I don't know if this would be actually enough to interpret axioms of $$ZF$$, but I guess so. Matters depend on whether Reinhardt proof of interpretability of axioms of $$ZF$$ in Ackermann's make essential use of Class comprehension schema, which I suspect it doesn't. That said what remains is the consistency question of this theory, and an answer as to whether class comprehension axiom matters.