Can Ackermann theory minus foundation minus class comprehension permit allowing every proper subclass of $V$ to be a set?

Question

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?

Answer 1

The theory is inconsistent. Let ZG(x) be the formula ∀u∈x∀v∈u(v∈x)∧∀t(x∈t→∃s∈t(∀𝑦∈s(y∉t)))∧(∀t(∃s∈x(s∉t)→∃y∈x(y∉t∧∀u∈(x-t)(u∉y))))∧∀t∈x∃s∀v(v∈s↔(v=tνv=x))∧∀u∈x∃t∈x∀s(s∈t↔(s∈u∧∃r(r∈s)))∧∃t∀s(s∈t↔(s∈x∧∃r(r∈s))) (That is x is transitive; if x is in t, then t has an ∈-minimal element; if x is not contained in t, then there is an ∈-minimal element of x which is not in t; if t is in x then the pair {t,x} exists; if t is in x then t-{0} exists; and x-{0} exists.)

We note some simple properties of x for which ZG(x) holds: If x∈𝑉 then x⊆𝑉.(If x were not contained in V, then there is an ∈-minimal element m of x which is not in V. If m is not V, then by Set-hood m is in V. If m=V,then {V,x} exists and has no ∈-minimal element.) If x∈𝑉, then there is a W∈𝑉 such that t is in W iff t is contained in x.(By the above property x⊆𝑉, and so anything contained in x is properly contained in V since x is not in
x. Therefore by Set construction such a W must exist.) We will denote such a W by Px. If x∈𝑉 then ZG(Px).

Suppose that ZG(x) implies x∈𝑉. By Set construction, there is a z∈𝑉 which consists of all x for which ZG(x) holds. Let y be the union of z. Then y is in 𝑉 and ZG(y). Therefore ZG(Py).
Then Py is contained in y. But then y∈y which is impossible. Therefore there must be an x such that ZG(x) and x∉𝑉. If x is contained in 𝑉 then x-{0} is in 𝑉 and so x is in 𝑉. If x is not contained in 𝑉 then there is an ∈-minimal element m of x which is not in 𝑉. But then m-{0} is in 𝑉 and so m is in 𝑉.