From [Ackermann's set theory equals ZF](https://www.sciencedirect.com/science/article/pii/0003484370900112) (1970) by William N. Reinhardt:

Let **A** be the theory determined by the following axioms:

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

2. **Class construction**: $\exists x \forall y (y \in x \leftrightarrow y \in V \land \phi)$ where $x$ is not free in $\phi$

3. **Strong completeness of V**: $(y \in x \lor y \subseteq x) \land x \in V \to y \in V$

4. **Ackermann's schema**: $a, b \in V \to (\forall y)(\phi \to y \in V) \to (\exists x \in V) (\forall y) (y \in x \leftrightarrow \phi)$ where $\phi$ does not mention $V$ and its only free variables are $y,a,b$

On page 191, the paper says

> In the paper introducing the theory **A** [1], Ackermann showed that the relativizations to V of all the axioms of Zermelo's theory (**Z**) (except for the regularity axiom) could be proved in **A**. He also stated that the replacement schema of **ZF** (relativized to V) can be proved in **A**. An error in Ackermann's proof of this assertion was found by Levy [8], however, and the question remained open.

On page 192-193, it says

> Recall that Ackermann originally stated that the replacement schema of **ZF** (relativized to V) can be proved in **A**. Notice that (2) establishes this with **A** replaced by **A***, i.e. assuming the axiom of regularity for sets. Since **A*** is consistent relative to **A** in much the same way that **ZF** is consistent relative to **ZF** without regularity, this shows that Ackermann's statement is correct in the universe of "regular" sets and classes. I do not know whether the assumption of regularity is necessary.

Does **A** alone prove replacement (relativized to V)?