Is this form of replacement suitable for ZF - Powerset + well-ordering principle?

Question

The following scheme can be understood as a form of replacement. Axiomatizing $\sf ZF$ with it instead of the usual replacement schema renders it immune to removal of extensionality; see here.

In an article by Gitman/Hamkins/Johnstone, it was shown that the usual version of replacement is in some sense problematic if power set axiom is removed, while at the same time having the well-ordering principle.

Would the following version of replacement work as collection does if powerset is removed while having the well-ordering principle?

Replacement$^*$: if $\varphi$ is a formula in which $B$ is not free, then:

$$\forall A \exists B \forall y \, (y \in B \leftrightarrow \exists x \in A: \forall z (z \in y \leftrightarrow \varphi(x,z)))$$

If we have extensionality, then this is:

$$\forall A \exists B \forall y \, (y \in B \leftrightarrow \exists x \in A: y= \{z \mid \varphi(x,z)\})$$

Answer 1

The answer is no.

Your version of replacement is a weakening of ordinary replacement. To see this assume that replacement holds (but perhaps not power set or collection), and then observe that for any set $A$ and any formula $\varphi$, we can first restrict to the subset of $x\in A$ for which $\varphi(x,z)$ defines a set $y$, and then apply replacement to gather these $y$s together into a set.

Since replacement does not imply collection, by the results in the paper you mention, it follows that your principle Replacement* also does not imply collection.