(This is from a problem set given years ago that I am self-studying.)
Let $S$ denote the successor function given by $S(x) = x \cup \{x\}$. Given a set $A$, prove that the class $\{S(x) : x \in A\}$ exists as a set, without using replacement.
Hint: consider $\{y \in B : y = S(x) \text{ for some }x\in A\}$ for a sufficiently large set $B$
Does the hint not solve the problem by Separation, i.e., since $A$ is a set, there is a unique (by Extension) $B$ such that: $B=\{x\in A:\phi (x)\}=\{y \in B : y = S(x) \text{ for some }x\in A\}$.
If by chance this is correct, what constitutes a "sufficiently large" set $B$.
Thanks