Minimum-rank selection and Collection

Instantiate one $c\in C$ and minimize the ranks attained in $C$ among ordinals at most $\mathrm{rank}(c)$. Separation on $\mathrm{rank}(c)+1$ and ordinal well-ordering yield a least such $\alpha$, which is globally least. All members of rank $\alpha$ lie in $V_{\alpha +1}$, so Separation inside that stage forms the asserted nonempty set.

Given the Collection premise, for each $x\in a$ the class of witnesses has a unique least rank ${\alpha }_x$ by step 1.1. Replacement collects these ordinals; set $\gamma =\sup \{{\alpha }_x+1:x\in a\}$. The stage $V_{\gamma }$ contains at least one witness for each $x$, because the witnesses of its minimum rank lie in $V_{{\alpha }_x+1}\subseteq V_{\gamma }$. Thus $b=V_{\gamma }$ suffices. For $a=\mathrm{\varnothing }$, take $\gamma =0$.

Conversely assume Separation and Collection, and let $\varphi$ be functional on the set $a$. Collection supplies a set $b$ containing a witness for each $x\in a$. Separation gives $\{y\in b:\mathrm{\exists }x\in a\varphi (x,y)\}$. Uniqueness ensures every value of $\varphi$ occurs in this set and that every member is such a value, which is the Replacement conclusion. This direction does not use rank or a prior application of Replacement.