Countable completeness and transitive collapse

Assume U countably complete. If a nonempty set S of Scott classes has no E-minimal member, choose an initial a_0 in S. By F4 choose, for each a in S, a predecessor s(a) in S; recursion gives $a_{n+1}=s(a_n)$. By F2 and F4 choose representative functions f_n from the nonempty Scott sets a_n. Each $A_n=\{i:f_{n+1}(i)\in f_n(i)\}$ belongs to U. Countable completeness makes their intersection a member of U, hence nonempty. At any i in it, the set $\{f_n(i):n\in \omega \}$ has no membership-minimal member, contrary to Foundation. Thus every nonempty set of classes has an E-minimal member. This also suffices for definable subclasses: for any chosen member, its finite predecessor closure is a set by set-likeness and Replacement, and an E-minimal member of its intersection with the subclass is minimal in that subclass.

Conversely, if U is not countably complete, take A_n in U with intersection A not in U. Set $B_n=(I\setminus A)\cap {\bigcap }_{k\le n}{A}_k$. Each B_n is in U and their intersection is empty. For every i let e(i) be the least n with i not in B_n. Then $\{i:e(i)>n\}=B_n$. Put $g_m(i)=\max (e(i)-m,0)$, viewed as a finite von Neumann ordinal. On B_m, g_(m+1)(i) is strictly smaller than g_m(i), hence belongs to it. Consequently $[g_{m+1}{]}_{U}E[g_m{]}_U$ for all m. Their range is a nonempty set with no E-minimal member, so the ultrapower is not well-founded.

In the complete case, F1 gives extensionality, F2 gives set-likeness and step 1.1 gives well-foundedness. Apply F3 to obtain a definable collapse pi onto transitive M. Composing pi with the constant map yields a definable elementary j by F1. For each ordinal alpha, elementarity says j(alpha) is an ordinal in M; transitivity makes this an actual ordinal. The map on ordinals is strictly increasing since membership is preserved. Induction gives $j(\alpha )\ge \alpha$: j(alpha) is above all j(beta) for beta below alpha, hence above or equal to their required supremum alpha. Given any ordinal gamma, j(gamma+1) belongs to M and is larger than gamma, so transitivity puts gamma in M. Restrictions of pi and j to sets are sets by Replacement; no Global Choice is involved.