A countably incomplete ultrapower need not be well-founded

A free ultrafilter on omega contains no finite set: if a finite union of singletons belonged to U, the ultrafilter complement decision and finite intersections would force one singleton to belong, making it principal. Hence each cofinite tail $B_m=\{n:n>m\}$ belongs to U. Their countable intersection is empty, explicitly witnessing failed countable completeness.

For n>m, $g_m(n)=n-m>0$ and $g_{m+1}(n)=n-m-1<g_m(n)$. Since these are finite von Neumann ordinals, this is exactly $g_{m+1}(n)\in g_m(n)$. F1 therefore gives $[g_{m+1}{]}_{U}E[g_m{]}_U$. Also every g_m(n) belongs to omega, so every displayed class belongs to [c_omega], the internal natural-number set. For n<=m both functions in the comparison are zero, so membership fails there; the truth set is exactly B_m, not just an unspecified large set.

Replacement forms the nonempty set $\{[g_m{]}_U:m\in \omega \}$. Each member has its next displayed class as an E-predecessor in this set, so the set has no E-minimal member and the relation is not well-founded. All functions and classes were explicitly defined; no choice of representatives and no additional AC were used.