Scott coding and set-likeness of ultrapower membership

U-equivalence is reflexive and symmetric, and transitive because the intersection of two coordinate equality sets is contained in the third. The function f itself witnesses a possible representative rank; minimize within rank(f)+1 among ranks attained by equivalent functions. The least rank rho is attained, and Separation in $V_{\rho +1}$ forms all equivalent functions of rank rho, a nonempty set. Equivalent f,g have the same equivalence class and hence the same minimum-rank set. Conversely equal Scott sets have a common representative, so f and g are equivalent by transitivity.

If f,f-prime and g,g-prime are respectively equivalent, their coordinate membership truth sets agree on the intersection of their two equality sets, which is in U. A truth set agreeing there with a U-large set is U-large by intersection and upward closure; this implication is symmetric. Thus E does not depend on the selected functions representing either Scott set.

Fix g and put $A=\bigcup \mathrm{ran}(g)\cup \{\mathrm{\varnothing }\}$. If [f] E [g], replace f by $h(i)=f(i)$ when $f(i)\in g(i)$ and by empty otherwise. Then h maps I to the set A and is U-equivalent to f. All predecessors are consequently among $\{[h{]}_U:h\in {}^{I}A\}$, a set by Replacement on the set of functions. Separate those satisfying E with [g] to get exactly the predecessor collection. The fallback is fixed empty, so this bounding argument uses no AC.