BPI and the set ultrafilter lemma are equivalent over ZF

Assume BPI and let $\mathcal{F}$ be a proper filter on a set $S$; if $S=\mathrm{\varnothing }$ there is no such filter. Put $I_{\mathcal{F}}=\{S\setminus A:A\in \mathcal{F}\}$. Filter closure makes this an ideal of $\mathcal{P}(S)$, and it is proper because $S\in I_{\mathcal{F}}$ would mean $\mathrm{\varnothing }\in \mathcal{F}$. Thus the quotient $\mathcal{P}(S)\mathrm{/}{I}_{\mathcal{F}}$ is a nontrivial Boolean algebra.

Conversely assume UFL, and let $B$ be a nontrivial Boolean algebra. Let $X$ be the set of all homomorphisms $p:A_p\to \mathbf{2}$ whose domains are finite Boolean subalgebras of $B$, and put $D_b=\{p\in X:b\in A_p\}$. The set $X$ is nonempty, and every finite intersection $D_{b_1}\cap \cdots \cap D_{b_m}$ is nonempty: apply F4 from the unique map on $\{0,1\}$ to the subalgebra generated by the listed elements.

By BPI choose a prime ideal $P$ of the quotient from step 1.1, and let $J=\{A\subseteq S:[A{]}_{I_{\mathcal{F}}}\in P\}$. The quotient map shows that $J$ is a prime ideal of $\mathcal{P}(S)$ containing $I_{\mathcal{F}}$.

The finite-intersection property from step 1.2 makes the supersets of finite intersections of the $D_b$ a proper filter $\mathcal{G}$ on $X$. By UFL extend it to an ultrafilter $\mathcal{V}$. For each $b$, the two disjoint sets $D_{b,0}=\{p\in D_b:p(b)=0\}$ and $D_{b,1}=\{p\in D_b:p(b)=1\}$ partition $D_b\in \mathcal{V}$, so exactly one lies in $\mathcal{V}$.

Define $\mathcal{U}=\{A\subseteq S:S\setminus A\in J\}$. It is a proper filter, contains $\mathcal{F}$, and decides every $A\subseteq S$: primality applied to $A\cap (S\setminus A)=\mathrm{\varnothing }\in J$ puts $A$ or its complement in $J$, while properness prevents both. Any proper filter strictly extending $\mathcal{U}$ would contain some $A\mathrm{\notin }\mathcal{U}$ as well as $S\setminus A\in \mathcal{U}$, hence $\mathrm{\varnothing }$; therefore $\mathcal{U}$ is maximal and is an ultrafilter.

Define $h(b)=i$ when $D_{b,i}\in \mathcal{V}$. For any finite Boolean equation among elements of $B$, the intersection of their deciding sets lies in $\mathcal{V}$ and every partial homomorphism in it obeys that equation. If the selected bits violated it, intersecting the corresponding value cells would give the empty set in $\mathcal{V}$. Hence $h$ preserves $0,1,\mathrm{\neg },\wedge ,\vee$ and is a homomorphism $B\to \mathbf{2}$.

Its zero fibre $h^{-1}(0)$ is a proper ideal, and $h(a\wedge b)=h(a)h(b)=0$ implies one factor is zero, so the ideal is prime. Thus UFL implies BPI.

Steps 1.1–3.1 prove BPI implies UFL, and steps 1.2, 2.2, 3.2, and 4.1 prove UFL implies BPI, all in ZF. The empty-set UFL instance is vacuous and the trivial Boolean algebra is excluded exactly as in F1.