AC implies BPI

Let $P$ consist of the proper filters $G\supseteq F$, ordered by inclusion. This is a set of subsets of $B$ and inclusion is reflexive, antisymmetric and transitive. It is nonempty because $F\in P$. The empty chain has upper bound $F$.

For a nonempty chain $\mathcal{C}\subseteq P$, put $G=\bigcup \mathcal{C}$. It contains $F$ and $1$ and excludes $0$. If $a\in G$ and $a\le b$, a member containing $a$ also contains $b$. If $a,b\in G$, two chain members witnessing this are comparable, so one contains both and their meet. Thus $G\in P$ and bounds the chain.

Apply F2, with AC supplied by F1 and its poset hypotheses checked in steps 1.1 and 2.1, to obtain a maximal $U\in P$. Every proper filter extending $U$ also extends $F$ and belongs to $P$, so maximality in $P$ is maximal properness in $B$. This application of Zorn is the use of AC.

For a nontrivial $B$, start with the proper filter $\{1\}$; step 3.1 gives an ultrafilter whose complement is a prime ideal by F3. This is BPI by F4. For the trivial algebra there is no proper initial filter and BPI makes no existence demand. QED.