A Suslin tree has a Suslin regular-open algebra

By F3-F4, $B$ is a complete Boolean algebra and each $0<b\in B$ has some tree condition $t$ with $0<e(t)\le b$. Since $P$ is nonempty, its underlying space is nonempty, so the regular-open bounds $0=\mathrm{\varnothing }$ and $1=P$ are distinct. Thus $B$ is nontrivial.

Let $0<b\in B$ and choose $t$ with $e(t)\le b$. Splitting supplies two distinct immediate tree successors $s,u$ of $t$. They are incompatible, so F3 gives nonzero disjoint elements $e(s),e(u)\le e(t)\le b$. Therefore $0<e(s)<b$, proving atomlessness.

Let $X\subseteq B^{+}$ be pairwise disjoint. By A1 and density of $e$, choose $t_b\in P$ with $e(t_b)\le b$ for every $b\in X$. If $b\mathrm{\ne }c$, compatibility of $t_b,t_c$ would make $e(t_b)\wedge e(t_c)$ nonzero by F3, while it lies below $b\wedge c=0$. Thus the $t_b$ form a forcing antichain; they are distinct and F2 makes $X$ countable. Hence $B$ is ccc.

Fix a double sequence $(b_{n,m}{)}_{n,m<\omega }$ in $B$ and put $c={\bigwedge }_n{\bigvee }_m{b}_{n,m}$ and $r={\bigvee }_{f\in {\omega }^{\omega }}{\bigwedge }_n{b}_{n,f(n)}$. Always $r\le c$. In any complete Boolean algebra, $a\wedge \bigvee X={\bigvee }_{x\in X}(a\wedge x)$: the right side is below $a\wedge \bigvee X$, and if $y$ bounds all $a\wedge x$, then $\mathrm{\neg }a\vee y$ bounds $X$, giving $a\wedge \bigvee X\le y$. Suppose $0<d=c\wedge \mathrm{\neg }r$ and choose $p_0$ with $e(p_0)\le d$. For each $n$, let $D_n$ contain the conditions $q$ such that either $q$ is incompatible with $p_0$, or $e(q)\le b_{n,m}$ for some $m$. This set is open. It is dense: if $q$ is compatible with $p_0$, take $q^{\mathrm{\prime }}\le q,p_0$; since $e(q^{\mathrm{\prime }})\le c\le {\bigvee }_m{b}_{n,m}$, the just-proved distributive identity makes some $e(q^{\mathrm{\prime }})\wedge b_{n,m}$ nonzero, and density of $e$ plus compatibility reflection gives an actual $s\le q^{\mathrm{\prime }}$ with $e(s)\le b_{n,m}$. By F2 choose $q\le p_0$ in every $D_n$. Then incompatibility with $p_0$ is impossible. Let $f(n)$ be the least $m$ with $e(q)\le b_{n,m}$. Completeness gives $e(q)\le {\bigwedge }_n{b}_{n,f(n)}\le r$, while $e(q)\le e(p_0)\le \mathrm{\neg }r$, a contradiction. Therefore $d=0$, so $c\le r$ and the exact diagonal law holds.

Steps 1.1-2.3 give nontriviality, completeness, atomlessness, ccc, and precisely the identity in F1. Therefore $B$ is a Suslin algebra. AC is used only at step 2.2 for a set-indexed simultaneous selection and through the already declared distributivity supplier; the regular-open construction itself is choice free.