Completeness, extremal disconnectedness, and regular-open clopens

Assume $B$ is complete, and let $G\subseteq X$ be open. Set $A=\{a\in B:[a]\subseteq G\}$ and $b=\bigvee A$. The basis property of F3 gives $G={\bigcup }_{a\in A}[a]$. Hence $G\subseteq [b]$, and since $[b]$ is closed, $\bar{G}\subseteq [b]$. If $[b]\setminus \bar{G}$ were nonempty, it would be open and F3 would supply $0<d\in B$ with $[d]\subseteq [b]\setminus \bar{G}$. Order reflection gives $d\le b$ and $d\wedge a=0$ for every $a\in A$. Thus $b\wedge \mathrm{\neg }d$ would be an upper bound of $A$ strictly below $b$, impossible. Therefore $[b]=\bar{G}$, which is open. This proves extremal disconnectedness, including $G=\mathrm{\varnothing }$, where $A=\{0\}$, $b=0$ and $\bar{G}=\mathrm{\varnothing }$.

Assume $X$ is extremally disconnected. Each clopen $K$ satisfies $K=\mathrm{int}\bar{K}$, so it is regular open. Conversely for regular open $U$, the closure $\bar{U}$ is open by the assumption. Hence $U=\mathrm{int}\bar{U}=\bar{U}$, so $U$ is also closed. This proves equality of the two sets of subsets.

Assume $\mathrm{Clop}(X)=\mathrm{RO}(X)$. On clopens, F2's binary meet, complement and binary join reduce respectively to intersection, set complement and union, since the result is already clopen; the bounds are the same as well. Thus the equality is an equality of Boolean algebras. Completeness of F2 transfers through the isomorphism F3 to $B$: for $A\subseteq B$, take the regular-open supremum of $\{[a]:a\in A\}$, which by the assumed equality is a clopen $[b]$; order reflection makes $b$ exactly the least upper bound of $A$. Empty suprema are included. This proves completeness of $B$ and closes the three-implication cycle with steps 1.1 and 1.2. If $B$ is trivial, $X$ is empty and all three claims hold with the sole subset $\mathrm{\varnothing }$; the operation comparison still applies. Arbitrary unions of clopens need not themselves be clopen: the common algebra's arbitrary join is the regularized union from F2. QED.