Setting: Let $H$ be a compact Hausdorff space and suppose the collection $\{ a_i \mid i \in I \}$ is a subbasis for the topology on $H$. Moreover, suppose each of the $a_i$ is regular open.
Let $\mathbf{I}u$ and $\mathbf{C}u$ denote the interior and closure of a set $u$. It is known that the collection of all regular open sets on a compact Hausdorff space forms a Boolean algebra with join $\bigvee u_j = \mathbf{IC}(\bigcup u_j)$, meet $\bigwedge u_j = \mathbf{I}(\bigcap u_j)$ and negation $\neg u = \mathbf{I}(H \setminus u)$. Call this Boolean algebra $\operatorname{RO}(H)$.
Let $A$ be the collection of subsets of $H$ generated by the $a_i$ and closed under the intersection, union and negation ($\bigwedge, \bigvee, \neg$) as described in the previous alinea. (That is, the ones used to make the collection of regular opens a Boolean algebra.) Then $A$ carries a Boolean algebra structure.
Question: Is $A = \operatorname{RO}(H)$?
Ideas: It is obvious that $A$ is a sub-Boolean algebra of $\operatorname{RO}(H)$. I suspect the converse might be true because the $a_i$ generate the topology on $H$, but I have not found a proof yet.