Let $A$ be a Boolean algebra, and let $A^+$ denote the set of non-zero elements of $A$. A cut $U \subseteq A^+$ is a set such that if $q\in U$, then $p\le q \implies p \in U$, for all $p \in A^+$. A cut $U$ is regular if whenever non-zero $p \not\in U$, there exists $q \le p$ such that $U_q \cap U = \emptyset$, where $U_q := \{ x \in A^+ \mid x \le q\}$.
Since $A^+$ is a regular cut, and any intersection of regular cuts is regular, it follows that any cut $U$ is contained in a least regular cut $\overline{U}$. I want to show that $U_{p+q} = \overline{U_p \cup U_q}$. One direction is obvious, since clearly $U_p, U_q \subseteq U_{p+q} \implies U_p \cup U_q \subseteq U_{p+q} \implies \overline{U_p \cup U_q} \subseteq U_{p+q}$ by leastness. But I am having trouble showing inclusion in the other direction. Any help would be appreciated.