I am looking for a proof of the fact which is formulated at the bottom of this post. The property of regular convex sets which the fact expresses seems to be true to me, yet I have not been able to demonstrate it in a rigorous way. Any hints will by very appreciated.

Let $\mathrm{r}\mathscr{O}$ be the family of open domains (regular open sets) of a topological Cartesian space $\langle\mathbf{R}^2,\mathscr{O}\rangle$ (with the standard topology), that is: $$A\in\mathrm{r}\mathscr{O}\iff A=\mathrm{int(\mathrm{cl(A)})}.$$

It is well know that $\langle\mathrm{r}\mathscr{O},+,\cdot,-,0,1\rangle$ is a complete Boolean algebra, where: $A+B=\mathrm{int(\mathrm{cl(A\cup B)})}$, $A\cdot B=A\cap B$ and $-A=\mathrm{int}(\complement A)$ (with $\complement$ the standard set-theoretical complement operation).

In this algebra define the disjoint sum operation in the usual way: $$A\oplus B=(A-B)+(B-A)\,.$$

By *a convex* set $A$ I understand, standardly, a regular open set for which it is the case that for any two points $x,y\in A$: $[x,y]\subseteq A$. The fact I would like to prove is (the order $\leqslant$ is standard) the following kind of irreducibility:

Let $A,B,C\in\mathrm{r}\mathscr{O}$ be convex: $$A\leqslant B\oplus C\ \mbox{and}\ B\cdot C\neq 0\Longrightarrow A\leqslant B-C\ \mbox{or}\ A\leqslant C-B\,.$$