In my research I stumbled across a following result:
Let $G = (V, E)$ be a multigraph with three chosen vertices $a, b, c \in V$. We color its edges into red and blue colors: $E = R \sqcup S$. Events we consider refer to $H = (V, R)$, so for example ``$ab|c$'' is the event that $a$ and $b$ are connected via red edges and there is a blue cut between them and $c$. Then
- If $a|b|c$ holds, then $E$ can be split as $E_1 \sqcup E_2 \sqcup E_3$ so that for any coloring of $G$ coinciding with our coloring on $E_1$ ($E_2$ and $E_3$ respectively) we have $\neg a|bc$ ($\neg ac|b$ and $\neg ab|c$ respectively).
- Same is true if $abc$ holds.
The statement is interesting even if all the edges are colored red as in the picture. Different shades of red represent $E_1$, $E_2$ and $E_3$.
So if we use $A \square B$ to denote that $A$ and $B$ have disjoint witnesses (as in the van den Berg-Kesten inequality), one can write it as $$abc \cup a|b|c = (\neg ab|c) \square (\neg ac|b) \square (\neg a|bc).$$
The proof I have is rather tedious. I would like to get some references to the literature concerned with this type of question. It would be really helpful to know that we can have two events in different colorings of the same graph, but the witnesses are still disjoint.
