BK inequality for increasing and decreasing events. Let $\Omega = \{0,1\}^N$. Given two increasing events $A, B \subset \Omega$, and let $\overline{A}$ denote the event where we flip the $0$'s and $1$'s inside each configuration $\omega\in A$. (So $\overline{A}$ is a decreasing event.) Then is it true that
$$\mathbb{P}(A\circ B) \leq \mathbb{P}(\overline{A}\circ B)$$
where $\circ$ denotes the disjoint occurrence of the two events.
The intuition behind this is that in a perfect world if the numbers of $0$'s and $1$'s are the same, then it seems to cost more for $A\circ B$ to occur than $\overline{A}\circ B$.
 A: It is indeed true, and the proof is applying the lemma behind Reimer's inequality to the left-hand side, and then noticing that $\bar A\circ B = A \cap B$ to rewrite the right-hand side.  More precisely:

*

*Underlying Reimer's inequality is the following fact, which can be
found as Lemma 4.1 in the paper cited below.  For
$\omega\in\{0,1\}^N$ and $S\subseteq\{1,2,\dots,N\}$, define the
following cylinder set:
$$[\omega]_S = \{\omega'\in\{0,1\}^N:\, \omega'_i = \omega_i \text{
for all $i\in S$}\}.$$
Then given any $X\subseteq\{0,1\}^N$ and any choice of
$S(\omega)\subseteq\{1,2,\dots,N\}$ for each $\omega\in X$, we have
$$|X| \leq \Big|\overline{\bigcup_{\omega\in
X}[\omega]_{S(\omega)}}\, \cap\, \bigcup_{\omega\in
X}[\omega]_{S(\omega)^c}\Big|.$$
Now let's apply this inequality to $X = A\circ B$. By definition, for each $\omega\in A\circ B$ we
can choose some $S(\omega)$ such that
$[\omega]_{S(\omega)} \subseteq A$ and $[\omega]_{S(\omega)^c}
\subseteq B$. So the first union on the right-hand side is contained
in $A$, and the second union is contained in $B$.  Hence
$$|A\circ B| \leq |\bar A \circ B|.$$


*To complete the proof, we just need to observe that $\bar A\circ B =
\bar A \cap B$, which is true because $\bar A$ is decreasing while
$B$ is increasing.  Indeed, the $\subseteq$ direction is trivial,
and for the other direction, we make the following argument.  Fix
$\omega\in \bar A \cap B$, and let $S = \{i:\, \omega_i = 0\}$. Then
$[\omega]_S \subseteq \bar A$ because every $\omega'\in[\omega]_S$
can only have more zeros, meaning $\omega'\leq\omega$. Similarly, $[\omega]_{S^c} \subseteq
B$ because every $\omega'\in[\omega]_{S^c}$ satisfies
$\omega'\geq\omega$.  So we have shown $\omega\in\bar A\circ B$.
Borgs, C.; Chayes, J. T.; Randall, D., The van den Berg-Kesten-Reimer inequality: a review., Bramson, Maury (ed.) et al., Perplexing problems in probability. Festschrift in honor of Harry Kesten. Boston: Birkhäuser (ISBN 0-8176-4093-2). Prog. Probab. 44, 159-173 (1999). ZBL1115.82320.
