Let $\mathfrak{A}$ and $\mathfrak{B}$ be (fixed) boolean lattices (with lattice operations denoted $\sqcup$ and $\sqcap$, bottom element $\bot$ and top element $\top$).
I call a boolean funcoid a pair $(\alpha;\beta)$ of functions $\alpha:\mathfrak{A}\rightarrow\mathfrak{B}$, $\beta:\mathfrak{B}\rightarrow\mathfrak{A}$ such that (for every $X\in\mathfrak{A}$, $Y\in\mathfrak{B}$) $$Y\sqcap^{\mathfrak{B}}\alpha(X)\ne\bot^{\mathfrak{B}} \Leftrightarrow X\sqcap^{\mathfrak{A}}\beta(Y)\ne\bot^{\mathfrak{A}}.$$
(Boolean funcoids are a special case of pointfree funcoids as defined in my free ebook.)
Order boolean funcoids by the formula $$(\alpha_0;\beta_0)\le (\alpha_1;\beta_1) \Leftrightarrow \forall X\in\mathfrak{A}: \alpha_0(X)\le\alpha_1(X) \land \forall Y\in\mathfrak{B}: \beta_0(Y)\le\beta_1(Y).$$
Conjecture The set of boolean funcoids with above defined order is itself a boolean lattice.
If this conjecture does not hold in general, does it hold for: a. atomic boolean lattices? b. atomistic boolean lattices? c. complete boolean lattices?
For the special case when $\mathfrak{A}$ and $\mathfrak{B}$ are complete atomic boolean lattices, the conjecture easily follows from this my answer to my own question.