A formula in complete lattice: $\bigwedge\{ a\vee b : a\in A , b\in B\}\leq (\bigwedge_{a\in A}a)\vee(\bigwedge_{b\in B}b) $ Let $(\Gamma,\leq)$ be a complete lattice. Assume $A$ and $B$ are two subsets of $\Gamma$  with $a\wedge b=0$ for every $a\in A$ and $b\in B$.
Q. True or false:
 $$\bigwedge\{ a\vee b : a\in A , b\in B\}\leq (\bigwedge_{a\in A}a)\vee(\bigwedge_{b\in B}b) $$
 A: As stated, the property is false in general: for example, let $\Gamma$ be the diamond lattice $M_3$ with top element $1$, bottom element $0$, and middle elements $x,y,z$. Put $A=\{x,y\}$ and $B=\{z\}$. Then the LHS evaluates to $1$, but the RHS to $z$. EDIT: In response to the comments, we can slightly modify this example by inserting in the lattice an extra point $w$ such that $x,y<w<1$ (thus, $w=x\lor y$). The inequality still fails, while the additional hypotheses $\bigl(\bigvee A\bigr)\land\bigl(\bigvee B\bigr)=0$ and $\bigvee(A\cup B)=1$ hold. (Notice also that like $M_3$, the resulting lattice remains modular.)
On the other hand, the inequality is true if $\Gamma$ is distributive. First, if $A$ or $B$ is empty, then both sides evaluate to $1$, hence we may assume $A\ne\varnothing\ne B$. Fix $a_0\in A$, and $b_0\in B$. Let
$$c=\bigwedge_{\substack{a\in A\\b\in B}}(a\lor b)$$
denote the left-hand side. Using distributivity and the assumptions on $A$ and $B$, we have
$$\begin{align}
a_0\land c&=\bigwedge_{\substack{a\in A\\b\in B}}(a_0\land(a\lor b))\\
&=\bigwedge_{\substack{a\in A\\b\in B}}((a_0\land a)\lor(a_0\land b))\\
&=\bigwedge_{\substack{a\in A\\b\in B}}(a_0\land a)\\
&=\bigwedge_{a\in A}a,
\end{align}$$
and symmetrically,
$$b_0\land c=\bigwedge_{b\in B}b.$$
Thus, using distributivity again, we have
$$c=c\land(a_0\lor b_0)=(c\land a_0)\lor(c\land b_0)=\bigwedge_{a\in A}a\lor\bigwedge_{b\in B}b.$$
On the third hand, the property is weaker than distributivity. For example, let $\Gamma_1$ be an arbitrary complete lattice (in particular, non-distributive), and let $\Gamma$ be $\Gamma_1$ with a new least element $0$ attached. Then $\Gamma$ satisfies the property for the simple reason that the assumption $a\land b=0$ for all $a\in A$ and $b\in B$ holds only if one of the sets $A$, $B$ is $\varnothing$ or $\{0\}$, and the inequality is easy to verify in these cases.
