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Let $(\Gamma,\leq)$ be a complete lattice and put $\bigwedge_{x\in \Gamma}x =0$. 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: $$\bigwedge\{ a\vee b : a\in A , b\in B\}\leq (\bigwedge_{a\in A}a)\vee(\bigwedge_{b\in B}b) $$

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    $\begingroup$ I doubt this stands a chance without assuming some form of distributivity. For a simple counterexample, let $\Gamma$ be the diamond lattice with “middle elements” $x,y,z$, and put $A=\{x,y\}$, $B=\{z\}$. Then the left-hand side is $1$, but the right-hand side is $z$. $\endgroup$ – Emil Jeřábek Jul 3 '17 at 12:14
  • $\begingroup$ It does, in fact, hold for distributive lattices. (I thought at first that one would need an infinitary distributive law, but plain distributivity actually suffices.) $\endgroup$ – Emil Jeřábek Jul 3 '17 at 12:40
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    $\begingroup$ Asked also on Math.SE: A formula in complete lattice. $\endgroup$ – Martin Sleziak Jul 3 '17 at 13:37
  • $\begingroup$ I answered on math.SE . $\endgroup$ – Emil Jeřábek Jul 3 '17 at 14:00

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