distributive inequalities

Let $L$ be a lattice. Then for $a,b,c\\in L$ , we have the following inequalities:

1.
$a\\vee(b\\wedge c)\\leq(a\\vee b)\\wedge(a\\vee c)$ ,
2.
$(a\\wedge b)\\vee(a\\wedge c)\\leq a\\wedge(b\\vee c)$ .

Proof.

Since $a\\leq a\\vee b$ and $a\\leq a\\vee c$ , $a\\leq(a\\vee b)\\wedge(a\\vee c)$ .Similarly, $b\\wedge c\\leq b\\leq a\\vee b$ and $b\\wedge c\\leq c\\leq a\\vee c$ imply $b\\wedge c\\leq(a\\vee b)\\wedge(a\\vee c)$ .Together, we have $a\\vee(b\\wedge c)\\leq(a\\vee b)\\wedge(a\\vee c)$ .

The second inequality is the dual of the first one.∎

The two inequalities above are called the distributive inequalities.

Proposition A lattice $L$ is a distributive lattice if one of the following inequalities holds:

1.
$(a\\vee b)\\wedge(a\\vee c)\\leq a\\vee(b\\wedge c)$ ,
2.
$a\\wedge(b\\vee c)\\leq(a\\wedge b)\\vee(a\\wedge c)$ .

Proof.

By the distributive inequalities, all we need to show is that 1. implies 2. (that 2. implies 1. is just the dual statement). So suppose 1. holds. Then

$\\displaystyle(a\\wedge b)\\vee(a\\wedge c)$	$\\displaystyle\\geq((a\\wedge b)\\vee a)\\wedge((a\\wedge b)\\vee c)$	$\\displaystyle\\qquad\\mbox{by assumption}$
	$\\displaystyle=a\\wedge((a\\wedge b)\\vee c)$	$\\displaystyle\\qquad\\mbox{by absorption}$
	$\\displaystyle\\geq a\\wedge((c\\vee a)\\wedge(c\\vee b))$	$\\displaystyle\\qquad\\mbox{by assumption}$
	$\\displaystyle=(a\\wedge(c\\vee a))\\wedge(c\\vee b)$	$\\displaystyle\\qquad\\mbox{meet associativity}$
	$\\displaystyle=a\\wedge(c\\vee b).$	$\\displaystyle\\qquad\\mbox{by absorption}$

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