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

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

∎

Title	distributive inequalities
Canonical name	DistributiveInequalities
Date of creation	2013-03-22 16:37:48
Last modified on	2013-03-22 16:37:48
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	5
Author	CWoo (3771)
Entry type	Derivation
Classification	msc 06D99
Related topic	ModularInequality