modular inequality

In any lattice (http://planetmath.org/lattice) the self-dual modular inequality is true: if $x\le z$ then $x\vee (y\wedge z)\le (x\vee y)\wedge z$.

Proof.

$x\le x\vee y$ and we are given that $x\le z$, so $x\le (x\vee y)\wedge z$. Also, $y\wedge z\le y\le x\vee y$ and $y\wedge z\le z$ imply that $y\wedge z\le (x\vee y)\wedge z$. Therefore, $x\vee (y\wedge z)\le (x\vee y)\wedge z$. ∎

Title	modular inequality
Canonical name	ModularInequality
Date of creation	2014-02-01 1:48:21
Last modified on	2014-02-01 1:48:21
Owner	ixionid (16766)
Last modified by	ixionid (16766)
Numerical id	10
Author	ixionid (16766)
Entry type	Theorem
Classification	msc 06C05
Related topic	ModularLattice
Related topic	DistributiveInequalities
Defines	modular inequality