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, $x\vee (y\wedge z)\le (x\vee y)\wedge z$.∎