# Definition:Distributive Lattice

## Definition

Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.

### Definition 1

Then $\struct {S, \vee, \wedge, \preceq}$ is distributive if and only if $\struct {S, \vee, \wedge, \preceq}$ satisfies one of the distributive lattice axioms:

 $(1)$ $:$ $\ds \forall x, y, z \in S:$ $\ds x \wedge \paren {y \vee z} = \paren {x \wedge y} \vee \paren {x \wedge z}$ $(1')$ $:$ $\ds \forall x, y, z \in S:$ $\ds \paren {x \vee y} \wedge z = \paren {x \wedge z} \vee \paren {y \wedge z}$ $(2)$ $:$ $\ds \forall x, y, z \in S:$ $\ds x \vee \paren {y \wedge z} = \paren {x \vee y} \wedge \paren {x \vee z}$ $(2')$ $:$ $\ds \forall x, y, z \in S:$ $\ds \paren {x \wedge y} \vee z = \paren {x \vee z} \wedge \paren {y \vee z}$

### Definition 2

Then $\struct {S, \vee, \wedge, \preceq}$ is distributive if and only if $\struct {S, \vee, \wedge, \preceq}$ satisfies all of the distributive lattice axioms:

 $(1)$ $:$ $\ds \forall x, y, z \in S:$ $\ds x \wedge \paren {y \vee z} = \paren {x \wedge y} \vee \paren {x \wedge z}$ $(1')$ $:$ $\ds \forall x, y, z \in S:$ $\ds \paren {x \vee y} \wedge z = \paren {x \wedge z} \vee \paren {y \wedge z}$ $(2)$ $:$ $\ds \forall x, y, z \in S:$ $\ds x \vee \paren {y \wedge z} = \paren {x \vee y} \wedge \paren {x \vee z}$ $(2')$ $:$ $\ds \forall x, y, z \in S:$ $\ds \paren {x \wedge y} \vee z = \paren {x \vee z} \wedge \paren {y \vee z}$

That these statements are in fact equivalent is shown on Equivalence of Definitions of Distributive Lattice.

Hence, $\struct {S, \vee, \wedge, \preceq}$ is distributive if and only if $\wedge$ and $\vee$ distribute over each other.

## Also see

• Results about distributive lattices can be found here.