# Equivalence of Definitions of Distributive Lattice

## Theorem

The following definitions of the concept of Distributive Lattice are equivalent:

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

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

## Proof

### 1 is equivalent to 1'

By applying Meet is Commutative several times, we have:

 $\ds x \wedge \paren {y \vee z}$ $=$ $\ds \paren {x \wedge y} \vee \paren {x \wedge z}$ $\ds \leadstoandfrom \ \$ $\ds \paren {y \vee z} \wedge x$ $=$ $\ds \paren {y \wedge x} \vee \paren {z \wedge x}$

which (after renaming variables as appropriate) establishes the equivalence.

$\Box$

### 2 is equivalent to 2'

By applying Join is Commutative several times, we have:

 $\ds x \vee \paren {y \wedge z}$ $=$ $\ds \paren {x \vee y} \wedge \paren {x \vee z}$ $\ds \leadstoandfrom \ \$ $\ds \paren {y \wedge z} \vee x$ $=$ $\ds \paren {y \vee x} \wedge \paren {z \vee x}$

which (after renaming variables as appropriate) establishes the equivalence.

$\Box$

### 1 implies 2

Suppose that $(1)$ holds, and hence $(1')$ as well.

 $\ds \paren {x \vee y} \wedge \paren {x \vee z}$ $=$ $\ds \paren {\paren {x \vee y} \wedge x} \vee \paren {\paren {x \vee y} \wedge z}$ by $(1)$ $\ds$ $=$ $\ds x \vee \paren {\paren {x \vee y} \wedge z}$ $\wedge$ absorbs $\vee$ $\ds$ $=$ $\ds x \vee \paren {\paren {x \wedge z} \vee \paren {y \wedge z} }$ by $(1')$ $\ds$ $=$ $\ds \paren {x \vee \paren {x \wedge z} } \vee \paren {y \wedge z}$ $\vee$ is associative $\ds$ $=$ $\ds x \vee \paren {y \wedge z}$ $\vee$ absorbs $\wedge$

$\Box$

### 2 implies 1

By inspection, aided by Dual Pairs (Order Theory), we see that $(2)$ is dual to $(1)$.

Thus by Global Duality, $(2)$ implies $(1)$ as soon as $(1)$ implies $(2)$.

That direction was already established above.

$\blacksquare$