# Equivalence of Definitions of Distributive Lattice

## Theorem

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

Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice.

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

## Proof

### 1 is equivalent to 1'

By applying Meet is Commutative several times, we have:

 $\displaystyle x \wedge \left({y \vee z}\right)$ $=$ $\displaystyle \left({x \wedge y}\right) \vee \left({x \wedge z}\right)$ $\displaystyle \iff \ \$ $\displaystyle \left({y \vee z}\right) \wedge x$ $=$ $\displaystyle \left({y \wedge x}\right) \vee \left({z \wedge x}\right)$

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

$\Box$

### 2 is equivalent to 2'

By applying Join is Commutative several times, we have:

 $\displaystyle x \vee \left({y \wedge z}\right)$ $=$ $\displaystyle \left({x \vee y}\right) \wedge \left({x \vee z}\right)$ $\displaystyle \iff \ \$ $\displaystyle \left({y \wedge z}\right) \vee x$ $=$ $\displaystyle \left({y \vee x}\right) \wedge \left({z \vee x}\right)$

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

$\Box$

### 1 implies 2

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

 $\displaystyle \left({x \vee y}\right) \wedge \left({x \vee z}\right)$ $=$ $\displaystyle \left({\left({x \vee y}\right) \wedge x}\right) \vee \left({\left({x \vee y}\right) \wedge z}\right)$ by $(1)$ $\displaystyle$ $=$ $\displaystyle x \vee \left({\left({x \vee y}\right) \wedge z}\right)$ $\wedge$ absorbs $\vee$ $\displaystyle$ $=$ $\displaystyle x \vee \left({\left({x \wedge z}\right) \vee \left({y \wedge z}\right)}\right)$ by $(1')$ $\displaystyle$ $=$ $\displaystyle \left({x \vee \left({x \wedge z}\right)}\right) \vee \left({y \wedge z}\right)$ $\vee$ is associative $\displaystyle$ $=$ $\displaystyle x \vee \left({y \wedge z}\right)$ $\vee$ 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.

$\blacksquare$