# Equivalence of Definitions of Distributive Lattice

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## Contents

## 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.

$\blacksquare$