Dual Distributive Lattice is Distributive

Theorem

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

Then

$L$ is a distributive lattice

if and only if

$L^{-1}$ is a distributive lattice

where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.

Proof

Sufficient Condition

Let $L$ be a distributive lattice.

By Dual of Lattice Ordering is Lattice Ordering:

$L^{-1}$ is lattice.

Let $x, y, z \in S$.

$\vee'$ and $\wedge'$ denotes join and meet in $L^{-1}$.

Thus

\(\ds x \wedge' \paren {y \vee' z}\)

\(=\)

\(\ds x \wedge' \paren {y \wedge z}\)

Join is Dual to Meet

\(\ds \)

\(=\)

\(\ds x \vee \paren {y \wedge z}\)

Join is Dual to Meet

\(\ds \)

\(=\)

\(\ds \paren {x \vee y} \wedge \paren {x \vee z}\)

Definition of Distributive Lattice

\(\ds \)

\(=\)

\(\ds \paren {x \wedge' y} \wedge \paren {x \wedge' z}\)

Join is Dual to Meet

\(\ds \)

\(=\)

\(\ds \paren {x \wedge' y} \vee' \paren {x \wedge' z}\)

Join is Dual to Meet

$\Box$

Necessary Condition

This follows by mutatis mutandis.

$\blacksquare$

Sources

• Mizar article YELLOW_7:25