Dual Distributive Lattice is Distributive
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Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Then
- $L$ is a distributive lattice
- $L^{-1}$ 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