Equivalence of Definitions of Distributive Lattice

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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.

$\blacksquare$


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