Equivalence of Definitions of Distributive Lattice
Theorem
The following definitions of the concept of Distributive Lattice are equivalent: Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Definition 1
Then $\struct {S, \vee, \wedge, \preceq}$ is distributive if and only if $\struct {S, \vee, \wedge, \preceq}$ satisfies one of the distributive lattice axioms:
\((1)\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds x \wedge \paren {y \vee z} = \paren {x \wedge y} \vee \paren {x \wedge z} \) | ||||||
\((1')\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds \paren {x \vee y} \wedge z = \paren {x \wedge z} \vee \paren {y \wedge z} \) | ||||||
\((2)\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds x \vee \paren {y \wedge z} = \paren {x \vee y} \wedge \paren {x \vee z} \) | ||||||
\((2')\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds \paren {x \wedge y} \vee z = \paren {x \vee z} \wedge \paren {y \vee z} \) |
Definition 2
Then $\struct {S, \vee, \wedge, \preceq}$ is distributive if and only if $\struct {S, \vee, \wedge, \preceq}$ satisfies all of the distributive lattice axioms:
\((1)\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds x \wedge \paren {y \vee z} = \paren {x \wedge y} \vee \paren {x \wedge z} \) | ||||||
\((1')\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds \paren {x \vee y} \wedge z = \paren {x \wedge z} \vee \paren {y \wedge z} \) | ||||||
\((2)\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds x \vee \paren {y \wedge z} = \paren {x \vee y} \wedge \paren {x \vee z} \) | ||||||
\((2')\) | $:$ | \(\ds \forall x, y, z \in S:\) | \(\ds \paren {x \wedge y} \vee z = \paren {x \vee z} \wedge \paren {y \vee z} \) |
Proof
In what follows it is shown that if one of the axioms distributive lattice axioms holds then they all hold.
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.
$\blacksquare$