Category:Definitions/Distributive Lattices

This category contains definitions related to Distributive Lattices.
Related results can be found in Category:Distributive Lattices.

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

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} $

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} $

Subcategories

This category has only the following subcategory.

The following 4 pages are in this category, out of 4 total.

\((1)\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds x \wedge \paren {y \vee z} = \paren {x \wedge y} \vee \paren {x \wedge z} \)
\((1')\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds \paren {x \vee y} \wedge z = \paren {x \wedge z} \vee \paren {y \wedge z} \)
\((2)\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds x \vee \paren {y \wedge z} = \paren {x \vee y} \wedge \paren {x \vee z} \)
\((2')\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds \paren {x \wedge y} \vee z = \paren {x \vee z} \wedge \paren {y \vee z} \)

\((1)\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds x \wedge \paren {y \vee z} = \paren {x \wedge y} \vee \paren {x \wedge z} \)
\((1')\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds \paren {x \vee y} \wedge z = \paren {x \wedge z} \vee \paren {y \wedge z} \)
\((2)\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds x \vee \paren {y \wedge z} = \paren {x \vee y} \wedge \paren {x \vee z} \)
\((2')\)	$:$	\(\ds \forall x, y, z \in S:\)	\(\ds \paren {x \wedge y} \vee z = \paren {x \vee z} \wedge \paren {y \vee z} \)