\begin{definition}[Definition:Distributive Lattice]
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice.
Then $\left({S, \vee, \wedge, \preceq}\right)$ is '''distributive''' {{iff}} one of the following statements holds:
{{begin-axiom}}
{{axiom | n = 1
        | q = \forall x, y, z \in S
        | m = x \wedge \left({y \vee z}\right) = \left({x \wedge y}\right) \vee \left({x \wedge z}\right)
}}
{{axiom | n = 1'
        | q = \forall x, y, z \in S
        | m = \left({x \vee y}\right) \wedge z = \left({x \wedge z}\right) \vee \left({y \wedge z}\right)
}}
{{axiom | n = 2
        | q = \forall x, y, z \in S
        | m = x \vee \left({y \wedge z}\right) = \left({x \vee y}\right) \wedge \left({x \vee z}\right)
}}
{{axiom | n = 2'
        | q = \forall x, y, z \in S
        | m = \left({x \wedge y}\right) \vee z = \left({x \vee z}\right) \wedge \left({y \vee z}\right)
}}
{{end-axiom}}
That these statements are in fact equivalent is shown on Equivalence of Definitions of Distributive Lattice.
Hence, $\left({S, \vee, \wedge, \preceq}\right)$ is '''distributive''' {{iff}} $\wedge$ and $\vee$ distribute over each other.
\end{definition}