Set of All Completely Irreducible Elements is Smallest Order Generating

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.

Then $\map {\operatorname {Irr} } L$ is order generating and
 * $\forall X \subseteq S: X$ is order generating $\implies \map {\operatorname {Irr} } L \subseteq X$

where $\map {\operatorname {Irr} } L$ denotes the set of all completely irreducible elements of $L$.

Proof
By Not Preceding implies Exists Completely Irreducible Element in Algebraic Lattice:
 * $\forall x, y \in S: y \npreceq x \implies \exists p \in \map {\operatorname {Irr} } L: p \preceq x \land p \npreceq y$

Thus by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding:
 * $\map {\operatorname {Irr} } L$ is order generating.

Let $X \subseteq S$ such that
 * $X$ is order generating.

Let $x \in \map {\operatorname {Irr} } L$.

By definition of $\operatorname {Irr}$:
 * $x$ is completely irreducible.

Thus by Order Generating Subset Includes Completely Irreducible Elements:
 * $x \in X$