Set of All Completely Irreducible Elements is Smallest Order Generating

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

Then $\mathit{Irr}\left({L}\right)$ is order generating and
 * $\forall X \subseteq S: X$ is order generating $\implies \mathit{Irr}\left({L}\right) \subseteq X$

where $\mathit{Irr}\left({L}\right)$ denotes the set of all completely irreducible elements of $L$.

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

Thus by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding:
 * $\mathit{Irr}\left({L}\right)$ is order generating.

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

Let $x \in \mathit{Irr}\left({L}\right)$.

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

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