Order Generating Includes Completely Irreducible Elements

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $X \subseteq S$ be an order generating subset of $S$.

Let $X \in S$ be a completely irreducible element of $S$.

Then $x \in X$.

Proof
By definition of order generating:
 * $x^\succeq \cap X$ admits an infimum and $x = \inf\left({x^\succeq \cap X}\right)$

By Completely Irreducible and Subset Admits Infimum Equals Element implies Element Belongs to Subset:
 * $x \in x^\succeq \cap X$

Thus by definition of intersection:
 * $x \in X$