Intersection of Upper Section with Directed Set is Directed Set

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

Let $A, B$ be subsets of $S$ such that
 * $A \cap B \ne \varnothing$

and
 * $A$ is an upper set,
 * $B$ is a directed set.

Then $A \cap B$ is a directed set.

Proof
Thus by assumption:
 * $A \cap B$ is a non-empty set.

Let $x, y \in A \cap B$.

By definition of intersection:
 * $x, y \in A$ and $x, y \in B$

By definition of directed subset:
 * $\exists z \in B: x \preceq z \land y \preceq z$

By definition of upper set:
 * $z \in A$

Thus by definition of intersection:
 * $x \in A \cap B$

Thus
 * $x \preceq z$ and $y \preceq z$