Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets

Theorem
Let $\left({S, \preceq}\right)$ be an up-complete meet semilattice.

Let $\left({S \times S, \precsim}\right)$ be the Cartesian product of $\left({S, \preceq}\right)$ and $\left({S, \preceq}\right)$.

Let $f: S \times S \to S$ be a mapping such that
 * $\forall s, t \in S: f\left({s, t}\right) = s \wedge t$

and
 * $f$ preserves directed suprema.

Let $D_1, D_2$ be directed subsets of $S$.

Then
 * $\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$

Proof
By Up-Complete Product:
 * $\left({S \times S, \precsim}\right)$ is up-complete.

By Up-Complete Product/Lemma 1:
 * $D_1 \times D_2$ is directed subsets of $S \times S$

By definition of mapping preserves directed suprema:
 * $f$ preserves the supremum of $D_1 \times D_2$

By definition of up-complete:
 * $D_1 \times D_2$ admits a supremum

and $D_1$ and $D_2$ admit suprema.

Thus