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

Theorem
Let $\struct {S, \preceq}$ be an up-complete meet semilattice.

Let $\struct {S \times S, \precsim}$ be the Cartesian product of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.

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

and
 * $f$ preserves directed suprema.

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

Then
 * $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {x \wedge y: x \in D_1, y \in D_2}$

Proof
By Up-Complete Product:
 * $\struct {S \times S, \precsim}$ 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