Meet Preserves Directed Suprema

Theorem
Let $\mathscr S = \left({S, \preceq}\right)$ be an up-complete meet semilattice such that
 * $\forall x \in S$, a directed subset $D$ of $S$: $x \preceq \sup D \implies x \preceq \left\{ {x \wedge y: y \in D}\right\}$

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

Then:
 * $f$ preserves directed suprema as a mapping from Cartesian product $\left({S \times S, \precsim}\right)$ of $\mathscr S$ and $\mathscr S$ into $\mathscr S$.

Proof
Let $X$ be a directed subset of $S \times S$ such that
 * $X$ admits a supremum.

By Up-Complete Product:
 * the Cartesian product of $\mathscr S$ and $\mathscr S$ is up-complete.

By Up-Complete Product/Lemma 2:
 * $X_1 := \operatorname{pr}_1^\to\left({X}\right)$ is directed

and
 * $X_2 := \operatorname{pr}_2^\to\left({X}\right)$ is directed

where
 * $\operatorname{pr}_1$ denotes the first projection on $S \times S$
 * $\operatorname{pr}_2$ denotes the second projection on $S \times S$
 * $\operatorname{pr}_1^\to\left({X}\right)$ denotes the image of $X$

We will prove that
 * $\left\{ {x \wedge \sup X_2: x \in X_1}\right\} = \left\{ {\sup \left\{ {x \wedge y: y \in X_2}\right\}: x \in X_1}\right\}$

first inclusion:

Let $a \in \left\{ {x \wedge \sup X_2: x \in X_1}\right\}$.

Then
 * $\exists x \in X: a = x \wedge \sup X_2$

By Meet Precedes Operands:
 * $x \wedge \sup X_2 \preceq \sup X_2$

By assumption:
 * $x \wedge \sup X_2 \preceq \sup\left\{ {x \wedge \sup X_2 \wedge y: y \in X_2}\right\}$

By definition of supremum:
 * $\forall y \in X_2: y \preceq \sup X_2$

By Preceding iff Meet equals Less Operand:
 * $\forall y \in X_2: \sup X_2 \wedge y = y$

By Meet is Associative:
 * $x \wedge \sup X_2 \preceq \sup\left\{ {x \wedge y: y \in X_2}\right\}$

By Meet Semilattice is Ordered Structure:
 * $\forall y \in X_2: x \wedge y \preceq x \wedge \sup X_2$

By definition:
 * $x \wedge \sup X_2$ is upper bound for $\left\{ {x \wedge y: y \in X_2}\right\}$

By definition of supremum:
 * $\sup \left\{ {x \wedge y: y \in X_2}\right\} \preceq x \wedge \sup X_2$

By definition of antisymmetry:
 * $a = \sup \left\{ {x \wedge y: y \in X_2}\right\}$

Thus
 * $a \in \left\{ {\sup \left\{ {x \wedge y: y \in X_2}\right\}: x \in X_1}\right\}$

second inclusion

Let $a \in \left\{ {\sup \left\{ {x \wedge y: y \in X_2}\right\}: x \in X_1}\right\}$.

Then
 * $\exists x \in X_1: a = \sup \left\{ {x \wedge y: y \in X_2}\right\}$

Analogically to first inclusion
 * $a = x \wedge \sup X_2$

Thus
 * $a \in \left\{ {x \wedge \sup X_2: x \in X_1}\right\}$