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
 * $(1): \quad \left\{ {x \wedge \sup X_2: x \in X_1}\right\} = \left\{ {\sup \left\{ {x \wedge y: y \in X_2}\right\}: x \in X_1}\right\}$

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

We will prove as lemma that
 * $\left({f^\to \left({X}\right)}\right)$ is directed.

Let $x, y \in \left({f^\to \left({X}\right)}\right)$.

By image of set:
 * $\exists \left({a, b}\right) \in X: x = f\left({a, b}\right)$

and
 * $\exists \left({c, d}\right) \in X: y = f\left({c, d}\right)$

By definition of $f$:
 * $x = a \wedge b$ and $y = c \wedge d$

By definition of directed subset:
 * $\exists \left({g, h}\right) \in X: \left({a, b}\right) \precsim \left({g, h}\right) \land \left({c, d}\right) \precsim \left({g, h}\right)$

By Cartesian product of ordered sets:
 * $a \preceq g$, $b \preceq h$, $c \preceq g$, and $d \preceq h$

By Meet Semilattice is Ordered Structure:
 * $x \preceq g \wedge h$ and $y \preceq g \wedge h$

By definition of image of set:
 * $g \wedge h = f\left({g, h}\right) \in f^\to \left({X}\right)$

Thus
 * $\exists z \in f^\to \left({X}\right): x \preceq z \land y \preceq z$

This ends the proof of lemma.

Thus by definition of up-complete:
 * $\left({f^\to \left({X}\right)}\right)$ admits a supremum.

Thus