Meet Preserves Directed Suprema

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

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

Then:
 * $f$ preserves directed suprema as a mapping from simple order product $\struct{S \times S, \precsim}$ of $\mathscr S$ and $\mathscr S$ into $\mathscr S$.

Lemma 2
Let $X$ be a directed subset of $S \times S$ such that
 * $X$ admits a supremum.

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

By Up-Complete Product/Lemma 2:
 * $X_1 := \map {\pr_1^\to} X$ is directed

and
 * $X_2 := \map {\pr_2^\to} X$ is directed

where
 * $\pr_1$ denotes the first projection on $S \times S$
 * $\pr_2$ denotes the second projection on $S \times S$
 * $\map {\pr_1^\to} X$ denotes the image of $X$

We will prove that
 * $(1): \quad \set {x \wedge \sup X_2: x \in X_1} = \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$

First inclusion:

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

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:
 * $x \wedge \sup X_2 \preceq \sup \set {x \wedge \sup X_2 \wedge y: y \in X_2}$

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:
 * $x \wedge \sup X_2 \preceq \sup \set {x \wedge y: y \in X_2}$

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 $\set {x \wedge y: y \in X_2}$

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

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

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

Second inclusion

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

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

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

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

We will prove as lemma that:
 * $\paren {\map {f^\to} X}$ is directed.

Let $x, y \in \paren {\map {f^\to} X}$.

By image of set:
 * $\exists \tuple {a, b} \in X: x = \map f {a, b}$

and
 * $\exists \tuple {c, d} \in X: y = \map f {c, d}$

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

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

By simple order product:
 * $a \preceq g$, $b \preceq h$, $c \preceq g$, 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:
 * $g \wedge h = \map f {g, h} \in \map {f^\to} X$

Thus:
 * $\exists z \in \map {f^\to} X: x \preceq z \land y \preceq z$

This ends the proof of lemma.

Thus by definition of up-complete:
 * $\paren {\map {f^\to} X}$ admits a supremum.

Thus