Supremum of Meet Image of Directed Set

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

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

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

Then:
 * $\sup \left({f^\to\left({D}\right)}\right) = \sup \left\{ {x \wedge y: x \in \operatorname{pr}_1^\to\left({D}\right), y \in \operatorname{pr}_2^\to\left({D}\right)}\right\}$

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

Proof
By definition of image of set:
 * $f^\to\left({D}\right) = \left\{ {x \wedge y: \left({x, y}\right) \in D}\right\}$

By definition of subset:
 * $f^\to\left({D}\right) \subseteq \left\{ {x \wedge y: x \in \operatorname{pr}_1^\to\left({D}\right), y \in \operatorname{pr}_2^\to\left({D}\right)}\right\}$

By Up-Complete Product/Lemma 2:
 * $D_1 := \operatorname{pr}_1^\to\left({D}\right)$ is directed

and
 * $D_2 := \operatorname{pr}_2^\to\left({D}\right)$ is directed.

By Meet of Directed Subsets is Directed:
 * $\left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$ is directed.

By definition of up-complete:
 * $\left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$ admits a supremum.

By Meet is Increasing:
 * $f$ is increasing mapping.

By Image of Directed Subset under Increasing Mapping is Directed:
 * $f^\to\left({D}\right)$ is directed.

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

By Supremum of Subset:
 * $\sup \left({f^\to\left({D}\right)}\right) \preceq \sup \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$

We will prove that
 * $\left\{ {x \wedge y: x \in D_1, y \in D_2}\right\} \subseteq \left({f^\to\left({D}\right)}\right)^\preceq$

where
 * $\left({f^\to\left({D}\right)}\right)^\preceq$ denotes the lower closure of set of $f^\to\left({D}\right)$.

Let $z \in \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$.

Then
 * $\exists x \in D_1, y \in D_2: z = x \wedge y$

By definition of image of set:
 * $\exists \left({a, b}\right) \in D: \operatorname{pr}_1\left({a, b}\right) = x$

and
 * $\exists \left({c, d}\right) \in D: \operatorname{pr}_2\left({c, d}\right) = y$

By definition of first projection and second projection:
 * $a = x$ and $d = y$

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

By definition of Cartesian product of ordered sets:
 * $x \preceq g$ and $y \preceq h$

By Meet Semilattice is Ordered Structure:
 * $x \wedge y \preceq g \wedge h \in \left\{ {x \wedge y: \left({x, y}\right) \in D}\right\}$

Thus by definition of lower closure of set:
 * $z \in \left({f^\to\left({D}\right)}\right)^\preceq$

By Supremum of Lower Closure of Set:
 * $\left({f^\to\left({D}\right)}\right)^\preceq$ admits a supremum

and
 * $\sup \left({f^\to\left({D}\right)}\right)^\preceq = \sup \left({f^\to\left({D}\right)}\right)$

By Supremum of Subset:
 * $\sup \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\} \preceq \sup \left({f^\to\left({D}\right)}\right)$

Thus by definition of antisymmetry:
 * $\sup \left({f^\to\left({D}\right)}\right) = \sup \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$