Supremum by Suprema of Directed Set in Simple Order Product

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

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

Let $D$ be a directed subset of $S \times S$.

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

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 Up-Complete Product:
 * $\left({S \times S, \precsim}\right)$ is up-complete.

By definition of up-complete:
 * $D$ admits a supremum.

By definition of Cartesian product:
 * $\exists d_1, d_2 \in S: \sup D = \left({d_1, d_2}\right)$

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 definition of up-complete:
 * $D_1$ admits a supremum

and
 * $D_2$ admits a supremum

We will prove that
 * $d_2$ is upper bound for $D_2$

Let $x \in D_2$.

By definition of image of set:
 * $\exists \left({a, b}\right) \in D: \operatorname{pr}_2\left({a, b}\right) = x$

By definition of second projection: $b = x$

By definition of supremum:
 * $\left({d_1, d_2}\right)$ is upper bound for $D$.

By definition of upper bound:
 * $\left({a, x}\right) \precsim \left({d_1, d_2}\right)$

Thus by definition of Cartesian product of ordered sets:
 * $x \preceq d_2$

Analogically we have that
 * $d_1$ is upper bound for $D_1$

By definition of supremum:
 * $\sup D_1 \preceq d_1$ and $\sup D_2 \preceq d_2$

By definition of Cartesian product of ordered sets:
 * $\left({\sup D_1, \sup D_2}\right) \precsim \sup D$

By Up-Complete Product/Lemma 1:
 * $D_1 \times D_2$ is directed.

By definition of up-complete:
 * $D_1 \times D_2$ admits a supremum.

By definition of subset:
 * $D \subseteq D_1 \times D_2$

By Supremum of Subset:
 * $\sup D \precsim \sup\left({D_1 \times D_2}\right)$

By Supremum of Cartesian Product:
 * $\sup D \precsim \left({\sup D_1, \sup D_2}\right)$

Thus by definition of antisymmetry:
 * $\sup D = \left({\sup D_1, \sup D_2}\right)$