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:
 * $\exists d_1, d_2 \in S: \sup D = \left({d_1, d_2}\right)$