Meet Preserves Directed Suprema/Lemma 2

Theorem
Let $x$ be an element of $S$, $D$ be a directed subset of $S$.

Then:
 * $\paren {\sup D} \wedge x = \sup \set {d \wedge x: d \in D}$

Proof
By Meet Precedes Operands:
 * $\paren {\sup D} \wedge x \preceq \sup D$

By assumption:
 * $\paren {\sup D} \wedge x \preceq \sup \set {\paren {\sup D} \wedge x \wedge d: d \in D}$

By definition of supremum:
 * $\forall d \in D: d \preceq \sup D$

By Preceding iff Meet equals Less Operand:
 * $\forall d \in D: d \wedge \sup D = d$

By Meet is Associative and Meet is Commutative:
 * $\paren {\sup D} \wedge x \preceq \sup \set {d \wedge x: d \in D}$

By Meet Semilattice is Ordered Structure:
 * $\forall d \in D: d \wedge x \preceq \paren {\sup D} \wedge x$

By definition:
 * $\paren {\sup D} \wedge x$ is upper bound for $\set {d \wedge x: d \in D}$

By definition of supremum:
 * $\sup \set {d \wedge x: d \in D} \preceq \paren {\sup D} \wedge x$

if $\set {d \wedge x: d \in D}$ admits a supremum

We will prove that:
 * $\set {d \wedge x: d \in D}$ is directed

Let $y, z \in \set {d \wedge x: d \in D}$.

Then
 * $\exists d_1 \in D: y = d_1 \wedge x$

and
 * $\exists d_2 \in D: z = d_2 \wedge x$

By definition of directed subset:
 * $\exists d \in D: d_1 \preceq d \land d_2 \preceq d$

By Meet Semilattice is Ordered Structure:
 * $y \preceq d \wedge x$ and $z \preceq d \wedge x$

Thus
 * $d \wedge x \in \set {d \wedge x: d \in D}$

Hence by definition:
 * $\set {d \wedge x: d \in D}$ is directed.

By definition of up-complete:
 * $\set {d \wedge x: d \in D}$ admits a supremum

Thus by definition of antisymmetry:
 * $\paren {\sup D} \wedge x = \sup \set {d \wedge x: d \in D}$