Lower Closure of Meet of Lower Closures

Theorem
Let $\left({S, \preceq}\right)$ be a meet semilattice.

Let $D_1, D_2$ be subsets of $S$.

Then
 * $\left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq = \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$

where
 * $D_1^\preceq$ denotes the lower closure of $D_1$.

Proof
By Lower Closure is Closure Operator:
 * $D_1 \subseteq D_1^\preceq$ and $D_2 \subseteq D_2^\preceq$

Then
 * $\left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\} \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}$

By Lower Closure is Closure Operator:
 * $\left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq$

By definition of set equality it remains to prove that
 * $\left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$

Let $x \in \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq$.

By definition of lower closure of set:
 * $\exists y \in \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\} \land x \preceq y$

Then:
 * $\exists i_1 \in D_1^\preceq, i_2 \in D_2^\preceq: y = i_1 \wedge i_2$

By definition of lower closure of set:
 * $\exists d_1 \in D_1: i_1 \preceq d_1$

and
 * $\exists d_2 \in D_2: i_2 \preceq d_2$

By Meet Semilattice is Ordered Structure:
 * $y \preceq d_1 \wedge d_2$

By definition of transitivity:
 * $x \preceq d_1 \wedge d_2$


 * $d_1 \wedge d_2 \in \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}$

Thus by definition of lower closure of set:
 * $x \in \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$

Thus the result.