Meet-Continuous iff Ideal Supremum is Meet Preserving

Theorem
Let $\mathscr S = \left({S, \vee, \wedge, \preceq}\right)$ be an up-complete lattice.

Let $f: {\it Ids}\left({\mathscr S}\right) \to S$ be a mapping such that
 * $\forall I \in {\it Ids}\left({\mathscr S}\right): f\left({I}\right) = \sup_{\mathscr S} I$

where
 * ${\it Ids}\left({\mathscr S}\right)$ denotes the set of all ideals in $\mathscr S$

Then
 * $\mathscr S$ is meet-continuous
 * $f$ preserves meet as a mapping from $\left({ {\it Ids}\left({\mathscr S}\right), \subseteq}\right)$ into $\mathscr S$

Sufficient Condition
Assume that
 * $\mathscr S$ is meet-continuous.

We will prove that
 * for every directed subsets $D_1, D_2$ of $S$: $\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}\right\}$

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

By Meet-Continuous implies Element Precedes Supremum of Directed Subset:
 * for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \left\{ {x \wedge d: d \in D}\right\}$