Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving

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

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 $\mathscr S$

Let
 * $\forall I_1, I_2 \in {\it Ids}\left({\mathscr S}\right): \left({\sup I_1}\right) \wedge \left({\sup I_2}\right) = \sup \left\{ {i \wedge j: i \in I_1, j \in I_2}\right\}$

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

Proof
Let $I, J \in {\it Ids}\left({\mathscr S}\right)$ such that
 * $\left\{ {I, J}\right\}$ admits an infimum in $\left({ {\it Ids}\left({\mathscr S}\right), \subseteq}\right)$.

By definition of image of set:
 * $f^\to\left({\left\{ {I, J}\right\} }\right) = \left\{ {f\left({I}\right), f\left({J}\right)}\right\}$

Thus by definition of meet semilattice:
 * $f^\to\left({\left\{ {I, J}\right\} }\right)$ admits an infimum in $\mathscr S$.

Thus