Down Mapping is Generated by Approximating Relation

Theorem
Let $L = \left({S, \wedge, \preceq}\right)$ be a bounded below meet-continuous meet semilattice.

Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$.

Let $I$ be an ideal in $L$.

Let $f: S \to \mathit{Ids}\left({L}\right)$ be a mapping such that
 * $\forall x \in S: x \preceq \sup I \implies f\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$

and
 * $\forall x \in S: x \npreceq \sup I \implies f\left({x}\right) = x^\preceq$

where $x^\preceq$ denotes the lower closure of $x$.

Then
 * there exists an auxiliary approximating relation $\mathcal R$ on $S$ such that
 * $\forall s \in S: f\left({s}\right) = s^{\mathcal R}$

where $s^{\mathcal R}$ denotes the $\mathcal R$-segment of $s$.

Proof
Define relation $\mathcal R$ on $S$:
 * $\forall x, y \in S: \left({x, y}\right) \in \mathcal R \iff x \in f\left({y}\right)$

By definition of $\mathcal R$-segment:
 * $\forall x \in S: f\left({x}\right) = x^{\mathcal R}$

We will prove that
 * $\mathcal R$ is an approximating relation.

Let $x \in S$.

Suppose the case holds that:
 * $x \preceq \sup I$

Thus:

Suppose the case holds that
 * $x \npreceq \sup I$

Thus