Intersection of Relation Segments of Approximating Relations equals Way Below Closure

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below meet-continuous lattice.

Let $\map {\operatorname {App} } L$ be the set of all auxiliary approximating relations on $S$.

Let $x \in S$.

Then
 * $\ds \bigcap \set {x^\RR: \RR \in \map {\operatorname {App} } L} = x^\ll$

Proof
By Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element:
 * $\ds \bigcap \set {I \in \operatorname {Ids}: x \preceq \sup I} = x^\ll$

where $\operatorname {Ids}$ denotes the set of all ideals in $L$.

For all $I \in \operatorname {Ids}$ define a mapping $m_I: S \to \operatorname {Ids}$:
 * $\forall x \in S: x \preceq \sup I \implies \map {m_I} x = \set {x \wedge i: i \in I}$

and
 * $\forall x \in S: x \npreceq \sup I \implies \map {m_I} x = x^\preceq$

By Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element:
 * $\ds \forall x \in S: \bigcap \set {\map {m_I} x: I \in \operatorname {Ids} } = x^\ll$

We will prove that
 * $\set {\map {m_I} x: I \in \operatorname {Ids} } \subseteq \set {x^{\RR}: \RR \in \map {\operatorname {App} } L}$

Let $a \in \set {\map {m_I} x: I \in \operatorname {Ids} }$

Then
 * $\exists I \in \operatorname {Ids}: a = \map {m_I} x$

By Down Mapping is Generated by Approximating Relation:
 * $\exists \RR \in \map {\operatorname {App} } L: \forall s \in S: \map {m_I} s = s^\RR$

Then
 * $a = x^\RR$

Thus
 * $a \in \set {x^\RR: \RR \in \map {\operatorname {App} } L}$

By Intersection of Family is Subset of Intersection of Subset of Family:
 * $\ds \bigcap \set {x^\RR: \RR \in \map {\operatorname {App} } L} \subseteq x^\ll$

We will prove that
 * $\set {x^\RR: \RR R \in \map {\operatorname {App} } L} \subseteq \set {I \in \operatorname {Ids}: x \preceq \sup I}$

Let $a \in \set {x^\RR: \RR \in \map {\operatorname {App} } L}$

Then
 * $\exists \RR \in \map {\operatorname {App} } L: a = x^\RR$

By definition of approximating relation:
 * $x = \sup a$

By Relation Segment of Auxiliary Relation is Ideal:
 * $a \in \operatorname {Ids}$

Thus by definition of reflexivity:
 * $a \in \set {I \in \operatorname {Ids}: x \preceq \sup I}$

By Intersection of Family is Subset of Intersection of Subset of Family:
 * $\ds x^\ll \subseteq \bigcap \set {x^\RR: \RR \in \map {\operatorname {App} } L}$

Hence the result by definition of set equality.