Auxiliary Approximating Relation has Interpolation Property

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.

Let $x, z \in S$ such that
 * $x \ll z \land x \ne z$

Let $\mathcal R$ be an auxiliary approximating relation on $S$.

Then
 * $\exists y \in S: \left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \land x \ne y$

Proof
Define $I := \left\{ {u \in S: \exists y \in S: \left({u, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R}\right\}$

By definition of auxiliary relation:
 * $\left({\bot, \bot}\right) \in \mathcal R$ and $\left({\bot, z}\right) \in \mathcal R$

where $\bot$ denotes the smallest element in $L$.

Then
 * $\bot \in I$

By definition:
 * $I$ is a non-empty set.

We will prove that
 * $I$ is a lower set.

Let $a \in I, b \in S$ such that
 * $b \preceq a$

By definition of $I$:
 * $\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$

By definitions of auxiliary relation and reflexivity:
 * $\left({b, s}\right) \in \mathcal R$

Thus
 * $b \in I$

We will prove that
 * $I$ is directed.

Let $a, b \in I$.

By definition of $I$:
 * $\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$

and
 * $\exists t \in S: \left({b, t}\right) \in \mathcal R \land \left({t, z}\right) \in \mathcal R$

By Auxiliary Relation is Congruent:
 * $\left({a \vee b, s \vee t}\right) \in \mathcal R$

By definition of auxiliary relation:
 * $\left({s \vee t, z}\right) \in \mathcal R$

Thus by definition of $I$:
 * $a \vee b \in I$

Thus by Join Succeeds Operands:
 * $a \preceq a \vee b$ and $b \preceq a \vee b$

By definition:
 * $I$ is ideal in $L$.

We will prove that
 * $I \subseteq z^{\mathcal R}$

Let $a \in I$.

By definition of $I$:
 * $\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$

By definition of auxiliary relation:
 * $a \preceq s$

Again by definition of auxiliary relation and reflexivity:
 * $\left({a, z}\right) \in \mathcal R$

Thus by definition of $\mathcal R$-segment:
 * $a \in z^{\mathcal R}$

By Supremum of Subset and definition of approximating relation:
 * $\sup I \preceq \sup \left({z^{\mathcal R} }\right) = z$

We will prove that
 * $\sup I = z$

Aiming for a contradiction suppose that
 * $\sup I \ne z$

By definition of $\prec$:
 * $\sup I \prec z$

Then by definition of antisymmetry:
 * $z \npreceq \sup I$

By Not Preceding implies Approximating Relation and not Preceding:
 * $\exists y \in S: \left({y, z}\right) \in \mathcal R \land y \npreceq \sup I$

Again by Not Preceding implies Approximating Relation and not Preceding:
 * $\exists u \in S: \left({u, y}\right) \in \mathcal R \land u \npreceq \sup I$

By definition of $I$:
 * $u \in I$

This contradicts $u \preceq \sup I$ by definition of supremum.

By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $x \in I$

By definition of $I$:
 * $\exists s \in S: \left({x, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$

By definition of auxiliary relation:
 * $x \preceq s$

By definition of auxiliary relation and reflexivity:
 * $\left({x, z}\right) \in \mathcal R$

By Auxiliary Approximating Relation has Quasi Interpolation Property:
 * $\exists y \in S: x \preceq y \land \left({y, z}\right) \in \mathcal R \land x \ne y$

By definition of $\prec$:
 * $ x \prec y$

Define $Y := s \vee y$

By Join Succeeds Operands:
 * $s \preceq Y$ and $y \preceq Y$

Then
 * $x \ne Y$

By definition of reflexivity:
 * $x \preceq x$

By definition of auxiliary relation:
 * $\left({x, Y}\right) \in \mathcal R$

Again by definition of auxiliary relation:
 * $\left({Y, z}\right) \in \mathcal R$

Hence
 * $\exists y \in S: \left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \land x \ne y$