# Auxiliary Approximating Relation has Interpolation Property

## Theorem

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.

Let $x, z \in S$ such that:

$x \ll z \land x \ne z$

Let $\RR$ be an auxiliary approximating relation on $S$.

Then

$\exists y \in S: \tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \land x \ne y$

## Proof

Define $I := \set {u \in S: \exists y \in S: \tuple {u, y} \in \RR \land \tuple {y, z} \in \RR}$

By definition of auxiliary relation:

$\tuple {\bot, \bot} \in \RR$ and $\tuple {\bot, z} \in \RR$

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 section.

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

$b \preceq a$

By definition of $I$:

$\exists s \in S: \tuple {a, s} \in \RR \land \tuple {s, z} \in \RR$

By definitions of auxiliary relation and reflexivity:

$\tuple {b, s} \in \RR$

Thus

$b \in I$

$\Box$

We will prove that

$I$ is directed.

Let $a, b \in I$.

By definition of $I$:

$\exists s \in S: \tuple {a, s} \in \RR \land \tuple {s, z} \in \RR$

and

$\exists t \in S: \tuple {b, t} \in \RR \land \tuple {t, z} \in \RR$
$\tuple {a \vee b, s \vee t} \in \RR$

By definition of auxiliary relation:

$\tuple {s \vee t, z} \in \RR$

Thus by definition of $I$:

$a \vee b \in I$

Thus by Join Succeeds Operands:

$a \preceq a \vee b$ and $b \preceq a \vee b$

$\Box$

By definition:

$I$ is ideal in $L$.

We will prove that

$I \subseteq z^{\RR}$

Let $a \in I$.

By definition of $I$:

$\exists s \in S: \tuple {a, s} \in \RR \land \tuple {s, z} \in \RR$

By definition of auxiliary relation:

$a \preceq s$

Again by definition of auxiliary relation and reflexivity:

$\tuple {a, z} \in \RR$

Thus by definition of $\RR$-segment:

$a \in z^{\RR}$

$\Box$

By Supremum of Subset and definition of approximating relation:

$\sup I \preceq \sup \left({z^{\RR} }\right) = z$

We will prove that

$\sup I = z$

$\sup I \ne z$

By definition of $\prec$:

$\sup I \prec z$

Then by definition of antisymmetry:

$z \npreceq \sup I$
$\exists y \in S: \tuple {y, z} \in \RR \land y \npreceq \sup I$
$\exists u \in S: \tuple {u, y} \in \RR \land u \npreceq \sup I$

By definition of $I$:

$u \in I$

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

$\Box$

$x \in I$

By definition of $I$:

$\exists s \in S: \tuple {x, s} \in \RR \land \tuple {s, z} \in \RR$

By definition of auxiliary relation:

$x \preceq s$

By definition of auxiliary relation and reflexivity:

$\tuple {x, z} \in \RR$
$\exists y \in S: x \preceq y \land \tuple {y, z} \in \RR \land x \ne y$

By definition of $\prec$:

$x \prec y$

Define $Y := s \vee y$

$s \preceq Y$ and $y \preceq Y$

Then

$x \ne Y$

By definition of reflexivity:

$x \preceq x$

By definition of auxiliary relation:

$\tuple {x, Y} \in \RR$

Again by definition of auxiliary relation:

$\tuple {Y, z} \in \RR$

Hence

$\exists y \in S: \tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \land x \ne y$

$\blacksquare$