Intersection of Auxiliary Relations is Auxiliary Relation

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Theorem

Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.

Let $\FF$ be a non-empty set of auxiliary relations on $S$.


Then

$\ds \bigcap \FF$ is auxiliary relation.


Proof

By definition of non-empty set:

$\exists \RR: \RR \in \FF$

We will prove that:

$\ds \forall x, y \in S: \tuple {x, y} \in \bigcap \FF \implies x \preceq y$

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

$\ds \tuple {x, y} \in \bigcap \FF$

By definition of intersection:

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

Thus by definition of auxiliary relation:

$x \preceq y$

$\Box$


We will prove that:

$\ds \forall x, y, z, u \in S: x \preceq y \land \tuple {y, z} \in \bigcap \FF \land z \preceq u \implies \tuple {x, u} \in \bigcap \FF$

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

$\ds x \preceq y \land \tuple {y, z} \in \bigcap \FF \land z \preceq u$

By definition of intersection:

$\forall r \in \FF: \tuple {y, z} \in r$

By definition of auxiliary relation:

$\forall r \in \FF: \tuple {x, u} \in r$

Thus by definition of intersection:

$\ds \tuple {x, u} \in \bigcap \FF$

$\Box$


We will prove that:

$\ds \forall x, y, z \in S: \tuple {x, z} \in \bigcap \FF \land \tuple {y, z} \in \bigcap \FF \implies \tuple {x \vee y, z} \in \bigcap \FF$

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

$\ds \tuple {x, z} \in \bigcap \FF \land \tuple {y, z} \in \bigcap \FF$

By definition of intersection:

$\forall r \in \FF: \tuple {x, z} \in r \land \tuple {y, z} \in r$

By definition of auxiliary relation:

$\forall r \in \FF: \tuple {x \vee y, z} \in r$

Thus by definition of intersection:

$\tuple {x \vee y, z} \in \bigcap \FF$

$\Box$


By definition of auxiliary relation:

$\forall x \in S: \forall r \in \FF: \tuple {\bot, x} \in r$

Thus by definition of intersection:

$\ds \forall x \in S: \tuple {\bot, x} \in \bigcap \FF$

Thus by definition:

$\ds \bigcap \FF$ is auxiliary relation.

$\blacksquare$


Sources