Intersection of Auxiliary Relations is Auxiliary Relation

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:
 * $x \preceq y$

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

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

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.