Multiplicative Auxiliary Relation iff Images are Filtered

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

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

Then $\RR$ is multiplicative :
 * for all $x \in S$: $\map \RR x$ is filtered

where $\map \RR x$ denotes the $\RR$-image of $x$.

Sufficient Condition
Let $\RR$ be multiplicative.

Let $x \in S$.

Let $a, b \in \map \RR x$.

By definition of $\RR$-image of element:
 * $\tuple {x, a}, \tuple {x, b} \in \RR$

By definition of multiplicative relation:
 * $\tuple {x, a \wedge b} \in \RR$

By definition of $\RR$-image of element:
 * $a \wedge b \in \map \RR x$

By Meet Precedes Operands:
 * $a \wedge b \preceq a$ and $a \wedge b \preceq b$

Thus
 * $\exists c \in \map \RR x: c \preceq a \land c \preceq b$

Hence $\map \RR x$ is filtered.

Necessary Condition
Suppose that:
 * for all $x \in S$: $\map \RR x$ is filtered

Let $a, x, y \in S$ such that:
 * $\tuple {a, x}, \tuple {a, y} \in \RR$

By definition of $\RR$-image of element:
 * $x, y \in \map \RR a$

By Auxiliary Relation Image of Element is Upper Set:
 * $\map \RR a$ is upper set.

By assumption:
 * $\map \RR a$ is filtered.

By Filtered in Meet Semilattice:
 * $x \wedge y \in \map \RR a$

Thus by definition of $\RR$-image of element:
 * $\tuple {a, x \wedge y} \in \RR$