Multiplicative Auxiliary Relation iff Images are Filtered

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

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

Then $\mathcal R$ is multiplicative
 * for all $x \in S$: $^{\mathcal R}x$ is filtered

where $^{\mathcal R}x$ denotes the $\mathcal R$-cosegment of $x$.

Sufficient Condition
Let $\mathcal R$ be multiplicative.

Let $x \in S$.

Let $a, b \in {}^{\mathcal R}x$.

By definition of $\mathcal R$-cosegment:
 * $\left({x, a}\right), \left({x, b}\right) \in \mathcal R$

By definition of multiplicative relation:
 * $\left({x, a \wedge b}\right) \in \mathcal R$

By definition of $\mathcal R$-cosegment:
 * $a \wedge b \in {}^{\mathcal R}x$

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

Thus
 * $\exists c \in {}^{\mathcal R}x: c \preceq a \land c \preceq b$

Hence $^{\mathcal R}x$ is filtered.

Necessary Condition
Suppose that
 * for all $x \in S$: $^{\mathcal R}x$ is filtered

Let $a, x, y \in S$ such that
 * $\left({a, x}\right), \left({a, y}\right) \in \mathcal R$

By definition of $\mathcal R$-cosegment:
 * $x, y \in {}^{\mathcal R}a$

By Auxiliary Relation Cosegment is Upper Set:
 * $^{\mathcal R}a$ is upper set.

By assumption:
 * $^{\mathcal R}a$ is filtered.

By Filtered in Meet Semilattice:
 * $x \wedge y \in {}^{\mathcal R}a$

Thus by definition of $\mathcal R$-cosegment:
 * $\left({a, x \wedge y}\right) \in \mathcal R$