Arithmetic iff Way Below Relation is Multiplicative in Algebraic Lattice

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

Then $L$ is arithmetic
 * $\ll$ is a multiplicative relation

where $\ll$ denotes the way below relation of $L$.

Sufficient Condition
Let $L$ be arithmetic.

Let $a, x, y \in S$ such that
 * $a \ll x$ and $a \ll y$

By Algebraic iff Continuous and For Every Way Below Exists Compact Between:
 * $\exists c \in K\left({L}\right): a \preceq c \preceq x$

and
 * $\exists k \in K\left({L}\right): a \preceq k \preceq y$

where $K\left({L}\right)$ denotes the compact subset of $L$.

By :
 * $c \wedge k \preceq x \wedge y$

By definition of arithmetic:
 * $K\left({L}\right)$ is meet closed.

By definition of meet closed:
 * $c \wedge k \in K\left({L}\right)$

By definition of compact subset:
 * $c \wedge k$ is compact.

By definition of compact:
 * $c \wedge k \ll c \wedge k$

By Meet is Idempotent:
 * $a \wedge a = a$

By :
 * $a \preceq c \wedge k$

Thus by Preceding and Way Below implies Way Below
 * $a \ll x \wedge y$

Necessary Condition
Let $\ll$ be a multiplicative relation.

Thus $L$ is algebraic.

It remains to prove that
 * $K\left({L}\right)$ is meet closed.

Let $x, y \in K\left({L}\right)$.

By definition of compact subset:
 * $x$ and $y$ are compact.

By definition of compact:
 * $x \ll x$ and $y \ll y$

By Way Below Relation is Auxiliary Relation:
 * $\ll$ is auxiliary relation.

By Multiplicative Auxiliary Relation iff Congruent:
 * $x \wedge y \ll x \wedge y$

By definition:
 * $x \wedge y$ is compact.

Thus by definition of compact subset:
 * $x \wedge y \in K\left({L}\right)$