Way Below iff Second Operand Preceding Supremum of Prime Ideal implies First Operand is Element of Ideal

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

Let $x, y \in S$.

Then $x \ll y$
 * for every prime ideal $P$ in $L$: $y \preceq \sup P \implies x \in P$

Sufficient Condition
The result follows by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal.

Necessary Condition
Suppose
 * for every prime ideal $P$ in $L$: $y \preceq \sup P \implies x \in P$

We will prove that
 * for every ideal $I$ in $L$: $y \preceq \sup I \implies x \in I$

Let $I$ be an ideal in $L$ such that
 * $y \preceq \sup I$

Aiming for a contradiction suppose that
 * $x \notin I$

By If Element Does Not Belong to Ideal then There Exists Prime Ideal Including Ideal and Excluding Element:
 * there exists a prime ideal $P$ in $L$: $I \subseteq P$ and $x \notin P$

By Supremum of Subset:
 * $\sup I \preceq \sup P$

By definition of transitivity:
 * $y \preceq \sup P$

By assumption:
 * $x \in P$

Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $x \ll y$