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

Theorem
Let $\mathscr S = \struct {S, \preceq}$ be an up-complete ordered set.

Let $x, y \in S$.

Then $x \ll y$
 * $\forall I \in \map {\operatorname {Ids} } {\mathscr S}: y \preceq \sup I \implies x \in I$

where
 * $\ll$ denotes the way below relation,
 * $\map {\operatorname {Ids} } {\mathscr S}$ denotes the set of all ideals in $\mathscr S$.

Sufficient Condition
Let $x \ll y$

Let $I \in \map {\operatorname {Ids} } {\mathscr S}$ such that
 * $y \preceq \sup I$

By definition of ideal:
 * $I$ is directed and lower.

By definition of up-complete:
 * $I$ admits a supremum.

By definition of way below relation:
 * $\exists i \in I: x \preceq i$

Thus by definition of lower section:
 * $x \in I$

Necessary Condition
Assume that
 * $\forall I \in \map {\operatorname {Ids} } {\mathscr S}: y \preceq \sup I \implies x \in I$

Let $D$ be a directed subset of $S$ such that:
 * $D$ admits a supremum and $y \preceq \sup D$

By Supremum of Lower Closure of Set:
 * $D^\preceq$ admits a supremum and $\sup D^\preceq = \sup D$

where $D^\preceq$ denotes the lower closure of $D$.

By Lower Closure of Directed Subset is Ideal:
 * $D^\preceq$ is an ideal.

By assumption:
 * $x \in D^\preceq$

Thus by definition of lower closure of subset:
 * $\exists d \in D: x \preceq d$

Thus by definition of way below relation:
 * $x \ll y$