Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element

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

Let $x \in S$.

Then:
 * $\ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I} = x^\ll$

where $\mathit {Ids}$ denotes the set of all ideals in $L$.

Proof
By Supremum of Lower Closure of Element:
 * $\map \sup {x^\preceq} = x$

By Lower Closure of Element is Ideal:
 * $x^\preceq \in \mathit {Ids}$

Then by definition of reflexivity:
 * $x^\preceq \in \set {I \in \mathit {Ids}: x \preceq \sup I}$

We will prove that:
 * $\ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I} \subseteq x^\ll$

Let $z \in \ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I}$

By definition of intersection:
 * $\forall I \in \mathit {Ids}: x \preceq \sup I \implies z \in I$

By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $z \ll x$

Thus by definition of way below closure:
 * $z \in x^\ll$

By definition of set equality, it remains to prove the pposite inclusion.

Let $z \in x^\ll$

By definition of way below closure:
 * $z \ll x$

We will prove that
 * $\forall Y \in \set {I \in \mathit {Ids}: x \preceq \sup I}: z \in Y$

Let $Y \in \set {I \in \mathit {Ids}: x \preceq \sup I}$

Then
 * $Y \in \mathit {Ids}$ and $x \preceq \sup Y$

Thus Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $z \in Y$

Thus by definition of intersection:
 * $z \in \ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I}$