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

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

Let $x \in S$.

Then
 * $\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} = x^\ll$

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

Proof
By Supremum of Lower Closure of Element:
 * $\sup \left({x^\preceq}\right) = x$

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

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

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

Let $z \in \displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$

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:
 * $z \in x^\ll$

It by definition of set equality remains to prove opposite inclusion.

Let $z \in x^\ll$

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

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

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

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:
 * $z \in \displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$