If Element Does Not Belong to Ideal then There Exists Prime Ideal Including Ideal and Excluding Element

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

Let $I$ be an ideal in $L$.

Let $x$ be an element of $S$.

Suppose $x \notin I$

Then there exists a prime ideal $P$ in $L$: $I \subseteq P$ and $x \notin P$

Proof
By Upper Closure of Element is Filter:
 * $x^\succeq$ is a filter.

Wr will prove that
 * $I \cap x^\succeq = \varnothing$

Aiming for a contradiction suppose that
 * $\exists y: y \in I \cap x^\succeq$

By definition of intersection:
 * $y \in I$ and $y \in x^\succeq$

By definition of upper closure of element:
 * $x \preceq y$

By definition of lower set:
 * $x \in I$

This contradicts $x \notin I$

Thus by If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter:
 * there exists prime ideal $P$ in $L$: $I \subseteq P$ and $P \cap x^\succeq = \varnothing$

By definition of reflexivity:
 * $x \preceq x$

By definition of upper closure of element:
 * $x \in x^\succeq$

Thus by definitions of intersection and empty set:
 * $x \notin P$