If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter

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

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

Let $F$ be a filter on $L$ such that
 * $I \cap F = \varnothing$

Then there exists a prime ideal $P$ in $L$: $I \subseteq P$ and $P \cap F = \varnothing$

Proof
Define $X := \left\{ {P \in \mathit{Ids}\left({L}\right): I \subseteq P \land P \cap F = \varnothing}\right\}$

where $\mathit{Ids}\left({L}\right)$ denotes set of all ideals in $L$.

By Set is Subset of Itself:
 * $I \in X$

We will prove that
 * $\forall Z: Z \ne \varnothing \land Z \subseteq X \land \left({\forall Y_1, Y_2 \in Z: Y_1 \subseteq Y_2 \lor Y_2 \subseteq Y_1}\right) \implies \bigcup Z \in X$

By Zorn's Lemma:
 * there exists $Y \in X$: $Y$ is maximal set of $X$.

Then by definition of $X$:
 * $Y \in \mathit{Ids}\left({L}\right)$ and $I \subseteq Y$ and $Y \cap F = \varnothing$

We will prove that
 * $Y$ is a prime ideal.

Let $x, y \in S$ such that
 * $x \wedge y \in Y$

Aiming for a contradiction suppose that
 * $x \notin Y$ and $y \notin Y$