Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding

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

Let $X$ be a subset of $S$.

Then
 * $X$ is order generating


 * $\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}\right)$

Sufficient Condition
Let $X$ be order generating.

Let $x, y \in S$ such that
 * $y \npreceq x$

By Order Generating iff Every Element is Infimum:
 * $\exists P \subseteq X: x = \inf P$

By definition of infimum:
 * $y$ is lower bound for $P \implies y \preceq x$

By definition of lower bound:
 * $\exists p \in P: y \npreceq p$

Thus by definition of subset:
 * $p \in X$

By definition of infimum:
 * $x$ is lower bound for $P$.

Thus by definition of lower bound:
 * $x \preceq p$

Thus
 * $y \npreceq p$

Necessary Condition
Suppose that
 * $\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}\right)$

Let $x \in S$.

Thus by definition of complete lattice:
 * $x^\succeq \cap X$ admits an infimum.

Define $y := \inf \left({x^\succeq \cap X}\right)$

By definition of infimum:
 * $y$ is lower bound for $x^\succeq \cap X$

We will prove that
 * $x$ is lower bound for $x^\succeq \cap X$

Let $a \in x^\succeq \cap X$

By definition of intersection:
 * $a \in x^\succeq$

Thus by definition of upper closure of element:
 * $x \preceq a$

By definition of infimum:
 * $x \preceq y$

By definition of antisymmetry:
 * $y \nprec x$

It remains to prove that
 * $y = x$

Aiming for a contradiction suppose that
 * $y \ne x$

By definition of $\prec$
 * $y \npreceq x$ or $y = x$

Case $y \npreceq x$

By assumption:
 * $\exists p \in X: x \preceq p \land y \npreceq p$

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

By definition of intersection:
 * $p \in x^\succeq \cap X$

By definition of lower bound:
 * $y \preceq p$

Thus this contradicts $y \npreceq p$

Case $y = x$

Thus this contradicts $y \ne x$