Maximal implies Difference equals Intersection

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $x, y \in S$ such that
 * $x$ is maximal in $S \setminus y^\succeq$

Then $x^\succeq \setminus \left\{ {x}\right\} = x^\succeq \cap y^\succeq$

Proof
By definition of maximal:
 * $x \in S \setminus y^\succeq$

By definition of difference:
 * $x \notin y^\succeq$

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

We will prove that
 * $x^\succeq \setminus \left\{ {x}\right\} \subseteq x^\succeq \cap y^\succeq$

Let $a \in x^\succeq \setminus \left\{ {x}\right\}$

By definition of difference:
 * $a \in x^\succeq$ and $a \notin \left\{ {x}\right\}$

By definitions of upper closure of element and singleton:
 * $x \preceq a$ and $a \ne x$

By definition of strictly precede:
 * $x \prec a$

By definition of maximal:
 * $a \notin S \setminus y^\succeq$

By definition of difference:
 * $a \in y^\succeq$

Thus by definition of intersection:
 * $a \in x^\succeq \cap y^\succeq$

We will prove that
 * $x^\succeq \cap y^\succeq \subseteq x^\succeq \setminus \left\{ {x}\right\}$

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

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

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

Then
 * $a \ne x$

By definition of singleton:
 * $a \notin \left\{ {x}\right\}$

Thus by definition of difference:
 * $a \in x^\succeq \setminus \left\{ {x}\right\}$

Thus by definition of set equality:
 * $x^\succeq \setminus \left\{ {x}\right\} = x^\succeq \cap y^\succeq$