Correctness of Definition of Increasing Mappings Satisfying Inclusion in Lower Closure

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

Let ${\it Ids}\left({R}\right)$ be the set of all ideals in $R$.

Let $L = \left({ {\it Ids}\left({R}\right), \precsim}\right)$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{ {\it Ids}\left({R}\right) \times {\it Ids}\left({R}\right)}$

Let
 * $M = \left({F, \preccurlyeq}\right)$

where
 * $F = \left\{ {f: S \to {\it Ids}\left({R}\right): f}\right.$ is increasing mapping $\left.{\land \forall x \in S: f\left({x}\right) \subseteq x^\preceq}\right\}$

and
 * $\preccurlyeq$ is ordering on mappings generated by $\precsim$

where $x^\preceq$ denotes the lower closure of $x$.

Then
 * $M$ is an ordered set.

Reflexivity
Let $f \in F$.

By definition of reflexivity:
 * $\forall x \in S: f\left({x}\right) \precsim f\left({x}\right)$

Thus by definition of ordering on mappings:
 * $f \preccurlyeq f$

Transitivity
Let $f, g, h \in F$ such that
 * $f \preccurlyeq g \preccurlyeq h$

By definition of ordering on mappings:
 * $\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \precsim h\left({x}\right)$

By definition of transitivity:
 * $\forall x \in S: f\left({x}\right) \precsim h\left({x}\right)$

Thus by definition of ordering on mappings:
 * $f \preccurlyeq h$

Antisymmetry
Let $f, g \in F$ such that
 * $f \preccurlyeq g$ and $g \preccurlyeq f$

By definition of ordering on mappings:
 * $\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \land g\left({x}\right) \precsim f\left({x}\right)$

By definition of antisymmetry:
 * $\forall x \in S: f\left({x}\right) = g\left({x}\right)$

Thus by equality of mappings:
 * $f = g$

Thus by definition:
 * $M$ is an ordered set.