Segment of Auxiliary Relation Mapping is Increasing

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 $r$ be an auxiliary relation on $S$.

Let $f: S \to {\it Ids}\left({R}\right)$ be a mapping such that
 * $\forall x \in S: f\left({x}\right) = x^r$

where $x^r$ denotes the $r$-segment of $x$.

Then
 * $f$ is increasing mapping.

Proof
$f$ is well-defined because by Relation Segment of Auxiliary Relation is Ideal:
 * $\forall x \in S: x^r$ is ideal in $L$

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

By Preceding implies Inclusion of Segments of Auxiliary Relation:
 * $x^r \subseteq y^r$

Thus by definitions of $\precsim$ and $f$:
 * $f\left({x}\right) \precsim f\left({y}\right)$

Thus by definition:
 * $f$ is increasing mapping.