Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure

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

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

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

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

Let $M = \left({F, \preccurlyeq}\right)$ be the ordered set of increasing mappings $g$ satisfying $\forall x \in S: g\left({x}\right) \subseteq x^\preceq$.

Let $f: S \to \mathit{Ids}\left({R}\right)$ 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:


 * $(1): \quad f \in F$

Let $h: S \to \mathit{Ids}\left({R}\right): x \mapsto x^\preceq$

Then:


 * $(2): \quad h \in F \land f \preccurlyeq h$

Let $k: S \to \mathit{Ids}\left({R}\right): x \mapsto \left\{ {\bot}\right\}$

where $\bot$ denotes the smallest element in $L$.

Then:


 * $(3): \quad k \in F$ and $k \preccurlyeq f$

Condition $(1)$
By Segment of Auxiliary Relation Mapping is Increasing:
 * $f$ is an increasing mapping.

By Segment of Auxiliary Relation is Subset of Lower Closure:
 * $\forall x \in S: x^r \subseteq x^\preceq$

By definition of $f$:
 * $\forall x \in S: f\left({x}\right) \subseteq x^\preceq$

Thus
 * $f \in F$

Condition $(2)$
By Lower Closure is Increasing:
 * $\forall x, y \in S: x \preceq y \implies x^\preceq \subseteq y^\preceq$

By definitions of $h$ and $\precsim$:
 * $\forall x, y \in S: x \preceq y \implies h\left({x}\right) \precsim h\left({y}\right)$

By definition:
 * $h$ is an increasing mapping.

By definition of $h$:
 * $\forall x \in S: h\left({x}\right) \subseteq x^\preceq$

Thus
 * $h \in F$

By Segment of Auxiliary Relation is Subset of Lower Closure:
 * $\forall x \in S: x^r \subseteq x^\preceq$

By definitions of $f$ and $h$:
 * $\forall x \in S: f\left({x}\right) \subseteq h\left({x}\right)$

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

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

Condition $(3)$
$k$ is well-defined because by Singleton of Bottom is Ideal:
 * $\left\{ {\bot}\right\}$ is an ideal in $R$.

By Mapping is Constant iff Increasing and Decreasing:
 * $k$ is an increasing mapping.

By definition of smallest element:
 * $\forall x \in S: \bot \preceq x$

By definition of lower closure of element:
 * $\forall x \in S: \bot \in x^\preceq$

By definition of subset:
 * $\forall x \in S: \left\{ {\bot}\right\} \subseteq x^\preceq$

By definition of $k$:
 * $\forall x \in S: k\left({x}\right) \subseteq x^\preceq$

Thus
 * $k \in F$

By definition of auxiliary relation:
 * $\forall x \in S: \left({\bot, x}\right) \in r$

By definition of $r$-segment:
 * $\forall x \in S: \bot \in x^r$

By definition of subset:
 * $\forall x \in S: \left\{ {\bot}\right\} \subseteq x^r$

By definitions of $k$ and $f$:
 * $\forall x \in S: k\left({x}\right) \subseteq f\left({x}\right)$

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

Thus by definition of ordering of $M$:
 * $k \preccurlyeq f$