Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping is Increasing

Theorem
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be lattices.

Let $f: S \to T$ be a mapping such that:
 * for all directed set $\struct {D, \precsim}$ and Moore-Smith sequence $N:D \to S$ in $S$: $\map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$

Then $f$ is an increasing mapping.

Proof
Let $a, b \in S$ such that
 * $a \preceq_1 b$

Define $M = \struct {\N, \le}$ being an ordered set.

We will prove that:
 * $M$ is a directed set.

Let $x, y \in \N$.

Thus by definition of max operation:
 * $\max \set {x, y} \in \N$

Thus by definition of max operation:
 * $x \le \max \set {x, y}$ and $y \le \max \set {x, y}$

Define $g = \sequence {c_i}_{i \mathop \in \N} = \tuple {a, b, a, b, \dots}: \N \to S$ being a Moore-Smith sequence in $S$.

By Limit Inferior of Repetition Moore-Smith Sequence:
 * $\liminf \sequence {c_i}_{i \mathop \in \N} = a \wedge_1 b$

By Preceding iff Meet equals Less Operand:
 * $\map f {\liminf \sequence {c_i}_{i \mathop \in \N} } = \map f a$

By definition of composition of mappings:
 * $f \circ g = \sequence {\map f {c_i} }_{i \mathop \in \N} = \tuple {\map f a, \map f b, \map f a, \map f b, \dots}$

By Limit Inferior of Repetition Moore-Smith Sequence:
 * $\map \liminf {f \circ g} = \map f a \wedge_2 \map f b$

By assumption:
 * $\map f a \preceq_2 \map f a \wedge_2 \map f b$

By Meet Precedes Operands:
 * $\map f a \wedge_2 \map f b \preceq_2 \map f a$

By definition of antisymmetry:
 * $\map f a = \map f a \wedge_2 \map f b$

Thus by Preceding iff Meet equals Less Operand:
 * $\map f a \preceq_2 \map f b$