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

Theorem
Let $\left({S, \vee_1, \wedge_1, \preceq_1}\right)$ and $\left({T, \vee_2, \wedge_2, \preceq_2}\right)$ be lattices.

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

Then $f$ is an increasing mapping.

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

Define $M = \left({\N, \le}\right)$ being an ordered set.

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

Let $x, y \in \N$.

Thus by definition of max operation:
 * $\max\left({x, y}\right) \in \N$

Thus by definition of max operation:
 * $x \le \max\left({x, y}\right)$ and $y \le \max\left({x, y}\right)$

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

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

By Preceding iff Meet equals Less Operand:
 * $f\left({\liminf \left({c_i}\right)_{i \in \N} }\right) = f\left({a}\right)$

By definition of composition of mappings:
 * $f \circ g = \left({f\left({c_i}\right)}\right)_{i \in \N} = \left({f\left({a}\right), f\left({b}\right), f\left({a}\right), f\left({b}\right), \dots}\right)$

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

By assumption:
 * $f\left({a}\right) \preceq_2 f\left({a}\right) \wedge_2 f\left({b}\right)$

By Meet Precedes Operands:
 * $f\left({a}\right) \wedge_2 f\left({b}\right) \preceq_2 f\left({a}\right)$

By definition of antisymmetry:
 * $f\left({a}\right) = f\left({a}\right) \wedge_2 f\left({b}\right)$

Thus by Preceding iff Meet equals Less Operand:
 * $f\left({a}\right) \preceq_2 f\left({b}\right)$