Ordered Set of All Mappings is Ordered Set

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

Let $X$ be a set.

Then $L^X$ is also an ordered set.

Proof
By definition of ordered set of all mappings:
 * $L^X = \left({S^X, \precsim}\right)$

where
 * $\forall f, g \in S^X: f \precsim g \iff f \preceq g$
 * $\preceq$ denotes the ordering on mappings,
 * $S^X$ denotes the set of all mappings from $X$ into $S$.

Reflexivity
Let $f \in S^X$.

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

By definition of ordering on mappings:
 * $f \preceq f$

Thus by definition of $\precsim$:
 * $f \precsim f$

Transitivity
Let $f, g, h \in S^X$ such that
 * $f \precsim g$ and $g \precsim h$

By definition of $\precsim$:
 * $f \preceq g$ and $g \preceq h$

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

and
 * $\forall x \in X: g\left({x}\right) \preceq h\left({x}\right)$

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

By definition of ordering on mappings:
 * $f \preceq h$

Thus by definition of $\precsim$:
 * $f \precsim h$

Antisymmetry
Let $f, g \in S^X$ such that
 * $f \precsim g$ and $g \precsim f$

By definition of $\precsim$:
 * $f \preceq g$ and $g \preceq f$

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

and
 * $\forall x \in X: g\left({x}\right) \preceq f\left({x}\right)$

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

Thus by Equality of Mappings:
 * $f = g$