Identity Mapping is Order Isomorphism/Proof 1

Proof
By definition:
 * $\forall x \in S: I_S \left({x}\right) = x$

So:
 * $x \preceq y \implies I_S \left({x}\right) \preceq I_S \left({y}\right)$

As $I_S$ is a bijection, we also have:
 * $I_S^{-1} \left({x}\right) = x$

So:
 * $x \preceq y \implies I_S^{-1} \left({x}\right) \preceq I_S^{-1} \left({y}\right)$