Identity Mapping is Order Isomorphism/Proof 1

Proof
By definition:
 * $\forall x \in S: \map {I_S} x = x$

So:
 * $x \preceq y \implies \map {I_S} x \preceq \map {I_S} y$

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

So:
 * $x \preceq y \implies \map {I_S^{-1} } x \preceq \map {I_S^{-1} } y$