Identity Mapping is Order Isomorphism

Theorem
Let $$\left({S; \preceq}\right)$$ be a poset.

The identity mapping $$I_S$$ is an order isomorphism from $$\left({S; \preceq}\right)$$ to itself.

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