Order Isomorphism forms Galois Connection

Theorem
Let $L_1 = \left({S_1, \preceq_1}\right)$, $L_2 = \left({S_2, \preceq_2}\right)$ be ordered sets.

Let $f:S_1 \to S_2$ be an order isomorphism between $L_1$ and $L_2$.

Then $\left({f, f^{-1} }\right)$ is a Galois connection.

Proof
Let $t \in S_2$, $s \in S_1$.

We will prove that
 * $t \preceq_2 f\left({s}\right) \implies f^{-1}\left({t}\right) \preceq_1 s$

Assume that
 * $t \preceq_2 f\left({s}\right)$

By Inverse of Order Isomorphism is Order Isomorphism:
 * $f^{-1}$ is an order isomorphism.

By definition of order isomorphism:
 * $f^{-1}$ is an order embedding.

By definition of order embedding:
 * $f^{-1}\left({t}\right) \preceq_1 f^{-1}\left({f\left({s}\right)}\right)$

Thus by definition of bijection:
 * $f^{-1}\left({t}\right) \preceq_1 s$

We will prove that
 * $f^{-1}\left({t}\right) \preceq_1 s \implies t \preceq_2 f\left({s}\right)$

Assume that
 * $f^{-1}\left({t}\right) \preceq_1 s$

By definition of order embedding:
 * $f\left({f^{-1}\left({t}\right)}\right) \preceq_2 f\left({s}\right)$

Thus by definition of bijection:
 * $t \preceq_2 f\left({s}\right)$