Inverse of Order Isomorphism is Order Isomorphism

Theorem
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.

Let $\phi$ be a bijection from $\struct {S, \preceq_1}$ to $\struct {T, \preceq_2}$.

Then:
 * $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$

is an order isomorphism :
 * $\phi^{-1}: \struct {T, \preceq_2} \to \struct {S, \preceq_1}$

is also an order isomorphism.

Proof
Follows directly from the definition of order isomorphism.

Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an order isomorphism.

Then:
 * $\forall x, y, \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$

and:


 * $\forall \map \phi x, \map \phi y \in T: \map \phi x \preceq_2 \map \phi y \implies \map {\phi^{-1} } {\map \phi x} \preceq_1 \map {\phi^{-1} } {\map \phi y}$

That is:
 * $\forall \map \phi x, \map \phi y \in T: \map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$

Hence the result.