Induced Relation Generates Order Isomorphism

Theorem
Let $\left({ A_1, \preceq_1 }\right)$ be an ordered set.

Let $\phi : A_1 \to A_2$ be a bijection.

Let:
 * $\preceq_2 \mathop{:=} \left\{ \left({\phi\left({x}\right), \phi\left({y}\right) }\right): x \in A_1 \land y \in A_1 \land x \mathop{\preceq_1} y \right\}$

Then $\phi : \left({ A_1, \preceq_1 }\right) \to \left({ A_2 , \preceq_2 }\right)$ is an order isomorphism.

Proof
Take any $x,y \in A_1$ such that $x \preceq_1 y$.

Since $x, y \in A_1$, it follows by the definition of a mapping that:
 * $\phi\left({x}\right), \phi\left({y}\right) \in A_2$

So $x \in A_1$ and $y \in A_1$ and $x \preceq_1 y$.

It follows from the definition of $\preceq_2$ that:
 * $\phi\left({x}\right) \preceq_2 \phi\left({y}\right)$

Conversely, suppose that:
 * $\phi\left({x}\right) \preceq_2 \phi\left({y}\right)$

By the definition of $\preceq_2$, it follows that:
 * $x \preceq_1 y$

Therefore, the biconditional holds:


 * $x \preceq_1 y \iff \phi\left({x}\right) \preceq_2 \phi\left({y}\right)$

By definition, it follows that:
 * $\phi: \left({A_1, \preceq_1}\right) \to \left({A_2, \preceq_2 }\right)$

is an order isomorphism.