Induced Relation Generates Order Isomorphism

Theorem
Let $\left({ A_1, \preceq_1 }\right)$ be a poset.

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

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

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 that $\phi\left({x}\right), \phi\left({y}\right) \in A_2$ by the definition of a mapping.

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 the definition of order isomorphism, it follows that $\phi : \left({ A_1, \preceq_1 }\right) \to \left({ A_2 , \preceq_2 }\right)$ is an order isomorphism.