Order Isomorphism Preserves Strictly Minimal Elements

Theorem
Let $A_1$ and $A_2$ be classes.

Let $\prec_1$ and $\prec_2$ be relations.

Let $\phi : A_1 \to A_2$ create an order isomorphism between $\struct {A_1, \prec_1}$ and $\struct {A_1, \prec_2}$.

Let $B \subseteq A_1$.

Then $\phi$ maps the strictly minimal elements under $\prec_1$ of $B$ to the strictly minimal elements under $\prec_2$ of $\phi \sqbrk B$.

Proof
Suppose $x$ is a strictly minimal element of $B$. That is:


 * $\neg \exists y \in B: x \prec_1 y$

This is equivalent to the statement:


 * $\neg \exists z \in \phi \sqbrk B: \map \phi x \prec_2 z$ since $z$ is of the form $\map \phi y$ by the definition of order isomorphism.

Thus, this statement is true $\map \phi x$ is a strictly minimal element of $B$.