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 $\left({ A_1, \prec_1 }\right)$ and $\left({ A_2 , \prec_2 }\right)$.

Let $B \subseteq A_1$.

Then $\phi$ maps the $\prec_1$-minimal elements (for strict orderings) of $B$ to the $\prec_2$-minimal element of $\phi \left({B}\right)$.

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


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

This is equivalent to the statement:


 * $\neg \exists z \in \phi\left({B}\right): \phi\left({x}\right) \prec_2 z$ since $z$ is of the form $\phi\left({y}\right)$ by the definition of order isomorphism.

Thus, this statement is true iff $\phi\left({x}\right)$ is a minimal element of $B$.