Order Isomorphism on Strictly Well-Founded Relation preserves Strictly Well-Founded Structure

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

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

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

Then $\struct {A_1, \prec_1}$ is a foundational structure $\struct {A_2, \prec_2}$ is also a foundational structure.

Proof
Take any nonempty subset $B \subseteq A_1$.

Then $x$ is a minimal element of $B$ $\map \phi x$ is a minimal element of $\phi \sqbrk B$ by the fact that an Order Isomorphism Preserves Minimal Elements.

By the definition of foundational relation, $\struct {A_1, \prec_1}$ is foundational $\struct {A_2, \prec_2}$ is foundational.