Equivalence of Definitions of Minimally Closed Class

Theorem
Let $A$ be a class.

Let $g$ be a mapping on $A$.

Proof
Let it be given that $A$ is closed under $g$.

$(1)$ implies $(2)$
Let $A$ be a minimally closed class under $g$ by definition 1.

Then by definition:
 * there exists $b \in A$ such that $A$ has no proper subclass $B$ such that:
 * $b \in B$
 * $B$ is closed under $g$.

Let $A$ have a subclass $C$ which is closed under $g$ such that $b \in C$.

Then by definition, $C$ is not a proper subclass of $A$.

Thus by definition of proper subclass:
 * $C = A$

By definition of subclass:
 * $A \subseteq C$

and:
 * $C \subseteq A$

That is:
 * $\forall x \in A: x \in C$

and:
 * $\forall x \in C: x \in A$

That is: $C$ contains all elements of $A$.

Thus $A$ is a minimally closed class under $g$ by definition 2.

$(2)$ implies $(1)$
Let $A$ be a minimally closed class under $g$ by definition 2.

Then by definition:
 * Every subclass of $A$ containing $b$ which is closed under $g$ contains all the elements of $A$.

Let $C$ be a subclass of $A$ which is closed under $g$ such that $b \in C$.

By definition of subclass:
 * $C \subseteq A$

But by hypothesis:
 * $\forall x \in A: x \in C$

That is:
 * $A \subseteq C$

By definition of equality of classes it follows that:
 * $A = C$

and so by definition $C$ cannot be a proper subclass of $A$.

Thus $A$ is a minimally closed class under $g$ by definition 1.