Superinductive Class under Strictly Progressing Mapping is Proper Class

Theorem
Let $A$ be a class.

Let $g: A \to A$ be a strictly progressing mapping on $A$.

Let $A$ be superinductive under $g$.

Then $A$ cannot be a set, and thus is a proper class.

Proof
$A$ is a set.

Then from Set which is Superinductive under Progressing Mapping has Fixed Point, $A$ has a fixed point.

However, we have that $g$ is a strictly progressing mapping on $A$.

Hence $g$ has no fixed point in $A$.

Hence the result by Proof by Contradiction.