Add a primitive a one place function symbol $c$ to represent "True cardinality" of a set, to the first order language of set theory.

Add the following schema:

If $\phi(x,y)$ is a formula in which $x,y$ occur free, and only occur free, then all closures of:

$\forall X,Y: \\\forall x \in X \exists! y \in Y (\phi(x,y)) \land \\\forall y \in Y \exists! x \in X (\phi(x,y)) \\ \to c(X)=c(Y)$

are axioms.

Add the following $\omega$-rule of inference:

if $\psi(X); \varphi(Y)$, are formulas in which $X,Y$ occur free and only free respectively, then:

From: $\big{[}$if $\phi(x,y)$ is a formula in which $x,y$ occur free, and only occur free, then all closures of the following formula are true:

$\forall X,Y (\psi(X) \land \varphi(Y) \to \\\neg [\forall x \in X \exists! y \in Y (\phi(x,y)) \land \\\forall y \in Y \exists! x \in X (\phi(x,y))]) \big{]}$

______________________we Infer

All closures of $\forall X,Y (\psi(X) \land \varphi(Y) \to c(X)\neq c(Y) )$ are true.

Now if a set theory T extended with the above, proves that:

$\exists X,Y: |X|\neq|Y| \land c(X)=c(Y)$

Then its guilty of committing cardinaity error of the first kind.

If it proves that:

$\exists X,Y: |X| = |Y| \land c(X) \neq c(Y)$

Then its guilty of committing cardinality error of the second kind.

Now NFU is an example of a set theory that commit cardinality error of the first kind, but this cannot occur in ZFC.

Can ZFC commit cardinality error of the second kind?

Based on comments with Monroe Eskew. The following question presents itself.

Is there a natural statement that the theory "ZFC + ZFC doesn't commit cardinality error of the second kind" can settle, that ZFC + V=L cannot?