 Proof. If \(D_f\) is in \(B\), then there is some \(y\) in \(A\) such that \(f(y) = D_f\). It now follows that \(y\) is in \(D_f\) iff it’s not. Contradiction. Note that when \(A = B\) and \(f\) is the identity map defined on \(A\), the result is (a form of) the theorem asserting that the set of non-self-membered sets does not exist.