The proof is an easy modification of the argument for Russell's Paradox, using the contradiction this time as a reductio. By Separation, let M0 be the subset of M consisting of those elements x of M such that  x ∉ x. Now either  M0 ∈ M0 or M0 ∉ M0. Assume that  M0 ∈ M0. Since  M0 is a subset of M, this tells us that  M0 ∈ M. But M0 is then a member of M which fails to satisfy the condition for belonging to M0, showing that  M0 ∉ M0, which is a contradiction. Hence, necessarily,  M0 ∉ M0. But now if we suppose that M0 were in M, then M0 itself is bound to be in M0 by the defining condition of this set. Hence,  M0 ∉ M on pain of contradiction. The argument for the Russell paradox is used here to constructive effect: one person's contradiction is another person's reductio. Zermelo comments: