The second point presents greater difficulties. We cannot deny that there are ranges of propositions, for we often wish to assert the logical product of such ranges; yet we cannot admit that there are more ranges than propositions. At first sight, the difficulty might be thought to be solved by the fact that there is a proposition associated with every range of propositions which is not null, namely the logical product of the propositions in the range^[141]; but this does not destroy Cantor’s proof that a range has more sub-ranges than members. Let us apply the proof by assuming a particular one-one relation, which associates every proposition p which is not a logical product with the range whose only member is p, while it associates the product of all propositions with the null-range of propositions, and associates every other logical product of propositions with the range of its own factors. Then the range w which, by the general principle of Cantor’s proof, is not correlated with any proposition, is the range of propositions which are logical products, but are not themselves factors of themselves. But, by the definition of the correlating relation, w ought to be correlated with the logical product of w. It will be found that the old contradiction breaks out afresh; for we can prove that the logical product of w both is and is not a member of w. This seems to show that there is no such range as w; but the doctrine of types does not show why there is no such range. It seems to follow that the Contradiction requires further subtleties for its solution; but what these are, I am at a loss to imagine.(§ 500 ¶ 1)

Let us state this new contradiction more fully. If m be a class of propositions, the proposition every m is true may or may not be itself an m. But there is a one-one relation of this proposition to m: if n be different from m, every n is true is not the same proposition as every m is true. Consider now the whole class of propositions of the form every m is true, and having the property of not being members of their respective m’s. Let this class be w, and let p be the proposition every w is true. If p is a w, it must possess the defining property of w; but this property demands that p should not be a w. On the other hand, if p be not a w, then p does possess the defining property of w, and therefore is a w. Thus the contradiction appears unavoidable.(§ 500 ¶ 2)

In order to deal with this contradiction, it is desirable to reopen the quesiton of the identity of equivalent propositional functions and of the nature of the logical product of two propositions. These questions arise as follows. If m be a class of propositions, their logical product is the proposition every m is true, which I shall denote by ^‘m. If we now consider the logical product of the class of propositions composed of m together with ^‘m, this is equivalent to Every m is true and every m is true, i.e. to every m is true i.e. to ^‘m. Thus the logical product of the new class of propositions is equivalent to a member of the new class, which is the same as the logical product of m. Thus if we identify equivalent propositional functions (^‘m being a propositional function of m), the proof of the above contradiction fails, since every proposition of the form ^‘m is the logical product both of a class of which it is a member and of a class of which it is not a member.(§ 500 ¶ 3)

But such an escape is, in reality, impracticable, for it is quite self-evident that equivalent propositional functions are often not identical. Who will maintain, for example, that x is an even prime number other than 2 is identical with x is one of Charles II.’s wise deeds or foolish sayings? Yet these are equivalent, if a well-known epitath is to be credited. The logical product of all the propositions of the class composed of m and ^‘m is Every proposition which either is an m or asserts that every m is true, is true; and this is not identical with every m is true, although the two are equivalent. Thus there seems no simple method of avoiding the contradiction in question.(§ 500 ¶ 4)

The close analogy of this contradiction with the one discussed in Chapter X strongly suggests that the two must have the same solution, or at least very similar solutions. It is possible, of course, to hold that propositions themselves are of various types, and that logical products must have propositions of only one type as factors. But this suggestion seems harsh and highly artificial.(§ 500 ¶ 5)

To sum up: it appears that the special contradiction of Chapter X is solved by the doctrine of types, but that there is at least one closely analogous contradiction which is probably not soluble by this doctrine. The totality of all logical objects, or of all propositions, involves, it would seem, a fundamental logical difficulty. What the complete solution ofthe difficulty may be, I have not succeeded in discovering; but as it affects the very foundations of reasoning, I earnestly commend the study of it to the attention of all students of logic.(§ 500 ¶ 6)