Any x is b,
whence Any x not u is b,
Any x not u is b not m.
But since
Any b not m is n,
whence Any x not u is n,
and therefore Any x not u is both a and n.
But since
Whatever is both a and n is v,
whence Any x not u is v.
Corollary 1.—This proposition readily extends itself to arithmetical addition.
Corollary 2.—The converse propositions produced by transposing the last two identities of Theorems VIII and IX are also true.
Corollary 3.—Theorems VI, VII, and IX hold also with arithmetical multiplication. This is sufficiently evident in the case of theorem VI, because by definition 7 we have an additional premise, namely, that a and b are independent, and an additional conclusion which is the same as that premise.
In order to show the extension of the other theorems, I shall begin with the following lemma. If a and b are independent, then corresponding to every pair of individuals, one of which is both a and b, there is just one pair of individuals one of which is a and the other b; and conversely, if the pairs of individuals so correspond, a and b are independent. For, suppose a and b independent, then, by definition 7, condition 3, every class (Am,Bn) is an individual. If then Aa denotes any Am which is a, and Bb any Bm which is b, by condition 6 (Aa,Bn) and (Am,Bb) both exist, and by conditions 4 and 5 the former is any individual a, and the latter any individual b. But given this pair of individuals, both of the pair (Aa,Bn) and (Am,Bb) exist by condition 6. But one individual of this pair is both a and b. Hence the pairs correspond, as stated above. Next, suppose a and b to be any two classes. Let the series of Am’s be a and not-a; and let the series of Bm’s be all individuals separately. Then the first five conditions can always be satisfied. Let us suppose, then, that the sixth alone cannot be satisfied. Then Ap and Bq may be taken such that (Ap,Bq) is nothing. Since Ap and Bq are supposed both to exist, there must be two individuals (Ap,Bn) and (Am,Bq) which exist. But there is no corresponding pair (Am,Bn) and (Ap,Bq). Hence, no case in which the sixth condition cannot be satisfied simultaneously with the first five is a case in which the pairs rightly correspond; or, in other words, every case in which the pairs correspond rightly is a case in which the sixth condition can be satisfied, provided the first five can be satisfied. But the first five can always be satisfied. Hence, if the pairs correspond as stated, the classes are independent.
In order to show that Theorem VII may be extended to arithmetical multiplication, we have to prove that if a and b, b and c, and a and (b, c), are independent, then (a, b) and c are independent. Let s denote any individual. Corresponding to every s with (a,b,c), there is an a and (b,c). Hence, corresponding to every s with s and with (a,b,c) (which is a particular case of that pair), there is an s with a and with (b,c). But for every s with (b,c) there is a b with c; hence, corresponding to every a with s and with (b,c), there is an a with b and with c. Hence, for every s with s and with (a,b,c) there is an a with b and with a For every a with b there is an s with (a,b); hence, for every a with b and with c, there is an s with (a,b) and a Hence, for every s with s and with (a,b,c) there is an s with (a,b) and with a Hence, for every s with (a,b,c) there is an (a,b) with c. The converse could be proved in the same way. Hence, &c.
Theorem IX holds with arithmetical addition of whichever sort the multiplication is. For we have the additional premise that “No m is n”; whence since “any u is m” and “any v is n,” “no u is v,” which is the additional conclusion.
Corollary 2, so far as it relates to Theorem IX, holds with arithmetical addition and multiplication. For, since no m is n, every pair, one of which is a and either m or n, is either a pair, one of which is a and m, or a pair, one of which is a and n, and is not both. Hence, since for every pair one of which is a and m, there is a pair one of which is a and the other m, and since for every pair one of which is a, n there is a pair one of which is a and the other n; for every pair one of which is a and either m or n, there is either a pair one of which is a and the other m, or a pair one of which is a and the other n, and not both; or, in other words, there is a pair one of which is a and the other either m or n.
[It would perhaps have been better to give this complicated proof in its full syllogistic form. But as my principal object is merely to show that the various theorems could be so proved, and as there can be little doubt that if this is true of those which relate to arithmetical addition it is true also of those which relate to arithmetical multiplication, I have thought the above proof (which is quite apodeictic) to be sufficient. The reader should be careful not to confound a proof which needs itself to be experienced with one which requires experience of the object of proof.]
X
If
This does not hold with logical, but does with arithmetical multiplication.
For if a is not identical with a′, it may be divided thus
if ā’ denotes not a’. Then
and by the definition of independence the last term does not vanish unless
XI
This is not true of arithmetical addition, for since by definition 7,
by Theorem IX
Whence