OF ARTS AND SCIENCES : SEPTEMBER 10, 1867. 407 



A m which is «, and B b any B m which is b, by condition 6 (A a , B n ) and 

 (A m , B b ) both exist, and by conditions 4 and 5 the former is any indi- 

 vidual a, and the latter any individual b. But given this pair of indi- 

 viduals, both of the pair (A a , B b ) and ( A m , B n ) exist by condition 6. 

 But one individual of this pair is both a and b. Hence the pairs cor- 

 respond, as stated above. Next, suppose a and b to be any two classes. 

 Let the series of A m 's be a and not-a ; and let the series of J5 m '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 A p and i? q may be taken such that ( A p , Z? q ) is noth- 

 ing. Since A p and 2? q are supposed both to exist, there must be two 

 individuals (A p , B n ) and (A m , i? q ) which exist. But there is no cor- 

 responding pair (^ m , B n ) and (A p , i? q ). 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 vn. 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 de- 

 note any individual. Corresponding to every s with (a, b, c), there is 

 an a and (5, 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 c. 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 c. Hence, 

 for every s with s and with (a, b, c) there is an s with (a, b) and with 

 c. 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 arithmet- 

 ical addition and multiplication. For, since no m is n, every pair, one 



