408 PROCEEDINGS OF THE AMERICAN ACADEMY 



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 ?i, 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 multiplica- 

 tion, 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 a b == c and a' b == c, then a == a', or no b exists. 



This does not hold with logical, but does with arithmetical multipli- 

 cation. 



For if a is not identical with a', it may be divided thus 



a == a, a' -\- a, a' 

 if a' denotes not a'. Then 



a , b = (a , a') , b -f {a , a 1 ) ,b 



and by the definition of independence the last term does not vanish 

 unless (a , a') == or all a is a' ; but since a,b = a' ,b= (a , a') , b -(- 

 (a , a') , b, this term does vanish, and, therefore, only a is a 1 , and in a 

 similar way it could be shown that only a 1 is a. 



XI. 



l + a=l. 



This is not true of arithmetical addition, for since by definition 7, 



1 X, 1 ===SG 1 



