254 PROCEEDINGS OF THE AMERICAN ACADEMY 



Developing x;y in the same way, we have * 



x;y==l;l,x,y-\-l;0,x,y-{-0;l,x,y-\-OiO,x,§. 

 So that, by (14), 

 (20.) 1;1=f1 1;0==[1;0] 0;1 = 0^0 = 1;. 



Boole gives (20), but not (19). 



In solving identities we must remember that 



(21.) (a-\rb) — b=za 



(22.) (a -r b) -{r b == a. 



From a -r b the value of h cannot be obtained. 



(23.) {a ,b) -^ b =z a 



(24.) a;b ,b=pa. 



From a ; b the value of b cannot be determined. 



Given the identity gj a; = 0. 



Required to eliminate x. 



cf (I) = X ,cf {I) + {I - x) ,cp {\) 



cf (0) =x ,cp {0) -^ {I - x) ,q> {0). 



Logically multiplying these identities, we get 



g. (1) ,9, (0) =^,9 (1)^9 (0) + (1 — x) ,9, (1) ,g, (0). 



For two terms disappear because of (17). 



But we have, by (18), 



(p 0) ,x-\- cf>{0),{l—x) = q>X = 0. 



Multiplying logically by x we get 



(p{\),x = Q 



and by (1 — x) we get 



q> (0) , (1 _ x) = 0. 



Substituting these values above, we have 



(25.) <P (1) J <? (0) = when (f>x = 0. 



* a;b ,c must always be taken as (a;b) , c, not as a ; (6 , c). 



