25-1 PROCEEDINGS OF THE AMERICAN ACADEMY 



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



x ;y == 1 ;l ,x ,y -{- I ;0 ,x ,§ -\- ;l , x ,y -\- ;0 ,x ,y. 

 So that, by (14), 

 (20.) 1 ; 1 = 1 1;0=f[1;0] 0;1 = 0,0 ==v. 



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



In solving identities we must remember that 



(21.) (a-\rb)—b = a 



(22.) (a -7- b) -\r b == a. 



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



(23.) (a,b) -~ b==a 



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



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



Given the identity qp x == 0. 



Required to eliminate x. 



g,(l) ==x,<p(l) + (1— aO,g,(l) 



9>(0)== x, 9,(0) + (1-30,9,(0). 



Logically multiplying these identities, we get 



9 (1) , <P (0) =F a , <P (1) , 9> (0) + (1 — *) , <p (1) , 9> (0). 

 For two terms disappear because of (17). 

 But we have, by (18), 



(p C 1 ) > x + 9 (°) j C 1 — ») =F «JP a == 0. 

 Multiplying logically by # we get 



»(1),*===0 

 and by (1 — a?) we get 

 9 (0) , (1 ■— *) = 0. 

 Substituting these values above, we have 

 (25.) 9, (1) , <p (0) == when 9, x = 0. 



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



