OF ARTS AND SCIENCES: MARCH 12, 1867. 253 



It is plain that if for the moment we allow a:b to denote the maxi- 

 mum value of a;b, then 



(16.) X == 1 — X == :x. 



So that 



(17.) x,(l — x) = x-jrO:x = l. 



The rules for the transformation of expressions involving logical 

 subtraction and division would be very complicated. The following 

 method is, therefore, resorted to. 



It is plain that any operations consisting solely of logical addition 

 and multiplication, being performed upon interpretable symbols, can 

 result in nothing uninterpretable. Hence, if (jd -}- X ^ signifies such 

 an operation performed upon symbols of which x is one, we have 



cp-{-Xx = a,x-\-b,(l — x) 



where a and b are interpretable. 



It is plain, also, that all four operations being performed in any 

 way upon any symbols, will, in general, give a result of which one 

 term is interpretable and another not ; although either of these terms 

 may disappear. "We have then 



cpx==i,x-{-J , (1 —x). 



We have seen that if either of these coefficients i and j is unin- 

 terpretable, the other factor of the same term is equal to nothing, or 

 else the whole expression is uninterpretable. But 



<P (1) =F * and cjp (0) =/ 

 Hence 

 (18.) q>x==<p{l),x-\-cp (0) ,(l—x) 



tp (x and y) = qp (1 and l),x ,y -\- cp (1 and 0) , x,y -\-(p (0 and 1) , x , y 



-}- 9 (0 and 0) ,x ,y. 



(18'.) cpx=(cpil)-\rx), (cp (0) + x) 



cp (x and y) == (qp (1 and 1) -\r x -\r §) , ((p (I and 0) -\rx-\ri/), 

 {<p (0 and 1) + :r -jr y) ,(9 (0 and 0) -jr a; -}r y). 



Developing by (18) a: — y, we have, 



x — y==(l — l),a;,y+(l — 0),x,§-{- (0 -r l),x,y+(0 -^ 0),x,y. 



So that, by (11), 



(19.) (l^l) = y 1-0 = 1 - 1===[0 -,-1] 0-0 = 0. 



