CHAP. IX.] METHODS OF ABBREVIATION. 147 



Let us represent the property A by x, B by y, C by z, D by 

 iv, E by v. The data are 



~xz = qv (yw + wy)\ (1) 



vxw = q (yz + y~z)', (2) 



a^ + #i># = t0J + ZMJ; (3) 



a? standing for 1 - #, &c., and ^ being an indefinite class symbol. 

 Eliminating q separately from the first and second equations, 

 and adding the results to the third equation reduced by (5), 

 Chap.'vm., we get 



xz (1 - vyw - vwy) + vxw (yz + zy) + (xy + xvy) (wz + wz) 



+ (wz + zw) (\-xy- xvy) = 0. (4) 



From this equation v must be eliminated, and the value of x 

 determined from the result. For effecting this object 9 it will 

 be convenient to employ the method of Prop. 3 of the present 

 chapter. 



Let then the result of elimination be represented by the 

 equation 



To find E make x = 1 in the first member of (4), we find 

 vw (yz + zy) + (y + vy) (wz + wz) + (wz + zw) vy. 

 Eliminating u, we have 



(wz -f wz) {w (ylz + zy) + y (wz + wz) + y (wz + zw) } ; 



which, on actual multiplication, in accordance with the conditions 

 ww = 0, zz = 0, &c., gives 



Next, to find E make x = in (4), we have 



z (1 - vyw - vyw) + wz + zw. . 

 whence, eliminating v, and reducing the result by Propositions 



1 and 2, we find 



E' = wz + zw + ywz; 



and, therefore, finally we have 



(wz + ywz)x + (wz + zw+ywz)x = Q; (5) 



from which 



L2* 



