Logical Spectrum. 289 
from which 
2—1 pnd 7—0; 
hence 
A, =Ieand, B,=0. 
For the second term abcdef’ we get 
aty=1, #2—y=0; 
hence y is complementary to w and also identical with 2, 
which is impossible unless the term abcdef’ is null. 
For the fifth term abcd’ef we get 
TOY et 
but y cannot be negative unless abcd’ef is null. 
For the seventh term, 
re — to 
but the latter equation is impossible unless the term vanish. 
For the eighth term, 
2ru—1, -0—O; 
hence wz and y are indeterminate, but complementary to one 
another. In the case of the 58th term, we get x and y both 
indeterminate but identical with one another. These are 
interpretations which are not conceived of by Boole. 
For the nineteenth term, 
2p t=1; 
hence Aj is 1 and By, is indeterminate, that is may have any 
value between 0 and 1. In a similar manner, by solving 
each of the sixty-four sets of equations the several coefficients 
are found. In the last case only are both coefficients inde- 
terminate and independent of one another. 
I have exhibited the solution obtained by this analysis in 
the accompanying spectrum (diagram 9). Hach null term is 
shaded ; each term which is wholly included is white; each 
term which is totally excluded is black ; and each term which 
is indeterminate is partly black and partly white. The two 
complementary indeterminates have complementary parts 
white, and the two identical indeterminates have the same 
parts white. 
The solution is verified by finding whether the az together 
with the by is identical with the c, and whether the dz, ex- 
cepting the ey, is identical with the /; and by testing whether 
any other solution not comprehended as a particular case of 
that obtained would satisfy the conditions. 
