40 Thus from what was said in (70), we find BOOK ONE N [(u A , x ) B , y . . . n] T = N (u . . . n) T - N (u . . . n) T -A-1 - N (u . . . n) +N (u . . . n) T -B-2 - N (u) A+ A -B-1 A T - A -1 × N (u . . . n - 1) T -B-1 . Corollary (76.) If A + A < B, then u and x would not reach the dimension B to- gether; rather the highest dimension they can reach is A+ A . The expression we just found would thus not be exact. But if we replace B by A + A , we obtain N [(u A , x ) B , y . . . n] T A T - A -1 = N (u . . . n) T - N (u . . . n) T -A-1 - N (u . . . n) T -A- A -2 +N (u . . . n) when A + A < B; this requires observing that N (u) -1 = 0, since in general N (u) A+ A -B-1 = A + A - B. Problem XIII We ask for the value of N ((u A , x ) B , y . . . n) T , when this polynomial satisfies the conditions given in (74). (77.) In this polynomial, all terms that may be divided by y , z , etc., are missing. But under our assumptions, the absence of terms that may be divided by y A +1 A +1 A +1 A A , for example, does not imply the absence of terms that A +1 may be divided by z ; this argument is valid for all other unknowns; thus we conclude, as we have already done in (59), that the number of missing terms in y is N (u . . . n) T - A -1 T - A -1 , that the number of missing terms in z, is N (u . . . n) , and so on. Therefore, and from what was said about the preceding problem, we conclude that N [(u A , x ) B , y . . . n] T = N (u . . . n) T - N (u . . . n) T -A-1 -N (u . . . n) T - A -1 T -B-2 A A T - A -1 T - A -1 - N (u . . . n) - N (u) - N (u . . . n) , etc. ; +N (u . . . n) A+ A -B-1 × N (u . . . n - 1) T -B-1 We abbreviate this expression by writing it as N (u . . . n) T - N (u . . . n) T -A-1 , etc. +N (u . . . n) T -B-2 - N (u) A+ A -B-1 × N (u . . . n - 1) T -B-1 .