942] ON SEMIN VARIANTS. 365 
where to explain the algorithm, I remark, that if 
A=(A 0 , A 1} A 2 ,...^x, y) 5 and D = (D 0 , D lf D 2 ,...\x, y) 3 , 
then 
(A, D)- = B 0 A, - 2A^i + D,A 0 , 
represented as above by 
/ a /)N2 _ f A » A, A) ice - c 3 , cf- c 2 d, df- cd?\ 
1 ’ V) \A a , A 0 , l]"! c , b , 1 }' 
The result is in every case given as df — c 4 ; in each case there is only a single 
term c.c 3 , = c 4 , and the term in c 4 certainly presents itself. In (A, AB) 4 there is a 
single term d.f, =df and in (A, _D) 2 a single term df, and thus the term df certainly 
presents itself: in (yl, E) 3 there are two terms d.f, =df and df and it is conceivable 
that, inserting the proper numerical coefficients, these might destroy each other : if 
this were so, the form instead of being df—c i would be e 2 — c 4 ; and similarly in 
(A, F) 5 , there are two terms df = df and df which it is conceivable might destroy 
each other, and the form would then be e 2 — c 4 . But in every case we have the term 
c 4 , and it thus appears that the form df — b 2 d 2 is not obtainable by mere derivation. 
The form in question is in fact obtained by a linear combination of df—c 4 and 
e 2 — c 4 , viz. writing down the leading coefficients of the covariants B' 2 and H, we have 
37/ - 2B 2 = 
df 
3 
3 
e 2 
0-2 
- 2 
bcf 
- 9 
- 9 
bde 
-15 +16 
+ 1 
c 2 e 
+ 30 - 12 
- 18 
cd 2 
- 12 
- 12 
b*f 
+ 6 
+ 6 
b 2 ce 
- 15 
- 15 
bW 2 
+ 42 - 32 
- 10 
bc 2 d 
- 48 +48 
0 
c 4 
+ 18 - 18 
0 
viz. the form in question df—b 2 d? is = Sdf— 2e 2 — ... — 10b 2 d 2 .