530 Proceedings of the Royal Society of Edinburgh. [Sess. 
case c — 1 in (10), and the corresponding statement derived from the second 
form of (1) ; for this last is 
g = ^ 2 a< c) Sa( c >K^ . . . . . (13) 
and the particular cases of (10) and (13) mentioned are 
p = ^a_ 1 Sa_ 1 Kp = ^a_ 1 Sa_ 1 K|0 . . . (14) 
[That (13) corresponds to the second form of (1) follows easily by noting 
that this second form clearly gives q = 'E?xa^ ) and deducing x from (8).] 
From (10) § 2 it now follows at once that «_ l5 a_ 2 , . . . . must be the same 
functions of a_ x , d_ 2 , .... that the latter are of the former. 
We will now show that af is the same function of d_ 1} d_ 2 , .... that 
a ( c is of «_!, a_ 2 , . . . . , and therefore that d (c) may be defined as the same 
product of d_ 1} d_ 2 , .... that a {c} is of a_ 1} a_ 2 , .... [This statement would 
not be true if the K had been omitted from the right of (7) or the left of 
(12), though it would if K were replaced by Q. Instead, we should have 
had to add that the sequence in d^ ] was the opposite of that in a ® ; that is, 
QaJ. c) would have been the same function of Qa_ l5 Q a_ 2 , .... as a { c c) is of 
a_i, a_ 2 , . . . .] 
Let a (c-1) stand for any product of c — 1 of the c fictors constituting a ,c) 
in such sequence that if a is the c th fictor 
$c(a ( c C -i ] a) = a?, 
and let a be the d- a which corresponds to a according to (12). If we can 
show that 
S c (d[ c JT 1 1 *a) = a,y 
it will follow that aj. c) is derived from a_ 1} a_ 2 .... just as aj. c) is from 
a_i, a_ 2 , .... 
Now, putting for brevity 
we have from (7) 
Hence 
S M 1 a_ 1 a_ 2 . . . . a_ n =£, 
K ^ 1) | S - C +i aa(n " C ^ 
S e (KaKagn 1) ) = ( - ) c - 1 S c (S n _ 1 a< c -^a<— ^.S w _ c+1 aa<- c ^) 
= ( _ ) n(n-^2 Sc ( Sji _ ia (c-a(n- C , <Sn _ c+iaa («-c,) 
[by transposing ^ with S n _ c+1 ( ) and noticing that = scalar] 
= ( - ) n(n - c, ^ 2 S n a (c - 1, a (n - c) a.S n _ c a (M - c) [(10) § 5]. 
