442 Proceedings of Eoyal Society of Edinburgh, [july 18 , 
and for a(l, 2, q , r - 2) by considering 
f 0, y + S + 1 f 1, /? + y + S 
„ _ , ! l,y + S along with each I 
eacn of the duads < ' < 
of the triads | 
S, y + 1 1^1, ^ — 1, /3-hy/ + S — <y + 3. 
There is more trouble here in comparing the number of dovetail- 
ing^ in the case of a( 0, 1, q, r) with the first set in the case of 
a( 0, 2, q, r- 1), but the result is the same as before, viz. x" and 
x" — 1 ; so that the coefficient is again 1. 
And thus the theorem is established. 
The case where q = r — 1 is important, as it furnishes a result 
similar to those of the first two theorems. Putting <? = 2 + s and 
r = 3 + s we obtain 
ct( 0, 1, 2 + s, 3 + s) = + [] + n(0, 2, 2 + s, 2 -f- s) — ct(l, 2, 2 + s, 1 + s) 
— + ] -f- 0 + ct(l 5 2, 1 + s, 2 + s) 
= [5a s 6 s + ] + abed . a(0, 1, s, 1 + s) ; 
and as the a( ) on the right is the same function of s, as the 
a( ) on the left is of s + 2, we see that by repeated application 
of the result, the full expansion of a(0, 1, 2 + s, 3 + 5 ) in terms of 
symmetric functions is obtainable. We have, in fact, 
a(0, 1, 2 + s, 3 + s) 
- + ] + abcd{\%a s ~ 2 b s ~ 2 + ] + abed . a(0, l,s-2, s-1)} 
= + ] + [^a s_1 5 s-1 c^ + ] + \^a s ~ 2 b s ~ 2 c 2 d 2 + ] + .... (iv.) 
For example, 
a(0, 1, 6. 7 ) = \%a 4 b 4 + ] + \_^a 3 bhd + ] + \^a 2 b 2 c 2 d 2 + ], 
= 2a 4 b 4 + 'Zct 4 b 3 c + %a 4 b 2 c 2 + 2a 4 b 2 cd + 2 a 3 b 3 c 2 + 2,a 3 b 3 cd + ~2a 3 b 2 c 2 d + %a 2 b 2 c 2 d 2 \ 
+ 2a 3 b 3 cd + ~Za 3 b 2 c 2 d + '2 i d?b 2 (?d 2 
+ 3,a 2 b 2 c 2 d 2 , J 
= 2a 4 b 4 + 2a 4 b 3 c + 2 a 4 b 2 c 2 + 2 a 4 b 2 cd + %(t 3 b 3 c 2 + 2 '2a 3 b 3 cd + 2'Xa 3 b 2 c 2 d + 3 2a 2 b 2 c 2 
The fifth theorem is, in symbols, 
a(0, 2, 2+s, 3 + s) - a(l, 2, 1 + s, 3 + s) = \%a s b s c + ] . . (v.) 
The only point of the proof which requires care is where the 
dovetailings of the 
