Here again the two methods diverge. The WZ method operates without a hitch as has already been shown by Zeilberger and his computer .
Namely, if a+f=-m and b+c+d+e=m+1, then
The WZ-method produces the certificate functionand it is then immediate that and consequently by summing (4.3) from r=0 to m+1 we find
In contrast, Pfaff again is unable to prove the required result by itself. In this instance, one is forced to prove a much less known companion identity due to A. Lakin .
Namely if a+f=-m and b+c+d+e=m, thenwhere is the third elementary symmetric function of a,b,c,d,e and f.
The Pfaff method now proceeds as before. In this case,and To conclude the simultaneous proof of Dougall's (4.1) and Lakin's (4.5) theorems, one verifies that the right-hand sides satisfy (4.6) and (4.7) also. In this section then, the WZ method has been fast and elegant. The Pfaff method has required the unearthing of an almost forgotten result . It has produced more but with twice the effort. If Lakin's result had been totally forgotten, considerable computer time would have been required to find it.
It should be pointed out here that Pfaff's method has been able to act with simpler recurrences in Sections 3 and 4 because it relies on shifts of parameters in addition to the shift of the index m.