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From: feldman@hermod.cs.cornell.edu (Ronen Feldman)
Newsgroups: comp.lang.prolog,comp.lang.functional,comp.lang.lisp
Subject: Correction to mutual recursion elimination query
Message-ID: <43273@cornell.UUCP>
Date: 13 Jul 90 02:26:12 GMT
Sender: nobody@cornell.UUCP
Reply-To: feldman@cs.cornell.edu (Ronen Feldman)
Organization: Cornell Univ. CS Dept, Ithaca NY
Lines: 24

In my former message I claimed that program2 was equivalent to program1 for 
every X .
As Brendan Mahony pointed out to me if for some X :
f1(X) and f2(X) are true but f3(X) is false then 
a(X) is true but a1(X) is false.

I have to change the declaration of a1 to be:  

a1(X) :- f1(X), f2(X).
a1(s(X)) :- b1(X).

b1(X) :- f1(X), f2(X), f3(X).
b1(s(X)) :- b1(X).

I think that now a1(X) = a(X) for every X. If we want a1(X) to be 
equivalent to b(X) or c(X) the first clause will have to be changed 
to a1(X) :- f2(X), f3(X) or a1(X) :- f3(X), f1(X) accordingly.

Sorry about that

The query still holds, can anybody suggest an automatic way to get
the new declaration?

Ronen (feldman@cs.cornell.edu)