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Narrowed it down a bit to an aggregate using ground versus non-ground count-equals.
foo(1) :- 1=#count{ na_1 : a; nb_1 : b }.
:- not foo(1).
na_1 :- not a.
a :- not na_1.
nb_1 :- not b.
b :- not nb_1.
The program above erroneously yields 1 answer set, while the variant below yields the correct 2 answer sets. The major difference is 1=#count{..} versus X=#count{..}.
foo(X) :- X=#count{ na_1 : a; nb_1 : b }.
:- not foo(1).
na_1 :- not a.
a :- not na_1.
nb_1 :- not b.
b :- not nb_1.
The following program unexpectedly has Alpha report 2 answer sets while there should be three.
Commandline:
-n 0 -str [above program]
Output:
Expected: a third answer set
{ a, nb_1, nc_1}
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