Why this program answers False in SWI-PROLOG?
sor(x, y):- sorted(y), perm(x, y).
sorted([]).
sorted([x, []]).
sorted([x, y, z]):- mi(x, y), sorted([y, z]).
perm([], []).
perm([x,y],[u,v]):- delete(u,[x,u],z), perm(z,v).
delete(x,[x,y],y].
delete(x, [y, z], [y, w]):- delete(x,z,w).
mi(0, x).
mi(s(x), s(y)):- mi(x, y).
for the query ?-
sor([s(s(s(s(s(0))))), s(s(s(s(s(s(0)))))), s(s(s(0))), s(s(0)), []], y).
This is an adaptation to SWIProlog of an inefficient sorting-program used as example in the book Foundations of Logic Programming, by Loyd (you can find the original SLOWSORT program example in this pdf, on page 9)
SWI Prolog is a standard Prolog, isn't it?
Edit
Now I have tried to correct the program (looking a little to the lists syntax in Prolog)
sor(X, Y):- perm(X, Y), sorted(Y).
sorted([]).
sorted([X|[]]).
sorted([X|[Y|Z]]):- mi(X, Y), sorted([Y|Z]).
perm([], []).
perm([X|Y],[U|V]):- delete(U,[X|Y],Z), perm(Z, V).
delete(X,[X|Y],Y).
delete(X, [Y|Z], [Y|W]):- delete(X, Z, W).
mi(0, X).
mi(s(X), s(Y)):- mi(X, Y).
and changing the query in
sor([s(s(s(s(s(0)))))|[ s(s(s(s(s(s(0))))))|[s(s(s(0)))|[ s(s(0))|[]]]]], Y).
Well, Prolog now gives success, but it gives this substitution
Y = [s(s(0)), s(s(s(0))), s(s(s(s(s(0))))), s(s(s(s(s(s(...))))))]
and I don't understand the meaning of (...): Why not (0)?
Edit2
I notice that after giving the command swipl -s slowsort.pl I obtain this error message
Warning: /home/navigazione/Scrivania/slowsort.pl:3:
Singleton variables: [X]
Warning: /home/navigazione/Scrivania/slowsort.pl:9:
Singleton variables: [X]
It seems to refer to 3th and 9th rows of the program, but I don't understand what it means.