Solving CNF using Prolog

Question

While learning Prolog, I tried to write a program solving CNF problem (the performance is not an issue), so I ended up with the following code to solve (!x||y||!z)&&(x||!y||z)&&(x||y||z)&&(!x||!y||z):

vx(t).
vx(f).
vy(t).
vy(f).
vz(t).
vz(f).

x(X) :- X=t; \+ X=f.
y(Y) :- Y=t; \+ Y=f.
z(Z) :- Z=t; \+ Z=f.
nx(X) :- X=f; \+ X=t.
ny(Y) :- Y=f; \+ Y=t.
nz(Z) :- Z=f; \+ Z=t.

cnf :-
   (nx(X); y(Y); nz(Z)),
   (x(X); ny(Y); z(Z)),
   (x(X); y(Y); z(Z)),
   (nx(X); ny(Y); z(Z)),
   write(X), write(Y), write(Z).

Is there any simpler and more direct way to solve CNF using this declarative language?

Answer 1

Consider using the built-in predicates true/0 and false/0 directly, and use the toplevel to display results (independently, instead of several subsequent write/1 calls, consider using format/2):

boolean(true).
boolean(false).

cnf(X, Y, Z) :-
        maplist(boolean, [X,Y,Z]),
        (\+ X; Y ; \+ Z),
        (   X ; \+ Y ; Z),
        (   X ; Y ; Z),
        (   \+ X ; \+ Y ; Z).

Example:

?- cnf(X, Y, Z).
X = true,
Y = true,
Z = true .

EDIT: As explained by @repeat, also take a serious look at CLP(B): Constraint Solving over Booleans.

With CLP(B), you can write the whole program above as:

:- use_module(library(clpb)).

cnf(X, Y, Z) :-
        sat(~X + Y + ~Z),
        sat(X + ~Y + Z),
        sat(X + Y + Z),
        sat(~X + ~Y + Z).

Please see the answer by @repeat for more information about this.