Prolog First Order Logic - Printing a Truth Table

Question

I have to write program that prints a truth table of expressions. So, I wrote the following function:

bool(true).
bool(fail).

tableBody(A,B,E) :-
    bool(A),
    bool(B) ,
    write(A) ,
    write('    '),
    write(B),
    write('    '),
    write(E),nl, fail.

My problem is that E (wich is expression that contains A and B) is not evaluated, but printed as is. For example:

296 ?- table(A,B,and(A,B)).
A    B    expr(A,B)
true    true    and(true, true)
true    fail    and(true, fail)
fail    true    and(fail, true)
fail    fail    and(fail, fail)
false.

I am interested to write the evaluated value of and(true, true) ("and(X,Y)" is a functor I defined earlier) instead of what is currently displayed. I thought about writing an eval functor, but would not it have the same effect? How can I solve this?

I am using SWI-Prolog 5.8. Thank you.

Answer 1

Here's one way to do it:

and(A, B) :- A, B.

evaluate(E, true) :- E, !.
evaluate(_, false).

bool(true).
bool(false).

tableBody(A,B,E) :-
  bool(A),
  bool(B),
  write(A),
  write(' \t '),
  write(B),
  write(' \t '),
  evaluate(E, Result),
  write(Result),nl, fail.

Produces:

?- tableBody(A,B,and(A,B)).
true    true    true
true    false   false
false   true    false
false   false   false
false.

Answer 2

As usual, one-liner here

?- forall((member(A,[true,false]),member(B,[true,false]),(A,B->C=true;C=false)),format('~w|~w|~w~n',[A,B,C])).
true|true|true
true|false|false
false|true|false
false|false|false