CODEWITHSUNDEEP
prolog
Qwerto
Qwerto

Reputation: 275

Define AND, OR, NOT operators in Prolog

I have to define a prolog program which gives the truth table for a logic formula like this:

(a or non (b and c))

where the logic variables can only have true or false value, and the only operators are AND,OR and NOT. The program should behave like this:

table(a and (b or non a)).

[a, b]
[v, v] v
[v, f] f
[f, v] f
[f, f] f
yes

What i did is define the 3 operators ,but i dont know how to continue. Can you help me ?

:- op(30,fx,non).
:- op(100,xfy,or).
:- op(100,xfy,and).

Upvotes: 0

Views: 10583

Answers (1)

coredump
coredump

Reputation: 38967

Not going for the full solution, but here are some hints.

Basic approach

% fact: truth value "v" is satisfiable in all environments.
sat(v,_).

% rule: and(X,Y) is satisfiable in environment E iff both X and Y are sat in E
sat(and(X,Y),E) :- sat(X,E), sat(Y,E).

Bindings

sat(Var, E) :- 
  (member(Var:Value,E) -> 
    Value = v 
  ; throw(unknown_variable(Var,E))).

Example:

[eclipse 6]: sat(o,[o:v]).

Yes (0.00s cpu)
[eclipse 7]: sat(o,[o:f]).

No (0.00s cpu)
[eclipse 8]: sat(o,[u:v]).
uncaught exception in throw(unknown_variable(o, [u : v]))
Abort

Enumeration

Define a rule (binding) which binds a variable to a value non-deterministically, and another one (bindings) which binds a list of symbols (atoms) to a list of bindings.

% Two different solution possible when binding Var
binding(Var, Var:v).
binding(Var, Var:f).

% Lists of bindings
bindings([],[]).
bindings([V|VL],[B|BL]) :-
  binding(V,B), 
  bindings(VL,BL).

For example:

[eclipse 9]: bindings([a,b,c],L).

L = [a : v, b : v, c : v]
Yes (0.00s cpu, solution 1, maybe more) ? ;

L = [a : v, b : v, c : f]
Yes (0.00s cpu, solution 2, maybe more) ? ;

L = [a : v, b : f, c : v]
Yes (0.00s cpu, solution 3, maybe more) ? ;

L = [a : v, b : f, c : f]
Yes (0.00s cpu, solution 4, maybe more) ? ;

L = [a : f, b : v, c : v]
Yes (0.00s cpu, solution 5, maybe more) ? ;

L = [a : f, b : v, c : f]
Yes (0.00s cpu, solution 6, maybe more) ? ;

L = [a : f, b : f, c : v]
Yes (0.00s cpu, solution 7, maybe more) ? ;

L = [a : f, b : f, c : f]
Yes (0.00s cpu, solution 8)

Operators

First, you could declare the following and predicate:

and(0,0,0).
and(1,0,0).
and(0,1,0).
and(1,1,1).

The rule can be applied as and(X,Y,R), and R is the result of the and operation. The same can be done with or, etc.

Your declaration:

:- op(100,xfy,and).

... allows to write X and Y instead of and(X,Y), but notice how there is no third argument here. With the ECLiPSe environment, the operator notation is also used in conjunction with is/2 to evaluate arithmetic expressions. Since the above add predicate deals with numbers, the following works:

X is 0 and 1.

The above unifies X with 0.

Upvotes: 1

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