Can prolog be used to determine invalid inference? - prolog

Question

If I have two premises as follows:
a -> c (a implies c)
b -> c (b implies c)
and a derived conclusion:
a -> b (a therefore implies b),
then the conclusion can be shown to be invalid because:
a -> c is valid for statement #1 when a is true and c is true, and
b -> c is valid for statement #2 when b is false and c is true. This leads to a -> b when a is true and b is false, a direct contradiction of statement #3.
Or, per proof with a truth table that contains a line for where the premises are true but the conclusion is false:
Truth Table with true premises and false conclusion
My question is: "Is there a way to use prolog to show that the assertions of statements #1 and #2 contradict the conclusion of statement #3? If so, what is a clear and concise way to do so?"

Answer 1

#coder has already given a very good answer, using clpb constraints.
I would like to show a slightly different way to show that the conclusion does not follow from the premises, also using CLP(B).
The main difference is that I post individual sat/1 constraints for each of the premises, and then use taut/2 to see whether the conclusion follows from the premises.
The first premise is:
a → c
In CLP(B), you can express this as:
sat(A =< C)
The second premise, i.e., b &rightarrow; c, becomes:
sat(B =< C)
If the a &rightarrow; b followed from these premises, then taut/2 would succeed with T = 1 in:
?- sat(A =< C), sat(B =< C), taut(A =< B, T).
false.
Since it doesn't, we know that the conclusion does not follow from the premises.
We can ask CLP(B) to show a counterexample, i.e., an assignment of truth values to variables where a &rightarrow; c and b &rightarrow; c both hold, and a &rightarrow; b does not hold:
?- sat(A =< C), sat(B =< C), sat(~(A =< B)).
A = C, C = 1,
B = 0.
Simply posting the constraints suffices to show the unique counterexample that exists in this case. If the counterexample were not unique, we could use labeling/1 to produce ground instances, for example: labeling([A,B,C]).
For comparison, consider for example:
?- sat(A =< B), sat(B =< C), taut(A =< C, T).
T = 1,
sat(A=:=A*B),
sat(B=:=B*C).
This shows that a &rightarrow; c follows from a &rightarrow; b ∧ b &rightarrow; c.

Answer 2

You could use library(clpb):
Firstly assign to a variable Expr your expression:
Expr = ((A + ~C)*(B + ~C)+(~(A + ~B)).
Note that:
'+' stands for logical OR
'*' for logical AND
'~' for logical NOT
Also A->B is equivalent with A+(~B). So the above expression is equivalent with ((A->C),(B->C))-> (A->C) where we wrote '->' using +,~ and ',' with *.
Now if we query:
?- use_module(library(clpb)).
true.
?- Expr=((A + ~C)*(B + ~C))+(~(A + ~B)),taut(Expr,T).
false.
Predicate taut/2 takes as input an clpb Expression and returns T=1 if it tautology, T=0 if expression cannot be satisfied and fails in any other case. So the fact that the above query failed means that Expr was nor a tautology neither could not be satisfied, it meant that it could be satisfied.
Also by querying:
?- Expr=((A + ~C)*(B + ~C))+(~(A + ~B)),sat(Expr).
Expr = (A+ ~C)* (B+ ~C)+ ~ (A+ ~B),
sat(1#C#C*B),
sat(A=:=A).
Predicate sat/1 returns True iff the Boolean expression is satisfiable and gives the answer that the above Expression is satisfiable when:
sat(1#C#C*B),
sat(A=:=A).
where '#' is the exclusive OR which means that your expression (we know from taut/2 that is satisfiable) is satisfied when 1#C#C*B is satisfied.
Another solution without using libraries could be:
truth(X):-member(X,[true,false]).
test_truth(A,B,C):-
truth(A),
truth(B),
truth(C),
((A->C),(B->C)->(A->C)).
Example:
?- test_truth(A,B,C).
A = B, B = C, C = true ;
false.
Also if I understood correctly from your comment, to collect all possible solutions you could write:
?- findall([A,B,C],test_truth(A,B,C),L).
L = [[true, true, true]].
Which gives a list of lists, where the inner lists have the form [true,true,true] in the above example which means a solution is A=true,B=true,C=true and in the above case it has only one solution.
To find all contradictions you could write:
truth(X):-member(X,[true,false]).
test_truth(A,B,C):-
truth(A),
truth(B),
truth(C),
not( (\+ ((\+A; C),(\+B ; C)) ; (\+A ; B)) ).
last line could also been written as:
not( ( (A->C;true),(B->C;true) ) -> (A->B;true) ;true ).
Example:
?- findall([A,B,C],test_truth(A,B,C),L).
L = [[true, false, true]].