Artificial Intelligence and Soft Computing: Behavioral ... - Arteimi.info

Definition 5.10: A proof process is called complete, if for any inference
α, that follows logically from a given set of axioms S, i..e., S α, the proof
procedure can prove α, i.e., Sα.
Theorem 5.1: The resolution theorem is sound.
Proof: Given a set of clauses S and a goal α. Suppose we derived α from S
by the resolution theorem. By our usual notation, we thus have S α. We
want to prove that the derivation is logically sound, i.e., S α. Let us prove
the theorem by the method of contradiction. So, we presume that the
consequent S α is false, which in other words means S ¬ α. Thus ¬ α is
satisfiable. To satisfy it, we assign truth values (true / fa se) to all
propositions that are used in α. We now claim that for such assignment,
resolution of any two clauses from S will be true. Thus the resulting clause
even after exhaustion of all clauses through resolution will not be false. Thus
S α is a contradiction. Hence, the assumption S ¬ α is false, and
consequently S α is true. This is all about the proof [5].
Theorem 5.2: The resolution theorem is complete.
Proof: Let α be a formula, such that from a given set of clauses S, we have
S α, i.e., α can be logically proved from S. We have to show there exists a
proof procedure for α, i.e., S α.
We shall prove it by the method of contradiction, i.e. let S α not
follow, i.e., S ¬α. In words α is not derivable by a proof procedure from S.
Therefore, S1 S ∪ α is unsatisfiable. We now use an important theorem,
called the ground resolution theorem, that states “if a set of ground clauses
(clauses with no variables) is unsatisfiable, then the resolution closure of those
clauses contains the ‘false’ clause. Thus as S1 is unsatisfiable, the resolution
closure of S1 yields the null clause, which causes a contradiction to S α.
Thus the assumption is wrong and hence S α is true.
We now prove the ground resolution theorem, stated below.
Theorem 5.3: If a set of ground clauses S is unsatisfiable, then the
resolution closure T of those clauses contains the false clause.
Proof: We prove the theorem by the method of contradiction. So, we
presume that resolution closure T does not contain false clause and will
terminate the proof by showing that S is satisfiable.