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Bulletin of the Section of Logic Volume 39:3/4 (2010), pp. 147–152 Gábor Sági∗ A SHORT PROOF FOR THE COMPLETENESS OF PARAMODULACION Abstract In this note we provide a short and elementary proof for the well known fact that first order resulution extended with paramodulation is a sound and refutation complete calculus. Keywords. Automated theorem proving, resolution, paramodulation. 1. Introduction The resolution calculus is a classical tool for deciding if a given set of first order clauses are satisfiable or not. We refer to [2] as a standard monograph of the subject. Although the resolution calculus is comprehensive, elegant and complete, in its original form does not handle the equation symbol in an efficient way. In order to increase it’s efficiency in this respect, Robinson and Wos extended it by a new inference rule called paramodulation. In [4] they proved that this extended calculus is sound and refutation complete over the so called E-interpretations (E refers to the special treatment of the equality symbol; see Chapter 8.2 of [2] for the details). In the present paper we provide a short and elementary proof for completeness of first order paramodulation, see Theorem 2.7. Although the result is well known, we believe our new proof may provide some additional insight into the notions involved: it is essentially shorter than the ones in [3] and in [1], moreover, the main steps of usual completeness proofs (for other logics) can be recognized in it. ∗ Supported by Hungarian National Foundation for Scientific Research grant K68262. 148 Gábor Sági In order to keep this note as short as possible, here we do not deal with extensions of paramodulation which make it more efficient. In particular, we do not analyze how to eliminate the functionally reflexive axioms. We assume that the reader is familiar with the theory of resolution calculus. For completeness, we close this section by fixing some notation. Throughout, ↑ denotes “true” and ↓ denotes “false”. As usual, a clause containing exactly one literal is called a unit clause. Atomic formulas are also called positive literals and a literal will be called negative iff it is a negation of an atomic formula. • Σ always denotes a set of first order clauses; Σ∗ = Σ ∪ {x = x}; • Σ(A) is the set of ground instances of Σ over the set A; • H(Σ) is the Herbrand Universe of Σ (if Σ is clear from the context, we omit it); • `P means “can be derived by paramodulation (and resolution)”. 2. Completeness of First Order Paramodulation We start by recalling the notion of paramodulation. Definition 2.1. Suppose L∨C1 and [e = f ]∨C2 are ground clauses. Then L[e ← f ] ∨ C1 ∨ C2 is called the binary paramodulant of the second clause into the first one, where L[e ← f ] is obtained from L by replacing an occurrence of e by f in it. If C1 , C2 and C3 are first order clauses such that C3 is the binary paramodulant of suitable factors of C1 and C2 over their Herbrand universe, then C3 is defined to be a first order paramodulant of C1 and C2 . Our goal in the rest of this paper is to provide a short proof that first order resolution combined with first order paramodulation yields a sound and refutation complete calculus over E-interpretations (we assume the reader is familiar with this notion, for the details see Chapter 8.2 of [2]). To achieve this goal, we need some further preparations. Definition 2.2. Let Γ be a set of ground clauses. Then ded(Γ) = {C : Γ `p C} A Short Proof for the Completeness of Paramodulacion 149 is the set of ground clauses derivable from Γ by paramodulation and resolution. Γ is defined to be closed iff Γ = ded(Γ); Γ called consistent iff the empty clause is not an element of Γ. We say, that Γ is maximal iff it is closed, consistent and there does not exists a closed consistent set Γ0 of ground clauses with Γ ⊂ Γ0 , Γ 6= Γ0 . Lemma 2.3. Suppose Γ+ is a maximal (closed, consistent) set of ground clauses and C ∈ Γ+ . Then there is a literal L in C with L ∈ Γ+ . Proof: Suppose, seeking a contradiction, that the set E = {C ∈ Γ+ : C does not have a literal in Γ+ } is not empty. Choose C ∈ E such that the number of literals of C is as small as possible (that is, every element of E contains at least as many literals as C does). Observe, that C is not the empty clause, because Γ+ is consistent. Moreover C is not a unit clause , because the statement of the lemma trivially holds for unit clauses. So C has at least two different literals, let L be one of them. Let Γ++ = ded(Γ+ ∪ {L}). Since C ∈ E, it follows, that Γ+ ⊂ Γ++ and + Γ 6= Γ++ . Clearly, Γ++ is closed. Hence, by maximality of Γ+ , we have that Γ++ is inconsistent. Consequently, the empty clause may be derived from Γ+ ∪ {L}; let p0 , ...pn be such a derivation. Define the clauses p0i by recursion on i as follows: • if pi ∈ Γ+ then let p0i = pi ; • if pi = L then let p0i = C; • if pi is obtained by resolution or by paramodulation from pj and pk (k, j < i), then let p0i be the appropriate resolvent or paramodulant of p0j and p0k . Identifying clauses with the set of their literals, it is easy to show by induction on i that (∗) pi ⊆ p0i and p0i − pi ⊆ C − {L}. Therefore the third stipulation in the construction of p0i really makes sense. By construction, p00 , ..., p0n is a derivation from Γ+ and p0n ⊆ C − {L}. But then, closedness of Γ+ implies p0n ∈ Γ+ . In addition, (∗) implies that p0n is strictly shorter than C, hence p0n ∈ E. This contradicts to the choice of C. 150 Gábor Sági Theorem 2.4. If {x = x}(H) ⊆ Γ is a consistent set of ground clauses then Γ is E-satisfiable. Proof: Clearly, ded(Γ) is closed and consistent, hence by Zorn’s lemma there exists a maximal set Γ+ of ground clauses with Γ ⊆ Γ+ . Now we define an interpretation I as follows. Let P be an arbitrary atomic formula. • If P ∈ Γ+ then let I(P ) =↑; • if ¬P ∈ Γ+ , then let I(P ) =↓; • if P, ¬P 6∈ Γ+ then let I(P ) =↓. Since Γ+ is consistent, there is no atomic formula P for which P, ¬P ∈ Γ , hence I is well-defined. We have to show that I is an E-interpretation satisfying Γ+ (and consequently satisfying Γ, as well). Let C ∈ Γ+ be arbitrary. Then by Lemma 2.3, there exists L ∈ C with L ∈ Γ+ . So, according to the first two stipulations of definition of I, we have I |= C. Since C was an arbitrary element of Γ+ , it follows, that I |= Γ+ . Finally we show that I is an E-interpretation. Let t, s, r be ground terms in the Herbrand Universe and let L be an arbitrary literal. Assume, in addition, that L0 may be obtained from L by replacing an occurrence of t by s. Since {x = x}(H) ⊆ Γ ⊆ Γ+ , we have I(t = t) =↑. Moreover, if I(t = s) =↑, then t = s ∈ Γ+ , because if a positive literal is true in I then it should be an element of Γ+ . Since {t = s, t = t} `p s = t, closedness of Γ+ implies s = t ∈ Γ+ so I(s = t) =↑. Similarly, if I(t = s) = I(s = r) =↑ then t = s, s = r ∈ Γ+ and {t = s, s = r} `p t = r, that is, by closedness of Γ+ we have t = r ∈ Γ+ . So I(t = r) =↑. If I(t = s) = I(L) =↑ and L ∈ Γ+ then t = s, L ∈ Γ+ . Moreover, {t = s, L} `p L0 , so closedness of Γ+ implies L0 ∈ Γ+ . Therefore I(L0 ) =↑. Finally assume I(t = s) = I(L) =↑, but L 6∈ Γ+ . Then, by construction of I, ¬L 6∈ Γ+ also holds. Hence I(L) =↑ implies that L is a negated atomic formula. Observe, that if L0 ∈ Γ+ (respectively, if ¬L0 ∈ Γ+ ) then Γ+ `p L (respectively Γ+ `p ¬L) would follow. Hence, closedness of Γ+ implies L0 , ¬L0 6∈ Γ+ . Hence I makes true L0 which contains a negation symbol. So I(L0 ) =↑ holds in this case, as well. + A Short Proof for the Completeness of Paramodulacion 151 Theorem 2.5. (Completeness of Ground Paramodulation.) Let Σ be a set of first order clauses. Then the following statements are equivalent. (1) Σ∗ (H) is not E-satisfiable. (2) The empty clause may be derived from Σ∗ (H) by paramodulation and resolution. Proof: (2) ⇒ (1) is the “soundness” part; this may be established by a straightforward induction on the length of the derivation supposed to exist in (2). The converse implication (1) ⇒ (2) is a reformulation of Theorem 2.4. Definition 2.6. Let Σ be a set of first order clauses. Then Σ∗∗ is defined to be the following set of first order clauses Σ∗∗ = Σ∗ ∪{ f (x1 , ..., xn ) = f (x1 , ..., xn ) : f is an n-ary function symbol }. Here x1 , ..., xn are individuum variables; n may be equal to 0, in this case the equality of the corresponding constant symbols should have been taken into Σ∗∗ . In [2], the second part of the above union in Definition 2.6 is called functionally reflexive axioms. Theorem 2.7. (Completeness of First Order paramodulation.) Let Σ be a set of first order clauses. Then the following statements are equivalent. (1) Σ∗ is not E-satisfiable. (2) The empty clause can be derived from Σ∗∗ by first order resolution and paramodulation. Proof: The implication (2) ⇒ (1) is the “soundness” part; it may be easily established by observing Σ∗ |= Σ∗∗ . We turn to show the converse implication (1) ⇒ (2). Let t be an arbitrary ground term over H. One can show by a straightforward induction on the complexity of t, that t = t can be derived from Σ∗∗ by first order paramodulation. In addition, we claim, that every ground instance (over H) of any element of Σ∗ can be derived from Σ∗∗ by first order paramodulation. Indeed, assume C ∈ Σ∗ (H). Then C is a ground instance over 152 Gábor Sági H of some C 0 ∈ Σ∗ . For every ground term t occurring in C we have Σ∗∗ `P t = t, paramodulating these terms consecutively into the appropriate terms of C 0 we get C. If, according to (1), Σ∗ is not E-satisfiable, then by Theorem 2.5 the empty clause can be derived from Σ∗ (H) by paramodulation and resolution. But then, (2) follows from the previous paragraph. References [1] L. Bachmair, H. Ganzinger, Rewrite-based equational theorem proving with selection and simplification, J. Logic and Computation 4 (1994), pp. 217–247. [2] C. L. Chang, R. C. Lee, Symbolic Logic and Mechanical Theorem Proving, Academic Press, London, 1973. [3] G. Pais, G. E. Peterson, Using Forcing to Prove Completeness of Resolution and Paramodulation, J. Symb. Comput. 11 (1991), pp. 3–19. [4] G. Robinson, L. Wos, Paramodulation and Theorem Proving in First Order Logic with Equality, [in:] Machine Intelligence 4 (1969) pp. 135–150. BUTE Department of Algebra Budapest, Egry J. u. 1 H-1111 Hungary e-mail: [email protected] Alfréd Rényi Institute of Mathematics Hungarian Academy of Sciences Budapest Pf. 127 H-1364 Hungary e-mail:[email protected]