Download A SHORT PROOF FOR THE COMPLETENESS OF

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.
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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.
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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
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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]