In this section we present our calculus for labeled clauses LPSup. We continue building on the idea that labeled clauses represent standard clauses from the “ground level” of deciding the existence of (K, L)-models, and show how to “lift” the operation of PSup, a sound and complete calculus for the ground proof tasks, and abbreviate it into a single saturation process on the level of labeled clauses.
We make sure the new calculus retains the valuable features of its originator including ordering restrictions for inferences and an abstract redundancy criterion for justifying concrete reductions. We show that the inference rules of the new calculus are sound and that the proposed reductions are instances of the new redundancy criterion. The com- pleteness result for LPSup is postponed till Section 2.3.4, because it relies on properties of saturated clause sets, which we develop in Section 2.3.3.
Ordering and label merge
The calculus LPSup and the corresponding version of PSup which it lifts from the ground level are parametrized by a common ordering on the infinite signature Σ∗, which
is uniformly derived from a given ordering on Σ.
Definition 2.11. Given an ordering < over Σ, its temporal extension over Σ∗ =S
i∈NΣ(i),
denoted again by <, is defined by
p(i) < q(j) if and only if i < j, or i = j and p < q.
For the rest of the presentation we fix an ordering < over Σ along with its temporal extension. We use the standard extension of this ordering to compare literals in clauses (recall Section 2.2.1).
Example 2.5. Assume the signature Σ = {a, b} is totally ordered by a < b. Then Σ∗
is totally ordered by the corresponding temporal extension as a(0) < b(0) < a(1) <
b(1)< a(2) < . . . and the corresponding standard extension orders the literals over Σ∗ as
a(0)<¬a(0)< b(0)<¬b(0) < a(1)<¬a(1) < b(1) <¬b(1)< a(2) < . . .
When two labeled clauses participate in an inference the label of the conclusion is computed by the following operation from the labels of the premises.
Definition 2.12. The merge of two labels (b1, k1) and (b2, k2) is the label (b, k) defined
imperatively as follows:
• if b1 =∗ and b2 =∗ then b ← ∗ else b ← 0,
• if k1 =∗ then k ← k2 else if k2 =∗ then k ← k1 else if k1 = k2 then k← k1.
In the case when k1, k2 ∈ N and k1 6= k2, the merge operation is undefined.
This idea behind label merge is that the labeled premises only interact when they represent standard clauses that interact on the ground level. Moreover, the resulting labeled conclusion represents exactly all the conclusions of the corresponding inferences from the ground level.
Ordered Resolution:
I (b1, k1)|| C ∨ a (b(b, k) 2, k2)|| D ∨ ¬a
|| C ∨ D ,
where the atom a is maximal in C, its complement ¬a is maximal in D, and the merge of labels (b1, k1) and (b2, k2) is defined and equal to (b, k).
Temporal Shift:
I ((∗, k) || C ∗, k0)|| (C)0 ,
where C is a clause over Σ, and • k = ∗ and k0 =∗, or • k ∈ N and k0 = k + 1. Leap:
I {(b, u + i · v) || C}i∈N derivable from the current clause set N
(b, u− v) || C ,
where u≥ v > 0 are integers and C an arbitrary standard clause (see the main text for details concerning the derivability condition).
Figure 2.4: Inference rules of LPSup.
Lemma 2.4. Let (b, k) be the merge of the labels (b1, k1) and (b2, k2), and (K, L) any
rank. Then
R(K,L)(b, k) = R(K,L)(b1, k1)∩ R(K,L)(b2, k2).
Proof. The proof is straightforward from the definitions. We check by case analysis that (b1 6= ∗ ⇒ t = 0) and (b2 6= ∗ ⇒ t = 0)
is equivalent to (b6= ∗ ⇒ t = 0), and also that
(k1 6= ∗ ⇒ ∃s ∈ N . t + k = K + s · L) and (k2 6= ∗ ⇒ ∃s ∈ N . t + k = K + s · L)
is equivalent to (k 6= ∗ ⇒ ∃s ∈ N . t + k = K + s · L), under the condition that (k1=∗ or k2 =∗ or k1 = k2).
Inference rules and their soundness
The inference rules of LPSup are presented in Figure 2.4. While the Ordered Resolution rule constitutes a labeled analogue of the corresponding rule of PSup, Temporal Shift and Leap are “structural” in nature. We will show that these two latter rules only modify the syntactic form of the clauses, but the underlying set of the represented standard clauses remains the same.
It is important to note that each of the LPSup inference rules preserves the fact that the standard parts of the involved labeled clauses span only the signature Σ∪ Σ0. This
follows, in particular, from the restriction on the premise of the Temporal Shift inference to involve clauses with literals only over the signature Σ.
Example 2.6. The starting labeled clause set NT of our running example contains among
others also clauses (∗, ∗) || a ∨ b0 and (∗, 0) || ¬b. We can apply Temporal Shift to the
second clause to obtain (∗, 1) || ¬b0. Now b0 is the only literal over Σ0 in the first clause
and therefore maximal. So the first clause and the newly derived one can participate in Ordered Resolution inference with a conclusion (∗, 1) || a.
Further explanation is needed for the Leap rule. It is stated as an inference with infinitely many premises and so we only require their potential derivability from the current clause set N . The appeal to infinity is just a useful mathematical abstraction. When we discuss the saturation with LPSup in Section 2.3.3, we will show how to employ repetition detection and deduction replaying to make the Leap inference effective.
Soundness of the Ordered Resolution rule is derived from the same property of the corresponding PSup inference, as captured by the following lemma.
Lemma 2.5. Let (K, L) be a rank. Any standard clause represented in (K, L) by the con- clusion of the Ordered Resolution inference of LPSup can be derived by the corresponding PSup inference from some standard clauses represented in (K, L) by the premises of the inference.
Proof. Let (C ∨ D)(t) be a standard clause represented in (K, L) by the conclusion
(b, k)|| C ∨D of the Ordered Resolution inference of LPSup with premises (b1, k1)|| C ∨a
and (b2, k2)|| D ∨ ¬a. This means that t ∈ R(K,L)(b, k) and thus, by Lemma 2.4, also
t ∈ R(K,L)(b1, k1) and t ∈ R(K,L)(b2, k2). Consider the standard clauses (C∨ a)(t) and
(D∨ ¬a)(t) represented in (K, L) by the respective labeled premises. If follows from
Definition 2.11 that the atom a(t) is maximal in C(t) and its complement (¬a)(t) is
maximal in D(t). Thus the clauses (C∨ a)(t) and (D∨ ¬a)(t) are valid premises of the
Ordered Resolution inference of PSup with the conclusion (C∨ D)(t).
As already indicated, the Temporal Shift and Leap inferences are sound, because they do not introduce any new clauses to the ground level. The next two lemmas formalize this observation.
Lemma 2.6. Let (K, L) be a rank. Any standard clause represented in (K, L) by a conclusion of a Temporal Shift inference is represented in (K, L) by its premise.
Proof. Let (C0)(t)be a standard clause represented in (K, L) by a conclusion (∗, k0)
|| (C)0 of Temporal Shift inference. This means that t∈ R(K,L)(∗, k0). We either have k =∗ and
k0 =∗ or k ∈ N and k0 = k +1. In any case t+1∈ R(K,L)(∗, k), and thus C(t+1)= (C0)(t)
is represented in (K, L) by the premise (∗, k) || C of the inference.
Lemma 2.7. Let (K, L) be a rank. Any standard clause represented in (K, L) by a conclusion of a Leap inference is represented in (K, L) by one of its premises.
Proof. Let C(t) be a standard clause represented in (K, L) by a conclusion (b, u− v) || C
of a Leap inference. This means we have t ∈ R(K,L)(b, u− v). We need to show that
t∈S
i∈NR(K,L)(b, u+i·v) and thus C(t) is represented in (K, L) by one of the inference’s
premises. This is equivalent to showing that whenever t + (u− v) = K + s1· L
for some s1∈ N, we can find i, s2 ∈ N such that
t + (u + i· v) = K + s2· L.
This can be achieved by setting i = L− 1 and s2= s1+ v.
Theorem 2.2 (Soundness of LPSup). Let N be a set of labeled clauses and (b, k)|| C a labeled clause derivable from N by LPSup. Then for any rank (K, L) and any t ∈ R(K,L)(b, k) the standard clause C(t) is derivable from N(K,L) by PSup. Moreover, if an
empty labeled clause (b, k)|| ⊥ is derivable from N by LPSup such that R(K,L)(b, k)6= ∅, then N is not (K, L)-satisfiable.
Proof. The first part is proved by induction on the length of the derivation, using Lem- mas 2.5, 2.6, and 2.7. The second part then follows from the soundness of PSup.
Notice that in LPSup the fact that an empty labeled clause (b, k)|| ⊥ is derived does not necessarily mean that the whole clause set is unsatisfiable. It only rules out those (K, L)-models for which R(K,L)(b, k) is non-empty. This motivates the following notion,
which will be later used for the formulation of the completeness result.
Definition 2.13. An empty labeled clause (b, k)|| ⊥ is called conditional if b = 0 and k∈ N, and unconditional otherwise. A set of labeled clauses N is obviously contradictory if it contains an unconditional empty clause or if (0, k)|| ⊥ is in N for every k ∈ N. Lemma 2.8. Any obviously contradictory set of a labeled clauses is unsatisfiable. Redundancy and reductions
Abstract redundancy for LPSup lifts the corresponding notion (see Definition 2.1 on page 15) from the ground level to the level of labeled clauses.
Definition 2.14. A labeled clause (b, k)|| C is redundant with respect to a set of labeled clauses N , if for any rank (K, L) every standard clause represented by (b, k)|| C in (K, L) is redundant with respect to N(K,L).
We present two example reductions for LPSup in Figure 2.5. These are the labeled analogues of the tautology deletion and clause subsumption, respectively. To prove that they satisfy the above redundancy criterion, we need to show in both cases that the clause missing in the conclusion of the reduction is redundant in the presence of the remaining premises. This is trivial for Tautology deletion and covered by the following lemma for Subsumption.
Tautology deletion:
R (b, k)|| C ∨ l ∨ ∼l , where the literal ∼l is the complement of l.
Subsumption:
R (b1, k1)|| C (b2, k2)|| D (b1, k1)|| C
, where C is a strict subset of D and
the merge of labels (b1, k1) and (b2, k2) is defined and equal to (b2, k2).
Figure 2.5: Possible reduction rules of LPSup.
Lemma 2.9. Let (b1, k1)|| C and (b2, k2)|| D be the premises of the Subsumption reduc-
tion, i.e., C is a subset of D and the merge of labels (b1, k1) and (b2, k2) is defined and
equal to (b2, k2). Then (b2, k2)|| D is redundant with respect to {(b1, k1)|| C}.
Proof. Let (K, L) be a rank and let D(t) be a standard clause represented in (K, L)
by (b2, k2)|| D. This means that t ∈ R(K,L)(b2, k2) and, because the merge of labels
(b1, k1) and (b2, k2) is defined and equal to (b2, k2), we obtain from Lemma 2.4 that
R(K,L)(b2, k2) ⊆ R(K,L)(b1, k1), and therefore t ∈ R(K,L)(b1, k1). Thus the standard
clause C(t) is represented in (K, L) by (b
1, k1)|| C. Because C is a strict subset of D, we
obtain that C(t) <cD(t) and that C(t) |= D(t).
The power of abstract redundancy lies in the modularity with which further new reduction rules can be introduced to the calculus. As long as they fit into the framework prescribed by Definition 2.14, they are guaranteed to preserve completeness (to be shown in Section 2.3.4) and its underlying proof need not be revised in any way.