When two layers of clauses are equal up to reindexing the corresponding sets of goal vertexes (of respective orders) are necessarily also equal. We would like to establish under what conditions does also the opposite implication hold, i.e., when can we expect to obtain syntactically equal representations of semantically equivalent sets of vertexes. Similarly to Section 2.3.3 the argument we provide here is proof theoretic and focuses solely on Ordered Resolution as the inference (Temporal Shift is handled explicitly) and on Tautology Deletion and Subsumption as reductions. The proof would need to be reexamined and revised should additional reductions be added to LPSup.
Saturation is a non-deterministic process during which we are required to perform certain inferences and allowed to perform certain reductions. Here we will insist that even reductions are performed exhaustively, meaning that no opportunity remains for performing a reduction in the final saturated set. Obviously, we cannot expect to obtain syntactically equal sets of clauses if we on one side, e.g., perform a certain subsumption and on the other side leave the corresponding subsumed clause in place.
Formulating a theorem
We assume a fixed ordering < on Σ for constraining Ordered Resolution inferences. Let N be set of (∗, ∗)-clause that is saturated by LPSup. Further, let N1, N2 be two sets of
simple (∗, k)-clauses such that Gk
(N ∪N1) =G k
(N ∪N2). For i = 1, 2, let N 0
corresponding shifted versions of clauses of Ni, i.e.,
Ni0 ={(∗, k + 1) || (C)0 | (∗, k) || C ∈ Ni},
and let N∗
i denote a set of (∗, k+1)-clauses obtained from (N ∪Ni0) by exhaustive satura-
tion by the Ordered Resolution inference and the Tautology Deletion and Subsumption reductions. Finally, let N∗
i be the subset of simple clauses from N ∗
i. Our claim is that
despite the potentially different syntactic representations N1 and N2 of the same set
of goal vertexes of order k and despite the non-determinism involved in the saturation process, the sets N∗
1 and N2∗ are necessarily equal.
Derivability relation as a rewrite system
We split the proof of our theorem into two parts. In this first, preparatory part, we explain how to model LPSup derivations as a rewrite system (see, e.g., Baader and Nipkow, 1998, for an overview) and use the rewriting theory to obtain a normalization result for saturations, which will later on help us to focus on derivations of a particular shape. We structure this first part into several steps.
First, we define the derivability relation on sets of clauses. Two sets N1, N2 are in the
relation, written N1B N2, if the set N2 can be obtained from N1 by adding a conclusion
of the Ordered Resolution inference with premises in N1, by deleting a tautology in
N1, or by deleting a clause from N1 subsumed by another (different) clause in N1.
We can fix the details of the definition in such a way that the derivability relation B be terminating. This is achieved by stipulating, without the loss of generality, that tautologies and subsumed (or present) clauses cannot be (re-)derived.
Next, we show that the derivability relation is confluent. This means that for any sets N1, N2, N3 such that N1B∗N2 and N1B∗N3, where B∗ is the reflexive and transitive
closure of B, there is a set N4 such that N2B∗N4 and N3B∗N4. By Newman’s lemma
(Newman, 1942) and relying on the fact that our relation is terminating, it is sufficient to show only local confluence, a property which requires that for any sets N1, N2, N3
such that N1 B N2 and N1 B N3, there is N4 with N2B∗ N4 and N3 B∗ N4. This is
done by case analysis over the possible ways in which the sets N2 and N3 can be derived
(for local confluence in one step) from N1. As an example, we present here the most
interesting case, the combination of Ordered Resolution and subsumption, leaving the other cases to the reader.
We work on the level of PSup relying on the obvious correspondence to LPSup. Let us assume that the set N2is derived from N1 by adding the conclusion C∨ D of Ordered
Resolution inference with premises C∨ a and D ∨ ¬a. Moreover, assume that N3 arises
from N1 by deleting the premise C∨ a. (Note that we can choose C ∨ a over D ∨ ¬a
without loss of generality. Also note that when the deleted clause is not equal to one of the two premises, the case becomes trivial.) The premise C∨ a was deleted in N3,
because it is subsumed by a clause C1 ⊂ (C ∨ a). We consider two subcases:
• C1 ⊂ C. Then we set N4 = N3 and arrive from N2 to N4 in two subsumption
steps using C1 ⊂ (C ∨ a) to delete the said premise and C1 ⊂ (C ∨ D) to delete
Σ Σ′ N N′ i ∗ k + 1 N N0 i ∗ k + 1 N N∗ i ∗ k + 1 N N0 i ∗ k + 1 N N∗ i ∗ k + 1 N1 i N∗ i N∗ i Σ Σ′ Σ Σ′ N N′ i ∗ k + 1 N N∗ i ∗ k + 1 N∗ i
Figure 2.12: Illustrating the proof of the uniqueness of saturations.
• C1 is of the form C0∨ a and C0 ⊂ C. Because the literal a must be maximal in
C∨ a, we know it is also maximal in C0∨ a. Thus there is an Ordered Resolution
inference with C0 ∨ a and D ∨ ¬a as premises. We define N4 as N3 with the
conclusion C0∨ D of this inference added. We arrive from N2 to N4 in three steps.
First, we repeat the step by which N3 was derived from N1, i.e., we delete C∨ a,
which is subsumed by C1. Then we repeat the above Ordered Resolution inference
and add C0∨ D to the set, and, finally, we delete the old conclusion C ∨ D as it is
now subsumed by C0∨ D.
When a rewriting system is both confluent and terminating it is called convergent. In a convergent rewriting system every object has a unique normal form. For our case of the derivability relation B this means that for every set of clauses N1 there is a unique
set N2 such that 1) N1B∗N2 and 2) there is no N3 for which N2B N3. Informally, the
final result of the saturation process does not depend on the order in which inferences and reductions are performed. We will slightly generalize this result for our main proof. Definition 2.18. Let N1, N2 be two sets of clauses. We say that N1 subsumes N2, if
for every C ∈ N2 there is C0 ∈ N1 such that C0 subsumes C. We say that N1 and N2
are subsumption equivalent, if N1 subsumes N2 and N2 subsumes N1.
Lemma 2.14. Subsumption equivalent sets of clauses have identical normal form. Proof. Let the sets N2 and N3 be subsumption equivalent. Set N1 = (N2 ∪ N3) and
check that N1B∗N2 and N1B∗N3.
Proof tree analysis
Let us now focus back on the theorem about uniqueness of saturations. We start with a set N of (∗, ∗)-clause that is saturated by LPSup and two sets N1, N2 of simple (∗, k)-
clauses such thatGk
(N ∪N1)=G k
(N ∪N2). We then, for i = 1, 2, separately saturate (N∪N 0 i)
to obtain (N∪ N∗
i), where Ni0 contains the shifted versions of clauses from Ni, and Ni∗
denotes the (∗, k+1)-part of the saturated set (the set of (∗, ∗)-clauses N does not change anymore). We want to study the sets N∗
i of simple clauses from N ∗
i. See Figure 2.12 on
the left.
Because we now know that the result of saturation is unique, we can assume that the derivation (N∪ N0
i) B∗(N∪ Ni∗) is of a particular shape. We imagine a saturation
strategy that first performs all the resolutions and only then deletes all the tautologies and subsumed clauses. This way we split the saturation into two parts
(N∪ Ni0) B ∗ (N∪ N0 i) B ∗ (N∪ Ni∗),
where the first part (N∪ N0
i) B∗(N∪ Ni0) consist of additions only and the second part
(N∪ N0 i) B
∗(N
∪ Ni∗) consist only of deletions. This is illustrated in Figure 2.12 in the
middle column.
We now analyze parent-child relations between the clauses of N0
i. We say that a clause
C is a parent of a clause D (and D is a child of C) if C is a premise and D the conclusion of Ordered Resolution inference in our derivation. Note that some clauses from N0
i have
only one parent in N0
i, the other parent being a (∗, ∗)-clause from N. Clauses from Ni0
without any parent are those from the starting set N0 i ⊆ Ni0.
Of special interest to us are simple (∗, k + 1)-clauses of N0
i that do not have a simple
parent. Let us denote their set N1
i. Note that if a premise of Ordered Resolution is
simple, then so is the other premise and the conclusion. This means that (N∪ N1 i) B∗
(N∪ N∗
i), i.e., the set Ni1 already contains all the clauses needed to derive the simple
(∗, k + 1)-clauses N∗
i we are interested in (as indicated in Figure 2.12 on the right).
Our current plan is to show that the sets N1
1 and N21 are subsumption equivalent and
from that to conclude that the sets N∗
1 and N2∗ are equal. In what follows we assume
the literals of every labeled clauseD to be separated, as in D = (b, k) || Dl∨ (Du)0, into
a lower, non-primed part Dl and upper, primed part (Du)0, where where Dl and Du are
standard clauses over Σ.
Let us consider a clauseC ∈ N1
1 and let us follow the parent links fromC to construct
the whole proof tree for C with C as its root and some subsets M ⊆ N and M0 1 ⊆ N10
forming the leaves. We now observe that:
• The set of upper parts of clauses from M and M0
1 is unsatisfiable. Indeed, if we
just “forget” the lower parts of the clauses in the proof tree for C, we obtain a proof of the empty clause.
• If we, on the other hand, track these lower parts of the clauses in the proof tree for C, we realize they are being collected to constitute the standard part of C. Specifically, the simple labeled clause C is necessarily of the form (∗, k + 1) || C, where
C =_ {Dl | (∗, ∗) || Dl∨ (Dh)0∈ M}.
The clauses from M10 do not directly influence the form of C, because their lower parts are empty.
We now use the assumption that Gk (N ∪N1)=G k (N ∪N2) to discover a clauseC0 ∈ N 1 2 that
subsumesC. The details follow. • The assumption Gk
(N ∪N1)=G k
(N ∪N2)means that the sets (N∪N1) and (N∪N2) are
logically equivalent, where N stands for the simple (∗, ∗)-clauses from N. To focus on the main idea, let us first formulate our result for the case when the equivalence holds already between N1 and N2.
• Then, since the upper parts of clauses from (M ∪ M10) are unsatisfiable, so are
the upper parts of clauses from (M ∪ N0
2): we first strengthen (M ∪ M10) to (M∪
N10), which preserves unsatisfiability, and than exchange N10 with N20 using our equivalence.
• Because the set of the upper parts of the clauses (M ∪ N20) is unsatisfiable, by
completeness of PSup the empty clause can be derived from the set. In analogy to what we have seen above, there is a corresponding proof tree from the clauses (M∪ N0
2) as leaves. The root of the tree is formed by a clauseC0 = (∗, k + 1) || C0,
where
C0=_ {Dl | (∗, ∗) || Dl∨ (Dh)0 ∈ M0},
and where M0 is the subset those clauses of M that were actually used in the
derivation of the empty clause. From M0 ⊆ M it follows that C0 ⊆ C and we are
done.
• In the general case, where we assume equivalence of (N ∪ N1) and (N ∪ N2), we
can only obtain unsatisfiability of (M∪ N0∪ N0
2) instead of (M ∪ N20), where
N0 ={(∗, ∗) || (C)0 | (∗, ∗) || C ∈ N}
is the set of shifted versions of clauses from N . If a clause D ∈ N0 is not in N , it must be subsumed by another clause D0 ∈ N, because we assume N to be
saturated by LPSup and so, in particular, by Temporal Shift. This means each of the additional assumptions from N0 (or its stronger version) is available for the derivation of the empty clause. Moreover, these additional assumptions cannot influence the form of the clauseC0, because their lower parts are empty.
We have just shown that the set N1
1 is subsumed by N21. Because the argument is
symmetrical, we obtain that sets N1
1 and N21 are subsumption equivalent, and so the
sets N∗
1 and N2∗ are equal. This concludes the proof.
Final remarks
As a corollary of our theorem we obtain that under the suitable conditions imposed on the saturation process the exact syntactic form of the derived layers is uniquely determined by their semantics, i.e., by the set of goal vertexes the layer represents. The only layer which can have a “non-compatible” syntactic form is the 0-layer, which our decision procedure obtains as an input.
Example 2.12. We close this section with an example showing that subsumption is im- portant for the proof of our theorem to go through. The following three (∗, ∗)-clauses
(∗, ∗) || a ∨ ¬a0, (∗, ∗) || a ∨ b ∨ b0, (∗, ∗) || a ∨ ¬b ∨ ¬b0
are saturated by LPSup (we may assume that a < b in the ordering). If we now use the following (redundant) representation of the 0-layer
(∗, 0) || a ∨ ¬b (∗, 0) || a
and altogether refrain from applying subsumption, the subsequent saturation will gen- erate the following simple clauses in the 1-layer and 2-layer:
(∗, 1) || a ∨ b (∗, 1) || a
(∗, 2) || a ∨ ¬b (∗, 2) || a
We see that without subsumption the “syntactic period” of the saturation is 2 while the “semantic period” is just 1.