6.2 Refutability
8.2.1 The OLRR problem for resolution
In this sections, we discuss the problem of finding the optimal read-once refutation of a Horn formula.
Let α be an unsatisfiable Horn formula. From [Sze01] we know that α has a read- once refutation. The question whether α has a read-once resolution of length less than k is equivalent to the question whether α contains a minimal unsatisfiable formula consisting of at most k clauses.
Theorem 8.2.1. Let R denote an OLROR of Φ and let ΦR⊆ Φ be the set of clauses used
by R. R is also an optimal tree-like refutation of Φ and an optimal dag-like refutation of Φ.
Additionally, ΦR is a minimum unsatisfiable subset of Φ.
Proof. Sine R is a read-once refutation of Φ, R is also a tree-like and a dag-like refutation
of Φ.
Assume that T is a tree like refutation of Φ such that |T | < |R|. Let ΦT ⊆ Φ be the set
of clauses used by T . It follows that |ΦT| ≤ |T | + 1 < |R| + 1
By [Sze01], ΦT must have a read-once refutation, RT. However, |RT| ≤ |ΦT| − 1 < |R|.
This contradicts the fact that R is the optimal read-once refutation of Φ. Thus, R is also an optimal tree-like refutation of R. Similarly, R is an optimal dag-like refutation of R.
Let Φ0⊆ Φ be an unsatisfiable Horn formula such that Φ0< ΦR. By [Sze01], Φ0must
have a read-once refutation, R0. However, |R0| ≤ |Φ0| − 1 < |ΦR| − 1 = |R|. This contradicts
the fact that R is the optimal read-once refutation of Φ. Thus, ΦRis a minimum unsatisfiable
subset of R.
We conclude, that for unsatisfiable Horn formulas the length of the shortest resolu- tion refutation equals the length of the shortest read-once and the shortest tree resolution. Moreover, let k be the number of clauses of a minimal unsatisfiable subformula with mini-
CHAPTER 8. HORN CLAUSAL FORMULAS 75
mal number of clauses. Then the shortest resolution (read-once resolution, tree resolution) refutations is k − 1. Note that this only applies to regular resolution refutations.
Let α be a definite Horn formula without unit clauses, x a variable, U a set of variables (unit-clauses), and k ≥ 1.
Determining if there exists a subset K ⊆ U of at most k variables such that K ∧ α |= x is NP-complete. The proof is based on a reduction to the vertex cover problem. Note that this problem asks for the number of variables and not the number of clauses.
The following result is a corollary of Theorem 5.2 in [Lib08]. However, it is included here for completeness.
Theorem 8.2.2. The problem of deciding whether a Horn formula contains an unsatisfiable
sub-formula with at most k clauses isNP-complete.
Proof. We show this by a reduction from Vertex Cover. Let G = (V, E) be an undirected
graph where V = {v1, . . . , vn} and E = {e1, . . . , em}. We associate with G the Horn formula:
v1∧ . . . ∧ vn∧
^
1≤i≤m,ei=(vi1,vi2)
(vi1 → ei) ∧ (vi2 → ei) ∧ (¬e1∨ . . . ∨ ¬em)
Then there exists a subset V0⊆ V such that |V0| ≤ r and V0∩ eiis non-empty for every
1 ≤ i ≤ m if and only if the associated formula contains an unsatisfiable sub-formula with
at most (1 + m + r) clauses. (the negative clause, the clause (v → ei) for each 1 ≤ i ≤ m,
and r unit clauses).
Since the problem of finding a vertex cover of size at most k is NP-complete, the problem of finding a read-once resolution refutation of length at most k is NP-complete for Horn formulas.
8.2.2
The ROR problem for unit-resolution
In this section, we explore the problem of finding read-once unit resolution refutations for Horn formulas. If we restrict ourselves to unit resolution refutations then we are no
CHAPTER 8. HORN CLAUSAL FORMULAS 76
longer guaranteed read-once refutations. Example 35: Consider the Horn formula
(x1) ∧ (¬x1∨ x2) ∧ (¬x1∨ ¬x2∨ x3) ∧ (¬x3) This formula has the following read-once refutation:
(¬x1∨ x2) ∧ (¬x1∨ ¬x2∨ x3) |Res1 (¬x1∨ x3)
(x1) ∧ (¬x1∨ x3) | 1
Res (x3)
(x3) ∧ (¬x3) |Res1 t
However, we will now show that this formula does not have a read-once unit resolution refutation.
There are three possibilities for the final resolution step.
1. The final resolution step is (x1) ∧ (¬x1) |Res1 t: Note that in this case it is impossible
to generate the clause (¬x1) by unit resolution.
2. The final resolution step is (x2) ∧ (¬x2) |Res1 t: To generate the clause (x2) we need
to use the clause (x1). However, we also need to use this clause to generate (¬x2).
3. The final resolution step is (x3) ∧ (¬x3) |Res1 t: To generate the clause (¬x3) via unit
resolution, we need to use the clause (x1) twice. Once to resolve with (¬x1∨ x2) and
once to resolve with (¬x1, ¬x2, x3).
Thus, It is not always possible to find a read-once unit refutation.
It was shown in [KZ03] that the UROR problem for Horn formulas is NP-complete. We now provide an alternative proof that the unit ROR problem is NP-complete for Horn formulas. This is done by a reduction from the set packing problem.
CHAPTER 8. HORN CLAUSAL FORMULAS 77
Definition 8.2.1. The set packing problem is the following: Given a set S, m subsets S1, . . . , Smof S, and an integer k, does{S1, . . . , Sm} contain k mutually disjoint sets.
This problem is known to be NP-complete [Kar72].
Theorem 8.2.3. The unit ROR problem for Horn formulas is NP-complete.
Proof. Let us consider an instance of the set packing problem. We construct the Horn
formula Φ as follows.
1. For each xi∈ S, create the boolean variable xiand the clause (xi).
2. For j = 1 . . . k, create the boolean variable vj.
3. For each subset Sl, l = 1 . . . m create the clauses
(vj∨
_
xi∈Sl
¬xi) j= 1 . . . k.
4. Finally create the variable w and the clauses (w ∨ ¬v1∨ . . . ∨ ¬vk) and (¬w).
We now show that Φ has a read-once unit resolution refutation if and only if {S1, . . . , Sm} contains k mutually disjoint sets.
Suppose that {S1, . . . , Sm} does contain k mutually disjoint sets. Without loss of gener-
ality assume that these are the sets S1, . . . , Sk.
Let us consider the sets of clauses
Φj= {(vj∨
_
xi∈Sj
¬xi)} ∪ {(xi) | xi∈ Sj} j= 1 . . . k.
By the construction of Φ, Φj⊆ Φ for j = 1 . . . k. Since the sets S1, . . . , Sk are mutually
disjoint, so are the sets Φ1, . . . , Φk.
It it easy to see that the clause (vj) can be derived from the set Φj by read-once unit
CHAPTER 8. HORN CLAUSAL FORMULAS 78
ally disjoint, the set of clauses {(v1), . . . , (vk)} can be derived from Φ by read-once unit
resolution.
Together with the clause (w ∨ ¬v1∨ . . . ∨ ¬vk), this set of clauses has a read-once unit
derivation of the clause (w). Thus, Φ has a read-once unit derivation of the clause (w). Since Φ contains the clause (¬w), it follows that Φ has a read-once unit resolution refuta- tion.
Now suppose that Φ has a read-once unit resolution refutation R. Note that Φ/{(¬w)} can be satisfied by setting every variable to true. Thus, R must use the clause (¬w).
Let us consider the resolution step that involves the clause (¬w). By construction,
(¬w) must be resolved with the clause (w ∨ ¬v1∨ . . . ∨ ¬vk) or a clause derived from it.
We can assume without loss of generality that the clause resolved with (¬w) has the form (w ∨ ¬v1∨ . . . ∨ ¬vk0) where k0≤ k. To derive this clause, the clauses (vk0+1), . . . , (vk) must
have already been derived. Thus,
(¬w) ∧ (w ∨ ¬v1∨ . . . ∨ ¬vk0) |Res1 (¬v1∨ . . . ∨ ¬vk0).
To eliminate the clause (¬v1∨ . . . ∨ ¬vk0), R must either derive the clauses (v1), . . . , (vk0), or
reduce it to a unit clause and then resolve it with another clause. Without loss of generality
we can assume that this unit clause is (¬v1). In either case R must derive the clauses
(v2), . . . , (vk0).
Thus, we must derive the clauses (v2), . . . , (vk). Let us consider the clause (vj), 2 ≤ j ≤
k. By the construction of Φ, this clause must be derived from one of the clauses
(vj∨
_
xi∈Sl
¬xi) l= 1 . . . m.
To to this, we must use the set of clauses Ψlj = {(xi) | xi∈ Slj} for some lj≤ m.
Since the refutation is read-once, the sets Ψlj for j = 2 . . . k are mutually disjoint. Thus,
CHAPTER 8. HORN CLAUSAL FORMULAS 79
If R derives the clause (v1), then, by the construction of Φ, this clause must be derived
from one of the clauses
(v1∨
_
xi∈Sl
¬xi) l= 1 . . . m.
To do this, we must use the set of clauses Ψl1 = {(xi) | xi∈ Slj} for some l1≤ m.
If R reduces (¬v1∨ . . . ∨ ¬vk0) to the clause (¬v1), then we must resolve (¬v1) with one
of the clauses
(v1∨
_
xi∈Sl
¬xi) l= 1 . . . m.
This results in the clause (W
xi∈Sl1¬xi) for some l1≤ m. To eliminate this clause, we must
use the set of clauses Ψl1 = {(xi) | xi∈ Slj}.
Since R is read-once, the set Ψl1 does not share any clauses with the sets Ψlj, j = 2 . . . k.
Thus, the sets Slj for j = 1 . . . k are also mutually disjoint. This means that {S1, . . . , Sm}
contains k mutually disjoint sets.
Thus, Φ has a read-once unit resolution refutation if and only if {S1, . . . , Sm} contains
k mutually disjoint sets. As a result of this, the unit ROR problem for Horn formulas is
NP-complete.
Theorem 8.2.4. For a CNF formula Φ, the length of a read-once unit resolution refutation
is at most(m − 1), where m is number of clauses in the formula.
Proof. Recall that a read-once resolution step is equivalent to removing the two parent
clauses from the formula and adding the resolvent. Thus, each read-once resolution step effectively reduces the number of clauses in the formula by 1. Since Φ initially has m clauses, there can be at most (m − 1) such resolution steps.