maximal contractions in boolean algebras
TRANSCRIPT
. RESEARCH PAPER .
SCIENCE CHINAInformation Sciences
September 2012 Vol. 55 No. 9: 2044–2055
doi: 10.1007/s11432-012-4631-6
c© Science China Press and Springer-Verlag Berlin Heidelberg 2012 info.scichina.com www.springerlink.com
Maximal contractions in Boolean algebras
SHI HuiXian∗ & WANG GuoJun
Institute of Mathematics, Shaanxi Normal University, Xi’an 710062, China
Received July 4, 2011; accepted August 28, 2011; published online June 22, 2012
Abstract In the present paper, the concepts of deductive element and maximal contraction are introduced in
Boolean algebras, and corresponding theories of consistency and maximal contractions are studied. An algorithm
principle is proposed to compute all maximal contractions for a consistent set with respect to its refutation in
Boolean algebras. It is pointed out that the quotient algebra of the first-order language with respect to its
provable equivalence relation constitutes a Boolean algebra, and hence the computation of R-contractions for
closed formulas in first-order languages can be converted into the one in Boolean algebras proposed in this
paper. Furthermore, the concept of basic element is introduced in Boolean algebras, which contributes to the
definitions of clause and Horn clause transplanted from logic to a special type of Boolean algebras generated
by basic elements. It is also pointed out that the computation of R-contractions for clauses in the classical
propositional logic can be converted into the one in Boolean algebras generated by basic elements proposed in
this paper.
Keywords deductive element, consistency, maximal contraction, minimal subtraction, basic element, clause
Citation Shi H X, Wang G J. Maximal contractions in Boolean algebras. Sci China Inf Sci, 2012, 55: 2044–2055,
doi: 10.1007/s11432-012-4631-6
1 Introduction
Belief revision [1–4], mainly dealing with how belief sets change as agents gain new knowledge, plays
a significant role in the realm of artificial intelligence. When logical consequences of a scientific theory
contradict experimental results or natural phenomenon (we say that this theory is refuted by facts [5]),
our concerns are concentrated in the process of revising the original theory properly to eliminate such
contradictions, which is also one of the key problems for belief revision theory. In order to study formal
aspects of this process of revision, refs. [5,6] proposed a calculus system about logical connectives and
quantifier symbols, called the revision calculus or R-calculus for short, aiming at forming a new theory
by deleting part of the original one that contradicts its refutation by facts. Note that the process of
this deletion should remain the least to make sure that the remainder is close to the original theory as
much as possible. That is to say, we hope that the remainder of the original theory after deleting is a
maximal subset consistent with facts, which was called an R-contraction in [7,8]. Considering scientific
theories and facts could be formalized as sets of abstract formulas in logic, ref. [9] proposed an interactive
algorithm to enumerateR-contractions for Horn clauses in first-order languages, which provides a solution
to properly conduct the deletion above in the sense of logic to some degree.
∗Corresponding author (email: [email protected])
Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9 2045
It is worth noting that the algorithm in [9], which focuses on Horn clauses, special types of formulas in
first-order languages, is not suitable for usual formulas any more. As a result, a natural question arises,
i.e., how to expand the R-contraction theory in [9] to the one suitable for all formulas in first-order
languages, or more extensively, to the one applicable in some more general algebraic structure? This is
also one of the focuses of this paper. First of all, the concept of deductive element is introduced in Boolean
algebras in the present paper, and corresponding theory of consistency is considered. Secondly, the
concepts of refutation, maximal contraction and minimal subtraction are proposed for Boolean algebras
by generalizing the ones in [5–9], and a one-to-one corresponding relation is constructed between maximal
contractions and minimal subtractions. Besides, two algorithm principles are proposed to compute all
minimal inconsistent subsets for a given finite set as well as all minimal choices for finite given sets
in Boolean algebras, which consequently provides a method for finding out maximal contractions in the
sense of Boolean algebras. Moreover, it is pointed out that the quotient algebra of the first-order language
with respect to its provable equivalence relation constitutes a Boolean algebra, and the computation of
R-contractions for closed formulas in first-order languages can be converted into the one in Boolean
algebras proposed in this paper, since the theories of consistency and R-contractions under the scope of
closed formulas in first-order languages are in accordance with the ones in the sense of Boolean algebras.
Lastly, the concept of basic element is introduced in Boolean algebras, and a special type of Boolean
algebras generated by basic elements is defined. We point out that the quotient algebra of the classical
propositional logic with respect to its provable equivalence relation constitutes a special Boolean algebra
whose basic elements are actually the equivalence classes of all the atoms. Based on these, the definitions
of clause and Horn clause are consistently transplanted from logic to Boolean algebras generated by basic
elements, and we also point out that the computation of R-contractions for clauses, especially for Horn
clauses, in the classical propositional logic can be converted into the one in Boolean algebras generated
by basic elements proposed in this paper.
2 Consistent sets in Boolean algebras
Definition 1 (See [10,11]). Let (B,∨,∧, 0, 1) be a bounded lattice with respect to its order � satisfying
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c), (1)
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) (2)
for every a, b, c in B. If there is a mapping ′ : B −→ B satisfying
a ∨ a′ = 1, a ∧ a′ = 0, a ∈ B, (3)
then (B,∨,∧,′ , 0, 1) is called a Boolean algebra. The mapping ′ is said to be the complementary operator
on B, and we call a, a′ a complementary pair.
Refs. [10,11] proved that Eqs. (1) and (2) are equivalent in lattices, and only one of them is required
in Definition 1. If a lattice satisfies (1) or (2), then it is called a distributive lattice [10,11].
Proposition 1 (See [10,11]). Let (B,∨,∧,′ , 0, 1) be a Boolean algebra. Then the following statements
hold for every a, b and ai (i ∈ I) in B.
(i) 0′ = 1, 1′ = 0.
(ii) a′′ = a.
(iii) (∨i∈I ai)′ = ∧i∈I a′i, (∧i∈I ai)
′ = ∨i∈I a′i.(iv) a � b if and only if b′ � a′; if and only if a∧ b′ = 0 if and only if a′ ∨ b = 1; if and only if a∧ b = a
if and only if a ∨ b = b.
Definition 2. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra. Define a binary operator →: B2 −→ B as
a → b = a′ ∨ b, a, b ∈ B. (4)
Then → is called the implication operator on B.
2046 Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9
Definition 3. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M ⊆ B, x ∈ B. If there are finite elements
a1, . . . , an in M such that a1 ∧ · · · ∧ an � x, then x is called a deductive element of M . The set of all
deductive elements of M is denoted by D(M), that is,
D(M) = {x ∈ B | there are a1, . . . , an ∈ M such that a1 ∧ · · · ∧ an � x}. (5)
By (5), D can be seen as a self-mapping on P(B), that is, D :P(B) −→P(B), where P(B) is the
powerset of B. In this sense, D is actually a closure operator on B, which will be proved as follows.
Proposition 2. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, and let the mapping D :P(B) −→P(B) be
defined by (5). Then D is a closure operator [10,12] on B; that is, the following statements hold for every
M,M1,M2 in P(B).
(i) M ⊆ D(M).
(ii) If M1 ⊆ M2, then D(M1) ⊆ D(M2).
(iii) D(D(M)) = D(M).
Proof. (i) is trivial.
(ii) Let M1 ⊆ M2. Suppose x ∈ D(M1). Then there are elements a1, . . . , an ∈ M1 such that a1 ∧· · · ∧ an � x. Since M1 ⊆ M2, we have a1, . . . , an ∈ M2, and consequently x ∈ D(M2). Therefore,
D(M1) ⊆ D(M2).
(iii)D(M) ⊆ D(D(M)) can be obtained immediately by (i), and we only need to prove thatD(D(M)) ⊆D(M). Suppose x ∈ D(D(M)). Then there are elements a1, . . . , an ∈ D(M) such that a1 ∧ · · · ∧ an � x.
For every ai (1 � i � n), since ai ∈ D(M), there are elements b(i)1 , . . . , b
(i)ji
∈ M such that b(i)1 ∧· · ·∧b
(i)ji
�ai. As a result, (b
(1)1 ∧ · · · ∧ b
(1)j1
)∧ · · · ∧ (b(n)1 ∧ · · · ∧ b
(n)jn
)� a1 ∧ · · · ∧ an � x, and consequently x ∈ D(M)
since j1 + · · ·+ jn is finite. The proof is completed.
Proposition 3. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M ⊆ B.
(i) D(M) is an up-set [10]; that is, if x ∈ D(M) and x � y, then y ∈ D(M).
(ii) D(M) is closed under finite meet and arbitrary join.
(iii) D(M) is closed under Modus Ponens; that is, if x, x → y ∈ D(M), then y ∈ D(M).
(iv) If M is finite, then D(M) = {x ∈ B | ∧M � x}, where ∧M = ∧{x | x ∈ M}.Proof. (i) and (ii) are trivial.
(iii) Suppose x, x → y ∈ D(M). From (ii) and Eqs. (1), (3), (4), we have x ∧ y = (x ∧ x′) ∨ (x ∧ y) =
x ∧ (x′ ∨ y) = x ∧ (x → y) ∈ D(M), and then y ∈ D(M) by (i).
(iv) Suppose M = {a1, . . . , an}. On one hand, if x ∈ D(M), then there is a subsequence i1, . . . , ik of
1, . . . , n such that ai1 ∧ · · · ∧ aik � x. Therefore, ∧M = a1 ∧ · · · ∧ an � ai1 ∧ · · · ∧ aik � x. On the other
hand, if ∧M � x, then it is obvious that x ∈ D(M). The proof is completed.
Definition 4. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M ⊆ B. If 0 ∈ D(M), then M is called to be
inconsistent. Otherwise, M is called to be consistent.
Especially, we consider ∅ to be consistent.
The following corollary can be easily obtained by Propositions 2, 3 and Definition 4.
Corollary 1. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M ⊆ M1 ⊆ B.
(i) If there is an element a ∈ B such that a, a′ ∈ M , then M is inconsistent.
(ii) M is inconsistent if and only if D(M) = B.
(iii) Suppose M is finite. Then M is inconsistent if and only if ∧M = 0.
(iv) If M is inconsistent, then M1 is inconsistent as well.
Remark 1. The inverse of (i) and (iv) in Corollary 1 fails to hold, and counterexamples are as follows.
Let X = {a, b, c}, B = P(X) be the powerset of X . It is easy to infer that (B,∪,∩,′ , ∅, X) is a Boolean
algebra, where ∪,∩,′ are union, intersection and complementary operators of sets, respectively. Suppose
M = {{a}, {b}}. Then ∧M = {a} ∩ {b} = ∅, and M is inconsistent by Corollary 1 (iii). However, M
does not contain any complementary pair. Besides, let M1 = {{a}, {a, b}, {c}}. Then M1 is inconsistent
with a consistent subset {{a}, {a, b}}.
Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9 2047
Definition 5. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M ⊆ B. If M is inconsistent, and M − {a} is
consistent for every a ∈ M , then M is called a minimal inconsistent set. Furthermore, if M ⊆ M1 ⊆ B,
then M is called a minimal inconsistent subset of M1.
Example 1. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, a, b ∈ B such that {a, a′, b, b′} is an
antichain [10] in B. Let M1 = {a, a → b, b′}, M2 = {a, a′, b, b′}.(i) Since ∧M1 = a ∧ (a′ ∨ b) ∧ b′ = (a ∧ a′ ∧ b′) ∨ (a ∧ b ∧ b′) = 0, M1 is inconsistent. Assume
a ∧ b = 0. Then a′ ∧ b = (a′ ∧ b) ∨ (a ∧ b) = (a ∨ a′) ∧ b = b, and consequently b � a′ by Proposition 1,
which contradicts the statement that {a, a′, b, b′} is an antichain. As a result, we have a ∧ b = 0. Thus
a ∧ (a → b) = a ∧ (a′ ∨ b) = a ∧ b = 0, and consequently {a, a → b} is consistent. Similarly, it can be
proved that {a → b, b′} and {a, b′} are consistent as well. Thus, M1 is a minimal inconsistent set.
(ii) Since ∧M2 = a ∧ a′ ∧ b ∧ b′ = 0, M2 is inconsistent. Meanwhile, {a, a′, b} is also inconsistent,
which means that M2 is not a minimal inconsistent set. Furthermore, it can be proved that M2 has two
minimal inconsistent subsets, {a, a′} and {b, b′}.Remark 2. Boolean algebras in Example 1 do exist. In fact, let X = {x, y, z, w}, B =P(X) be the
powerset of X . Then (B,∪,∩,′ , ∅, X) is a Boolean algebra, where ∪,∩,′ are union, intersection and
complementary operators of sets, respectively. Suppose a = {x, y}, b = {y, z}. Then it is easy to obtain
that {a, a′, b, b′} is an antichain with respect to the inclusion order of sets.
Example 1 shows that the minimal inconsistent subsets of a given set in Boolean algebras are not
necessarily unique.
Proposition 4. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M ⊆ B. Then M is inconsistent if and only
if M has at least one minimal inconsistent subset.
Proof. The sufficiency can be easily obtained by Corollary 1, and we only need to prove the necessity.
Suppose M is inconsistent, and let M (k) = {x1 ∧ · · · ∧ xk | xi ∈ M, i = 1, . . . , k} (k = 1, 2, . . .). By
Corollary 1 (iv), if there is a number j (j � 1) such that 0 ∈ M (j), then 0 ∈ M (i) holds for arbitrary
i (i � j). Since M is inconsistent, 0 ∈ D(M); that is, there are elements x1, . . . , xn ∈ M such that
x1 ∧ · · · ∧ xn = 0, and consequently 0 ∈ M (n). Thus, for the sequence M (1), . . . ,M (n), . . ., there is a least
number k such that 0 ∈ M (k); that is, 0 /∈ M (i) holds for every i (1 � i < k). As a result, there are
elements y1, . . . , yk ∈ M such that y1∧· · · ∧yk = 0, and consequently {y1, . . . , yk} is inconsistent with its
all proper subsets being consistent. Therefore, {y1, . . . , yk} is a minimal inconsistent subset of M . The
proof is completed.
3 Maximal contractions in Boolean algebras
Definition 6. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M and R be finite consistent sets in B. If M
is inconsistent with R; that is, M ∪R is an inconsistent set, then R is called to be the refutation for M .
Definition 7. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, let M and R be finite consistent sets in B,
and let R be the refutation for M . If M0 (= ∅) is a maximal subset of M consistent with R, then M0 is
called a maximal contraction of M with respect to R.
Example 2. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra same as Example 1.
(i) Suppose M = {a, a → b, b}, R = {a, b′}. Then it is easy to show that both M and R are finite
consistent sets in B, and R is the refutation for M . Let M0 = {a}. Then M0 is a maximal subset of M
consistent with R, that is, a maximal contraction of M with respect to R. Furthermore, it can be proved
that M0 is the unique maximal contraction of M with respect to R.
(ii) Suppose M = {a, a → b}, R = {b′}. Then it is easy to show that both M and R are finite consistent
sets in B, and R is the refutation for M . Let M1 = {a}, M2 = {a → b}. Then it can be proved that M1
and M2 constitute all maximal contractions of M with respect to R.
(iii) Suppose M = {a}, R = {a → b, b′}. Then it is easy to show that both M and R are finite
consistent sets in B, and R is the refutation for M . However, there is no maximal contraction of M with
2048 Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9
respect to R.
Example 2 shows that the maximal contractions in Boolean algebras do not necessarily exist, nor they
are necessarily unique.
Proposition 5. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M and R be finite consistent sets in B, and
R be the refutation for M . Then the maximal contractions of M with respect to R exist if and only if
there is an element a ∈ M such that R ∪ {a} is consistent.
Proof. Suppose there is no maximal contraction of M with respect to R. Let M1 be arbitrary nonempty
subset of M . Assume that M1 is consistent with R. Since M1 is not maximal, there is a set M2 ⊆ M
such that M1 � M2 and M2 is consistent with R as well. Similarly, there is a set M3 ⊆ M such that
M2 � M3 and M3 is still consistent with R. This process could be carried on straight forward, and we
can finally get a sequence of nonempty subsets of M satisfying M1 � M2 � M3 � · · · . However, M is
finite, and contradiction arises. Therefore, M1 is inconsistent with R, and consequently any nonempty
subset of M is inconsistent with R. Especially, R∪ {a} is inconsistent for arbitrary a ∈ M , which proves
the sufficiency.
Meanwhile, suppose M0 is a maximal contraction of M with respect to R. Then M0 = ∅ and M0∪R is
consistent. Choose a ∈ M0 ⊆ M . Then R ∪ {a} is consistent by Corollary 1, which proves the necessity.
The proof is completed.
Definition 8. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M ⊆ B, Bi ⊆ B (i = 1, . . . ,m). If M ⊆B1 ∪ · · · ∪Bm, and M ∩Bi = ∅ holds for every i (1 � i � m), then M is called a choice of {B1, . . . , Bm}.Furthermore, if there is no other choice M0 of {B1, . . . , Bm} such that M0 � M , then M is called a
minimal choice of {B1, . . . , Bm}.Example 3. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, a, b, c ∈ B. Suppose B1 = {a, a′}, B2 = {a, b′, c},B3 = {b, b′}. Then it is easy to see that {a, b, c} is a choice, but not a minimal choice, of {B1, B2, B3},since its proper subset {a, b} is a choice of {B1, B2, B3} as well. Moreover, {a′, b′}, {a′, b, c} and {a, b}are minimal choices of {B1, B2, B3}.
Note that B1 ∪ · · · ∪ Bm is a choice of {B1, . . . , Bm}, which reveals the existence of choices of
{B1, . . . , Bm} in Boolean algebras. Furthermore, it can be seen from Example 3 that the minimal choices
of {B1, . . . , Bm} are not necessarily unique.
Proposition 6. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, Bi ⊆ B (i = 1, . . . ,m).
(i) If M ⊆ B1 ∪ · · · ∪Bm, and M ∩Bi (i = 1, . . . ,m) are all singletons, then M is a minimal choice of
{B1, . . . , Bm}.(ii) There is at least one minimal choice of {B1, . . . , Bm}.
Proof. (i) It is obvious that M is a choice of {B1, . . . , Bm}. Assume M is not minimal. Then there is
another choice M0 of {B1, . . . , Bm} such that M0 � M . Let a ∈ M −M0. Since M ⊆ B1 ∪ · · · ∪Bm and
M ∩Bi (i = 1, . . . ,m) are all singletons, there is a number k (1 � k � m) such that M ∩Bk = {a}. Sincea /∈ M0, we have M0 ∩ Bk = ∅, which contradicts the statement that M0 is a choice of {B1, . . . , Bm}.Therefore, M is a minimal choice of {B1, . . . , Bm}.
(ii) Choose a ∈ B1, and let M1 = {a}. For every k (i = 1, . . . ,m − 1), if Mk ∩ Bk+1 = ∅, then let
Mk+1 = Mk; if Mk ∩ Bk+1 = ∅ and Bk+1 − B1 ∪ · · · ∪ Bk = ∅, then choose b ∈ Bk+1 − B1 ∪ · · · ∪ Bk
and let Mk+1 = Mk ∪ {b}; if Mk ∩Bk+1 = ∅ and Bk+1 −B1 ∪ · · · ∪Bk = ∅, then Bk+1 ⊆ B1 ∪ · · · ∪Bk.
Choose c ∈ Bk+1. Then there is a subsequence i1, . . . , ij of 1, . . . , k such that c ∈ Bs (s = i1, . . . , ij) and
c /∈ Bt (t ∈ {1, . . . , k} − {i1, . . . , ij}). In this case, let Mk+1 = (Mk − Bi1 ∪ · · · ∪ Bij ) ∪ {c}. It will be
proved by induction that Mk is a minimal choice of {B1, . . . , Bk} (k = 1, . . . ,m) as follows.
Firstly, it is obvious thatM1 is a minimal choice of {B1}. Secondly, suppose thatMk is a minimal choice
of {B1, . . . , Bk}. If Mk ∩ Bk+1 = ∅, then Mk+1 = Mk is obviously a choice of {B1, . . . , Bk+1}. Assume
Mk+1 is not minimal. Then there is another choice G of {B1, . . . , Bk+1} such that G � Mk+1 = Mk. It
is easy to show that G is also a choice of {B1, . . . , Bk}, which contradicts the minimum of Mk.
IfMk∩Bk+1 = ∅ and Bk+1−B1∪· · ·∪Bk = ∅, then Mk+1 = Mk∪{b}, where b ∈ Bk+1−B1∪· · ·∪Bk. It
is obvious that Mk+1 is a choice of {B1, . . . , Bk+1}. Assume Mk+1 is not minimal. Then there is another
Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9 2049
choice H of {B1, . . . , Bk+1} such that H � Mk+1 = Mk ∪ {b}. Since H ∩Bk+1 = ∅ and b ∈ Mk+1 −Mk,
we have b ∈ H , and consequently H − {b} � Mk. It is easy to prove that H − {b} is also a choice of
{B1, . . . , Bk}, which contradicts the minimum of Mk.
If Mk ∩Bk+1 = ∅ and Bk+1 −B1 ∪ · · · ∪Bk = ∅, then Bk+1 ⊆ B1 ∪ · · · ∪Bk. Choose c ∈ Bk+1. Then
there is a subsequence i1, . . . , ij of 1, . . . , k such that c ∈ Bs (s = i1, . . . , ij) and c /∈ Bt (t ∈ {1, . . . , k} −{i1, . . . , ij}). If j = k, then c ∈ Bs (s = 1, . . . , k), and consequentlyMk+1 = (Mk−B1∪· · ·∪Bk)∪{c} = {c}is a minimal choice of {B1, . . . , Bk+1} by (i); if j < k, then Mk+1 = (Mk − Bi1 ∪ · · · ∪ Bij ) ∪ {c} is
obviously a choice of {B1, . . . , Bk+1}. Assume Mk+1 is not minimal. Then there is another choice Q of
{B1, . . . , Bk+1} such that Q � Mk+1. Since Q ∩ Bk+1 = ∅ and c ∈ Mk+1 − Mk, we have c ∈ Q. Let
Q1 = (Q−{c})∪(Mk∩Bi1 )∪· · ·∪(Mk∩Bij ), and choose d ∈ Mk+1−Q. Then d /∈ Mk∩Bs (s = i1, . . . , ij),
and d ∈ Mk−Q1 holds. ConsequentlyQ1 � Mk, andQ1 is also a choice of {B1, . . . , Bk}, which contradicts
the minimum of Mk. As a result, Mk+1 is a minimal choice of {B1, . . . , Bk+1}.From above, we know that Mm is a minimal choice of {B1, . . . , Bm}, and the proof is completed.
Remark 3. The inverse of (i) in Proposition 6 fails to hold, and a counterexample is as follows.
Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, a, b ∈ B. Suppose B1 = {a}, B2 = {a, b}, B3 = {b}. Then
{a, b} is a minimal choice of {B1, B2, B3}. However, {a, b} ∩B2 = {a, b} is not a singleton.
Definition 9. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, and M (⊆ B) be an inconsistent set with all
its minimal inconsistent subsets being B1, . . . , Bm. If G ⊆ B is a choice of {B1, . . . , Bm}, then G is called
a subtraction of M . Furthermore, if G is minimal, then G is called a minimal subtraction of M .
Example 4. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra same as Example 1. Suppose M = {a, a′, a →b, b, b′}. Then all minimal inconsistent subsets ofM are B1 = {a, a′}, B2 = {a, a → b, b′} and B3 = {b, b′}.Since {a, a → b, b} is a choice, but not a minimal choice, of {B1, B2, B3}, it is a subtraction, but
not a minimal subtraction, of M . Moreover, {a′, b′}, {a′, a → b, b} and {a, b} are minimal choices of
{B1, B2, B3} and consequently minimal subtractions of M .
In Example 4, let G1 = {a, a → b, b}, G2 = {a, b}. Then G1 is a subtraction of M , and G2 is a minimal
subtraction of M . Meanwhile, it is easy to infer that M −G1 = {a′, b′} is a consistent subset of M , and
M −G2 = {a′, a → b, b′} is a maximal consistent subset of M . More generally, the following proposition
holds.
Proposition 7. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, and M (⊆ B) be a finite inconsistent set
with all its minimal inconsistent subsets being B1, . . . , Bm. If G ⊆ B1 ∪ · · · ∪Bm, then
(i) M −G is consistent if and only if G is a subtraction of M .
(ii) M −G is a maximal consistent subset of M if and only if G is a minimal subtraction of M .
Proof. (i) Let M −G be a consistent set. Assume that G is not a subtraction of M ; that is, G is not a
choice of {B1, . . . , Bm}. Since G ⊆ B1∪· · ·∪Bm, there is a number k (1 � k � m) such that G∩Bk = ∅.Thus, Bk ⊆ M −G by Bk ⊆ M . Since Bk is inconsistent, M −G is inconsistent as well by Corollary 1,
and contradiction arises. This proves the necessity.
Let G be a subtraction of M . Assume that M −G is inconsistent. Then by Proposition 4, there is a
minimal inconsistent subset M0 of M − G. It is obvious that M0 is also a minimal inconsistent subset
of M . Consequently, there is a number j (1 � j � m) such that M0 = Bj . Since G is a subtraction
of M , we have G ∩ Bj = ∅. Choose a ∈ G ∩ Bj . Then a /∈ M − G, and consequently a /∈ Bj since
Bj = M0 ⊆ M −G. Contradiction arises, which proves the sufficiency.
(ii) Let M −G be a maximal consistent subset of M . By (i), G is a subtraction of M . Assume that
G is not minimal. Then there is another subtraction G1 of M such that G1 � G. By (i), M −G1 is a
consistent subset of M , and M −G1 � M −G, which contradicts the maximum of M −G. This proves
the necessity.
Let G be a minimal subtraction of M . By (i), M −G is a consistent subset of M . Assume that M −G
is not maximal. Then there is another consistent subset M1 of M such that M − G � M1. By (i),
M −M1 is a subtraction of M , and M −M1 � G, which contradicts the minimum of G. This proves the
sufficiency, and the proof is completed.
2050 Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9
Proposition 8. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, let M and R be finite consistent sets in B,
and let R be the refutation of M .
(i) If G is a minimal subtraction of M ∪ R with G ∩ R = ∅, then M −G is a maximal contraction of
M with respect to R.
(ii) If M1 is a maximal contraction of M with respect to R, then there is a minimal subtraction G of
M ∪R such that M1 = M −G with G ∩R = ∅.Proof. (i) Let G be a minimal subtraction of M ∪R with G∩R = ∅. By Proposition 7, (M −G)∪R =
(M ∪ R) − G is a maximal consistent subset of M ∪ R, and consequently M − G is consistent with R.
Assume that M −G is not a maximal contraction of M with respect to R. Then there is a set M0 ⊆ M
such that M0 ∪R is consistent and M −G � M0. Therefore, M0 ∪R � (M −G) ∪R, which contradicts
the maximum of (M −G) ∪R.
(ii) Let M1 be a maximal contraction of M with respect to R, and G = M −M1. Then it is obvious
that M1 = M −G, and we only need to prove that G is a minimal subtraction of M ∪R with G∩R = ∅.Since M1 is a maximal contraction of M with respect to R, M1∪R is consistent. Assume that M1 ∪R
is not a maximal consistent subset of M ∪R. Then there is a set M2 ⊆ M ∪R such that M2 is consistent
and M1∪R � M2. Consequently, (M ∩M2)∪R = M2 is consistent, and M ∩M2 � M1, which contradicts
the maximum of M1. Therefore, M1 ∪R is a maximal consistent subset of M ∪R.
Assume that G∩R = ∅, and choose a ∈ G∩R. Then M1 ∪ {a} ∪R = M1 ∪R is a maximal consistent
subset of M ∪ R. Since a ∈ G = M − M1, M1 ∪ {a} is a subset of M consistent with R. Note that
a /∈ M1. Then M1 ∪ {a} � M1, which contradicts the maximum of M1. Therefore, G ∩R = ∅.From above, since G ∩R = ∅, M1 ∪R = (M −G) ∪R = (M ∪R)−G is a maximal consistent subset
of M ∪R. By Proposition 7, G is a minimal subtraction of M ∪R. The proof is completed.
It is worth noting that Proposition 8 constructs a one-to-one corresponding relation in Boolean algebras
between the set of all maximal contractions for a given finite consistent setM with respect to its refutation
R and the set of all minimal subtractions of M ∪ R whose intersection with R is empty. As a result, in
order to find out all the maximal contractions for M with respect to R, we can equivalently compute all
the minimal subtractions of M ∪R whose intersection with R is empty by taking the following steps:
(i) Compute all the minimal inconsistent subsets, B1, . . . , Bm, of M ∪R.
(ii) Compute all the minimal choices of {B1, . . . , Bm} whose intersection with R is empty.
(iii) For every minimal choice G obtained in (ii), compute its complementary set M − G. All such
complementary sets constitute the set of all maximal contractions of M with respect to R.
Note that there are two key issues in the steps above. One is the computation of all the minimal
inconsistent subsets for a given finite inconsistent set, and the other is the computation of all the minimal
choices for finite given sets, which will be solved separately in the sequel.
Actually, ref. [9] proposed an algorithm to enumerate all minimal choices for finite given sets of formulas
in first-order languages. Similarly, Algorithm Principle I to compute all minimal choices for finite given
sets in Boolean algebras is outlined, and its correctness is proved as follows. Let (B,∨,∧,′ , 0, 1) be a
Boolean algebra, Bi ⊆ B (i = 1, . . . ,m). Algorithm Principle I considers {B1, . . . , Bm} as the input, and
Φm as the output, Φi (1 � i � m) storing all the minimal choices of {B1, . . . , Bi}.1) Let Φ0 = {∅}.2) For every Θ ∈ Φi−1.
(i) If Θ ∩ Bi = ∅, then for every a in Bi, check whether there is some set Λ in Φi such that Λ is a
subset of Θ ∪ {a}. If such Λ does not exist, then add Θ ∪ {a} into Φi.
(ii) If Θ ∩Bi = ∅, then check whether there is some set Γ in Φi such that Θ is a subset of Γ . If such
Γ exists, then delete Γ from Φi. After checking all the sets in Φi, add Θ into Φi.
Proposition 9. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, Bi ⊆ B (i = 1, . . . ,m). If the input of Algo-
rithm Principle I is {B1, . . . , Bm}, then its output Φm is the set of all minimal choices of {B1, . . . , Bm}.Proof. In the following, it will be proved by induction that Φk is the set of all minimal choices of
{B1, . . . , Bk} (1 � k � m).
Firstly, since Φ0 = {∅}, there is only one element Θ = ∅ in Φ0, and Θ ∩ B1 = ∅. For every a ∈ B1,
Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9 2051
since Θ ∪ {a} = {a}, and there is no subset of {a} in Φ1, such {a}’s are all added into Φ1. Therefore,
Φ1 = {{a} | a ∈ B1}, and it is easy to infer that Φ1 is the set of all minimal choices of {B1}.Secondly, suppose Φk−1 is the set of all minimal choices of {B1, . . . , Bk−1}. LetΘ ∈ Φk−1. IfΘ∩Bk = ∅,
then for every a in Bk, it is obvious that Θ ∪ {a} is a choice of {B1, . . . , Bk}. Note that Θ ∪ {a} will be
added into Φk only in the case where there is no subset of Θ ∪ {a} in Φk. If Θ ∩ Bk = ∅, then it can be
proved that Θ is also a minimal choice of {B1, . . . , Bk}, and Θ will be added into Φk only after deleting
all the supersets of Θ in Φk. As a result, Θ1 � Θ2 always fails to hold for arbitrary sets Θ1, Θ2 in Φk.
Moreover, when kth loop is finished, only the minimal ones of the following set
{Θ | Θ ∈ Φk−1, Θ ∩Bk = ∅} ∪ {Θ ∪ {a} | Θ ∈ Φk−1, a ∈ Bk, Θ ∩Bk = ∅}
will remain in Φk. Therefore, Φk is the set of all minimal choices of {B1, . . . , Bk}.From above, Φm is the set of all minimal choices of {B1, . . . , Bm}. The proof is completed.
Proposition 10. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, let M and R be finite consistent sets in
B, and let R be the refutation of M . Suppose that all the minimal inconsistent subsets of M ∪ R are
B1, . . . , Bm. If Algorithm Principle I considers {B1 − R, . . . , Bm − R} as its input, then the output Φm
is the set of all minimal subtractions of M ∪R whose intersection with R is empty.
Proof. By Proposition 9, if the input of Algorithm Principle I is {B1−R, . . . , Bm−R}, then its output Φm
is the set of all minimal choices of {B1−R, . . . , Bm−R}. Let Θ ∈ Φm. Then Θ ⊆ (B1−R)∪· · ·∪(Bm−R)
and Θ ∩ (Bi − R) = ∅ (1 � i � m). Consequently, Θ ⊆ B1 ∪ · · · ∪ Bm, Θ ∩ Bi = ∅ (1 � i � m) and
Θ ∩ R = ∅. Therefore, Θ is a choice of {B1, . . . , Bm} with Θ ∩ R = ∅; that is, Θ is a subtraction of
M ∪ R with Θ ∩ R = ∅. Assume that Θ is not a minimal subtraction of M ∪ R with Θ ∩ R = ∅. Then
there is another choice Θ1 of {B1, . . . , Bm} such that Θ1 ∩ R = ∅ and Θ1 � Θ. It is easy to prove that
Θ1 is also a choice of {B1 − R, . . . , Bm − R}, which contradicts Θ ∈ Φm. Therefore, Θ is a minimal
subtraction of M ∪R with Θ ∩R = ∅. Similarly, if Θ is a minimal subtraction of M ∪R with Θ ∩R = ∅,then Θ is a minimal choice of {B1 − R, . . . , Bm − R}. Since Φm contains all the minimal choices of
{B1 −R, . . . , Bm −R}, Φm is the set of all minimal subtractions of M ∪R whose intersection with R are
empty. The proof is completed.
It can be seen from above that the concepts of R-refutation, R-contraction and minimal subtraction
in first-order languages discussed in [9] have been consistently transplanted into Boolean algebras, where
part of similar results (e.g. Propositions 7–10) still hold. Note that, the algorithm for computing all
minimal inconsistent subsets for a given set of Horn clauses proposed in [9] is mainly based on the SLD-
resolution [9,13] for Horn clauses, which owns special properties that can not be easily transplanted into
Boolean algebras any more. However, a method can be given to find out all minimal inconsistent subsets
for a given set in Boolean algebras from the perspective of characteristics of Boolean algebras themselves
in the sequel.
Proposition 11. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M = {x1, . . . , xn} ⊆ B. Then M is a
minimal inconsistent set if and only if x1∧· · ·∧xn = 0, and xi1 ∧· · ·∧xik = 0 holds for every subsequence
i1, . . . , ik of 1, . . . , n.
Proof. We only prove the necessity, since the sufficiency can be similarly obtained. Suppose M is a
minimal inconsistent set. Then by Corollary 1, we have ∧M = x1 ∧ · · · ∧ xn = 0. Since M − {a} is
consistent for every a ∈ M , x1 ∧ · · · ∧ xj−1 ∧ xj+1 ∧ · · · ∧ xn > 0 holds for every j (1 � j � n). Thus,
for every subsequence i1, . . . , ik of 1, . . . , n, we can suppose j /∈ {i1, . . . , ik} (1 � j � n) without loss of
generality, and then xi1 ∧ · · · ∧ xik � x1 ∧ · · · ∧ xj−1 ∧ xj+1 ∧ · · · ∧ xn > 0. The proof is completed.
Based on Proposition 11, Algorithm Principle II to compute all minimal inconsistent subsets for a
given finite inconsistent set in Boolean algebras is outlined, and its correctness is proved as follows. Let
(B,∨,∧,′ , 0, 1) be a Boolean algebra, and let M = {x1, . . . , xn} be an inconsistent set in B. Algorithm
Principle II considers M = {x1, . . . , xn} as the input, and Φ as the output, Φi (0 � i � n) storing all the
subsets of M with power i.
(i) Let Φ0 = {∅}, Φ = ∅.
2052 Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9
(ii) For every Θ ∈ Φi−1, check whether each a ∈ M is contained in Θ. If a /∈ Θ, then add Θ ∪ {a} into
Φi.
(iii) For every Θ ∈ Φi, if ∧Θ = 0, then check whether there is a set Λ in Φ such that Λ is a subset of
Θ. If such Λ does not exist, then add Θ into Φ.
Proposition 12. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra, M = {x1, . . . , xn} be an inconsistent set in
B. If the input of Algorithm Principle II is M , then the output Φ is the set of all minimal inconsistent
subsets of M .
Proof. First of all, it is easy to prove that Φi (0 � i � n) is the set of all subsets of M with power
i. Moreover, let Θ ∈ Φ. Then Θ is inconsistent since ∧Θ = 0. By Proposition 4, there is a minimal
inconsistent subset Λ of Θ. Suppose |Θ| = k, |Λ| = j. Then Θ ∈ Φk, Λ ∈ Φj , and j � k. Assume that
j < k. Then Λ is a proper subset of Θ. Since Λ is a minimal inconsistent set, by Proposition 11 we have
∧Λ = 0, and ∧Λ1 = 0 for every proper subset Λ1 of Λ. Consequently, Λ ∈ Φ. By Algorithm Principle II,
Θ1 � Θ2 always fails to hold for arbitrary sets Θ1, Θ2 in Φ, which contradicts Λ � Θ. Therefore, k = j,
and Θ = Λ is a minimal inconsistent set. As a result, all the elements in Φ are minimal inconsistent
subsets of M . Besides, since Algorithm Principle II considers all the subsets of M , Φ is the set of all
minimal inconsistent subsets of M . The proof is completed.
Example 5. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra same as Example 1, and M = {a, a′, b, b′} be the
input of Algorithm Principle II.
Since ∅ is the unique element in Φ0, and x /∈ ∅ holds for every x ∈ M , such {x}’s are all added into
Φ1. Therefore, Φ1 = {{a}, {a′}, {b}, {b′}}. Besides, ∧Θ = 0 holds for every Θ ∈ Φ1, and consequently Φ
remains ∅.Let Θ = {a} ∈ Φ1. After checking all the elements in M , {a, a′}, {a, b}, {a, b′} are added into Φ2.
Similar discussion can be made for other sets in Φ1, and finally Φ2={{a, a′}, {a, b}, {a, b′}, {a′, b}, {a′, b′},{b, b′}}. As for {a, a′} ∈ Φ2, since a ∧ a′ = 0, and there is no subset of {a, a′} in Φ, {a, a′} is added into
Φ. After making similar discussion for other sets in Φ2, we can obtain Φ = {{a, a′}, {b, b′}}.Since every set in Φ3, Φ4 is a superset of some element in Φ, there will be no more new element added
into Φ. Finally, the output is Φ = {{a, a′}, {b, b′}}. By Proposition 12, all the minimal inconsistent
subsets of M are {a, a′} and {b, b′}.Example 6. In the first-order language KL (see [11,13] for details), the set of all formulas is denoted
by FL, and ∼ is the provable equivalence relation on FL, that is,
A ∼ B if and only if A → B,B → A are theorems. (6)
It can be proved that ∼ is a congruence relation [10,11] on FL of type (¬,∨,∧). Consequently, we obtaina quotient class of FL by ∼, denoted by [FL], that is, [FL] = FL/ ∼. Define relation � and operators
∨,∧,′ on [FL] as
[A] � [B] if and only if A → B is a theorem, (7)
[A] ∨ [B] = [A ∨B], [A] ∧ [B] = [A ∧B], [A]′ = [¬A]. (8)
Then it can be inferred that ([FL],∨,∧,′ , [¬�], [�]) constitutes a Boolean algebra, where � represents a
theorem.
Suppose Γ and Δ are finite consistent [9,11] sets of closed formulas [11] in FL. If Γ ∪Δ is inconsistent,
especially when Δ is composed of atoms or their negations, then Δ is said to be the R-refutation of Γ in
refs. [7–9]. In this case, if there is a maximal subset Γ1 of Γ such that Γ1∪Δ is consistent, then Γ1 is said
to be an R-contraction of Γ with respect to Δ in refs. [7-9]. Assume that there are no formulas provably
equivalent to each other in Γ ∪Δ. Let M = {[A] | A ∈ Γ}, R = {[A] | A ∈ Δ}. Note that the inference
rule Gen (brief notation for Generalization) has no real effect under the scope of closed formulas. In fact,
suppose that A(x) is a closed formula in FL containing the variable x, then it can be proved that A(x)
and (∀x)A(x) are logically equivalent as well as provably equivalent with each other. As a result, since Γ
and Δ consist of closed formulas, it can be proved that M and R are finite consistent sets in the Boolean
Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9 2053
algebra [FL] in the sense of Definition 4, and M ∪ R is inconsistent; that is, R is the refutation of M
in the sense of Definition 6. Since [FL] is a Boolean algebra, we can use Algorithm Principles I, II and
Propositions 8, 10 to compute all the maximal contractions of M with respect to R. Supposing Λ is one
of the maximal contractions of M with respect to R, without loss of generality, we can further assume
that Λ = {[A1], . . . , [An]} with Ai ∈ Γ (i = 1, . . . , n). Let Θ = {A1, . . . , An}. Then Θ ⊆ Γ , and Θ is an
R-contraction of Γ with respect to Δ. As a result, all the R-contractions of Γ with respect to Δ can be
found out by this process.
From above, the problem of computing all R-contractions for a given finite set of closed formulas in
first-order languages can be equivalently converted into the one in Boolean algebras proposed in this
paper.
Remark 4. Since ([FL],∨,∧,′ , [¬�], [�]) is a Boolean algebra by Example 6, some results in first-order
languages can be generalized to usual Boolean algebras, which contributes to the foundation of our work
above. However, ([FL],∨,∧,′ , [¬�], [�]) owns many special properties, and some concepts, such as literals
and Horn clauses, can not be directly transplanted in usual Boolean algebras. In order to modify this
limitation, the concept of basic element is introduced for a special type of Boolean algebras in the next
section.
4 Maximal contractions in Boolean algebras generated by basic elements
Definition 10. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra. Suppose that there is a set E ⊆ B satisfying
the following conditions:
(i) E ∪ E′ is an antichain, where E′ = {x′ | x ∈ E}.(ii) For finite arbitrary elements a1, . . . , an in E ∪E′, a1 ∧ · · · ∧ an = 0 if and only if there are numbers
i, j (1 � i, j � n) such that ai = a′j .(iii) B is a free algebra [11,14] of type (∨,∧,′ ) generated by E.
Then each element in E is called a basic element of B, and B is called to be a Boolean algebra generated
by E.
Example 7. (i) Let X = {a, b}, and B =P(X) be the powerset of X . Then (B,∪,∩,′ , ∅, X) is a
Boolean algebra, where ∪,∩,′ are union, intersection and complementary operators of sets, respectively.
Suppose E = {{a}}. Then it is easy to infer that E is the set of basic elements of B, and B is a Boolean
algebra generated by E.
(ii) In the classical propositional logic L (see refs. [11,13,15] for details), S = {p1, p2, . . .} is the set
of atoms, F (S) is the set of all formulas, and ∼ is the provable equivalence relation on F (S); that
is, ∼ is defined by (6). Ref. [11] proved that ∼ is a congruence relation on F (S), and obtained a
quotient class of F (S) by ∼, denoted by [F ], that is, [F ] = F (S)/ ∼. Define relation � and operators
∨,∧,′ on [F ] by (7) and (8), respectively. Ref. [11] proved that ([F ],∨,∧,′ , [¬�], [�]) constitutes a
Boolean algebra, where � represents a theorem. Let E = {[p] | p ∈ S}. Then it is easy to infer that
E ∪ E′ = {[p], [¬p] | p ∈ S} is an antichain in [F ], and for finite arbitrary elements [A1], . . . , [An] in
E ∪ E′, [A1] ∧ · · · ∧ [An] = [A1 ∧ · · · ∧ An] = [¬�] if and only if there are numbers i, j (1 � i, j � n)
such that Ai ∼ ¬Aj , that is, [Ai] = [Aj ]′. Besides, [F ] is a free algebra of type (∨,∧,′ ) generated by E.
Therefore, ([F ],∨,∧,′ , [¬�], [�]) is actually a Boolean algebra generated by E.
Definition 11. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra generated by the set E of basic elements,
x ∈ B. If there are elements a1, . . . , an ∈ E ∪ E′ such that x = a1 ∨ · · · ∨ an, then x is called a clause
in B, denoted by x = 〈a1, . . . , an〉. Furthermore, if at most one of a1, . . . , an is contained in E, then
x = 〈a1, . . . , an〉 is called a Horn clause in B.
In the classical propositional logic L, atoms and their negations are said to be positive literals and
negative literals, respectively, and the disjunction of finite literals is said to be a clause. Furthermore, a
clause is said to be a Horn clause if it has at most one positive literal [9,11]. Suppose A is a clause (or
Horn clause) in L. Then it can be inferred that [A] is a clause (or Horn clause) in the Boolean algebra
2054 Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9
[F ] constructed in Example 7 (ii) in the sense of Definition 11. As a result, the concepts of clause and
Horn clause proposed in the present paper are consistent with the ones in propositional logic.
Definition 12. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra generated by the set E of basic elements, and
let xi = 〈a(i)1 , a(i)2 , . . . , a
(i)ki〉 (i = 1, . . . , n) be clauses in B. Define Cartesian product x1 × · · · × xn of
x1, . . . , xn as
x1 × · · · × xn = {(a(1)j1, a
(2)j2
, . . . , a(n)jn
) | 1 � ji � ki (i = 1, · · · , n)}. (9)
Proposition 13. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra generated by the set E of basic elements,
and let M = {x1, . . . , xn} be a set of clauses in B. Then M is a minimal inconsistent set if and only
if each vector in x1 × · · · × xn contains a complementary pair, and there is a vector in xi1 × · · · × xim
containing no complementary pair for every subsequence i1, . . . , im of 1, . . . , n.
Proof. Suppose xi = 〈a(i)1 , . . . , a(i)ki〉 (i = 1, . . . , n). If M is a minimal inconsistent set, then ∧M =
x1 ∧ · · · ∧ xn = 0. By Eqs. (1) and (2), we have x1 ∧ · · · ∧ xn = (a(1)1 ∨ · · · ∨ a
(1)k1
) ∧ · · · ∧ (a(n)1 ∨
· · · ∨ a(n)kn
) = ∨{b1 ∧ · · · ∧ bn | (b1, . . . , bn) ∈ x1 × · · · × xn} = 0. Therefore, b1 ∧ · · · ∧ bn = 0 holds for
every vector (b1, . . . , bn) in x1 × · · · × xn, and consequently b1, . . . , bn contains a complementary pair by
Definition 11. Besides, since M is minimal, M − {xj} is consistent for every j (1 � j � n), and
consequently x1 ∧ · · · ∧ xj−1 ∧ xj+1 ∧ · · · ∧ xn > 0. Let i1, . . . , im be an arbitrary subsequence of 1, . . . , n.
Without loss of generality, we can suppose j /∈ {i1, . . . , im} (1 � j � n). Then xi1 ∧ · · · ∧xim � x1 ∧ · · · ∧xj−1∧xj+1∧· · ·∧xn > 0. Since xi1∧· · ·∧xim = ∨{bi1∧· · ·∧bim | (bi1 , . . . , bim) ∈ xi1×· · ·×xim} > 0, there
is some vector (bi1 , . . . , bim) in xi1 × · · · × xim such that bi1 ∧ · · · ∧ bim > 0, and consequently bi1 , . . . , bimcontains no complementary pair. This proves the necessity, and the sufficiency can be obtained similarly.
The proof is completed.
Based on Proposition 13, similar to Algorithm Principle II, Algorithm Principle III to compute all
minimal inconsistent subsets for a given finite inconsistent set of clauses in Boolean algebras generated
by basic elements is outlined as follows. Let (B,∨,∧,′ , 0, 1) be a Boolean algebra generated by the set E
of basic elements, and let M = {x1, . . . , xn} be an inconsistent set of clauses in B. Algorithm Principle
III considers M = {x1, . . . , xn} as the input, and Φ as the output, Φi (0 � i � n) storing all the subsets
of M with power i.
(i) Let Φ0 = {∅}, Φ = ∅.(ii) For every Θ ∈ Φi−1, check whether each a ∈ M is contained in Θ. If a /∈ Θ, then add Θ ∪ {a} into
Φi.
(iii) For every Θ ∈ Φi, suppose Θ = {xj1 , . . . , xji}. Check all the vectors in xj1 × · · · × xji . If each
vector contains some complementary pair, then check whether there is a set Λ in Φ such that Λ is a subset
of Θ. If such Λ does not exist, then add Θ into Φ.
By Propositions 12,13, it can be easily proved that the output Φ of Algorithm Principle III is the set
of all minimal inconsistent subsets of M , and the proof is omitted.
Example 8. In the classical propositional logic L [11], S = {p1, p2, . . .} is the set of all atoms, and
F (S) is the set of all formulas. Like [9], this paper defines the concepts of R-refutation and R-contraction
in L. Suppose Γ and Δ are finite consistent [11] sets of clauses in L. If Γ ∪Δ is inconsistent, especially
when Δ is composed of literals, then Δ can be said to be the R-refutation of Γ . In this case, if there
is a maximal subset Γ1 of Γ such that Γ1 ∪Δ is consistent, then Γ1 can be said to be an R-contraction
of Γ with respect to Δ. Assume that there are no formulas provably equivalent to each other in Γ ∪Δ.
Let M = {[A] | A ∈ Γ}, R = {[A] | A ∈ Δ}. Then it can be proved that M and R are finite consistent
sets in the Boolean algebra [F ] constructed in Example 7 (ii) in the sense of Definition 4, and M ∪ R is
inconsistent; that is, R is the refutation of M in the sense of Definition 6. Since [F ] is a Boolean algebra
generated by E = {[p] | p ∈ S}, we can use Algorithm Principles II, III and Propositions 8,10 to compute
all the maximal contractions of M with respect to R. Supposing Λ is one of the maximal contractions
of M with respect to R, without loss of generality, we can further assume that Λ = {[A1], . . . , [An]} with
Ai ∈ Γ (i = 1, . . . , n). Let Θ = {A1, . . . , An}. Then Θ ⊆ Γ, and Θ is an R-contraction of Γ with respect
to Δ. As a result, all the R-contractions of Γ with respect to Δ can be found out by this process.
Shi H X, et al. Sci China Inf Sci September 2012 Vol. 55 No. 9 2055
From above, the problem of computing all R-contractions for a given finite set of clauses, especially
Horn clauses, in the classical propositional logic can be equivalently converted into the one in Boolean
algebras generated by basic elements proposed in the present paper.
5 Conclusions
In the present paper, the concept of deductive element is introduced in Boolean algebras, and correspond-
ing theory of consistency is concerned firstly. Secondly, the concepts of refutation, maximal contraction
and minimal subtraction are proposed for Boolean algebras by generalizing the ones in [5–9], and a one-
to-one corresponding relation is constructed between maximal contractions and minimal subtractions,
similarly. Two algorithm principles are proposed to compute all minimal inconsistent subsets for a given
finite inconsistent set as well as all minimal choices for finite given sets in Boolean algebras, respectively,
which consequently provide a method to find out all maximal contractions in the sense of Boolean alge-
bras. Moreover, it is pointed out that the quotient algebra of the first-order language with respect to its
provable equivalence relation constitutes a Boolean algebra, and the computation of R-contractions for
closed formulas in the first-order languages can be converted into the one in Boolean algebras. Lastly, the
concept of basic element is introduced in Boolean algebras, which contributes to the definitions of clause
and Horn clause transplanted from logic to a special type of Boolean algebras generated by basic ele-
ments. It is pointed out that the computation of R-contractions for clauses in the classical propositional
logic can be converted into the one in Boolean algebras generated by basic elements. How to introduce
SLD-resolution in Boolean algebras, and how to expand the theory of maximal contractions suitable for
a more extensive structure will be investigated in a forthcoming paper.
Acknowledgements
This work was supported by National Natural Science Foundation of China (Grant Nos. 11171200, 61005046,
61103133) and Fundamental Research Funds for the Central Universities (Grant No. GK201004006). The authors
would like to express their sincere thanks to the reviewers for their valuable comments and suggestions.
References
1 Alchourron C, Gardenfors P, Makinson D. On the logic of theory change: Partial meet functions for contraction and
revision. J Symb Log, 1985, 50: 510–530
2 Alchourron C, Makinson D. On the logic of theory change: Safe contraction. Studia Logica, 1985, 44: 405–422
3 Katsuno H, Mendelzon A O. Propositional knowledge base revision and minimal change. Artif Intell, 1991, 52: 263–294
4 Darwixhe A, Pearl J. On the logic of iterated belief revision. Artif Intell, 1997, 89: 1–29
5 Li W. Logical verification of scientific discovery. Sci China Inf Sci, 2010, 53: 677–684
6 Li W. A logical framework for evolution of specifications. In: Programming Languages and Systems–ESOP’94.
LNCS788. Berlin: Springer-Verlag, 1994. 394–408
7 Li W. R-calculus: an inference system for belief revision. Comput J, 2007, 50: 378–390
8 Li W. Mathematical Logic: Foundations for Information Science. Basel: Birkhaeuser Publish, 2009. 139–164
9 Luo J, Li W. An algorithm to compute maximal contractions for Horn clauses. Sci China Inf Sci, 2011, 54: 244–257
10 Davey B A, Priestley H A. Introduction to Lattices and Order. New York: Cambridge University Press, 1990. 130–158
11 Wang G J, Zhou H J. Introduction to Mathematical Logic and Resolution Principle. Beijing: Science Press, Oxford:
Alpha Science International Limited, 2009. 5–96
12 Xiong J C. Lecture Notes on Point Set Topology (in Chinese). 3nd ed. Beijing: Higher Education Press, 2003. 63–73
13 Gallier J H. Logic for Computer Science: Foundations of Automatic Theorem Proving. New York: Harper & Row,
1986. 28–218
14 Wang G J. Non-classical Mathematical Logic and Approximate Reasoning (in Chinese). 2nd ed. Beijing: Science
Press, 2008. 1–15
15 Wang G J, Fu L, Song J S. Theory of truth degrees of formulas in two-valued propositional logic. Sci China Ser
A-Math, 2002, 45: 998–1008