A Tree Construction of the Preferable Answer Sets for Prioritized Basic Disjunctive Logic Programs

Transcription

1 A Tree Construction of the Preferable Answer Sets for Prioritized Basic Disjunctive Logic Programs Zaiyue Zhang 1, Yuefei Sui 2, and Cungen Cao 2 1 Department of Computer Science, Jiangsu University of Science and Technology njzzy@yzcn.net 2 Key Laboratory of Intelligent Information Processing, Institute of Computing Technology, Chinese Academy of Sciences suiyyff@hotmail.com, cgcao@ict.ac.cn Abstract. One of the most important works in the investigation of logic programming is to define the semantics of the logic programs and to find the preferable answer set of them. There are so far three methods can be used to establish the semantics of the logic programs, i.e., the means of model, fixpoint and proof theory. According to the form of the rules contained in a logic program, different logic program classes can be defined. Although well-defined semantics exist for some restricted classes of programs like Horn and stratified Programs, the declarative semantics of the more general program classes are still the subject of research. In this paper, the properties of the basic disjunctive logic programming are studied, and the notion of double prioritization is introduced, that is, the prioritization on both literals and clauses, by which the most preferable answer set of a basic disjunctive logic program is defined. In order to obtain the most preferable answer set of a basic disjunctive logic program explicitly, a recursion-theoretic construction called tree method is given. Keywords: logic programs, double priority, tree method. 1 Introduction The study of the logic programming originates in the mid-seventies last century ([1]). As a tool, logic programming has been widely used in the areas of computer science such as knowledge representation, knowledge reasoning, deductive databases, artificial intelligence etc. The important work in areas of logic programming is to define the semantics of a logic program and find the preferable answer set of the program. There are so far three methods can be used to establish the semantics of the logic programs, i.e., the means of model, fixpoint This work is supported by the National Natural Science Foundation (grant no , ), and the National 973 Programme (grants no. 2003CB and G ). J.-Y. Cai, S.B. Cooper, and A. Li (Eds.): TAMC 2006, LNCS 3959, pp , c Springer-Verlag Berlin Heidelberg 2006

2 A Tree Construction of the Preferable Answer Sets 589 and proof theory. These methods are summarized in [2]. According to the form of the rules contained in a logic program, different logic program classes can be defined. Extended logic programs and extended disjunctive logic programs use both classical negation and negation as failure. Their semantics is based on the method of stable models, which are extensively studied in paper [3]. The simplest class of the logic programs is Horn logic programs, which have been proved to be r.e. (recursively enumerable) complete. Analogical property also holds for stratified programs ([4], [5], [6], [7]), that is, for every natural number n, the n-stratified logic programs are Σ 0 n -complete. The semantics of logic programs are the logical consequences of the literals inferred from the logic programs. Answer set is usually used as the semantic model of a logic program. Although well-defined semantics exist for some restricted classes of programs like Horn and stratified Programs, the declarative semantics of the more general program classes are still the subject of research. Hence, it will be interesting to find an efficient process used to obtain an answer set of a general logic program. In this paper the properties of the basic disjunctive logic programming are discussed, and the notion of the double prioritization on both literals and clauses is introduced. By using prioritization, the most preferable answer set of a basic disjunctive logic program is defined. In order to find the most preferable answer set of a given basic disjunctive logic program, a mechanism called tree method is introduced. The paper is organized as follows. In section 2, There are some terminologies and notations used in the paper, and also in which the notion of the double prioritized basic disjunctive logic program is introduced, and the preferable answer set based on double prioritization is discussed. In section 3, the basic notions of the tree method are introduced, and the most preferable answer set of a basic disjunctive logic program is defined. In section 4, a recursion-theoretic construction for the most preferable answer set of a double prioritized basic disjunctive logic program is established, and an example is given to show how the construction works. Section 5 is verification. The last section is for short concluding remarks. 2 Preliminaries We begin with a nonempty set S of symbols called atoms. The choice of S determines the language of the programs. An atoms is also called a positive literal; a negative literal is an atom preceded by the classical negation symbol. A literal is a positive literal or a negative literal. The set of literals will be denoted by Lit S,orsimplyLit. For any literal l, the literal l and l are said to be complementary. A set of literals is inconsistent if it contains a complementary pair, and consistent otherwise. A logic program based on set S consists of clauses, or rules, of the form: l 1,..., l k,not l k+1,..., not l s l s+1,..., l m,not l m+1,..., not l n where n m s k 0, l i Lit S for all 1 i n and not is negation as failure.

3 590 Z. Zhang, Y. Sui, and C. Cao Definition 1. Aclauseissaidtobebasic disjunctive if every literal in the clause is ground, i.e., contains no variables, and there is no not in it. A logic program Π is called basic disjunctive logic program (BDLP) if the clauses contained in the program are all basic disjunctive. If π is a clause in a BDLP then π has the form: l 1,..., l n l 1,..., l m where l i,l j for 1 i n, 1 j m are literals. The set {l 1,..., l n }, denoted by head(π), is called the head of π; andtheset{l 1,..., l m }, denoted by body(π), is called the body of π. Aclauseπ is called afactif body(π) =. Definition 2. The semantics of BDLG is the answer set semantics. Let Π be a BDLP and A Lit Π,whereLit Π is the set of literals occur in Π. A is said to be an answer set of Π iff it satisfies the following conditions: (2.1) For each clause l 1,..., l n l 1,..., l m of Π, {l 1,..., l m} A implies l i A for some i, 1 i n. (2.2) A does not contain both l and l for any literal l. An answer set A of Π is said to be minimal if no B A satisfies both (2.1) and (2.2). An answer set of a logic program can be regarded as a set of consequences of the logic program. A BDLP is called consistent if it has an answer set. There are usually more than one answer set for a BDLP Π. Usually, it has a high computational complexity to select a preferable answer set among the answer sets of a BDLP. One way to reduce the computational complexity is to introduce the notion of the priority. The priorities in logic programs are discussed in many papers (see [8],[9] etc,). The priority introduced in a logic program can reduce the number of the answer sets among which the most preferable one is selected. For a non-disjunctive logic programs, the unique answer set can be fixed if the clauseprioritization is introduced. For a normal disjunctive logic program, i.e., with no classical negation in it, the same thing can be done if the literal-prioritization is used. However, to fix an answer set for a BDLP, the double prioritization on both literals and clauses must be considered. We look at the following example. Example 1. let Π be a logic program containing the following clauses: One can verify that the following sets are all answer sets of Π. π 1 : wetgrass π 2 : rain sprink wetgrass π 3 : dryroad π 4 : rain undertree dryroad. S 1 = {wetgrass, rain, dryroad, undertree}, S 2 = {wetgrass, dryroad, rain, sprink}, S 3 = {wetgrass, sprink, dryroad, undertree},

4 A Tree Construction of the Preferable Answer Sets 591 If there is only a priority order between the clauses in Π, say π 1 <π 2 < π 3 <π 4, then one can hardly to determine among S 1,S 2 and S 3 that which one would be a preferable answer set of Π. However, if the double priority on both clauses and literals is considered, say π 1 <π 2 <π 3 <π 4 together with that rain sprink and rain undertree, then the set S 1 = {wetgrass, rain, dryroad, undertree} maybechosenasananswersetofπ. The reason is that to build the semantics of the Π according to the double priority given as above, the clause π 1 will be considered firstly and the literal wetgrass is selected, and then for π 2, the literal rain is selected since rain has a higher priority than sprink does, for π 3 the literal dryroad is selected and for π 4 only undretree can be selected since π 2 has a higher priority than π 4 does and should be satisfied firstly. If the priority relation on the clauses is changed, say π 1 <π 4 <π 3 <π 2,and the priority on the literals is also rain sprink and rain undertree, then we prefer to choose S 2 = {wetgrass, dryroad, rain, sprink} as an answer set of Π, since this time π 4 has a higher priority than π 2 does and should be satisfied before π 2 is. The example shows that to choose a suitable answer set for a disjunctive logical program, the double priority on both clauses and literals must be considered. It is usually based on the practical application to build a double priority for a DBLS, which, however, will not be investigated in this paper. What we considered is how to find a suitable answer set for a double prioritized DBLP. In the following of this paper, we shall introduce a method that can be used to construct a suitable answer set for a double prioritized DBLP. 3 The Tree Method and the Preferable Answer Set In this section, we shall introduce the notion of the tree method and define the preferable answer set of a BDLP. Let be a linear ordering on the set Π = {π i } i<ω. Without loss of generality, we may assume that π i π j for i j. Ifπ i π j, i.e., π i π j and i<jthen we say that π i has a higher priority than π j. The same symbol is also used to establish the priority between the literals in a head of a clause. If head(π) = {l 1,,l n } for some clause π then we may assume that l 1 l 2 l n.if l i l j then we say that l i has a higher priority than l j,orl i is less than l j. Let Π = {π i } i ω be a BDLP with the double priority defined as above. We shall construct an answer set of Π by stages. At stage s, weconstructa set A s such that A s will do its best to satisfy the clauses so far as we have scanned. We say a clause π i requires attention at stage s if body(π i ) A s and head(π i ) A s =. Whenaclause,π say, requires attention at stage s we shall take some actions to modify the set A s, i.e., either to put in A s some literal of

5 592 Z. Zhang, Y. Sui, and C. Cao head(π) or extract some literal of body(π) froma s. We say that a literal l is consistent with a set of literals L if l L. The tree method is a useful tool in computability theory (See[10]) to construct an object that satisfies some requirements. Now it will be generalized in this paper to establish a mechanism to obtain a preferable answer set for a BDLP. The tree used to construct the most preferable answer set of the Π is called a priority tree of Π and will be defined as follows: Definition 3. Let Λ be the set of all the literals occurring in the heads of the clauses of Π together with a special letter o. The priority tree of Π, denoted by T Π, is a subset of Λ <ω, the finite sequences of the elements of Λ, such that (i) There is a unique node called the root node, denoted by o, at the 0-th level of T Π. The direct siblings of o are the literals of head(π 1 ) with the ordering such that if l, l head(π 1 )andl l then l is to the left of l. The 0-th level of T Π is called the π 1 -level. (ii) Every node at the π i -level (for i 1) of the tree is a literal l. Ifl is at the π i -level of the tree then its direct siblings are the literals of head(π i+1 )withthe ordering such that if l, l head(π i+1 )andl l then l is to the left of l. The i-th level of T Π is called the π i+1 -level. A path through the priority tree is a string of the literals l 0 l 1 l n,where l 0 = o, such that l i is a direct sibling of l i 1 for every 1 i n. We use the Greek letters α, β, γ,... to denote the paths. If α = l 0 l 1 l n then α(i) =l i for every 0 i n. Let α denote the length of α. Let λ be the empty string. Let αˆβ be the concatenation of α followed by string β. Definition 4. Let α, β T Π, i.e., α and β are paths through T Π. We say that α is to the left of β, denoted by α< L β, if l 1,l 2 Λ γ T Π (γˆl 1 α & γˆl 2 β & l 1 l 2 ). By the definition of T Π, we know that every level of the tree is associated with aclauseπ i which can be construed as a requirement. For each π i Π, there is an α-strategy which is designed for an attempt to satisfy the requirement π i, where α is a path of T Π such that α = i 1. If α is a strategy of π i then we shall replace π i by π α for convenience. We need the notion of the infinite paths through T Π to express our last result. η is an infinite path through T Π if η n T Π for all n, whereη n is the restriction of η to n. We can generalize the relation < L over the infinite paths such that for any infinite paths δ, η through T Π, δ< L η if there are paths α, β in T Π,α δ and β η such that α< L β. Let Π = {π α } α<ω be a BDLP. Then following proposition shows the relation between the answer sets of Π and the infinite paths through T Π. Proposition 1. Every answer set of Π is corresponding to at least an infinite path through T Π. Proof. Let A be an answer set of Π. We define along the tree T Π the infinite path η as follows: For any α η, ifbody(π α ) A then we define αˆl η,

6 A Tree Construction of the Preferable Answer Sets 593 where l is the literal such that l head(π α ) A, else choose the least literal l head(π α ) and define αˆl η. By Proposition 1 we know that every answer set A of a BDLP is corresponding to some infinite paths through the tree. Under the relation < L, the least such infinite path through the tree is called the true path of A and is denoted by trp(a). Definition 5. Let A be an answer set of a BDLP Π and its true path through T Π is trp(a). A is called the most preferable answer set of Π if for any answer set B of Π, trp(a) < L trp(b). 4 Constructing the Preferable Answer Set Let Π = {Π α } α<ω be a double prioritized BDLP contains infinitely many clauses. We shall give an efficient process to obtain the most preferable answer set of Π. In order to describe our main idea of the construction, we assume that each literal can appear at most once in all clause heads of the program. During the construction, the true path δ Λ ω through T Π is what we need at last. If δ is the true path on T Π then there will be an answer set A of Π such that for any α δ, the requirement π α is satisfied, i.e., either body(π α ) A or head(π α ) A. Our construction on T Π will produce a true path δ, which is the leftmost one on which every node is visited infinitely often, that is, for any α T, if α< L δ then α cannot be extended to be a true path. The leftmost true path produced in our construction will be constructed by stages. At any stage s, we define a recursive approximation δ s to δ, where δ s T Π and δ s s, such that δ = lim δ s. s Let A s be the approximation to the answer set A at stage s, and δ s be the approximation of δ. We say that π α requires attention at stage s if (4.1) α δ s, body(π α ) A s, and head(π α ) A = ; or (4.2) α δ s, body(π α ) A s, head(π α ) A s. When a clause π α requires attention at stage s, the α-strategy attempts to put some literal of head(π α )intoa s. If l is such a literal then we say that l is put in A s by the α-strategy. During the construction, some node on the tree will be initialized. To initialize α at stage s means to move from A s the literals which has been put in A s by the α-strategy at some previous stages. Following is the construction of A: Stage s =0:setA 0 =,δ 0 = λ. Stage s + 1 : If there is no α δ s requiring attention then define δ s+1 to be the leftmost path such that δ s+1 = δ s + 1 and go to the next stage. Otherwise, let α be the minimal γ δ s which requires attention, we do our work according to the following cases:

7 594 Z. Zhang, Y. Sui, and C. Cao Case 1. (4.1) holds. If α = δ s and there is a literal l head(π α+1 ) such that l is consistent with A s then we choose the least such l and put it in A s+1. Define δ s+1 = δ sˆl and go to the next stage; if there is no such literal then we take the modifying action to give the definition of δ s+1. Case 2. (4.1) holds and that α δ s. Let l be the literal such that αˆl δ s. If l is consistent with A s then put l in A s+1 and define δ s+1 = δ s. If l is not consistent with A s then we go to the modifying action. Case 3. (4.2) holds. Let l be the literal such that l head(π α ) A s. Extract l out of A s and define δ s+1 = δ s, go to the next stage. The modifying action: Let τ =max{γ δ s : l(γˆl δ s, l A s )}. Define δ s+1 = τˆl,wherel is the least literal in head(π τ ) such that l l. Initialize all α< L δ s+1, and go to the next stage. This completes the construction of A. The following example give us an intuition that the priority tree works. Example 2. Let Π = {π 1,π 2,..., π 5 } be a basic disjunctive logic program given below: π 1 : p 1 π 2 : p 2,p 3 p 1 π 3 : p 4,p 5 p 2,p 1 π 4 : p 4, p 2 p 1,p 6 π 5 : p 6, p 2 p 1. The priority relation is that π 1 π 2 π 3 π 4 π 5, and p 2 p 3,p 4 p 5, p 4 p 2 and p 6 p 2. The priority tree of Π is given as follows: p 6 p 2 p 6 p 2 p 6 p 2 p 6 p 2 p 6 p 2 p 6 p 2 p 6 p 2 p 6 p 2 p 4 p 2 p 4 p 2 p 4 p 2 p 4 p 2 π 5,α 4 p 4 p 5 p 4 π 4,α 3 p 2 p 5 p 3 π 3,α 2 p 1 π 2,α 1 π 1,α 0 Fig. 1. The priority tree of Π = {π 1,π 2,..., π n} The construction of the most preferable answer set of Π : Stage s =0: δ 0 = λ, A 0 =. Stage s =1: π 1 requires attention. The α 0 -strategy puts p 1 in A 1. Hence, A 1 = {p 1 } and δ 1 = op 1.

8 A Tree Construction of the Preferable Answer Sets 595 Stage s =2: π 2 requires attention. The α 1 -strategy puts p 2 in A 2. Hence, A 2 = {p 1,p 2 } and δ 2 = op 1 p 2. Stage s =3: π 3 requires attention. The α 2 -strategy puts p 4 in A 3. Hence, A 3 = {p 1,p 2,p 4 } and δ 3 = op 1 p 2 p 4. Stage s = 4 : There is no α δ s requiring attention. Hence, A 4 = {p 1,p 2,p 4 } and δ 4 = op 1 p 2 p 4 p 4. Stage s =5: π 5 requires attention. The α 4 -strategy puts p 6 in A 5. Hence, A 5 = {p 1,p 2,p 4,p 6 } and δ 5 = op 1 p 2 p 4 p 4 p 6. Stage s =6: π 4 requires attention. α 3ˆ p 4 δ 5 and p 4 is not consistent to A 5. We take the modifying action: firstly we find τ = α 4, α 4ˆp 6 δ 5 and p 6 A 5. Define δ 6 = op 1 p 2 p 4 p 4 p 2, and initialize α 4 i.e., extract p 6 from A 5, hence, A 6 = {p 1,p 2,p 4 }. Stage s =7: π 5 requires attention, and α 4ˆ p 2 δ 6 and p 2 is not consistent to A 6. We tack the modifying action: find τ = α 2, τˆp 4 δ 6 and p 4 A 6. Define δ 7 = op 1 p 2 p 5, and extract p 4 from A 6, hence A 7 = {p 1,p 2 }. Stage s =8: π 3 requires attention and α 2ˆp 5 δ 7, then put p 5 in A 7. Hence A 8 = {p 1,p 2,p 5 } and δ 8 = op 1 p 2 p 5. Stage s =9:Noα δ 8 requires attention. A 9 = A 8 and δ 9 = op 1 p 2 p 5 p 4. Stage s =10:Noα δ 9 requires attention. A 10 =A 9 and δ 10 =op 1 p 2 p 5 p 4 p 6. Stage s =11: π 5 requires attention, the α 4 -strategy puts p 6 in A 11, δ 11 = op 1 p 2 p 5 p 4 p 6, and A 11 = {p 1,p 2,p 5,p 6 }. Stage s =12: π 4 requires attention, and α 3 -strategy p 4 in A 12,wehave A 12 = {p 1,p 2,p 5,p 6, p 4 } and δ 12 = op 1 p 2 p 5 p 4 p 6. Stage t 13 : no clause requires attention again. This ends the construction of A, where δ = lim s δ s = δ 12 = op 1 p 2 p 5 p 4 p 6 isthetruepathoft Π, and A = lim s A s = A 12 = {p 1,p 2,p 5,p 6, p 4 } is the most preferable answer set of Π. 5 The Verification of the Construction Lemma 2. If Π is consistent then δ = lim s δ s exists. Proof. We prove the lemma by induction on n to show that for any n there exists a stage s such that δ n δ t for any t s. If n = 1 then we do nothing since δ 1 δ s for any s. Assume the result holds for n, i.e., there exists a stage s 0, choose the least one, such that δ n δ t for all t s 0. Let α = δ n. Noticethatα-strategy works for π α, thus the outputs of α are finite. Assume that the set of the outputs of α is {l 1,..., l n } =head(π α ). Then it is sufficient to show that there is a stage s s 0 and a literal l i head(π α ) such that αˆl i δ t for all t s. Without lose of generality, we may assume that the outputs of α are just l 1 l 2,andat s 0 we have that αˆl 1 δ s0.ifα-strategy dose not put l 1 in A s for any s s 0 then α can never be initialized after stage s 0, hence for any s s 0 we have that αˆl 1 δ s. Let s s 0 be the stage at which l 1 be put in A s +1 by α-strategy.ifα never be initialized after s then αˆl 1 δ s for any s s.lets s be the least stage at which α is initialized. Then at s we extract l 1 from A s

9 596 Z. Zhang, Y. Sui, and C. Cao and define δ s = αˆl 2. The initialization of α at stage s means that every path extended the path αˆl 1 can not produce a true path. If the same thing happens on the path αˆl 2 then any path through α can not produce a true path. Since Π is consistent, there must be an answer set of Π. To cope with this, our modifying action will initialize some node β α and make the approximation of true path go right below the node α, which contradicts the choice of the stage s 0. During the construction, we shall choose for each clause π α a literal in head(π α ) as an evidence such that whenever body(π α ) A s at some stage s, the evidence will be put into A s+1. The evidence used to satisfy the requirement π α is also called a candidate of π α. Lemma 3. Assume Π is consistent and π α Π is a clause such that l 1,..., l n head(π α ) with the priority l 1... l n. During the construction if there is a stage s such that α δ and l i turns to be a candidate of π α then for any j<i,l j can never be a candidate of π α again, and furthermore, for any j<i,αˆl j proves to be to the left of the true path. Proof. Without loss of generality, we assume that the literals in head(π α )are just l 1 l 2, and l 1 is used as a candidate of π α at first. Let s 0 be the least stage at which we have that α δ and αˆl 1 δ s0+1. By Lemma 2 s 0 exists if Π is consistent. Notice that the modifying action at stage s + 1 during the construction is to find the maximum α δ s such that α-strategy has put a literal in A t at some previous stage t, hence if l 1 A s then, as a candidate of π α, l 1 can not be changed at stage s +1. Thus we may also assume that l 1 A s0+1. Let s 1 >s 0 be the least stage and the modifying action make l 2 to be a candidate of π α at s Then the following conditions hold: (i) There is a β δ s1 such that body(π β ) A s1, head(π β ) A s1 = and every literal in head(π β ) is not consistence to A s1. (ii) For any γ, α γ δ s1, no literal is put into A s1 by γ-strategy. Bythechoiceofs 0,eachγ α can never be initialized after s 0 +1,andthe change of the candidate of π α from l 1 to l 2, means that any extension of the path αˆl 1 can not produce a model of Π. Hence, after stage s all approximation of the true path will always go along the path αˆl 2, thus l 1 cannot be a candidate of π α any more. More precisely, by (ii) above, we know that each element of A s1 is offered by some strategy γ α at some previous stage. Since body(π α ) A s1, all literals of body(π α ) are put into A s1 by the strategies below α. Notice that for any strategy γ α, the candidate of γ will hold the line forever, thus if we hold l 1 as a candidate of α after l 1 is extracted from A s1 then l 1 must be put in A s again at some later stage s, and the condition (i) will happen again. Lemma 4. If clause π α Π is a fact such that body(π α )= and literals in head(π α )arel 1... l n, then there exists a literal l i head(π α ) and a stage s such that αˆl i δ t and l i A t for all t s. Proof. This is straightforward from Lemma 3 and the assumption of π α.

10 A Tree Construction of the Preferable Answer Sets 597 If π α is a fact then the literals in head(π α ) are called termination element. Each fact contributes one and only one termination element to the model of Π during the construction. By Lemma 4 one can easily prove that for any terminal q there exists a stage s such that either q A t for all t s or q A t for all t s. There are dependent relations between the literals in the program, thus we have the following definition. Definition 6. A literal l 1 is said to be dependent on l 2 if there is a clause π such that l 1 head(π) andl 2 body(π). A sequence of literals l 1...l n is called a dependence chain if for any 1 i n 1thereisaclauseπ i such that l i head(π i )andl i+1 body(π i ). A dependence chain is complete if the last literal of the chain does not depend on any literal. During our construction we consider only those literals, whether or not they can be the consequences of the logic program. Thus every literal in a dependence chain is in a head of some clause. Therefore if a finite dependence chain is complete then the last literal in the chain must be a termination element. We use φ, ψ, etc, to express dependence chain, and φ denote the length of φ, the number of the literals in π. Dec(l) is the set of all dependence chains with l as being the first literal. Lemma 5. Let φ = l 1...l n be a dependence chain and l i be a literal in the chain. If there is a stage s such that l i A t for all t s then there exists a stage s s such that l 1 A t for all t s. Proof. Let π α be the clause such that l i 1 head(π α )andl i body(π α ). Notice that any literal can be offered in the answer set by a unique clause, thus if l i A t for all t s then body(π α ) A t for all t s. Ifl i 1 A s then it can not be put in A t for any t s; ifl i 1 A s then it will be extracted from A s at some stage s s and can never be put in A t any more at any stage t s.by induction on j<iwe can prove at last that l 1 A. Lemma 6. Let Dec(l) be the set of all dependence chains with l as the first literal. If there is φ Dec(l) such that φ = then the literal l A s for all s. Proof. Since at any stage s there are only finitely many clauses can be dealt with, hence there must be a literal l i in φ such that l i A t for any t s, thus by Lemma 5 we have that l A s. Lemma 7. Let l be a literal. If l is put in A s0 at stage s 0,then φ s 0 for any φ Dec(l). Proof. Let φ Dec(l). If l is put in A at stage s 0 then by Lemma 5 all literals in φ have been in A at stage s 0, and each of them is put in by a strategy α δ s0, hence φ δ s0 s 0. Lemma 8. There is no literal l can not be put in or extracted from A infinitely often, that is for any literal l, there exists a stage s such that either l A t for all t s or l A t for all t s.

11 598 Z. Zhang, Y. Sui, and C. Cao Proof. Let l be a literal such that l head(π α ) for some clause π α,andα be the strategy for π α. Choose the least stage s 0 such that αˆl δ t for all t s 0.If l A t for all t s 0 then there is nothing to do, hence we may also assume that l A s0. By Lemma 7, each φ Dec(l) has the length φ s 0 and ends with a termination element. Let C be the set of all termination elements appear in Dec(l). It is obvious that C is finite. By Lemma 4, there are two conditions for the elements in C. (i) There exists a stage s s 0, C A t for all t s; (ii) There is a q C and a stage s s 0 such that q A t for all t s. If (i) holds then we shall reach to a stage at which all literals appear in Dec(l) are in A, andl will be kept in A forever. If (ii) holds, let φ Dec(l) be the dependence chain ends with terminal q, then by Lemma 5 there will be a stage at which l must be extracted form A and can not be put in A again. Lemma 9. Let δ = lim s δ s bethetruepathanda be the set obtained by our construction. For any clause π α,ifbody(π α ) A then there exists a literal l head(π α ) such that αˆl δ and l A. Proof. By Lemma 8 we shall reach to a stage s 0 such that body(π α ) A s for all s s 0. Then at stage s 0 +1, π α will require attention if head(π α ) A s0 =, and the α-strategy will put a literal of head(π α )intoa s0+1. If the literal is such that αˆl δ then l will be kept in A forever. Lemma 10. The set A is consistent. Proof. Assume l is a literal such that l A. Then there will be a stage s, l A t for all t s, hence l can not be put into A t for any t s, and thus l A. Lemma 11. For any B A, B is not an answer set of Π. Proof. Let l be the literal such that l A and l B. Consider the set Dec(l), and let C be the set of all termination elements appear in Dec(l), hence C A. If there is a termination element q C such that q B then there must be a fact π α such that q head(π α )andπ α offers a q head(π α ), q q, inb. It is obvious that q A, which contradicts that B A. Thus we assume that C B. Let D be the set such that D = {π α : l ( l appears in Dec(l) andl head(π α ))}. Notice that every literal appears in Dec(l) isina and is offered by aclauseind. SinceC B, we know that some clauses in D will offer some literals in B. IfallclausesinD offer the same literals as them do for A then we shall have l B, which contradicts the assumption that l B. Ifthereisa clause π α D such that it offers a literal different from what it offers to A then we shall have that B A, also a contradiction. Theorem 12. The set A is the most preferable answer set of the logic program Π.

12 A Tree Construction of the Preferable Answer Sets 599 Proof. A is an answer set of Π can be proved immediately by Lemma 9, Lemma 10 and Lemma 11. To show that A is most preferable, it is sufficient to show that δ is the true path of A, because on the tree T Π every path α< L δ can not be extended to be a true path. By Proposition 1 and Lemma 9, it is obvious that trp(a) =δ. 6 Conclusion and Further Works Given a basic disjunctive logic program Π, there are many minimal Herbrand models of Π. In terms of the prioritization on the clauses and literals, the construction selects the unique minimal Herbrand model of Π, say M(Π). M(Π) can be taken as the canonical model of Π, and it is assumed that Cn(Π) = M(Π), where Cn(Π) is the logical closure of Π. Since the construction is recursive in Π, M(Π) is Turing-reducible to the halting problem, that is, every such constructed M(Π) is recursively-approximatible, comparing to the Σ n+1 - completeness of the stratified logic programs. Further works include the following two aspects: One is to consider the completeness of the construction. Given a basic disjunctive logic program Π and its minimal Herbrand model M, whether or not there is a prioritization on the clauses and literals of Π under which the construction produces the unique model M(Π) ism. Another is to consider the ω-completeness of the basic disjunctive logic programs. We know that the Horn logic programs are recursively enumerable-complete, that is, given a Horn logic program Π, if Π is recursive then the least Herbrand model M (Π) ofπ is recursively enumerable, and conversely, given a recursively enumerable subset X HB, there is an Horn logic program Π such that M (Π) =X. We want to know that if the basic disjunctive logic programs are ω-recursively enumerable-complete. Acknowledgement. The authors are grateful to the anonymous referees of the paper for the useful suggestion. References 1. van Emden M H, Kowalski R A. The semantics of predicate logic as a programming language. Journal of Association for Computing Machinery, 1976, 23(4): Dahr M. Deductive Databases: Theory and Applications. International Thomson Computer Press, London, Gelfond M, Lifschitz V. Classical negation in logic programs and disjunctive databases. New generation computing, 1991,9: Chandra A, Harel D. Horn clause queries and generalizations. Journal of logic programming, 1985, 2(1): Apt K, Blair H, and Walker A. Towards a theory of declarative knowledge. In Foundations of deductive databases and logic programming, ed. Jack Minker San Mateo, CA: Morgan Kaufmann, Van Gelder A. Negation as failure using tight derivations for general logic programs. In Foundations of deductive databases and logic programming, ed. Jack Minker San Mateo, CA: Morgan Kaufmann, 1988.

13 600 Z. Zhang, Y. Sui, and C. Cao 7. Przymusinski T. On the declarative semantics of deductive databases and logic programs. In Foundations of deductive databases and logic programming, ed. Jack Minker San Mateo, CA: Morgan Kaufmann, Sakama C, Inoue K. Representing priorities in logic programs. 9. Zhang Y, Foo N. Answer sets for prioritized logic programs. In Logic Programming: Proc. of the Intl. Symposium (ILPS-97), ed. Maluszynski J MIT Press, Cambridge, MA, Soare R I. Recursively enumerable sets and degrees. Springer-Verlag, New York, 1987.