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A Logical Characterization of the Preferred Models of Logic Programs with Ordered Disjunction

A. CHARALAMBIDIS    P. RONDOGIANNIS    and A. TROUMPOUKIS
Department of Informatics and Telecommunications
  
National and Kapodistrian University of Athens
{a.charalambidis, prondo, antru}@di.uoa.gr
Abstract

Logic Programs with Ordered Disjunction (LPODs) extend classical logic programs with the capability of expressing alternatives with decreasing degrees of preference in the heads of program rules. Despite the fact that the operational meaning of ordered disjunction is clear, there exists an important open issue regarding its semantics. In particular, there does not exist a purely model-theoretic approach for determining the most preferred models of an LPOD. At present, the selection of the most preferred models is performed using a technique that is not based exclusively on the models of the program and in certain cases produces counterintuitive results. We provide a novel, model-theoretic semantics for LPODs, which uses an additional truth value in order to identify the most preferred models of a program. We demonstrate that the proposed approach overcomes the shortcomings of the traditional semantics of LPODs. Moreover, the new approach can be used to define the semantics of a natural class of logic programs that can have both ordered and classical disjunctions in the heads of clauses. This allows programs that can express not only strict levels of preferences but also alternatives that are equally preferred.
This work is under consideration for acceptance in TPLP.

keywords:
Ordered Disjunction, Answer Sets, Logic of Here-and-There, Preferences.

1 Introduction

Logic Programs with Ordered Disjunction (LPODs) extend classical logic programs with the capability of expressing ordered alternatives in the heads of program rules. In particular, LPODs allow the head of a program rule to be a formula C1××CnC_{1}\times\cdots\times C_{n}, where “×\times” is a propositional connective called “ordered disjunction” and the CiC_{i}’s are literals. The intuitive explanation of C1××CnC_{1}\times\cdots\times C_{n} is “I prefer C1C_{1}; however, if C1C_{1} is impossible, I can accept C2C_{2}; \cdots; if all of C1,,Cn1C_{1},\ldots,C_{n-1} are impossible, I can accept CnC_{n}”. Due to their simplicity and expressiveness, LPODs are a widely accepted formalism for preferential reasoning, both in logic programming and in artificial intelligence.

At present, the semantics of LPODs is defined [Brewka (2002), Brewka et al. (2004)] based on the answer set semantics, using a two-phase procedure. In the first phase, the answer sets of the LPOD are produced. This requires a modification of the standard definition of answer sets. In the second phase, the answer sets are “filtered”, and we obtain the set of “most preferred” answer sets, which are taken as the meaning of the initial program. Notice that both phases are not purely model-theoretic: the first one requires the construction of the reduct of the program and the second one is performed using the so-called “degree of satisfaction of rules”, a concept that relies on examining the rules of the program to justify the selection of the most preferred answer sets. Apart from its logical status, the current semantics of LPODs produces in certain cases counterintuitive most preferred answer sets. This discussion leads naturally to the question: “Is it possible to specify the semantics of LPODs in a purely model-theoretic way?”.

An important first step in this direction was performed by \citeNlpod-cabalar, who used Equilibrium Logic [Pearce (1996)] to logically characterize the answer sets produced in the first phase described above. However, to our knowledge, the second phase (namely the selection of the most preferred answer sets), has never been justified model-theoretically. We consider this as an important shortcoming in the theory of LPODs. Apart from its theoretical interest, this question also carries practical significance, because, as we are going to see, the present formalization of the second phase produces in certain cases counterintuitive (and in our opinion undesirable) results. The main contribution of the present paper is to provide a purely model-theoretic characterization of the semantics of LPODs. The more specific contributions are the following:

  • We propose a new semantics for LPODs which uses an additional truth value in order to select as most preferred models those in which a top preference fails only if it is impossible to be satisfied. We demonstrate that the proposed approach overcomes the shortcomings of the traditional semantics of LPODs. In this way, the most preferred models of an LPOD can be characterized by a preferential ordering of its models.

  • We demonstrate that our approach can be seamlessly extended to programs that allow both ordered and classical disjunctions in the heads of clauses. In particular, we define a natural class of such programs and demonstrate that all our results about LPODs transfer, with minimal modifications, to this new class. In this way we provide a clean semantics for a class of programs that can express not only strict levels of preference but also alternatives that are equally preferred.

Section 2 introduces LPODs and gives relevant background. Sections 3 and 4 describe the shortcomings of LPOD semantics and give an intuitive presentation of the proposed approach for overcoming these issues. The remaining sections give a technical exposition of our results. The proofs of all results have been moved to corresponding appendices.

2 Background on LPODs

Logic programs with ordered disjunction are an extension of the logic programs introduced by \citeNGL91, called extended logic programs, which support two types of negation: default (denoted by 𝑛𝑜𝑡\mathit{not}) and strong (denoted by ¬\neg). Strong negation is useful in applications but it is not very essential from a semantics point of view: a literal ¬A\neg A is semantically treated as an atom AA^{\prime}. For the basic notions regarding extended logic programs, we assume some familiarity with the work of \citeNGL91.

Definition 1

A (propositional) LPOD is a set of rules of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

where the Ci,AjC_{i},A_{j}, and BlB_{l} are ground literals.

The intuitive explanation of a formula C1××CnC_{1}\times\cdots\times C_{n} is “I prefer C1C_{1}; however, if C1C_{1} is impossible, I can accept C2C_{2}; \cdots; if all of C1,,Cn1C_{1},\ldots,C_{n-1} are impossible, I can accept CnC_{n}”.

An interpretation of an LPOD is a set of literals. An interpretation is called consistent if there does not exist any atom AA such that both AA and ¬A\neg A belong to II. The notion of model of an LPOD is defined as follows.

Definition 2

Let PP be an LPOD and MM an interpretation. Then, MM is a model of PP iff for every rule

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},\mathit{not}{\,}B_{1},\ldots,\mathit{not}{\,}B_{k}

if {A1,,Am}M\{A_{1},\ldots,A_{m}\}\subseteq M and {B1,,Bk}M=\{B_{1},\ldots,B_{k}\}\cap M=\emptyset then there exists CiMC_{i}\in M.

To obtain the preferred answer sets of an LPOD, a two-phase procedure was introduced by \citeNlpod-brewka. In the first phase, the answer sets of the LPOD are produced. This requires a modification of the standard definition of answer sets for extended logic programs. In the second phase, the answer sets are “filtered”, and we obtain the set of “most preferred” ones. The first phase is formally defined as follows.

Definition 3

Let PP be an LPOD. The ×\times-reduct of a rule RR of PP of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

with respect to a set of literals II, is denoted by R×IR_{\times}^{I} and is defined as follows:

R×I={CiA1,,AmCiI and I{C1,,Ci1,B1,,Bk}=}R_{\times}^{I}=\{C_{i}\leftarrow A_{1},\ldots,A_{m}\mid C_{i}\in I\mbox{ and }I\cap\{C_{1},\ldots,C_{i-1},B_{1},\ldots,B_{k}\}=\emptyset\}

The ×\times-reduct of PP with respect to II is denoted by P×IP_{\times}^{I} and is the union of the reducts R×IR_{\times}^{I} for all RR in PP.

Definition 4

A set MM of literals is an answer set of an LPOD PP if MM is a consistent model of PP and MM is the least model of P×MP_{\times}^{M}.

The second phase produces the “most preferred” answer sets using the notion of the degree of satisfaction of a rule. Formally:

Definition 5

Let MM be an answer set of an LPOD PP. Then, MM satisfies the rule:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}
  • to degree 11 if AjMA_{j}\not\in M, for some jj, or BiMB_{i}\in M, for some ii,

  • to degree ll, 1ln1\leq l\leq n, if all AjMA_{j}\in M, no BiMB_{i}\in M, and l=min{rCrM}l=\min\{r\mid C_{r}\in M\}

The degree of a rule RR in the answer set MM is denoted by 𝑑𝑒𝑔M(R)\mathit{deg}_{M}(R).

The satisfaction degrees of rules are then used to define a preference relation on the answer sets of a program. Given a set of literals MM, let Mi(P)={RP𝑑𝑒𝑔M(R)=i}M^{i}(P)=\{R\in P\mid\mathit{deg}_{M}(R)=i\}. The preference relation is defined as follows.

Definition 6

Let M1,M2M_{1},M_{2} be answer sets of an LPOD PP. Then, M1M_{1} is inclusion-preferred to M2M_{2} iff there exists k1k\geq 1 such that M2k(P)M1k(P)M_{2}^{k}(P)\subset M_{1}^{k}(P), and for all j<kj<k, M2j(P)=M1j(P)M_{2}^{j}(P)=M_{1}^{j}(P).

Example 1

Consider the program:

wine ×\times beer.

The above program has the answer sets {wine}\{\mbox{\tt wine}\} and {beer}\{\mbox{\tt beer}\}. The most preferred one is {wine}\{\mbox{\tt wine}\} because it satisfies the unique fact of the program with degree 1 (while the answer set {beer}\{\mbox{\tt beer}\} satisfies the fact with degree 2). Consider now the program:

wine × beer.¬wine.\begin{array}[]{l}\mbox{\tt wine $\times$ beer.}\\ \mbox{\tt$\neg$wine.}\end{array}

This has the unique answer set {beer}\{\mbox{\tt beer}\} (the set {wine,¬wine}\{\mbox{\tt wine,$\neg$wine}\} is rejected due to its inconsistency). Therefore, {beer}\{\mbox{\tt beer}\} is also the most preferred answer set.

Notice that \citeNlpod-brewka originally defined only the preference relation of Definition 6. In the follow-up paper [Brewka et al. (2004)] two more preference relations were introduced, namely the cardinality and the Pareto, in order to treat cases for which the inclusion preference did not return the expected results. All these relations do not rely exclusively on the models of the source program, and are therefore subject to similar criticism. For this reason, in this paper we focus attention on the inclusion preference relation.

3 Some Issues with the Semantics of LPODs

From a foundational point of view, the main issue with the semantics of LPODs is that, in its present form, it is not purely model-theoretic. Despite the simplicity and expressiveness of ordered disjunction, one can not characterize the meaning of a program by just looking at its set of models. Recall that this principle is one of the cornerstones of logic programming since its inception: the meaning of positive logic programs is captured by their minimum Herbrand model [van Emden and Kowalski (1976)] and the meaning of extended logic programs is captured by their equilibrium models [Pearce (1996)]. How can the most preferred models of LPODs be captured model-theoretically? The existing issues regarding the semantics of LPODs are illustrated by the following examples.

3.1 The Logical Status of LPODs

Consider the following two programs:

a ×\times b.

and:

b ×\times a.

According to Definition 2, both programs have exactly the same models, namely {a}\{\mbox{\tt a}\}, {b}\{\mbox{\tt b}\}, and {a,b}\{\mbox{\tt a$,$b}\}. Moreover, both have the same answer sets, namely {a}\{\mbox{\tt a}\} and {b}\{\mbox{\tt b}\}. However, there is no model-theoretic explanation (namely one based on just the sets of models of the programs) of why the most preferred model of the first program is {a}\{\mbox{\tt a}\} while the most preferred model of the second program is {b}\{\mbox{\tt b}\}. As a conclusion, in order for the semantics of LPODs to be properly specified, a model-based approach needs to be devised.

3.2 Inaccurate Preferential Ordering of Answer Sets

Apart from the fact that Definition 5 is not purely model-theoretic, in many cases it also gives inaccurate preferential orderings of answer sets. Such inaccurate orderings have already been reported in the literature [Balduccini and Mellarkod (2003)]. The following program illustrates one such simple case.

Example 2

Consider the following program111Our example is identical (up to variable renaming) to an example given by \citeNlpod-crprolog. This work was brought to our attention by one of the reviewers of the present paper. whose declarative reading is “I prefer to buy a Mercedes than a BMW. In case a Mercedes is available, I prefer a gas model to a diesel one. A gas model (of Mercedes) is not available”.

mercedes × bmw.gas_mercedes × diesel_mercedes  mercedes.¬gas_mercedes.\begin{array}[]{l}\mbox{\tt mercedes $\times$ bmw.}\\ \mbox{\tt gas\_mercedes $\times$ diesel\_mercedes $\leftarrow$ mercedes.}\\ \mbox{\tt$\neg$gas\_mercedes.}\end{array}

The program has two answers sets: M1={mercedes,diesel_mercedes,¬gas_mercedes}M_{1}=\{\mbox{\tt mercedes},\mbox{\tt diesel\_mercedes},\mbox{\tt$\neg$gas\_mercedes}\} and M2={bmw,¬gas_mercedes}M_{2}=\{\mbox{\tt bmw},\mbox{\tt$\neg$gas\_mercedes}\}. M1M_{1} satisfies the first rule with degree 1, the second rule with degree 2, and the third rule with degree 1. M2M_{2} satisfies the first rule with degree 2, the second rule with degree 1 (because the body of the rule evaluates to false), and the third rule with degree 1. According to Definition 6, the two answer sets are incomparable. However, it seems reasonable that the most preferred model is M1M_{1}: the first rule, which is a fact, specifies unconditionally a preference; the preferences of the second rule seem to be secondary, because they depend on the choice that will be made in the first rule.

The problems in the above example appear to be related to Definition 5: it assigns degree 1 in two cases that are apparently different: a rule that has a false body gets the same degree of satisfaction as a rule with a true body in whose head the first choice is satisfied. These are two different cases which, however, it is not obvious how to handle if we follow the satisfaction degree approach of Definition 5.

3.3 Unsatisfiable Better Options

It has been remarked [Brewka et al. (2004), discussion in page 342] that the inclusion-preference is sensitive to the existence of unsatisfiable better options. The following example, taken from the paper by \citeNlpod-BNS04, motivates this problem.

Example 3

Assume we want to book accommodation for a conference (in the post-COVID era). We prefer a 3-star hotel from a 2-star hotel. Moreover, we prefer to be in walking distance from the conference venue. This can be modeled by the program:

walking × ¬walking.3-stars × 2-stars.\begin{array}[]{l}\mbox{\tt walking $\times$ $\neg$walking.}\\ \mbox{\tt 3-stars $\times$ 2-stars.}\end{array}

Consider now the scenario where the only available 3-star hotel (say ℎ𝑜𝑡𝑒𝑙1\mathit{hotel}_{1}), is not in walking distance. Moreover, assume that the only available 2-star hotel (say ℎ𝑜𝑡𝑒𝑙2\mathit{hotel}_{2}), happens to be in walking distance. According to Definition 6, these two options are incomparable because ℎ𝑜𝑡𝑒𝑙1\mathit{hotel}_{1} satisfies the first rule to degree 2 and the second rule to degree 1, while ℎ𝑜𝑡𝑒𝑙2\mathit{hotel}_{2} satisfies the first rule to degree 1 and the second rule to degree 2.

Assume now that the above program is revised after learning that there also exists a 4-star hotel, which however is not an option for us (due to restrictions imposed by our funding agencies). The new program is:

walking × ¬walking.4-stars × 3-stars × 2-stars.¬4-stars.\begin{array}[]{l}\mbox{\tt walking $\times$ $\neg$walking.}\\ \mbox{\tt 4-stars $\times$ 3-stars $\times$ 2-stars.}\\ \mbox{\tt$\neg$4-stars.}\end{array}

In the new program, ℎ𝑜𝑡𝑒𝑙1\mathit{hotel}_{1} satisfies the first rule to degree 2 and the second rule to degree 2, while ℎ𝑜𝑡𝑒𝑙2\mathit{hotel}_{2} satisfies the first rule to degree 1 and the second rule to degree 3. According to Definition 6, ℎ𝑜𝑡𝑒𝑙2\mathit{hotel}_{2} is our preferred option.

The above example illustrates that under the “degree of satisfaction of rules” semantics, a small (and seemingly innocent) change in the program, can cause a radical change in the final preferred model. This sensitivity to changes is another undesirable consequence that stems from the fact that the second phase of the semantics of LPODs is not purely model-theoretic.

4 An Intuitive Overview of the Proposed Approach

The main purpose of this paper is to define a model-theoretic semantics for LPODs. In other words, we would like to be able to choose the most preferred answer sets of a program using preferential reasoning on the answer sets themselves. Actually, such an approach should also be applicable directly on the models of the source program, without the need to first construct the answer sets of the program. We would expect that such an approach would also provide solutions to the shortcomings of the previous section.

But on what grounds can we compare the answer sets (or, even better, the models) of a program and decide that some of them satisfy in a better way our preferences than the others? This seems like an impossible task because the answer sets (or the models) do not contain any information related to the ordered disjunction preferences of the program.

It turns out that we can introduce preferential information inside the answer sets of a program by slightly tweaking the underlying logic. The answer sets of extended logic programs are two-valued, ie., a literal is either TT (true) or FF (false). We argue that in order to properly define the semantics of LPODs, we need a third truth value, which we denote by FF^{*}. The intuitive reading of FF^{*} is “impossible to make true”.

To understand the need for FF^{*}, consider again the intuitive meaning of C1×C2C_{1}\times C_{2}: we prefer C2C_{2} only if it is impossible for us to get C1C_{1}. Impossible here means that if we try to make C1C_{1} true, then the interpretation will become inconsistent. Therefore, we seem to need two types of false, namely FF and FF^{*}: FF means “false by default” while FF^{*} means “impossible to make true”. The following example demonstrates these issues.

Example 4

Consider the program:

wine × beer.¬wine.\begin{array}[]{l}\mbox{\tt wine $\times$ beer.}\\ \mbox{\tt$\neg$wine.}\end{array}

As we are going to see in the coming sections, the most preferred answer set according to our approach is {(𝚠𝚒𝚗𝚎,F),(𝚋𝚎𝚎𝚛,T),(¬𝚠𝚒𝚗𝚎,T)}\{({\tt wine},F^{*}),({\tt beer},T),(\neg{\tt wine},T)\}. Notice that wine receives the value FF^{*} because if we tried to make wine equal to TT, the interpretation would become inconsistent (because ¬\negwine is TT). Notice also that, as we are going to see, the interpretation {(𝚠𝚒𝚗𝚎,F),(𝚋𝚎𝚎𝚛,T),(¬𝚠𝚒𝚗𝚎,T)}\{({\tt wine},F),({\tt beer},T),(\neg{\tt wine},T)\} is not a model of the program.

The above discussion suggests the following semantics for “×\times” in the proposed logic. Let u,v{F,F,T}u,v\in\{F,F^{*},T\}. Then:

u×v={v,if u=Fu,otherwiseu\times v=\left\{\begin{array}[]{ll}v,&\mbox{if $u=F^{*}$}\\ u,&\mbox{otherwise}\end{array}\right.

The intuition of the above definition is that we return the value vv only if it is impossible to satisfy uu; in all other cases, we return uu.

We now get to the issue of how we can use the value FF^{*} to distinguish the most preferred answer sets: we simply identify those answer sets that are minimal with respect to their sets of atoms that have the value FF^{*}. As we have mentioned, the value FF^{*} means “impossible to make true”. By minimizing with respect to the FF^{*} values, we only keep those answer sets in which a top preference fails only if it is impossible to be satisfied.

The above discussion gives a description of how we can select the “most preferred (three-valued) answer sets” of an LPOD. However, it is natural to wonder whether it is possible to also characterize the answer sets model-theoretically, completely circumventing the construction of the reduct. A similar question was considered by \citeNlpod-cabalar, who demonstrated that the (two-valued) answer sets of an LPOD coincide with the equilibrium models of the program. We adapt \citeANPlpod-cabalar’s characterization to fit in our setting. More specifically, we extend our three-valued logic to a four-valued one by adding a new truth value TT^{*}, whose intuitive meaning is “not false but its truth can not be established”. We then demonstrate that the three-valued answer sets of an LPOD PP are those models of PP that are minimal with respect to a simple ordering relation and do not contain any TT^{*} values. In this way we get a two-step, purely model-theoretic characterization of the most-preferred models of LPODs: in the first step we use the TT^{*} values as a yardstick to identify those models that correspond to answer-sets, and in the second step we select the most preferred ones, by minimizing with respect to the FF^{*} values.

Finally, we consider the problem of characterizing the semantics of logic programs that contain both disjunctions and ordered disjunctions in the heads of rules. This is especially useful in cases where some of our preferences are equally important. For example:

(wine  beer) × (soda  juice).\begin{array}[]{l}\mbox{\tt(wine $\vee$ beer) $\times$ (soda $\vee$ juice).}\end{array}

states that wine and beer are our top preferences (but we have no preference among them), and soda and juice are our secondary preferences. We consider the class of programs in which the heads of rules consist of ordered disjunctions where each ordered disjunct is an ordinary disjunction (as in the above program). We demonstrate that the theory of these programs is very similar to that of LPODs. All our results for LPODs transfer with minimal modifications to this extended class of programs. This suggests that this is a natural class of programs that possibly deserves further investigation both in theory and in practice.

5 Redefining the Answer Sets of LPODs

In this section we provide a new definition of the answer sets of LPODs. The new definition is based on a three-valued logic which allows us to discriminate the most preferred answer sets using a purely model-theoretic approach. In Section 6 we demonstrate that by extending the logic to a four-valued one, we can identify directly the most preferred models of a program (without first producing the answer sets).

Definition 7

Let Σ\Sigma be a nonempty set of propositional literals. The set of well-formed formulas is inductively defined as follows:

  • Every element of Σ\Sigma is a well-formed formula,

  • The 0-place connective FF^{*} is a well-formed formula,

  • If ϕ1\phi_{1} and ϕ2\phi_{2} are well-formed formulas, then (ϕ1ϕ2)(\phi_{1}\wedge\phi_{2}), (ϕ1ϕ2)(\phi_{1}\vee\phi_{2}), (𝑛𝑜𝑡ϕ1)(\mathit{not}{\,}\phi_{1}), (ϕ1ϕ2)(\phi_{1}\leftarrow\phi_{2}), and (ϕ1×ϕ2)(\phi_{1}\times\phi_{2}), are well-formed formulas.

The meaning of formulas is defined over the set of truth values {F,F,T}\{F,F^{*},T\} which are ordered as F<F<TF<F^{*}<T. Given two truth values v1,v2v_{1},v_{2}, we write v1v2v_{1}\leq v_{2} iff either v1<v2v_{1}<v_{2} or v1=v2v_{1}=v_{2}.

Definition 8

A (three-valued) interpretation II is a function from Σ\Sigma to the set {F,F,T}\{F,F^{*},T\}. We can extend II to apply to formulas, as follows:

I(F)=FI(𝑛𝑜𝑡ϕ)={T,if I(ϕ)FF,otherwiseI(ϕψ)={T,if I(ϕ)I(ψ)F,otherwiseI(ϕ1ϕ2)=min{I(ϕ1),I(ϕ2)}I(ϕ1ϕ2)=max{I(ϕ1),I(ϕ2)}I(ϕ1×ϕ2)={I(ϕ2),if I(ϕ1)=FI(ϕ1),otherwise\begin{array}[]{lll}I(F^{*})&=&F^{*}\\ I(\mathit{not}{\,}\phi)&=&\left\{\begin{array}[]{ll}T,&\mbox{if $I(\phi)\leq F^{*}$}\\ F,&\mbox{otherwise}\end{array}\right.\\ I(\phi\leftarrow\psi)&=&\left\{\begin{array}[]{ll}T,&\mbox{if $I(\phi)\geq I(\psi)$}\\ F,&\mbox{otherwise}\end{array}\right.\\ I(\phi_{1}\wedge\phi_{2})&=&\min\{I(\phi_{1}),I(\phi_{2})\}\\ I(\phi_{1}\vee\phi_{2})&=&\max\{I(\phi_{1}),I(\phi_{2})\}\\ I(\phi_{1}\times\phi_{2})&=&\left\{\begin{array}[]{ll}I(\phi_{2}),&\mbox{if $I(\phi_{1})=F^{*}$}\\ I(\phi_{1}),&\mbox{otherwise}\end{array}\right.\\ \end{array}

It is straightforward to see that the meanings of “\vee”, “\wedge”, and “×\times” are associative and therefore we can write I(ϕ1ϕn)I(\phi_{1}\vee\cdots\vee\phi_{n}), I(ϕ1ϕn)I(\phi_{1}\wedge\cdots\wedge\phi_{n}), and I(ϕ1××ϕn)I(\phi_{1}\times\cdots\times\phi_{n}) unambiguously (without the need of extra parentheses). Moreover, given literals C1,,CnC_{1},\ldots,C_{n} we will often write I(C1,,Cn)I(C_{1},\ldots,C_{n}) instead of I(C1Cn)I(C_{1}\wedge\cdots\wedge C_{n}).

The ordering << (respectively, \leq) on truth values extends in the standard way on interpretations: given interpretations I1,I2I_{1},I_{2} we write I1<I2I_{1}<I_{2} (respectively, I1I2I_{1}\leq I_{2}), if for all literals LΣL\in\Sigma, I1(L)<I2(L)I_{1}(L)<I_{2}(L) (respectively, I1(L)I2(L)I_{1}(L)\leq I_{2}(L)).

When we consider interpretations of an LPOD program, we assume that the underlying set Σ\Sigma is the set of literals of the program. The following definition will be needed.

Definition 9

An interpretation II is a model of an LPOD PP if every rule of PP evaluates to TT under II. An interpretation II of PP is called consistent if there do not exist literals AA and ¬A\neg A in PP such that I(A)=I(¬A)=TI(A)=I(\neg A)=T.

We can now give the new definitions for reduct and answer sets for LPODs.

Definition 10

Let PP be an LPOD. The ×\times-reduct of a rule RR of PP of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

with respect to an interpretation II, is denoted by R×IR_{\times}^{I} and is defined as follows:

  • If I(Bi)=TI(B_{i})=T for some ii, 1ik1\leq i\leq k, then R×IR_{\times}^{I} is the empty set.

  • If I(Bi)TI(B_{i})\neq T for all ii, 1ik1\leq i\leq k, then R×IR_{\times}^{I} is the set that contains the rules:

    C1F,A1,,AmCr1F,A1,,AmCrA1,,Am\begin{array}[]{lll}C_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ C_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ C_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

    where rr is the least index such that I(C1)==I(Cr1)=FI(C_{1})=\cdots=I(C_{r-1})=F^{*} and either r=nr=n or I(Cr)FI(C_{r})\neq F^{*}.

The ×\times-reduct of PP with respect to II is denoted by P×IP_{\times}^{I} and is the union of the reducts R×IR_{\times}^{I} for all RR in PP.

The major difference of the above definition from that of Definition 3, are the clauses of the form CiF,A1,,AmC_{i}\leftarrow F^{*},A_{1},\ldots,A_{m}. These clauses are included so as that the value FF^{*} can be produced for CiC_{i} when I(A1)==I(Am)=TI(A_{1})=\cdots=I(A_{m})=T. Notice that if these clauses did not exist, there would be no way for the value FF^{*} to be produced by the reduct.

Definition 11

Let PP be an LPOD and MM an interpretation of PP. We say that MM is a (three-valued) answer set of PP if MM is consistent and it is the \leq-least model of P×MP_{\times}^{M}.

Notice that the least model of P×MP_{\times}^{M} in the above definition, can be constructed using the following immediate consequence operator TP×M:(Σ{F,F,T})(Σ{F,F,T})T_{P_{\times}^{M}}:(\Sigma\rightarrow\{F,F^{*},T\})\rightarrow(\Sigma\rightarrow\{F,F^{*},T\}):

TP×M(I)(C)=max{I(B1,,Bn)(CB1,,Bn)P×M}T_{P_{\times}^{M}}(I)(C)=\max\{I(B_{1},\ldots,B_{n})\mid(C\leftarrow B_{1},\ldots,B_{n})\in P_{\times}^{M}\}

Notice that since the set {F,F,T}\{F,F^{*},T\} is a complete lattice under the ordering \leq, it is easy to see that the set of interpretations is also a complete lattice under the ordering \leq. Moreover, the operator TP×MT_{P_{\times}^{M}} is monotonic over the complete lattice of interpretations; this follows from the fact that the meanings of conjunction (namely, min\min) and that of disjunction (namely, max\max), are monotonic. Then, by Tarski’s fixed-point theorem, TP×MT_{P_{\times}^{M}} has a least fixed-point, which can be easily shown to be the \leq-least model of P×MP_{\times}^{M}.

The following lemma guarantees that our definition is a generalization of the well-known one for extended logic programs [Gelfond and Lifschitz (1991)].

Lemma 1

Let PP be a consistent extended logic program. Then the three-valued answer sets of PP coincide with the standard answer sets of PP.

The following lemmas, which hold for extended logic programs, also extend to LPODs.

Lemma 2

Let PP be an LPOD and let MM be an answer set of PP. Then, MM is a model of PP.

Lemma 3

Let MM be a model of an LPOD PP. Then, MM is a model of P×MP^{M}_{\times}.

The answer sets of extended logic programs are minimal with respect to the classical truth ordering F<TF<T. As it turns out, the answer sets of LPODs are minimal but with respect to an extended ordering, which is defined below.

Definition 12

The ordering \prec on truth values is defined as follows: FTF\prec T and FFF\prec F^{*}. For all u,v{F,F,T}u,v\in\{F,F^{*},T\}, we write uvu\preceq v if either uvu\prec v or u=vu=v. Given interpretations I1,I2I_{1},I_{2}, we write I1I2I_{1}\prec I_{2} (respectively, I1I2I_{1}\preceq I_{2}) if for all literals LΣL\in\Sigma^{*}, I1(L)I2(L)I_{1}(L)\prec I_{2}(L) (respectively, I1(L)I2(L)I_{1}(L)\preceq I_{2}(L)).

Lemma 4

Every (three-valued) answer set MM of an LPOD PP, is a \preceq-minimal model of PP.

As in the case of the original semantics of LPODs, we now need to define a preference relation over the answer sets of a program. Intuitively, we prefer those answer sets that maximize the prospect of satisfying our top choices in ordered disjunctions. This can be achieved by minimizing with respect to FF^{*} values. More formally, we define the following ordering:

Definition 13

Let PP be an LPOD and let M1,M2M_{1},M_{2} be answer sets of PP. Let M1M_{1}^{*} and M2M_{2}^{*} be the sets of literals in M1M_{1} and M2M_{2} respectively that have the value FF^{*}. We say that M1M_{1} is preferred to M2M_{2}, written M1M2M_{1}\sqsubset M_{2}, if M1M2M_{1}^{*}\subset M_{2}^{*}.

Definition 14

An answer set of an LPOD PP is called most preferred if it is minimal among all the answer sets of PP with respect to the \sqsubset relation.

The intuition behind the definition of \sqsubset is that we prefer those answer sets that minimize the need for FF^{*} values. In other words, an answer set will be most preferred if all the literals that get the value FF^{*}, do this because there is no other option: these literals must be false in order for the program to have a model. We now examine the examples of Section 3 under the new semantics introduced in this section.

Example 5

Consider again the two programs discussed in Subsection 3.1. Under the proposed approach, the first program has two answer sets, namely {(𝚊,T),(𝚋,F)}\{({\tt a},T),({\tt b},F)\} and {(𝚊,F),(𝚋,T)}\{({\tt a},F^{*}),({\tt b},T)\}, and the most preferred one (ie., the minimal with respect to \sqsubset) is {(𝚊,T),(𝚋,F)}\{({\tt a},T),({\tt b},F)\}. The second program also has two answer sets, namely {(𝚊,F),(𝚋,T)}\{({\tt a},F),({\tt b},T)\} and {(𝚊,T),(𝚋,F)}\{({\tt a},T),({\tt b},F^{*})\}, and the most preferred one is {(𝚊,F),(𝚋,T)}\{({\tt a},F),({\tt b},T)\}. Notice that now the two programs have different sets of models and different answer sets and therefore it is reasonable that they have different most preferred ones.

Example 6

Consider the “cars” program of Subsection 3.2. It is easy to see that it has two answer sets, namely:

M1={(𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,T),(𝚋𝚖𝚠,F),(𝚐𝚊𝚜_𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,F),(𝚍𝚒𝚎𝚜𝚎𝚕_𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,T),(¬𝚐𝚊𝚜_𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,T)}M2={(𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,F),(𝚋𝚖𝚠,T),(𝚐𝚊𝚜_𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,F),(𝚍𝚒𝚎𝚜𝚎𝚕_𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,F),(¬𝚐𝚊𝚜_𝚖𝚎𝚛𝚌𝚎𝚍𝚎𝚜,T)}\begin{array}[]{lll}M_{1}&=&\{({\tt mercedes},T),({\tt bmw},F),({\tt gas\_mercedes},F^{*}),\\ &&\,\,\,({\tt diesel\_mercedes},T),(\neg{\tt gas\_mercedes},T)\}\\ M_{2}&=&\{({\tt mercedes},F^{*}),({\tt bmw},T),({\tt gas\_mercedes},F^{*}),\\ &&\,\,\,({\tt diesel\_mercedes},F^{*}),(\neg{\tt gas\_mercedes},T)\}\end{array}

According to the \sqsubset ordering, the most preferred answer set is M1M_{1}.

Example 7

Consider the “hotels” program from Subsection 3.3. Under the restriction that there does not exist any 3-star hotel in walking distance and also there does not exist any 2-star hotel outside walking distance, we get the two incomparable answer sets:

M1={(𝚠𝚊𝚕𝚔𝚒𝚗𝚐,T),(¬𝚠𝚊𝚕𝚔𝚒𝚗𝚐,F),(3-stars,F),(2-stars,T)}M2={(𝚠𝚊𝚕𝚔𝚒𝚗𝚐,F),(¬𝚠𝚊𝚕𝚔𝚒𝚗𝚐,T),(3-stars,T),(2-stars,F)}\begin{array}[]{lll}M_{1}&=&\{({\tt walking},T),(\neg{\tt walking},F),(\mbox{\tt 3-stars},F^{*}),(\mbox{\tt 2-stars},T)\}\\ M_{2}&=&\{({\tt walking},F^{*}),(\neg{\tt walking},T),(\mbox{\tt 3-stars},T),(\mbox{\tt 2-stars},F)\}\end{array}

Consider now the modified program given in Subsection 3.3 (which contains the unsatisfiable better option of a 4-star hotel). Under the same restrictions as above, we get the two answer sets:

M1={(𝚠𝚊𝚕𝚔𝚒𝚗𝚐,T),(¬𝚠𝚊𝚕𝚔𝚒𝚗𝚐,F),(3-stars,F),(2-stars,T),(4-stars,F),(¬4-stars,T)}M2={(𝚠𝚊𝚕𝚔𝚒𝚗𝚐,F),(¬𝚠𝚊𝚕𝚔𝚒𝚗𝚐,T),(3-stars,T),(2-stars,F),(4-stars,F),(¬4-stars,T)}\begin{array}[]{lll}M^{\prime}_{1}&=&\{({\tt walking},T),(\neg{\tt walking},F),(\mbox{\tt 3-stars},F^{*}),(\mbox{\tt 2-stars},T),\\ &&\,\,\,(\mbox{\tt 4-stars},F^{*}),(\neg\mbox{\tt 4-stars},T)\}\\ M^{\prime}_{2}&=&\{({\tt walking},F^{*}),(\neg{\tt walking},T),(\mbox{\tt 3-stars},T),(\mbox{\tt 2-stars},F),\\ &&\,\,\,(\mbox{\tt 4-stars},F^{*}),(\neg\mbox{\tt 4-stars},T)\}\\ \end{array}

Under the proposed approach, the above two answer sets are also incomparable, and the problem identified in Subsection 3.3 no longer exists.

We close this section by stating a result that establishes a relationship between the answer sets produced by our approach (Definition 4) and those ones produced by the original formulation [Brewka (2002), Brewka et al. (2004)].

Definition 15

Let II be a three-valued interpretation of LPOD PP. We define 𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(I)\mathit{collapse}(I) to be the set of literals LL in PP such that I(L)=TI(L)=T.

Lemma 5

Let PP be an LPOD and MM be a three-valued answer set of PP. Then, 𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M)\mathit{collapse}(M) is an answer set of PP according to Definition 4.

Lemma 6

Let NN be an answer set of PP according to Definition 4. There exists a unique three-valued interpretation MM such that N=𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M)N=\mathit{collapse}(M) and MM is a three-valued answer set of PP.

In other words, there is a bijection between the answer sets produced by our approach and the original ones. Moreover, each three-valued answer set only differs from the corresponding two-valued one in that some literals of the former may have a FF^{*} value instead of an FF value. However, these FF^{*} values play an important role because they allow us to distinguish the most preferred answer sets.

6 A New Logical Characterization of the Answer Sets of LPODs

In this section we demonstrate that the answer sets of LPODs can be characterized in a purely logical way, namely without even the use of the reduct. In particular, we demonstrate that the answer sets of a given program PP coincide with a well-defined subclass of the minimal models of PP in a four-valued logic. This logic is an extension of the three-valued one introduced in Section 5 and minimality is defined with respect to a four-valued relation \preceq that extends the three-valued one of Definition 12. The new logic is based on four truth values, ordered as follows:

F<F<T<TF<F^{*}<T^{*}<T

The value TT^{*} can be read as “not false but its truth can not be established”. The connections of this logic with Equilibrium Logic [Pearce (1996)] are discussed in Section 8.

An interpretation is now a function from Σ\Sigma to the set {F,F,T,T}\{F,F^{*},T^{*},T\}. The semantics of formulas with respect to an interpretation II is defined identically as in Definition 8. The notions of interpretation, consistent interpretation, and model are defined as in Definition 9. Moreover, we extend the three-valued \preceq relation of Definition 12, as follows:

Definition 16

The (four-valued) ordering \prec is defined as follows: FF{F\prec F^{*}}, FT{F\prec T^{*}}, FT{F\prec T}, and TT{T^{*}\prec T}. Given two truth values v1,v2v_{1},v_{2}, we write v1v2{v_{1}\preceq v_{2}} if either v1v2{v_{1}\prec v_{2}} or v1=v2{v_{1}=v_{2}}. Given interpretations I1,I2I_{1},I_{2} of a program PP, we write I1I2I_{1}\prec I_{2} (respectively, I1I2I_{1}\preceq I_{2}) if for all literals LL in PP, I1(L)I2(L)I_{1}(L)\prec I_{2}(L) (respectively, I1(L)I2(L)I_{1}(L)\preceq I_{2}(L)).

The following special kind of interpretations plays an important role in our logical characterization.

Definition 17

An interpretation II of LPOD PP is called solid if for all literals LL in PP, it is I(L)TI(L)\neq T^{*}.

We can now state the logical characterization of the answer sets of an LPOD.

Theorem 1

Let PP be an LPOD. Then, MM is a three-valued answer set of PP iff MM is a consistent \preceq-minimal model of PP and MM is solid.

In conclusion, given an LPOD we can purely logically characterize its most preferred models by first taking its consistent \preceq-minimal models that are solid and then keeping the \sqsubset-minimal ones (see Definitions 13 and 14).

An extended study of the properties of the proposed four-valued logic is outside the scope of the present paper. Figure 1 lists some useful equivalences; the first column is for classical connectives, while the second column contains equivalences involving the “×\times” operator. We note the interaction of ×\times with \vee because this will be the central theme of the next section. It is easy to see that:

(ϕ1ϕ2)×ϕ3(ϕ1×ϕ3)(ϕ2×ϕ3)(\phi_{1}\vee\phi_{2})\times\phi_{3}\not\equiv(\phi_{1}\times\phi_{3})\vee(\phi_{2}\times\phi_{3})

We note that this equivalence does not hold, a fact which will be referenced in the next section.

ϕϕϕϕ1(ϕ2ϕ3)(ϕ1ϕ2)ϕ3𝑛𝑜𝑡(ϕ1ϕ2)𝑛𝑜𝑡(ϕ1)𝑛𝑜𝑡(ϕ2)𝑛𝑜𝑡(ϕ1ϕ2)𝑛𝑜𝑡(ϕ1)𝑛𝑜𝑡(ϕ2)ϕ1(ϕ2ϕ3)(ϕ1ϕ2)(ϕ1ϕ3)\begin{array}[]{rcl}\phi\vee\phi&\equiv&\phi\\ \phi_{1}\vee(\phi_{2}\vee\phi_{3})&\equiv&(\phi_{1}\vee\phi_{2})\vee\phi_{3}\\ \mathit{not}{\,}(\phi_{1}\vee\phi_{2})&\equiv&\mathit{not}{\,}(\phi_{1})\wedge\mathit{not}{\,}(\phi_{2})\\ \mathit{not}{\,}(\phi_{1}\wedge\phi_{2})&\equiv&\mathit{not}{\,}(\phi_{1})\vee\mathit{not}{\,}(\phi_{2})\\ \phi_{1}\vee(\phi_{2}\wedge\phi_{3})&\equiv&(\phi_{1}\vee\phi_{2})\wedge(\phi_{1}\vee\phi_{3})\end{array}
\figrule
ϕ×ϕϕϕ1×(ϕ2×ϕ3)(ϕ1×ϕ2)×ϕ3ϕ1×ϕ2×ϕ1ϕ1×ϕ2ϕ1×(ϕ2ϕ3)(ϕ1×ϕ2)(ϕ1×ϕ3)ϕ1×(ϕ2ϕ3)(ϕ1×ϕ2)(ϕ1×ϕ3)\begin{array}[]{rcl}\phi\times\phi&\equiv&\phi\\ \phi_{1}\times(\phi_{2}\times\phi_{3})&\equiv&(\phi_{1}\times\phi_{2})\times\phi_{3}\\ \phi_{1}\times\phi_{2}\times\phi_{1}&\equiv&\phi_{1}\times\phi_{2}\\ \phi_{1}\times(\phi_{2}\vee\phi_{3})&\equiv&(\phi_{1}\times\phi_{2})\vee(\phi_{1}\times\phi_{3})\\ \phi_{1}\times(\phi_{2}\wedge\phi_{3})&\equiv&(\phi_{1}\times\phi_{2})\wedge(\phi_{1}\times\phi_{3})\end{array}
Figure 1: Equivalences in the 4-valued logic
\figrule

7 Answer Sets of Disjunctive LPODs

We now extend the ideas of the previous sections to programs that also allow standard disjunctions in the heads of rules. The case of disjunctive LPODs (DLPODs), was initially considered by \citeNtowards-dlpods and reexamined by \citeNlpod-cabalar.

The main idea of using disjunctions in the heads of LPOD rules is described [Kärger et al. (2008)] as follows: “we use ordered disjunction to express preferences and disjunction to express indifferences”.

Example 8 (taken from the paper by \citeNtowards-dlpods)

The program:

pub×(cinema  tv).\mbox{\tt pub}\times\mbox{\tt(cinema $\vee$ tv).}

expresses the fact that our top choice is going to the pub; if this is not possible, then our secondary preference can be satisfied by either going to the cinema or watching tv.

\citeN

towards-dlpods consider rules whose heads are arbitrary combinations of atoms and the operators ×\times and \vee. A set of transformations is then used in order to bring the heads of rules into “Ordered Disjunctive Normal Form (ODNF)”. More specifically, each head is transformed into a formula of the form 𝒞1𝒞n{\cal C}_{1}\vee\cdots\vee{\cal C}_{n} where each 𝒞i{\cal C}_{i} is an ordered disjunction of literals. The resulting normalized rules are then used to obtain the preferred answer sets of the original program.

Example 9

The program of Example 8 is transformed to:

(pub×cinema)(pub × tv).\mbox{\tt(pub}\times\mbox{\tt cinema)}\vee\mbox{\tt(pub $\times$ tv).}

This program is then used to get the preferred answer sets of the original one.

However, as observed by \citeNlpod-cabalar, one of the transformations used by \citeNtowards-dlpods to obtain the ODNF, can not be logically justified: the formula (ϕ1ϕ2)×ϕ3(\phi_{1}\vee\phi_{2})\times\phi_{3} is not logically equivalent to the formula (ϕ1×ϕ3)(ϕ2×ϕ3)(\phi_{1}\times\phi_{3})\vee(\phi_{2}\times\phi_{3}) in terms of the logic of Here-and-There. As discussed at the end of Section 6, these two formulas are also not equivalent under our four-valued logic.

As an alternative approach to the semantics of DLPODs, \citeNlpod-cabalar proposes to use the logical characterization on rules with heads that are arbitrary combinations of disjunctions and ordered disjunctions. We could extend this approach to get a logical characterization of the most preferred models of arbitrary DLPODs: given such a program PP, we could at first consider all the \preceq-minimal models of PP that are solid, and then select the \sqsubset-minimal among them. Such an approach is certainly general. However, we believe that not every such program carries computational intuition. A good example of this is given in [Cabalar (2011), Example 2], where a program with both disjunction and ordered disjunction is given, and whose computational meaning is far from clear.

In the following, we define a class of programs which, as we claim, have a clear computational interpretation and at the same time retain all properties that we have identified for LPODs. Intuitively, we allow the head of a program rule to be a formula 𝒞1××𝒞n{\cal C}_{1}\times\cdots\times{\cal C}_{n} where each 𝒞i{\cal C}_{i} is an ordinary disjunction of literals. Notice that the program in Example 8 belongs to this class, while the program in Example 9, does not. We believe that the programs of this class have a clear preferential interpretation. Intuitively, the rule heads of the programs we consider, denote a hierarchy of preferences imposed by the ×\times operator; in each level of this hierarchy, we may have literals that have equal preference (this is expressed by standard disjunction).

It is important to stress that if we allowed arbitrary combinations of disjunctions and ordered disjunctions, the preferential intuition would be lost. To see this, consider for example the formula (a ×\times b) \vee (c ×\times d). This gives us the information that (a ×\times b) is at the same level of preference as (c ×\times d), and that a is more preferred than b and c is more preferred than d; however, for example, it gives us no information of whether a is more preferred than c. On the other hand, a formula of the fragment we consider, such as (a \vee b) ×\times (c \vee d) gives us a total order of a, b, c, and d.

Definition 18

A (propositional) DLPOD is a set of rules of the form:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

where the AjA_{j} and BlB_{l} are ground literals and each 𝒞i{\cal C}_{i} is a disjunction of ground literals.

As it turns out, the answer sets of such programs can be defined in an almost identical way as those of LPODs.

Definition 19

Let PP be a DLPOD. The ×\times-reduct of a rule RR of PP of the form:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

with respect to an interpretation II, is denoted by R×IR_{\times}^{I} and is defined as follows:

  • If I(Bi)=TI(B_{i})=T for some ii, 1ik1\leq i\leq k, then R×IR_{\times}^{I} is the empty set.

  • If I(Bi)TI(B_{i})\neq T for all ii, 1ik1\leq i\leq k, then R×IR_{\times}^{I} is the set that contains the rules:

    𝒞1F,A1,,Am𝒞r1F,A1,,Am𝒞rA1,,Am\begin{array}[]{lll}{\cal C}_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ {\cal C}_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ {\cal C}_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

    where rr is the least index such that I(𝒞1)==I(𝒞r1)=FI({\cal C}_{1})=\cdots=I({\cal C}_{r-1})=F^{*} and either r=nr=n or I(𝒞r)FI({\cal C}_{r})\neq F^{*}.

The ×\times-reduct of PP with respect to II is denoted by P×IP_{\times}^{I} and is the union of the reducts R×IR_{\times}^{I} for all RR in PP.

Definition 20

Let PP be a DLPOD and MM a (three-valued) interpretation of PP. Then, MM is an answer set of PP if MM is consistent and MM is a minimal model of the disjunctive program P×MP_{\times}^{M}.

As it turns out, all the results we have obtained in the previous section, hold for DLPODs. The proofs of these results are (surprisingly) almost identical (modulo some minor notational differences) to the proofs of the corresponding results for LPODs. For reasons of completeness, the corresponding proofs are given in the appendix. The extended results are stated below.

Lemma 7

Let PP be a consistent disjunctive extended logic program. Then, the answer sets of PP according to Definition 20, coincide with the standard answer sets of PP.

Lemma 8

Let PP be a DLPOD and let MM be an answer set of PP. Then, MM is a model of PP.

Lemma 9

Let MM be a model of a DLPOD PP. Then, MM is a model of P×MP^{M}_{\times}.

Lemma 10

Every answer set MM of a DLPOD PP, is a \preceq-minimal model of PP.

Theorem 2

Let PP be a DLPOD. Then, MM is an answer set of PP iff MM is a consistent \preceq-minimal model of PP and MM is solid.

The similarity of the definitions and of the theoretical results of DLPODs to those of standard LPODs, makes us believe that this is indeed an interesting class of programs that deserves further attention.

8 Related and Future Work

The work on LPODs is closely related to “Qualitative Choice Logic” (QCL) [Brewka et al. (2004)]. QCL is an extension of propositional logic with the preferential connective “×\times”, which has the same intuitive meaning as in LPODs: A×BA\times B is read “if possible AA, but if AA is impossible then at least BB”. Essentially, QCL is the propositional logic underlying LPODs. It is worth noting that the semantics of QCL is based on the “degree of satisfaction” of formulas, which is connected to the idea of the degree of satisfaction of the rules of LPODs (Definition 5). Moreover, as remarked by one of the reviewers of the present paper, the DLPODs introduced in Section 7 are closely connected to the “basic choice formulas” of QCL [Brewka et al. (2004), Section 3.1, Definition 8]. It would be interesting to investigate whether our four-valued logic can be used to provide an alternative semantics for QCL.

The work reported in this paper is closely connected to the work of \citeNlpod-cabalar, who first considered the problem of expressing logically the semantics of LPODs. The key difference between the two works is that ours provides a characterization of both phases of the production of the most preferred models of an LPOD, while Cabalar’s work concentrates on the first one.

It is important to stress here that both our work as well as the work of \citeNlpod-cabalar, are influenced by the work of \citeNPearce96 who first gave a logical characterization of the answer sets of extended logic programs, using Equilibrium Logic. This is a non-monotonic logic which is defined on top of the monotonic logic of Here-and-There [Heyting (1930)], using a model preference approach. The technique we have proposed in this paper, when applied to a consistent extended logic program PP, produces the standard answer sets of PP; this is a direct consequence of Theorem 1 and Lemma 1. Therefore, for extended logic programs, the Equilibrium Logic gives the same outcome as our approach which is based on a four-valued logic and \preceq-minimal models that are solid. We believe that a further investigation of the connections of our approach with that of Equilibrium Logic is a worthwhile topic.

Our work is the first to provide a purely model-theoretic characterization of the semantics of LPODs. To our knowledge, the four-valued logic we have utilized does not appear to be a well-known variant/extension of Here-and-There. However, some seemingly related logics have been used in the literature of answer set extensions. The original definition of Equilibrium Logic included a second constructive negation, which corresponds to Nelson’s strong negation [Nelson (1949)]. This gave rise to a five-valued extension of Here-and-There, called 𝒩5{\cal N}_{5}. Also, a logic called 𝒳5{\cal X}_{5}, that is closely connected to 𝒩5{\cal N}_{5}, was recently proposed by \citeNACF0PV19 in order to capture the semantics of arbitrary combinations of explicit negation with nested expressions. Both 𝒩5{\cal N}_{5} and 𝒳5{\cal X}_{5} appear to be connected to our four-valued logic due to the different notions of false and true that they employ in order to capture aspects that arise in answer set semantics. However, the ordering of the truth values and the semantics of the logical connectives are different, and the exact correspondence (if any) between these logics and the present one, is not straightforward to establish. This is certainly an interesting topic for further investigation.

Another promising topic for future work is the characterization of the notion of strong equivalence [Lifschitz et al. (2001)] for LPODs and DLPODs. When two logic programs are strongly equivalent, we can replace one for the other inside a bigger program without worrying that the semantics of the bigger program will be affected. Characterizations of strong equivalence for LPODs have already been obtained by \citeNFaberTW08. It would be interesting to investigate if the logical characterization of the semantics of LPODs and DLPODs developed in the present paper, can offer advantages compared with their work.

Acknowledgments

We would like to thank the three anonymous reviewers of our paper for their careful and insightful comments.

This research is co-financed by Greece and the European Union (European Social Fund- ESF) through the Operational Programme “Human Resources Development, Education and Lifelong Learning 2014- 2020” in the context of the project “Techniques for implementing qualitative preferences in deductive querying systems” (5048151).

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Appendix A Proofs of Section 5

\@begintheorem

Lemma1 Let PP be a consistent extended logic program. Then the three-valued answer sets of PP coincide with the standard answer sets of PP. \@endtheorem

Proof A.3.

By taking n=1n=1 in Definition 10, we get the standard definition of reduct for consistent extended logic programs.

\@begintheorem

Lemma2 Let PP be an LPOD and let MM be an answer set of PP. Then, MM is a model of PP. \@endtheorem

Proof A.4.

Consider any rule RR in PP of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

If R×M=R_{\times}^{M}=\emptyset, then M(Bi)=TM(B_{i})=T for some ii, 1ik1\leq i\leq k. But then, the body of the rule RR evaluates to FF under MM, and therefore MM satisfies RR. Consider now the case where R×MR_{\times}^{M} is nonempty and consists of the following rules:

C1F,A1,,AmCr1F,A1,,AmCrA1,,Am\begin{array}[]{lll}C_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ C_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ C_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

We distinguish cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}):

Case 1: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F. Then, for some ii, M(Ai)=FM(A_{i})=F. Then, rule RR is trivially satisfied by MM.

Case 2: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. This implies that M(Cr)FM(C_{r})\geq F^{*}. We distinguish two subcases. If r=nr=n then M(C1××Cn)=M(C1××Cr)FM({C_{1}\times\cdots\times C_{n}})=M({C_{1}\times\cdots\times C_{r}})\geq F^{*} because, by the definition of P×MP_{\times}^{M} it is M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and we also know that M(Cr)FM(C_{r})\geq F^{*}. Thus, in this subcase MM satisfies RR. If r<nr<n, then by the definition of P×MP_{\times}^{M}, M(Cr)FM(C_{r})\neq F^{*}; however, we know that M(Cr)FM(C_{r})\geq F^{*}, and thus M(Cr)=TM(C_{r})=T. Thus, in this subcase MM also satisfies RR.

Case 3: M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Then, for all ii, M(Ai)=TM(A_{i})=T. Since MM is a model of P×MP_{\times}^{M}, we have M(Cr)=TM(C_{r})=T. Moreover, by the definition of P×MP_{\times}^{M}, M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*}. This implies that M(C1××Cn)=TM(C_{1}\times\cdots\times C_{n})=T.

\@begintheorem

Lemma3 Let MM be a model of an LPOD PP. Then, MM is a model of P×MP^{M}_{\times}. \@endtheorem

Proof A.5.

Consider any rule RR in PP of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

and assume MM satisfies RR. If M(Bi)=TM(B_{i})=T for some ii, 1ik1\leq i\leq k, then no rule is created in P×MP^{M}_{\times} for RR. Assume therefore that M(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=TM({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=T. By the definition of P×MP^{M}_{\times} the following rules have been added to P×MP^{M}_{\times}:

C1F,A1,,AmCr1F,A1,,AmCrA1,,Am\begin{array}[]{lll}C_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ C_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ C_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

where rr is the least index such that M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and either r=nr=n or M(Cr)FM(C_{r})\neq F^{*}. Obviously, the first r1r-1 rules above are satisfied by MM. For the rule CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m} we distinguish two cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}). If M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F, then, the rule is trivially satisfied. If M(A1,,Am)>FM(A_{1},\ldots,A_{m})>F, then, since rule RR is satisfied by MM and M(Cr)FM(C_{r})\neq F^{*}, it has to be M(Cr)=TM(C_{r})=T. Therefore, the rule CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m} is satisfied by MM.

\@begintheorem

Lemma4 Every (three-valued) answer set MM of an LPOD PP, is a \preceq-minimal model of PP. \@endtheorem

Proof A.6.

Assume there exists a model NN of PP with NMN\preceq M. We will show that NN is also a model of P×MP_{\times}^{M}. Since NMN\preceq M, we also have NMN\leq M. Since MM is the \leq-least model of P×MP_{\times}^{M}, we will conclude that N=MN=M.

Consider any rule RR in PP of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

Assume that R×MR_{\times}^{M} is nonempty. This means that there exists some rr, 1rn1\leq r\leq n, such that M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and either r=nr=n or M(Cr)FM(C_{r})\neq F^{*}. Then, R×MR_{\times}^{M} consists of the following rules:

C1F,A1,,AmCr1F,A1,,AmCrA1,,Am\begin{array}[]{lll}C_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ C_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ C_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

We show that NN satisfies the above rules. We distinguish cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}):

Case 1: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F. Then, N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F and the above rules are trivially satisfied by NN.

Case 2: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. Then, since NMN\preceq M, it is N(A1,,Am)FN(A_{1},\ldots,A_{m})\leq F^{*}. If N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F then NN trivially satisfies all the above rules. Assume therefore that N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F^{*}. Recall now that M(Ci)=FM(C_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1. Moreover, it has to be M(Cr)FM(C_{r})\geq F^{*}, because otherwise MM would not satisfy the rule RR. Since NMN\preceq M, it can only be N(Ci)=FN(C_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1 and N(Cr)FN(C_{r})\geq F^{*}, because otherwise NN would not be a model of PP. Therefore, NN satisfies the given rules of P×MP^{M}_{\times}.

Case 3: M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Then, since NMN\preceq M, it is either N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F or N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T. If N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F then NN trivially satisfies all the above rules. Assume therefore that N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T. Recall now that M(Ci)=FM(C_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1. Moreover, it has to be M(Cr)=TM(C_{r})=T, because otherwise MM would not satisfy the rule RR. Since NMN\preceq M, it can only be N(Ci)=FN(C_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1 and N(Cr)=TN(C_{r})=T, because otherwise NN would not be a model of PP. Therefore, NN satisfies the given rules of P×MP^{M}_{\times}.

In the proofs that follow, we will use the term Brewka-model to refer to that of Definition 2 and Brewka-reduct to refer to that of Definition 3 (although, to be precise, this definition of reduct was initially introduced in the paper by \citeNlpod-BNS04).

In order to establish Lemmas 5 and 6 we first show the following three propositions.

Proposition A.7.

Let PP be an LPOD and let MM be a three-valued model of PP. Then, N=𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M)N=\mathit{collapse}(M) is a Brewka-model of PP.

Proof A.8.

Consider any rule RR of PP of the form

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

If there exists AiNA_{i}\not\in N or there exists BjNB_{j}\in N then then NN trivially satisfies RR. Assume that {A1,,Am}N\{A_{1},\ldots,A_{m}\}\subseteq N and {B1,,Bk}N=\{B_{1},\ldots,B_{k}\}\cap N=\emptyset. By Definition 15 it follows that M(A1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=TM(A_{1},\ldots,A_{m},\mathit{not}{\,}B_{1},\ldots,\mathit{not}{\,}B_{k})=T. Since MM is a three-valued model of PP, it must satisfy RR and therefore M(C1××Cn)=TM(C_{1}\times\cdots\times C_{n})=T. Then, there exists rnr\leq n such that M(Cr)=TM(C_{r})=T and by Definition 15 we get that CrNC_{r}\in N. Therefore, NN satisfies rule RR.

Proposition A.9.

Let PP be an LPOD and MM be a Brewka-model of PP. Then, MM is also a model of the Brewka-reduct P×MP^{M}_{\times}.

Proof A.10.

Consider any rule RR in PP of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

and assume MM satisfies RR. If there exists BiMB_{i}\in M for some 1ik1\leq i\leq k, then no rule is created in the Brewka-reduct for RR. Moreover, if for all ini\leq n, CiMC_{i}\not\in M then also no rule is created in the Brewka-reduct. Assume therefore that {B1,,Bk}M=\{B_{1},\ldots,B_{k}\}\cap M=\emptyset and there exists rnr\leq n such that CrMC_{r}\in M and {C1,,Cr1}M=\{C_{1},\ldots,C_{r-1}\}\cap M=\emptyset. By the definition of P×MP^{M}_{\times} the only rule added to P×MP^{M}_{\times} because of RR is CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m}. Since CrMC_{r}\in M the rule is satisfied by MM.

Proposition A.11.

Let PP be an LPOD and let M1,M2M_{1},M_{2} be three-valued answer sets of PP such that 𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M1)=𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M2)\mathit{collapse}(M_{1})=\mathit{collapse}(M_{2}). Then, M1=M2M_{1}=M_{2}.

Proof A.12.

Assume, for the sake of contradiction, that M1M2M_{1}\neq M_{2}. We define:

M(A)={M1(A)if M1(A)=M2(A)FotherwiseM(A)=\left\{\begin{array}[]{ll}M_{1}(A)&\mbox{if $M_{1}(A)=M_{2}(A)$}\\ F&\text{otherwise}\end{array}\right.

It is MM1M\prec M_{1} and MM2M\prec M_{2}. We claim that MM is a model of PP. This will lead to contradiction because, by Lemma 4, M1M_{1} and M2M_{2} are \preceq-minimal models of PP.

Consider any rule RR in PP of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

If M(Bi)=TM(B_{i})=T for some ii, 1ik1\leq i\leq k, then MM satisfies the rule. Assume therefore that M(Bi)TM(B_{i})\neq T for all ii, 1ik1\leq i\leq k. We distinguish cases:

Case 1: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F. Then, obviously, MM satisfies RR.

Case 2: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. Then, M1(A1,,Am)=FM_{1}(A_{1},\ldots,A_{m})=F^{*} and M2(A1,,Am)=FM_{2}(A_{1},\ldots,A_{m})=F^{*}. Since, by Lemma 2, M1M_{1} and M2M_{2} are models of PP it follows that M1(C1××Cn)FM_{1}(C_{1}\times\cdots\times C_{n})\geq F^{*} and M2(C1××Cn)FM_{2}(C_{1}\times\cdots\times C_{n})\geq F^{*}. First assume that M1(C1××Cn)=TM_{1}(C_{1}\times\cdots\times C_{n})=T. This implies that there exists 1rn1\leq r\leq n such that M1(Cr)=TM_{1}(C_{r})=T and M1(Ci)=FM_{1}(C_{i})=F^{*} for all 1i<r1\leq i<r. Since, by assumption 𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M1)=𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M2)\mathit{collapse}(M_{1})=\mathit{collapse}(M_{2}) it follows that M2(Cr)=TM_{2}(C_{r})=T and therefore M(Cr)=TM(C_{r})=T. Moreover, it must be M2(Ci)=FM_{2}(C_{i})=F^{*} for all i<ri<r because we have already established that M2(C1××Cn)FM_{2}(C_{1}\times\cdots\times C_{n})\geq F^{*}. Therefore, M(Ci)=FM(C_{i})=F^{*} and M(C1××Cn)=TM(C_{1}\times\cdots\times C_{n})=T and MM satisfies the rule. Now assume that M1(C1××Cn)=FM_{1}(C_{1}\times\cdots\times C_{n})=F^{*}. It is easy to see that the only case is M1(Ci)=FM_{1}(C_{i})=F^{*} for all 1in1\leq i\leq n. Since M2M_{2} has the same collapse with M1M_{1} it follows that M2(Ci)FM_{2}(C_{i})\leq F^{*} and because M2(C1××Cn)FM_{2}(C_{1}\times\cdots\times C_{n})\geq F^{*} it also follows that M2(Ci)=FM_{2}(C_{i})=F^{*}. By definition of MM, M(Ci)=FM(C_{i})=F^{*} for all 1in1\leq i\leq n and M(C1××Cn)=FM(C_{1}\times\cdots\times C_{n})=F^{*}.

Case 3: M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Then, M1(A1,,Am)=TM_{1}(A_{1},\ldots,A_{m})=T and M2(A1,,Am)=TM_{2}(A_{1},\ldots,A_{m})=T and therefore M1(C1××Cn)=TM_{1}(C_{1}\times\cdots\times C_{n})=T and M2(C1××Cn)=TM_{2}(C_{1}\times\cdots\times C_{n})=T. This implies that there exists rr such that M1(C1)=M2(C1)=F,,M1(Cr1)=M2(Cr1)=FM_{1}(C_{1})=M_{2}(C_{1})=F^{*},\ldots,M_{1}(C_{r-1})=M_{2}(C_{r-1})=F^{*}, and M1(Cr)=M2(Cr)=TM_{1}(C_{r})=M_{2}(C_{r})=T. Therefore, M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and M(Cr)=TM(C_{r})=T, which implies that M(C1××Cn)=TM(C_{1}\times\cdots\times C_{n})=T, and therefore MM satisfies RR.

\@begintheorem

Lemma5 Let PP be an LPOD and MM be a three-valued answer set of PP. Then, 𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M)\mathit{collapse}(M) is an answer set of PP according to Definition 4. \@endtheorem

Proof A.13.

Since MM is an answer set of PP, then by Lemma 2, MM is also a model of PP. Moreover, by Proposition A.7, N=𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M)N=\mathit{collapse(M)} is a Brewka-model of PP. It also follows from Proposition A.9 that NN is a model of the Brewka-reduct PNP^{N}. It suffices to show that NN is also the minimum model of PNP^{N}. Assume there exists NN^{\prime} that is a model of PNP^{N} and NNN^{\prime}\subset N. We define MM^{\prime} as

M(A)={FAN and ANM(A)otherwiseM^{\prime}(A)=\left\{\begin{array}[]{ll}F^{*}&A\in N\text{ and }A\not\in N^{\prime}\\ M(A)&\text{otherwise}\end{array}\right.

It is easy to see that M<MM^{\prime}<M. We will show that MM^{\prime} is also model of P×MP^{M}_{\times} leading to contradiction because we assume that MM is the minimum model of P×MP^{M}_{\times}. Consider first a rule of the form CiF,A1,,AmC_{i}\leftarrow F^{*},A_{1},\ldots,A_{m}. Since MM is an answer set of PP it must be M(Ci)=FM(C_{i})=F^{*}. By the definition of MM^{\prime} it follows that M(Ci)FM^{\prime}(C_{i})\geq F^{*} and MM^{\prime} satisfies the rule. Now consider a rule of the form CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m}. We distinguish cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}).

Case 1: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F. Then, since M<MM^{\prime}<M it is M(A1,,Am)=FM^{\prime}(A_{1},\ldots,A_{m})=F and the rule is trivially satisfied.

Case 2: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. Then, M(Ai)FM(A_{i})\geq F^{*} and there exists AiA_{i} such that M(Ai)=FM(A_{i})=F^{*}. It follows that AiNA_{i}\not\in N and therefore M(Ai)=M(Ai)=FM^{\prime}(A_{i})=M(A_{i})=F^{*}. Moreover, by the construction of MM^{\prime}, for all AiA_{i} we have M(Ai)FM^{\prime}(A_{i})\geq F^{*} and therefore M(A1,,Am)=FM^{\prime}(A_{1},\ldots,A_{m})=F^{*}. Since MM is a model of P×MP^{M}_{\times}, M(Cr)FM(C_{r})\geq F^{*}. Again, by the construction of MM^{\prime} we have M(Cr)FM^{\prime}(C_{r})\geq F^{*} and the rule is satisfied.

Case 3: M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. By the construction of P×MP^{M}_{\times} the rule CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m} is a result of a rule in PP of the form

C1××Cr××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{r}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

and it must be M(Ci)=FM(C_{i})=F^{*} for all ir1i\leq r-1 and M(Bj)FM(B_{j})\leq F^{*} for all 1jk1\leq j\leq k. It follows that {C1,,Cr1}N=\{C_{1},\ldots,C_{r-1}\}\cap N=\emptyset and {B1,,Bk}N=\{B_{1},\ldots,B_{k}\}\cap N=\emptyset. Moreover, since MM is a model of P×MP^{M}_{\times} we get that M(Cr)=TM(C_{r})=T and it follows that CrNC_{r}\in N. By the construction of the Brewka-reduct, there exists a rule CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m} in PNP^{N}. We distinguish two cases. If {A1,,Am}N\{A_{1},\ldots,A_{m}\}\subseteq N^{\prime} then CrNC_{r}\in N^{\prime} because NN^{\prime} is a model of PNP^{N}. It follows by the construction of MM^{\prime} that M(Cr)=M(Cr)=TM^{\prime}(C_{r})=M(C_{r})=T and MM^{\prime} satisfies the rule. Otherwise, there exists ll, 1lm1\leq l\leq m such that AlNA_{l}\not\in N^{\prime}. Notice also that {A1,,Am}N\{A_{1},\ldots,A_{m}\}\subseteq N, so AlNA_{l}\in N. Therefore, M(Al)=FM^{\prime}(A_{l})=F^{*} and M(A1,,Am)FM^{\prime}(A_{1},\ldots,A_{m})\leq F^{*}. Moreover, since CrNC_{r}\in N, we have M(Cr)FM^{\prime}(C_{r})\geq F^{*} that satisfies the rule.

\@begintheorem

Lemma6 Let NN be an answer set of PP according to Definition 4. There exists a unique three-valued interpretation MM such that N=𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M)N=\mathit{collapse}(M) and MM is a three-valued answer set of PP. \@endtheorem

Proof A.14.

We construct iteratively a set of literals that must have the value FF^{*} in MM. Let n{\cal F}^{n} be the sequence:

0\displaystyle{\cal F}^{0} =\displaystyle=\emptyset
n+1\displaystyle{\cal F}^{n+1} ={Cj\displaystyle=\{C_{j}\mid (C1××CnA1,,Am,𝑛𝑜𝑡B1,𝑛𝑜𝑡Bk)P\displaystyle(C_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},\mathit{not}{\,}B_{1},\ldots\mathit{not}{\,}B_{k})\in P
and {B1,,Bk}N=\displaystyle\text{ and }\{B_{1},\ldots,B_{k}\}\cap N=\emptyset
and {C1,,Cj}N=\displaystyle\text{ and }\{C_{1},\ldots,C_{j}\}\cap N=\emptyset
and {A1,,Am}Nn}\displaystyle\text{ and }\{A_{1},\ldots,A_{m}\}\subseteq N\cup{\cal F}^{n}\}
ω\displaystyle{\cal F}^{\omega} =n<ωn\displaystyle=\hbox to0.0pt{$\cup_{n<\omega}{\cal F}^{n}$\hss}

We construct MM as

M(A)={FAN and AωFAN and AωTANM(A)=\left\{\begin{array}[]{ll}F&A\not\in N\text{ and }A\not\in{\cal F}^{\omega}\\ F^{*}&A\not\in N\text{ and }A\in{\cal F}^{\omega}\\ T&A\in N\end{array}\right.

First we prove that MM is a model of P×MP^{M}_{\times}. Consider first any rule of the form CiF,A1,,AmC_{i}\leftarrow F^{*},A_{1},\ldots,A_{m}. By the construction of P×MP^{M}_{\times}, such a rule exists because M(Ci)=FM(C_{i})=F^{*}; therefore MM satisfies this rule. Now consider any rule of the form CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m}. Such a rule was produced by a rule RR in PP of the form

C1××Cr××CnA1,,An,,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk.C_{1}\times\cdots\times C_{r}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{n},\ldots,\mathit{not}{\,}B_{1},\ldots,\mathit{not}{\,}B_{k}.

By the construction of P×MP^{M}_{\times} it follows that M(Ci)=FM(C_{i})=F^{*} for all i<ri<r. Therefore CiNC_{i}\not\in N and also CiωC_{i}\in{\cal F}^{\omega} for all i<ri<r. Moreover, it must be M(Bj)FM(B_{j})\leq F^{*} for all 1jk1\leq j\leq k, so {B1,,Bk}N=\{B_{1},\ldots,B_{k}\}\cap N=\emptyset. We distinguish cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}).

Case 1: If M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F then the rule is trivially satisfied by MM.

Case 2: If M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*} then for some AiA_{i}, M(Ai)=FM(A_{i})=F^{*}. By the construction of MM, it follows that AiωA_{i}\in{\cal F}^{\omega}. It follows by the definition of ω{\cal F}^{\omega} that CrωC_{r}\in{\cal F}^{\omega} and therefore M(Cr)FM(C_{r})\geq F^{*}.

Case 3: If M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T then {A1,,Am}N\{A_{1},\ldots,A_{m}\}\subseteq N and since NN is an answer set according to Definition 4 it follows that NN is a model of PP. It follows that there exists a least jnj\leq n such that CjNC_{j}\in N. Since we have already established that for all i<ri<r, CiNC_{i}\not\in N it must be rjnr\leq j\leq n. But, if r<jr<j then CrNC_{r}\not\in N and by the construction of MM it must be M(Cr)=FM(C_{r})=F^{*}. If M(Cr)=FM(C_{r})=F^{*}, then, by the construction of P×MP^{M}_{\times}, the rule for CrC_{r} should be of the form CrF,A1,,AmC_{r}\leftarrow F^{*},A_{1},\ldots,A_{m}. So, it must j=rj=r and CrNC_{r}\in N. Therefore, M(Cr)=TM(C_{r})=T and MM satisfies the rule.

Therefore, we have established that MM is a model of P×MP^{M}_{\times}. It remains to show that MM is the \leq-least model of P×MP^{M}_{\times}. Assume now that there exists MM^{\prime} that is a model of P×MP^{M}_{\times} and M<MM^{\prime}<M. Let N=𝑐𝑜𝑙𝑙𝑎𝑝𝑠𝑒(M)N^{\prime}=\mathit{collapse}(M^{\prime}). We distinguish two cases.

Case 1: N=NN^{\prime}=N and thus MM^{\prime} differs from MM only on some atoms CrC_{r} such that M(Cr)=FM^{\prime}(C_{r})=F and M(Cr)=FM(C_{r})=F^{*}. First, by the construction of MM, if M(Cr)=FM(C_{r})=F^{*} then CrωC_{r}\in{\cal F}^{\omega}. We show by induction on nn that for every CrnC_{r}\in{\cal F}^{n}, M(Cr)FM^{\prime}(C_{r})\geq F^{*}. This leads to contradiction and therefore MM is minimal.

Induction base: n=0n=0: the statement is satisfied vacuously.

Induction step: n=n0+1n=n_{0}+1: Every atom Crn0+1C_{r}\in{\cal F}^{n_{0}+1} must occur in a head of a rule in PP. such that {C1,,Cr1}N=\{C_{1},\ldots,C_{r-1}\}\cap N=\emptyset and therefore {C1,,Cr}n0+1\{C_{1},\ldots,C_{r}\}\subseteq{\cal F}^{n_{0}+1}. It follows then that M(Ci)=FM(C_{i})=F^{*} for 1ir1\leq i\leq r. By the construction of P×MP^{M}_{\times}, for every atom Crn0+1C_{r}\in{\cal F}^{n_{0}+1} there must be a rule in P×MP^{M}_{\times} either of the form CrF,A1,,AmC_{r}\leftarrow F^{*},A_{1},\ldots,A_{m} or of the form CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m}. Moreover, since Crn0+1C_{r}\in{\cal F}^{n_{0}+1} it follows that {A1,,Am}Nn0\{A_{1},\ldots,A_{m}\}\subseteq N\cup{\cal F}^{n_{0}}. Therefore, by the induction hypothesis, M(A1,,Am)=M(A1,,Am)FM(A_{1},\ldots,A_{m})=M^{\prime}(A_{1},\ldots,A_{m})\geq F^{*}. Since MM^{\prime} is also a model of P×MP^{M}_{\times} it must satisfy those rules thus M(Cr)FM^{\prime}(C_{r})\geq F^{*}.

Case 2: NNN^{\prime}\subset N. We show that NN^{\prime} is a model of PNP^{N} leading to contradiction because, by definition, NN is the minimum model of PNP^{N}. Consider a rule RR of the form CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m} in PNP^{N}. The rule RR has been produced by a rule in PP of the form:

C1××Cr××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{r}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

such that {C1,,Cr1}N=\{C_{1},\ldots,C_{r-1}\}\cap N=\emptyset and CrNC_{r}\in N.

If there exists AiNA_{i}\not\in N then also AiNA_{i}\not\in N^{\prime} and the rule is trivially satisfied by NN^{\prime}. Assume, on the other hand, that {A1,,An}N\{A_{1},\ldots,A_{n}\}\subseteq N. It follows, by the definition of MM, that M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T, M(Ci)=FM(C_{i})=F^{*} for i<ri<r and M(Cr)=TM(C_{r})=T. Therefore, there exist a rule in P×MP^{M}_{\times} of the form CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m}. If M(A1,,Am)=FM^{\prime}(A_{1},\ldots,A_{m})=F or M(A1,,Am)=FM^{\prime}(A_{1},\ldots,A_{m})=F^{*} then there exists AiNA_{i}\not\in N^{\prime} and NN^{\prime} again satisfies the rule. If M(A1,,Am)=TM^{\prime}(A_{1},\ldots,A_{m})=T then since MM^{\prime} is a model of P×MP^{M}_{\times} it follows that M(Cr)=TM^{\prime}(C_{r})=T. Since NN^{\prime} is the collapse of MM^{\prime} it is {A1,,Am}N\{A_{1},\ldots,A_{m}\}\subseteq N^{\prime} and CrNC_{r}\in N^{\prime}. Therefore, NN^{\prime} satisfies the rule RR in PNP^{N}.

The uniqueness of MM follows directly from Proposition A.11.

Appendix B Proofs of Section 6

In order to establish Theorem 1, we show two lemmas (which essentially establish the left-to-right and the right-to-left directions of the theorem, respectively).

Lemma B.15.

Let PP be an LPOD program and let MM be an answer set of PP. Then, MM is a \preceq-minimal model of PP and MM is solid.

Proof B.16.

Since MM is an answer set of PP, then, by Lemma 2, MM is a model of PP. Moreover, MM is solid because our definition of answer sets does not involve the value TT^{*}. It remains to show that it is minimal with respect to the \preceq ordering. Assume, for the sake of contradiction, that there exists a model NN of PP with NMN\prec M. By Lemma 4, MM is (three-valued) \preceq-minimal. Therefore, NN can not be solid. We first show that NN can not be a model of the reduct P×MP^{M}_{\times}. Assume for the sake of contradiction that NN is a model of P×MP^{M}_{\times}. We construct the following interpretation NN^{\prime}:

N(A)={F,if N(A)=TN(A),otherwiseN^{\prime}(A)=\left\{\begin{array}[]{ll}F^{*},&\mbox{if $N(A)=T^{*}$}\\ N(A),&\mbox{otherwise}\end{array}\right.

We claim that NN^{\prime} must also be a model of P×MP^{M}_{\times}. Consider first a rule of the form CF,A1,,AmC\leftarrow F^{*},A_{1},\ldots,A_{m}. Since NN is a model of P×MP^{M}_{\times}, it is N(C)FN(C)\geq F^{*}. By the definition of NN^{\prime}, it is N(C)FN(C)\geq F^{*} and therefore NN^{\prime} satisfies this rule. Consider now a rule of the form CA1,,AmC\leftarrow A_{1},\ldots,A_{m} in P×MP^{M}_{\times}. We show that NN^{\prime} also satisfies this rule. We perform a case analysis:

Case 1: N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F. Then, N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F and NN^{\prime} trivially satisfies the rule.

Case 2: N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F^{*}. Then, N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F^{*}. Moreover, N(C)FN(C)\geq F^{*} because NN is a model of P×MP^{M}_{\times}. By the definition of NN^{\prime}, it is N(C)FN^{\prime}(C)\geq F^{*}, and therefore NN^{\prime} satisfies the rule.

Case 3: N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T^{*}. Then, N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F^{*}. Moreover, N(C)TN(C)\geq T^{*} because NN is a model of P×MP^{M}_{\times}. By the definition of NN^{\prime}, it is N(C)FN^{\prime}(C)\geq F^{*}, and therefore NN^{\prime} satisfies the rule.

Case 4: N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T. Then, N(A1,,Am)=TN^{\prime}(A_{1},\ldots,A_{m})=T. Moreover, N(C)=TN(C)=T because NN is a model of P×MP^{M}_{\times}. By the definition of NN^{\prime}, it is N(C)=TN^{\prime}(C)=T, and therefore NN^{\prime} satisfies the rule.

Therefore, NN^{\prime} must also be a model of P×MP^{M}_{\times}. Moreover, by definition, NN^{\prime} is solid and N<MN^{\prime}<M. This contradicts the fact that, by construction, MM is the \leq-least model of P×MP^{M}_{\times}. In conclusion, NN can not be a model of P×MP^{M}_{\times}.

We now show that NN can not be a model of PP. As we showed above, NN is not a model of P×MP^{M}_{\times}, and consequently there exists a rule in P×MP^{M}_{\times} that is not satisfied by NN. Such a rule in P×MP^{M}_{\times} must have resulted due to a rule RR of the following form in PP:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

According to the definition of P×MP^{M}_{\times}, for all ii, 1ik1\leq i\leq k, M(𝑛𝑜𝑡Bi)=TM(\mathit{not}{\,}B_{i})=T, and since NMN\prec M, it is also N(𝑛𝑜𝑡Bi)=TN(\mathit{not}{\,}B_{i})=T. Moreover, there exists some rnr\leq n such that M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and either r=nr=n or M(Cr)FM(C_{r})\neq F^{*}. Since NMN\prec M, it is N(Ci)FN(C_{i})\leq F^{*} for all ii, 1ir11\leq i\leq r-1. Consider now the rule that is not satisfied by NN in P×MP^{M}_{\times}. If it is of the form CiF,A1,,AmC_{i}\leftarrow F^{*},A_{1},\ldots,A_{m}, ii, 1ir11\leq i\leq r-1, then N(A1,,Am)>FN(A_{1},\ldots,A_{m})>F and N(Ci)=FN(C_{i})=F. This implies that N(C1××Cn)=FN(C_{1}\times\cdots\times C_{n})=F and therefore NN does not satisfy the rule RR. If the rule that is not satisfied by NN in P×MP^{M}_{\times} is of the form CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m}, then N(Cr)<N(A1,,Am)N(C_{r})<N(A_{1},\ldots,A_{m}) and therefore, since N(Ci)FN(C_{i})\leq F^{*} for all ii, 1ir11\leq i\leq r-1, it is:

N(C1××Cn)<N(A1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)N(C_{1}\times\cdots\times C_{n})<N(A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})

Thus, NN is not a model of PP.

Lemma B.17.

Let PP be an LPOD program and let MM be a \preceq-minimal model of PP and MM is solid. Then, MM is an answer set of PP.

Proof B.18.

First observe that, by Lemma 3, MM is also a model of P×MP^{M}_{\times}. We demonstrate that MM is actually the \leq-least model of P×MP^{M}_{\times}. Assume, for the sake of contradiction, that NN is the \leq-least model of P×MP^{M}_{\times}. Then, NN will differ from MM in some atoms AA such that N(A)<M(A)N(A)<M(A). We distinguish two cases. In the first case all the atoms AA such that N(A)<M(A)N(A)<M(A) have M(A)FM(A)\leq F^{*}. In the second case there exist at least one atom AA such that M(A)>FM(A)>F^{*}.

In the first case it is easy to see that NMN\prec M. We demonstrate that NN is also model of PP leading to contradiction since MM is \preceq-minimal. Assume that NN is not a model of PP. Then, there exists in PP a rule RR of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

such that N(C1××Cn)<N(A1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)N(C_{1}\times\cdots\times C_{n})<N^{\prime}(A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}). Notice that this implies that N(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=M(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=TN({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=M({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=T. Therefore, N(C1××Cn)<N(A1,,Am)N({C_{1}\times\cdots\times C_{n}})<N(A_{1},\ldots,A_{m}). We distinguish cases based on the value of N(A1,,Am)N(A_{1},\ldots,A_{m}):

Case 1: N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F. This case leads immediately to contradiction because NN trivially satisfies RR.

Case 2: N(A1,,Am)>FN(A_{1},\ldots,A_{m})>F. Then, N(A1,,Am)=M(A1,,Am)N(A_{1},\ldots,A_{m})=M(A_{1},\ldots,A_{m}). Since MM is a model of PP, it is M(C1××Cn)M(A1,,Am)>FM(C_{1}\times\cdots\times C_{n})\geq M(A_{1},\ldots,A_{m})>F. This implies that there exists some rr, 1rn1\leq r\leq n, such that M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and M(Cr)FM(C_{r})\geq F^{*}. By the definition of the reduct, the rule CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m} exists in P×MP^{M}_{\times}. Since NN is a model of P×MP^{M}_{\times}, we get that N(Cr)>FN(C_{r})>F. Moreover, NN should also satisfy the rules CiF,A1,,AmC_{i}\leftarrow F^{*},A_{1},\ldots,A_{m} for 1ir11\leq i\leq r-1. Since N(Ci)M(Ci)N(C_{i})\leq M(C_{i}) and N(Cr)=M(Cr)N(C_{r})=M(C_{r}) we get N(C1)==N(Cr1)=FN(C_{1})=\cdots=N(C_{r-1})=F^{*} and N(Cr)=M(Cr)N(C_{r})=M(C_{r}). Therefore N(C1×Cn)=M(C1×Cn)N(C_{1}\times\cdots C_{n})=M(C_{1}\times\cdots C_{n}) and N(C1×Cn)N(A1,,Am)N(C_{1}\times\cdots C_{n})\geq N(A_{1},\ldots,A_{m}) (contradiction).

In the second case we construct the following interpretation NN^{\prime}:

N(A)={T,if M(A)=T and N(A){F,F}F,if M(A)=FN(A),otherwiseN^{\prime}(A)=\left\{\begin{array}[]{ll}T^{*},&\mbox{if $M(A)=T$ and $N(A)\in\{F,F^{*}\}$}\\ F^{*},&\mbox{if $M(A)=F^{*}$}\\ N(A),&\mbox{otherwise}\end{array}\right.

It is easy to see that NMN^{\prime}\prec M. We demonstrate that NN^{\prime} is a model of PP, which will lead to a contradiction (since we have assumed that MM is \preceq-minimal).

Assume NN^{\prime} is not a model of PP. Then, there exists in PP a rule RR of the form:

C1××CnA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡BkC_{1}\times\cdots\times C_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

such that N(C1××Cn)<N(A1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)N^{\prime}(C_{1}\times\cdots\times C_{n})<N^{\prime}(A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}). Notice that this implies that N(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=N(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=M(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=TN^{\prime}({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=N({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=M({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=T. Therefore, N(C1××Cn)<N(A1,,Am)N^{\prime}(C_{1}\times\cdots\times C_{n})<N^{\prime}(A_{1},\ldots,A_{m}). We distinguish cases based on the value of N(A1,,Am)N^{\prime}(A_{1},\ldots,A_{m}): Case 1: N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F. This case leads immediately to contradiction because NN^{\prime} trivially satisfies RR.

Case 2: N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F^{*}. Then, by the definition of NN^{\prime}, M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. Since MM is a model of PP, it is M(C1××Cn)FM(C_{1}\times\cdots\times C_{n})\geq F^{*}. This implies that either M(C1)==M(Cn)=FM(C_{1})=\cdots=M(C_{n})=F^{*} or there exists rnr\leq n such that M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and M(Cr)=TM(C_{r})=T. By the definition of NN^{\prime}, we get in both cases N(C1××Cn)FN^{\prime}(C_{1}\times\cdots\times C_{n})\geq F^{*} (contradiction).

Case 3: N(A1,,Am)=TN^{\prime}(A_{1},\ldots,A_{m})=T^{*}. Then, by the definition of NN^{\prime}, M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Since MM is a model of PP, it is M(C1××Cn)=TM(C_{1}\times\cdots\times C_{n})=T. This implies that there exists some rr, 1rn1\leq r\leq n, such that M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and M(Cr)=TM(C_{r})=T. By the definition of NN^{\prime}, we get that N(C1××Cn)TN^{\prime}(C_{1}\times\cdots\times C_{n})\geq T^{*} (contradiction).

Case 4: N(A1,,Am)=TN^{\prime}(A_{1},\ldots,A_{m})=T. Then, by the definition of NN^{\prime}, N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T and M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Since MM is a model of PP, it is M(C1××Cn)=TM(C_{1}\times\cdots\times C_{n})=T. This implies that there exists some rr, 1rn1\leq r\leq n, such that M(C1)==M(Cr1)=FM(C_{1})=\cdots=M(C_{r-1})=F^{*} and M(Cr)=TM(C_{r})=T. By the definition of the reduct, the rule CrA1,,AmC_{r}\leftarrow A_{1},\ldots,A_{m} exists in P×MP^{M}_{\times}. Since NN is a model of P×MP^{M}_{\times}, we get that N(Cr)=TN(C_{r})=T. Thus, N(C1)==N(Cr1)=FN^{\prime}(C_{1})=\cdots=N^{\prime}(C_{r-1})=F^{*} and N(Cr)=TN^{\prime}(C_{r})=T, and therefore N(C1××Cn)=TN^{\prime}(C_{1}\times\cdots\times C_{n})=T (contradiction).

\@begintheorem

Theorem1 Let PP be an LPOD. Then, MM is a three-valued answer set of PP iff MM is a consistent \preceq-minimal model of PP and MM is solid. \@endtheorem

Proof B.19.

Immediate from Lemma B.15 and Lemma B.17.

Appendix C Proofs of Section 7

\@begintheorem

Lemma7 Let PP be a consistent disjunctive extended logic program. Then, the answer sets of PP according to Definition 20, coincide with the standard answer sets of PP. \@endtheorem

Proof C.20.

By taking n=1n=1 in Definition 19, we get the standard definition of reduct for consistent disjunctive extended logic programs.

\@begintheorem

Lemma8 Let PP be a DLPOD program and let MM be an answer set of PP. Then, MM is a model of PP. \@endtheorem

Proof C.21.

Consider any rule RR in PP of the form:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

If R×M=R_{\times}^{M}=\emptyset, then M(Bi)=TM(B_{i})=T for some ii, 1ik1\leq i\leq k. But then, the body of the rule RR evaluates to FF under MM, and therefore MM satisfies RR. Consider now the case where R×MR_{\times}^{M} is nonempty and consists of the following rules:

𝒞1F,A1,,Am𝒞r1F,A1,,Am𝒞rA1,,Am\begin{array}[]{lll}{\cal C}_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ {\cal C}_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ {\cal C}_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

We distinguish cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}):

Case 1: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F. Then, for some ii, M(Ai)=FM(A_{i})=F. Then, rule RR is trivially satisfied by MM.

Case 2: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. This implies that M(𝒞r)FM({\cal C}_{r})\geq F^{*}. We distinguish two subcases. If r=nr=n then M(𝒞1××𝒞n)=M(𝒞1××𝒞r)FM({\cal C}_{1}\times\cdots\times{\cal C}_{n})=M({\cal C}_{1}\times\cdots\times{\cal C}_{r})\geq F^{*} because, by the definition of P×MP_{\times}^{M} it is M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and we also know that M(𝒞r)FM({\cal C}_{r})\geq F^{*}. Thus, in this subcase MM satisfies RR. If r<nr<n, then by the definition of P×MP_{\times}^{M}, M(𝒞r)FM({\cal C}_{r})\neq F^{*}; however, we know that M(𝒞r)FM({\cal C}_{r})\geq F^{*}, and thus M(𝒞r)=TM({\cal C}_{r})=T. Thus, in this subcase MM also satisfies RR.

Case 3: M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Then, for all ii, M(Ai)=TM(A_{i})=T. Since MM is a model of P×MP_{\times}^{M}, we have M(𝒞r)=TM({\cal C}_{r})=T. Moreover, by the definition of P×MP_{\times}^{M}, M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*}. This implies that M(𝒞1××𝒞n)=TM({\cal C}_{1}\times\cdots\times{\cal C}_{n})=T.

\@begintheorem

Lemma9 Let MM be a model of a DLPOD PP. Then, MM is a model of P×MP^{M}_{\times}. \@endtheorem

Proof C.22.

Consider any rule RR in PP of the form:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

and assume MM satisfies RR. If M(Bi)=TM(B_{i})=T for some ii, 1ik1\leq i\leq k, then no rule is created in P×MP^{M}_{\times} for RR. Assume therefore that M(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=TM({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=T. By the definition of P×MP^{M}_{\times} the following rules have been added to P×MP^{M}_{\times}:

𝒞1F,A1,,Am𝒞r1F,A1,,Am𝒞rA1,,Am\begin{array}[]{lll}{\cal C}_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ {\cal C}_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ {\cal C}_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

where rr is the least index such that M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and either r=nr=n or M(𝒞r)FM({\cal C}_{r})\neq F^{*}. Obviously, the first r1r-1 rules above are satisfied by MM. For the rule 𝒞rA1,,Am{\cal C}_{r}\leftarrow A_{1},\ldots,A_{m} we distinguish two cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}). If M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F, then, the rule is trivially satisfied. If M(A1,,Am)>FM(A_{1},\ldots,A_{m})>F, then, since rule RR is satisfied by MM and M(𝒞r)FM({\cal C}_{r})\neq F^{*}, it has to be M(𝒞r)=TM({\cal C}_{r})=T. Therefore, the rule 𝒞rA1,,Am{\cal C}_{r}\leftarrow A_{1},\ldots,A_{m} is satisfied by MM.

\@begintheorem

Lemma10 Every answer set MM of a DLPOD PP, is a \preceq-minimal model of PP. \@endtheorem

Proof C.23.

Assume there exists a model NN of PP with NMN\preceq M. We will show that NN is also a model of P×MP_{\times}^{M}. Since NMN\preceq M, we also have NMN\leq M. Since MM is the \leq-least model of P×MP_{\times}^{M}, we will conclude that N=MN=M.

Consider any rule RR in PP of the form:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

Assume that R×MR_{\times}^{M} is nonempty. This means that there exists some rr, 1rn1\leq r\leq n, such that M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and either r=nr=n or M(𝒞r)FM({\cal C}_{r})\neq F^{*}. Then, R×MR_{\times}^{M} consists of the following rules:

𝒞1F,A1,,Am𝒞r1F,A1,,Am𝒞rA1,,Am\begin{array}[]{lll}{\cal C}_{1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ &\cdots&\\ {\cal C}_{r-1}&\leftarrow&F^{*},A_{1},\ldots,A_{m}\\ {\cal C}_{r}&\leftarrow&A_{1},\ldots,A_{m}\end{array}

We show that NN satisfies the above rules. We distinguish cases based on the value of M(A1,,Am)M(A_{1},\ldots,A_{m}):

Case 1: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F. Then, N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F and the above rules are trivially satisfied by NN.

Case 2: M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. Then, since NMN\preceq M, it is N(A1,,Am)FN(A_{1},\ldots,A_{m})\leq F^{*}. If N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F then NN trivially satisfies all the above rules. Assume therefore that N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F^{*}. Recall now that M(𝒞i)=FM({\cal C}_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1. Moreover, it has to be M(𝒞r)FM({\cal C}_{r})\geq F^{*}, because otherwise MM would not satisfy the rule RR. Since NMN\preceq M, it can only be N(𝒞i)=FN({\cal C}_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1 and N(𝒞r)FN({\cal C}_{r})\geq F^{*}, because otherwise NN would not be a model of PP. Therefore, NN satisfies the given rules of P×MP^{M}_{\times}.

Case 3: M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Then, since NMN\preceq M, it is either N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F or N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T. If N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F then NN trivially satisfies all the above rules. Assume therefore that N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T. Recall now that M(𝒞i)=FM({\cal C}_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1. Moreover, it has to be M(𝒞r)=TM({\cal C}_{r})=T, because otherwise MM would not satisfy the rule RR. Since NMN\preceq M, it can only be N(𝒞i)=FN({\cal C}_{i})=F^{*} for all ii, 1ir11\leq i\leq r-1 and N(𝒞r)=TN({\cal C}_{r})=T, because otherwise NN would not be a model of PP. Therefore, NN satisfies the given rules of P×MP^{M}_{\times}.

\@begintheorem

Theorem2 Let PP be a DLPOD. Then, MM is an answer set of PP iff MM is a consistent \preceq-minimal model of PP and MM is solid. \@endtheorem The proof of the above theorem follows directly by the following two lemmas.

Lemma C.24.

Let PP be an DLPOD and let MM be an answer set of PP. Then, MM is a consistent \preceq-minimal model of PP and MM is solid.

Proof C.25.

Since MM is an answer set of PP, then, by Lemma 8, MM is a model of PP. Moreover, MM is solid because our definition of answer sets does not involve the value TT^{*}. It remains to show that it is minimal with respect to the \preceq ordering. Assume, for the sake of contradiction, that there exists a model NN of PP with NMN\prec M. By Lemma 10, MM is (three-valued) \preceq-minimal. Therefore, NN can not be solid. We first show that NN can not be a model of the reduct P×MP^{M}_{\times}. Assume for the sake of contradiction that NN is a model of P×MP^{M}_{\times}. We construct the following interpretation NN^{\prime}:

N(A)={F,if N(A)=TN(A),otherwiseN^{\prime}(A)=\left\{\begin{array}[]{ll}F^{*},&\mbox{if $N(A)=T^{*}$}\\ N(A),&\mbox{otherwise}\end{array}\right.

We claim that NN^{\prime} must also be a model of P×MP^{M}_{\times}. Consider first a rule of the form 𝒞F,A1,,Am{\cal C}\leftarrow F^{*},A_{1},\ldots,A_{m}. Since NN is a model of P×MP^{M}_{\times}, it is N(𝒞)FN({\cal C})\geq F^{*}. By the definition of NN^{\prime}, it is N(𝒞)FN({\cal C})\geq F^{*} and therefore NN^{\prime} satisfies this rule. Consider now a rule of the form 𝒞A1,,Am{\cal C}\leftarrow A_{1},\ldots,A_{m} in P×MP^{M}_{\times}. We show that NN^{\prime} also satisfies this rule. We perform a case analysis:

Case 1: N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F. Then, N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F and NN^{\prime} trivially satisfies the rule.

Case 2: N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F^{*}. Then, N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F^{*}. Moreover, N(𝒞)FN({\cal C})\geq F^{*} because NN is a model of P×MP^{M}_{\times}. By the definition of NN^{\prime}, it is N(𝒞)FN^{\prime}({\cal C})\geq F^{*}, and therefore NN^{\prime} satisfies the rule.

Case 3: N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T^{*}. Then, N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F^{*}. Moreover, N(𝒞)TN({\cal C})\geq T^{*} because NN is a model of P×MP^{M}_{\times}. By the definition of NN^{\prime}, it is N(𝒞)FN^{\prime}({\cal C})\geq F^{*}, and therefore NN^{\prime} satisfies the rule.

Case 4: N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T. Then, N(A1,,Am)=TN^{\prime}(A_{1},\ldots,A_{m})=T. Moreover, N(𝒞)=TN({\cal C})=T because NN is a model of P×MP^{M}_{\times}. By the definition of NN^{\prime}, it is N(𝒞)=TN^{\prime}({\cal C})=T, and therefore NN^{\prime} satisfies the rule.

Therefore, NN^{\prime} must also be a model of P×MP^{M}_{\times}. Moreover, by definition, NN^{\prime} is solid and N<MN^{\prime}<M. This contradicts the fact that, by construction, MM is the \leq-least model of P×MP^{M}_{\times}. In conclusion, NN can not be a model of P×MP^{M}_{\times}.

We now show that NN can not be a model of PP. As we showed above, NN is not a model of P×MP^{M}_{\times}, and consequently there exists a rule in P×MP^{M}_{\times} that is not satisfied by NN. Such a rule in P×MP^{M}_{\times} must have resulted due to a rule RR of the following form in PP:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

According to the definition of P×MP^{M}_{\times}, for all ii, 1ik1\leq i\leq k, M(𝑛𝑜𝑡Bi)=TM({\mathit{not}{\,}B_{i}})=T, and since NMN\prec M, it is also N(𝑛𝑜𝑡Bi)=TN({\mathit{not}{\,}B_{i}})=T. Moreover, there exists some rnr\leq n such that M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and either r=nr=n or M(𝒞r)FM({\cal C}_{r})\neq F^{*}. Since NMN\prec M, it is N(𝒞i)FN({\cal C}_{i})\leq F^{*} for all ii, 1ir11\leq i\leq r-1. Consider now the rule that is not satisfied by NN in P×MP^{M}_{\times}. If it is of the form 𝒞iF,A1,,Am{\cal C}_{i}\leftarrow F^{*},A_{1},\ldots,A_{m}, ii, 1ir11\leq i\leq r-1, then N(A1,,Am)>FN(A_{1},\ldots,A_{m})>F and N(𝒞i)=FN({\cal C}_{i})=F. This implies that N(𝒞1××𝒞n)=FN({\cal C}_{1}\times\cdots\times{\cal C}_{n})=F and therefore NN does not satisfy the rule RR. If the rule that is not satisfied by NN in P×MP^{M}_{\times} is of the form 𝒞rA1,,Am{\cal C}_{r}\leftarrow A_{1},\ldots,A_{m}, then N(𝒞r)<N(A1,,Am)N({\cal C}_{r})<N(A_{1},\ldots,A_{m}) and therefore, since N(𝒞i)FN({\cal C}_{i})\leq F^{*} for all ii, 1ir11\leq i\leq r-1, it is:

N(𝒞1××𝒞n)<N(A1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)N({\cal C}_{1}\times\cdots\times{\cal C}_{n})<N(A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})

Thus, NN is not a model of PP.

Lemma C.26.

Let PP be an DLPOD and let MM be a consistent \preceq-minimal model of PP and MM is solid. Then, MM is an answer set of PP.

Proof C.27.

First observe that, by Lemma 9, MM is also a model of P×MP^{M}_{\times}. We demonstrate that MM is actually the \leq-least model of P×MP^{M}_{\times}. Assume, for the sake of contradiction, that NN is the \leq-least model of P×MP^{M}_{\times}. Then, NN will differ from MM in some atoms AA such that N(A)<M(A)N(A)<M(A). We distinguish two cases. In the first case all the atoms AA such that N(A)<M(A)N(A)<M(A) have M(A)FM(A)\leq F^{*}. In the second case there exist at least one atom AA such that M(A)>FM(A)>F^{*}.

In the first case it is easy to see that NMN\prec M. We demonstrate that NN is also model of PP leading to contradiction since MM is \preceq-minimal. Assume that NN is not a model of PP. Then, there exists in PP a rule RR of the form:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

such that N(𝒞1××𝒞n)<N(A1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)N({\cal C}_{1}\times\cdots\times{\cal C}_{n})<N^{\prime}(A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}). Notice that this implies that N(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=M(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=TN({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=M({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=T. Therefore, N(𝒞1××𝒞n)<N(A1,,Am)N({\cal C}_{1}\times\cdots\times{\cal C}_{n})<N(A_{1},\ldots,A_{m}). We distinguish cases based on the value of N(A1,,Am)N(A_{1},\ldots,A_{m}):

Case 1: N(A1,,Am)=FN(A_{1},\ldots,A_{m})=F. This case leads immediately to contradiction because NN trivially satisfies RR.

Case 2: N(A1,,Am)>FN(A_{1},\ldots,A_{m})>F. Then, N(A1,,Am)=M(A1,,Am)N(A_{1},\ldots,A_{m})=M(A_{1},\ldots,A_{m}). Since MM is a model of PP, it is M(𝒞1××𝒞n)M(A1,,Am)>FM({\cal C}_{1}\times\cdots\times{\cal C}_{n})\geq M(A_{1},\ldots,A_{m})>F. This implies that there exists some rr, 1rn1\leq r\leq n, such that M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and M(𝒞r)FM({\cal C}_{r})\geq F^{*}. By the definition of the reduct, the rule 𝒞rA1,,Am{\cal C}_{r}\leftarrow A_{1},\ldots,A_{m} exists in P×MP^{M}_{\times}. Since NN is a model of P×MP^{M}_{\times}, we get that N(𝒞r)>FN({\cal C}_{r})>F. Moreover, NN should also satisfy the rules 𝒞iF,A1,,Am{\cal C}_{i}\leftarrow F^{*},A_{1},\ldots,A_{m} for 1ir11\leq i\leq r-1. Since N(𝒞i)M(𝒞i)N({\cal C}_{i})\leq M({\cal C}_{i}) and N(𝒞r)=M(𝒞r)N({\cal C}_{r})=M({\cal C}_{r}) we get N(𝒞1)==N(𝒞r1)=FN({\cal C}_{1})=\cdots=N({\cal C}_{r-1})=F^{*} and N(𝒞r)=M(𝒞r)N({\cal C}_{r})=M({\cal C}_{r}). Therefore N(𝒞1×𝒞n)=M(𝒞1×𝒞n)N({\cal C}_{1}\times\cdots{\cal C}_{n})=M({\cal C}_{1}\times\cdots{\cal C}_{n}) and N(𝒞1×𝒞n)N(A1,,Am)N({\cal C}_{1}\times\cdots{\cal C}_{n})\geq N(A_{1},\ldots,A_{m}) (contradiction).

In the second case we construct the following interpretation NN^{\prime}:

N(A)={T,if M(A)=T and N(A){F,F}F,if M(A)=FN(A),otherwiseN^{\prime}(A)=\left\{\begin{array}[]{ll}T^{*},&\mbox{if $M(A)=T$ and $N(A)\in\{F,F^{*}\}$}\\ F^{*},&\mbox{if $M(A)=F^{*}$}\\ N(A),&\mbox{otherwise}\end{array}\right.

It is easy to see that NMN^{\prime}\prec M. We demonstrate that NN^{\prime} is a model of PP, which will lead to a contradiction (since we have assumed that MM is \preceq-minimal).

Assume NN^{\prime} is not a model of PP. Then, there exists in PP a rule RR of the form:

𝒞1××𝒞nA1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk{\cal C}_{1}\times\cdots\times{\cal C}_{n}\leftarrow A_{1},\ldots,A_{m},{\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}}

such that N(𝒞1××𝒞n)<N(A1,,Am,𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)N^{\prime}({\cal C}_{1}\times\cdots\times{\cal C}_{n})<N^{\prime}(A_{1},\ldots,A_{m},\mathit{not}{\,}B_{1},\ldots,\mathit{not}{\,}B_{k}). Notice that this implies that N(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=N(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=M(𝑛𝑜𝑡B1,,𝑛𝑜𝑡Bk)=TN^{\prime}({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=N({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=M({\mathit{not}{\,}B_{1}},\ldots,{\mathit{not}{\,}B_{k}})=T. Therefore, N(𝒞1××𝒞n)<N(A1,,Am)N^{\prime}({\cal C}_{1}\times\cdots\times{\cal C}_{n})<N^{\prime}(A_{1},\ldots,A_{m}). We distinguish cases based on the value of N(A1,,Am)N^{\prime}(A_{1},\ldots,A_{m}):

Case 1: N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F. This case leads immediately to contradiction because NN^{\prime} trivially satisfies RR.

Case 2: N(A1,,Am)=FN^{\prime}(A_{1},\ldots,A_{m})=F^{*}. Then, by the definition of NN^{\prime}, M(A1,,Am)=FM(A_{1},\ldots,A_{m})=F^{*}. Since MM is a model of PP, it is M(𝒞1××𝒞n)FM({\cal C}_{1}\times\cdots\times{\cal C}_{n})\geq F^{*}. This implies that either M(𝒞1)==M(𝒞n)=FM({\cal C}_{1})=\cdots=M({\cal C}_{n})=F^{*} or there exists rnr\leq n such that M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and M(𝒞r)=TM({\cal C}_{r})=T. By the definition of NN^{\prime}, we get in both cases N(𝒞1××𝒞n)FN^{\prime}({\cal C}_{1}\times\cdots\times{\cal C}_{n})\geq F^{*} (contradiction).

Case 3: N(A1,,Am)=TN^{\prime}(A_{1},\ldots,A_{m})=T^{*}. Then, by the definition of NN^{\prime}, M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Since MM is a model of PP, it is M(𝒞1××𝒞n)=TM({\cal C}_{1}\times\cdots\times{\cal C}_{n})=T. This implies that there exists some rr, 1rn1\leq r\leq n, such that M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and M(𝒞r)=TM({\cal C}_{r})=T. By the definition of NN^{\prime}, we get that N(𝒞1××𝒞n)TN^{\prime}({\cal C}_{1}\times\cdots\times{\cal C}_{n})\geq T^{*} (contradiction).

Case 4: N(A1,,Am)=TN^{\prime}(A_{1},\ldots,A_{m})=T. Then, by the definition of NN^{\prime}, N(A1,,Am)=TN(A_{1},\ldots,A_{m})=T and M(A1,,Am)=TM(A_{1},\ldots,A_{m})=T. Since MM is a model of PP, it is M(𝒞1××𝒞n)=TM({\cal C}_{1}\times\cdots\times{\cal C}_{n})=T. This implies that there exists some rr, 1rn1\leq r\leq n, such that M(𝒞1)==M(𝒞r1)=FM({\cal C}_{1})=\cdots=M({\cal C}_{r-1})=F^{*} and M(𝒞r)=TM({\cal C}_{r})=T. By the definition of the reduct, the rule 𝒞rA1,,Am{\cal C}_{r}\leftarrow A_{1},\ldots,A_{m} exists in P×MP^{M}_{\times}. Since NN is a model of P×MP^{M}_{\times}, we get that N(𝒞r)=TN({\cal C}_{r})=T. Thus, N(𝒞1)==N(𝒞r1)=FN^{\prime}({\cal C}_{1})=\cdots=N^{\prime}({\cal C}_{r-1})=F^{*} and N(𝒞r)=TN^{\prime}({\cal C}_{r})=T, and therefore N(𝒞1××𝒞n)=TN^{\prime}({\cal C}_{1}\times\cdots\times{\cal C}_{n})=T (contradiction).