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Eliminating Disjunction from Propositional Logic Programs under Stable Model Preservation

Eliminating Disjunction from Propositional Logic Programs under Stable Model Preservation
Eliminating Disjunction from Propositional Logic Programs under Stable Model Preservation

Eliminating Disjunction from Propositional Logic

Programs under Stable Model Preservation

Thomas Eiter,Michael Fink,Hans Tompits,and Stefan Woltran

Institut f¨u r Informationssysteme184/3,Technische Universit¨a t Wien,

Favoritenstra?e9-11,A-1040Vienna,Austria

eiter,michael,tompits,stefan@kr.tuwien.ac.at Abstract.In general,disjunction is considered to add expressive power to propo-

sitional logic programs under stable model semantics,and to enlarge the range of

problems which can be expressed.However,from a semantical point of view,

disjunction is often not really needed,in that an equivalent program without dis-

junction can be given.We thus consider the question,given a disjunctive logic

program,does there exist an equivalent normal(i.e.,disjunction-free)logic

program?In fact,we consider this issue for different notions of equivalence,

namely for ordinary equivalence(regarding the collections of all stable models

of the programs)as well as for the more restrictive notions of strong and uniform

equivalence.We resolve the issue for propositional programs,and present a sim-

ple,appealing semantic criterion for the programs from which all disjunctions can

be eliminated under strong equivalence;testing this criterion is coNP-complete.

We also show that under ordinary and uniform equivalence,this elimination is

always possible.In all cases,there are constructive methods to achieve this.Our

results extend and complement recent results on simplifying logic programs un-

der different notions of equivalence,and add to the foundations of improving

implementations of Answer Set Solvers.

1Introduction

Disjunctive logic programming is an extension to normal logic programming which is generally considered to add expressive power to logic programs under stable model se-mantics,and to enlarge the range of problems which can be expressed.This view is supported by results on the expressiveness of disjunctive logic programs(DLPs)over ?nite structures,which show that properties at the second level of the Polynomial Hi-erarchy(PH)can be expressed by inference from function-free(Datalog)DLPs[11], while by normal logic programs only properties at the?rst level can be expressed[28]. However,from a semantical point of view,disjunction is often not really needed,in that an equivalent normal logic program(NLP,i.e.,without disjunction)can be given.For example,in[10],it was shown that in the presence of functions symbols,DLPs have over Herbrand models the same expressive power as NLPs,namely.

With the rise of Answer Set Programming as a program solving paradigm,in which solutions are computed in the answer sets resp.stable models of a logic program,at-tention has been directed to the expressiveness of logic programs in terms of the whole

152Thomas Eiter et al.

collection of their answer sets per se rather than their intersection(resp.union)as in cautious and brave reasoning,respectively),cf.[20];related to this is preliminary work on the expressiveness of formalisms such as default logic and circumscription[13,19].

In particular,equivalence of logic programs in terms of their collections of stable models has been considered,as well as the re?ned notions of strong equivalence,cf. [16,29,30,24,17,4],and uniform equivalence[7,8,25],which dates back to[27,18]. Two DLPs and are strongly equivalent(resp.,uniformly equivalent),if,for any set of rules(resp.,set of atoms),the programs and are equivalent

under the stable semantics,i.e.,have the same set of stable models.

Strong and uniform equivalence can be utilized for program optimization,cf.[30, 22,8],taking into account possible incompleteness of a program,where not all rules are known at the time of optimization,respectively varying input data given by atomic facts are respected.This is in particular helpful for optimizing components which are embedded into a more complex logic program.Note that as recently discussed by Pearce and Valverde[25],uniform and strong equivalence are essentially the only concepts of equivalence obtained by varying the logical form of the program extensions.

A natural issue in this context is the expressiveness of disjunction in rule heads, i.e.,whether it really adds expressive power.This is indeed the case,as can be seen on the simple example of the program:This program is not strongly equivalent to any normal logic program(cf.[30]).However,as easily seen is equivalent to the NLP since for both the stable models are and,and furthermore is also uniformly equivalent to (this is immediate from the result that rewriting a head-cycle free program to a normal logic program by standard shifting preserves uniform equivalence[7]).On the other hand,the enriched program is strongly equivalent to the program.

This raises the question of a criterion which tells when disjunctions can be elimi-nated,and a method for deciding,given a disjunctive logic program,does there exist an equivalent normal(i.e.,disjunction-free)program?We study this issue for propo-sitional programs,on which we focus here,and make the following contributions:

–We present a simple,appealing semantic characterization of the programs from which all disjunctions can be eliminated under strong equivalence.The charac-terization is based on the strong-equivalence models(SE-models)[29,30]which rephrase models in the more general logic of here-and-there[16]in logic program-ming terms.In fact,we show that this property holds for a program if and only if the collection of SE-models of is closed under here-intersection,i.e., whenever and are SE-models of,then also is an SE-model of.In more familiar terms,this condition is equivalent to the prop-erty that for each classical model of,the Gelfond-Lifschitz reduct of is semantically Horn if models not contained in are disregarded.

–We further show that under ordinary and uniform equivalence,this elimination is always possible.In all three cases,we obtain a constructive method to rewrite a DLP to an equivalent normal logic program.In general,the rewriting will be of exponential size(if it exists),but this will be unavoidable in practice.

–Finally,we show that testing whether for a given propositional DLP a strongly equivalent NLP exists is coNP-complete.

Eliminating Disjunction from Propositional Logic Programs153 Our results extend and complement recent results on simplifying logic programs un-der different notions of equivalence,cf.[22,30,8].They might be utilized for deciding whether a given disjunctive problem representation for a system such as such as DLV [6]or GnT[14]can,in principle,be replaced by an equivalent non-disjunctive repre-sentation,as well as for(automated)rewriting of disjunctive problem representations.

2Preliminaries

We deal with propositional disjunctive logic programs,containing rules of form

,where all are atoms from a?nite set of propositional atoms,, and denotes default negation.The head of is the set=,and the body of is the set=...,...,.We also use =and=.Moreover,for a set of atoms ,denotes.

A rule is normal,if;positive,if;and Horn,if it is normal and positive.If and,then is a constraint;if,is a fact, written as if,and as otherwise.

A disjunctive logic program(DLP)is a?nite set of rules.It is a normal logic program(NLP)(resp.,positive,Horn),if every is normal(resp.,positive,Horn).

We recall the stable model semantics for DLPs[12,26].Let be an interpretation, i.e.,a subset of.Then,satis?es a rule,denoted,iff whenever

,where iff for some,and iff(i)each

is true in,i.e.,,and(ii)each is false in,i.e.,. Furthermore,is a model of a program,denoted,iff,for all.

The reduct,,of a rule relative to a set of atoms is the positive rule such that and if,and is void otherwise.The Gelfond-Lifschitz reduct of a program is the positive program. An interpretation is a stable model of a program iff is a minimal model(under set inclusion)of.By we denote the set of all stable models of.

Lemma1.Let be a DLP and.Then,implies. The result is seen by the observation that implies.Thus, implies.In particular,for,implies,for any.

Several notions of equivalence for logic programs have been considered in the liter-ature(see,e.g.,[16,18,27]).Under stable semantics,two DLPs and are regarded as equivalent,denoted,iff.The more restrictive forms of strong equivalence[16]and uniform equivalence[27,18]are as follows:

De?nition1.Let and be two DLPs.Then,

(i)and are strongly equivalent,denoted,iff,for any set of rules,the

programs and are equivalent,i.e.,.

(ii)and are uniformly equivalent,denoted,iff,for any set of non-disjunctive facts,and are equivalent,i.e.,.

154Thomas Eiter et al.

Obviously,implies but not vice versa.Both notions of equivalence, however,enjoy interesting semantical characterizations.As shown in[16],strong equiv-alence is closely related to the non-classical logic of here-and-there,which was adapted to logic-programming terms by Turner[29,30]:

De?nition2.A pair with such that is called an SE-interpretation(over).By we denote the of all SE-interpretations over.An SE-interpretation is an SE-model of a DLP,if and.By we denote the set of all SE-models of.

Proposition1([29,30]).For every DLP and,iff.

SE-models also can be used to determine the stable models of a program. Proposition2([23,16]).Let be a DLP.Then,iff

and,for each,.

Recently,the following pendant to SE-models,characterizing uniform equivalence for(?nite)logic programs has been de?ned[7].

De?nition3.Let be a DLP and.Then,is UE-model of iff,for every,it holds that implies.By we denote the set of all UE-models of.

Proposition3([7];cf.also[25]).For any DLP and,iff.

This test can be reformulated as follows.

Proposition4.For DLPs and,iff and. Proof.From Proposition3,iff and. Clearly,holds for any DLP,which immediately gives the only-if direction.For the if direction,suppose.Hence,there exists an SE-interpretation ,such that either(i)and;or(ii)

and.For(i),we have two cases,by de?nition of UE-models. First,.But then,cannot hold.Second,there exists a set with such that.But since ,hence cannot hold.The argument for(ii)is analogous.

As a?nal result here,we characterize the set of SE-models of a disjunctive rule. Proposition5.Let be a disjunctive rule,and an SE-interpretation.Then, holds iff one of the following conditions is satis?ed:(i); (ii);or(iii)and.

Proof.By de?nition,iff,and.The former holds iff ,,or.The latter holds iff,, or.Hence,iff or

(1)

Eliminating Disjunction from Propositional Logic Programs155

Clearly,implies and,furthermore,implies .From this,it is easily veri?ed that satis?es(1)iff either,

(i.e.,(i)),or jointly and(i.e.,(iii)),holds.Hence, we have that iff either,,(i),or(iii)holds. Finally,or holds exactly iff(i.e.,(ii))holds.

3Strong Equivalence

We start with some informal discussion.Consider the following logic programs,each of them having as its only disjunctive rule.

Let us?rst compute the SE-models(over)of these programs:1

A good approximation to derive corresponding strongly equivalent normal logic pro-grams is to replace by the rules,i.e.,by the usual shifting technique.It is left to the reader to verify that this replacement works for, ,,and,but not for,,,,and.In fact,for the latter programs this replacement yields an additional SE-model.In some of these cases we can circumvent this problem by adding further rules.As is easily seen,adding to, ,and,respectively,solves this problem,since,and,for each ,holds().For and this does not work.As we will see soon,there is no normal logic program strongly equivalent to or.

Let us have a closer look at the difference between the SE-models of a disjunctive rule and its corresponding shifting rules.

Proposition6.For a disjunctive rule,,de?ne

;and

Then,.

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Proof.In what follows let denote that rule in with.

Clearly,we have and.In particular,the former relation can be seen by the fact that,for each,holds,since and.

It remains to show.Therefore,let

and suppose.We show that.Towards a contradiction, suppose.Since,we get by Proposition5that, ,and either or.We have two cases.

First,if,i.e.,,this clearly contradicts. Otherwise,if,we have,and we get,otherwise .Let and consider the rule.Obviously,. Moreover,we have.But then,,i.e.,, yields,and thus contradicts.

In other words,contains all SE-interpretations such that with and.Thus,and are disjoint,although,for each

,holds.

Hence,if we have a disjunctive program with a disjunctive rule;we can replace by but this may yield additional SE-models from.(The same observation can be found in[7],see Theorem4.3).In particular,this is the case if

.So,in addition to,we add the following rules which under certain circumstances eliminate all elements from but none from. Proposition7.For any sets,,de?ne the set of rules

Then,is SE-model of iff one of the following conditions holds:(1);(2);(3);or(4)and. Proof.Let and the corresponding rule in with.By Propo-sition5,iff one of the following conditions hold:(i)

(i.e.,);(ii)(i.e.,either or);or(iii)

and(i.e.,and).For any two rules,we have .Thus,iff either(ii),or,for any, (i)or(iii)holds.For the latter relations consider two cases.If,the relation holds iff for each,i.e.,for each.Hence,the relation holds iff.If and then from(iii)for each.Hence, has then to hold.

The rules play a central role and will be subsequently applied in several ways.However,we mention that may contain redundant rules,for instance if we have.It can be shown that then,which reduces the number of rules.However,for technical reasons we subsequently do not pay attention to this potential optimization.

De?nition4(here-intersection).2For any pair of SE-interpretations,and of,their here-intersection is the SE-interpretation.

Eliminating Disjunction from Propositional Logic Programs157 Recall our examples.The difference between and on the one side,and, ,and on the other side,is captured by the following property.

De?nition5.For any logic program,we say that is closed under here-intersection iff satis?es this property.

Lemma2.Each normal LP is closed under here-intersection.

Proof.Since is normal,is Horn.Then,and immediately implies since,each Horn program,satis?es the intersection property: and implies.

This leads us directly to our?rst central theorem.

Theorem1.Let be a DLP over.Then,there exists a NLP over,such that and are strongly equivalent iff is closed under here-intersection.

Proof.The only-if direction is immediate by Lemma2.For the if direction,assume is closed under here-intersection,let be disjunctive,let,and consider the logic program

with,

where

. Intuitively,collects,for each which is also SE-model of,the minimal SE-models of above(with?xed).Note that by de?nition of, for any,but implies existence of an with such that,since for,

holds(again by de?nition of).

We proceed with the proof and show.Clearly,if this holds,the above transformation sequentially applied to all disjunctive rules in yields a normal logic program strongly equivalent to.First observe that,by Proposition6,we have

The strategy for the remainder of the proof is as follows.We?rst show that

.Then,it remains to show.

We show.Let.If,we immediately have .Otherwise,.From above we know that then there exists

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a triple with.Thus,we have.We now show that.Assume the contrary,i.e.,. Then,by Proposition7one of the following conditions has to hold:(i),(ii) ,(iii),or(iv)and.(i)does not hold since we have ;(ii)since,which follows from;(iii+iv)do not hold trivially.Contradiction.

It remains to show that,i.e.,that

holds.Clearly,if is empty we are done.Hence,suppose.We show that, for each and each,holds. Towards a contradiction,let,,and suppose

.On the one hand,from,we have,which implies that(a);(b);and(c)

hold.On the other hand,by Proposition7,satis?es the following conditions for being a counter SE-model of:(1),(2),(3),and(4) or.

By assumption,.Hence,and.Moreover, by(3),holds.By Lemma1,we get and,and from(b) we get,and thus.Hence,and .Now is closed under here-intersection yielding,and

.We use(4)to distinguish between the following two cases.

:By(c),,and thus.On the other hand,from(1), we have.This implies.Hence,we have. Together with(a)and this clearly contradicts ,since is not minimal anymore.

:We have a similar argumentation.By(4),holds,yielding .Moreover,by(2)and by(c)hold.Thus,. Again,we have,and by(a)and,this contradicts.

Let us apply the construction of as used in the proof to the examples discussed in the beginning of the section(except and which are not closed under here-intersection).They all have as their only disjunctive rule.Hence, .Consider now.Clearly,if,is just replaced by.This applies to.For programs,

,and by the SE-models of the respective programs we get

and.Thus in and,exchange by, with(under assumption).This goes well conform from our informal analysis in the beginning of the section.Finally,for,we use instead of,yielding the following normal program

which is obviously strongly equivalent to.The latter has as its only SE-model and thus is strongly equivalent to.

4Uniform Equivalence

The intuitive problems in constructing a uniform equivalent normal logic program from a given disjunctive program are very similar to those observed in the case of strong

Eliminating Disjunction from Propositional Logic Programs159 equivalence.Consider a program having as its only disjunctive rule.Again,a good starting point is to replace by its corresponding shifting rules.But now,an SE-model from only comes into play,iff it is also UE-model of the rest program, i.e.,for each with,implies.Thus,if we want to eliminate such an SE-model,the problem of eliminating further SE-models, which should be retained,is less complicated compared to the case of strong equiva-lence.Roughly speaking,because of this difference we are always able to construct a uniformly equivalent normal program.For instance,all our example programs–except are uniformly equivalent to the program resulting from with replaced by.,however,matches exactly the case where is UE-model of the rest program.Adding rules as or(or both of them)circumvents the problem.Hence,in some(but less)cases we again have to add further rules,but as is seen in proof below,the difference to the construction of is very subtle.

Theorem2.For each DLP there exists a uniform equivalent NLP.

Proof.Again,we give a step-by-step transformation.So,let be a disjunctive rule in a DLP,,and consider the program

with,

where is de?ned as in the proof of Theorem1.Note that the only difference between and is that here we just consider those triples from where.To proceed with the proof,observe that analogously to the proof of Theorem1,we get

(2)

We show.By Proposition4,this holds iff both

and hold.

We?rst show,which clearly implies imme-diately.Note that if is empty we are done,since then

holds trivially.So consider.We show that,for each and each,holds.Towards a contradiction,let ,,and suppose.On the one hand,since,we have(a),(b),and(c)for each SE-interpretation with,implies.On the other hand,satis?es the following conditions for being a counter SE-model of ,by Proposition7:(i),(ii),(iii),and(iv)or .We use(iv)for distinguishing between the following two cases: First,assume.Clearly,holds and by(i)and by(iii) hold.We get.Moreover,holds,since. By Lemma1,we get and holds by(a),i.e.,since. Hence,which clearly is in contradiction to(c).

Second,assume.By(iv),then.Together with(iii)we have

.Moreover,holds by(ii).Since,holds.

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Lemma1yields,and since,we have with .Again this is in contradiction to(c).

It remains to show.In fact we show with

.By inspecting(2),it is seen that implies

which shows the claim since

holds trivially.To derive,we show,for any

,that either or.So,?x some

.We consider two cases.

Assume.Hence,there exists a set such that .We know that.Thus,.We already have shown,yielding.But then, ,since.

So assume and thus.However,we have

,since none of the following conditions from Proposition7is satis?ed: (i),(ii),(iii),or(iv)and.For(i+ii)this is seen by the fact that and thus.(iii+iv)trivially fail.Hence,.

As already discussed above,the only program from our examples–which is not uniformly equivalent after replacing by is.However,since is closed under here-intersection,we already know how to derive a strongly(and thus uniformly) equivalent normal logic program from the proof of Theorem1.In fact,one can ver-ify that.For an example program which is not closed under here-intersection and has,with disjunctive,consider

over.The SE-models of are given as follows:

.

Indeed,is not closed under here-intersection.Let.We have that is given by and as can be veri?ed,.Moreover,

Only the last triple,,is applied in the construction of.In fact,we have to add(which is given by)to.For the resulting program,we then have,but the “critical”SE-interpretation,,has been eliminated.In fact,

holds,since neither nor is UE-model of.

5Ordinary Equivalence

Finally,we discuss how to derive normal logic programs which are ordinary equiva-lent to given disjunctive ones.By Theorem2,such programs always exist and,for a given program,clearly does the job,since uniform equivalence implies ordinary equivalence.However,we give two further alternatives.The?rst is motivated by an enumeration of stable models,while the second one optimizes.To start with,we state the following result which is easily seen from Proposition7.

Eliminating Disjunction from Propositional Logic Programs161

Proposition8.or,for any. Theorem3.Let be DLP.Then,with. Proof.Let.Then,for each,either or.By Proposition8,.Towards a contradiction,suppose.By

Proposition2,there exists a such that.In particular,we must have.But by Proposition8,it is easily seen that this is contradicted by .Thus,.

Conversely,suppose and.That is,

for some.Then,for each.We have, for each by Proposition8.Thus,with.But this contradicts.

The following construction would be more“structure-preserving”and,to some ex-tend,“optimizes”the program.

Theorem4.Let be DLP and.Then,with

with

Proof.Let.Obviously,and,by Proposition6,. Moreover,,which implies,by Theorem3,.Thus,.It remains to show that no,yields an SE-model of.Towards a contradiction, suppose some such that.Clearly,,hence, since,must hold.Then,by Proposition6,. We have and,and thus get by construction.By Proposition8,which contradicts .Thus,no with exists.This means.

Conversely,let.This implies.Towards a contradiction, suppose,i.e.,there exists,such that.By Proposition6,.Since,we have, for each.By Proposition8,for each with .Hence,.By Proposition2,this contradicts.

To summarize,given a DLP with disjunctive,we are able to construct(via a replacement of by normal rules)a logic program which is ordinary equivalent to ;a program which is uniformly equivalent to;and,whenever is closed under here-intersection,a program which is strongly equivalent to.All these programs are of the form

for

By de?nition of,we furthermore can state that and.Hence, our method can be seen as a uniform framework for all important notions of equiva-lence.Moreover,our results extend and generalize methods based on shifting disjunc-tions,since the outcome of these methods coincides with the presented rewriting, whenever is empty.In particular,concerning equivalence in terms of stable mod-els,we present a semantic criterion(in contrast to the syntactic criterion of head-cycle free programs,cf.[1])which allows for shifting.Concerning equivalence in terms of UE-models,we generalize an observation made in[7](see Theorem4.3).

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6Complexity Issues

Finally,we deal with some related complexity issues.We can express the test for a DLP of being closed under here-intersection via the following normal logic program, which is linear in the size of.

De?nition6.Let be a DLP over atoms and let,,,,,,and be disjoint new atoms.De?ne

(3)

(4)

(5)

(6)

(7)

(8)

(9)

Intuitively,the program works as follows.Rules(3)guess an interpretation of and rules(5)check that is a model of.Similarly,rules(4)guess subsets and of such that both are models of,which is enforced by(6).Hence,(3)–(6)’compute’all pairs of SE-models and of.

Now,rules(7)compute the intersection and via rules(8)the new atom can be derived iff the intersection does not model,i.e.iff is no SE-model of.The constraint(9)kills all models of for which this is not the case, i.e.for which is an SE-model of.Thus,(7)–(9)ensure that has no stable model iff is closed under here-intersection.Formally,we have:

Theorem5.A DLP is closed under here-intersection iff has no stable model.

Based on this,we derive the following complexity result.

Theorem6.Let be a DLP.Then,checking whether there exists a normal logic pro-gram strongly equivalent to is complete for coNP.

Proof.By Theorem5,and the linear encoding from De?nition6we get that closed-ness under here-intersection is in coNP.By Theorem1,we immediately get the coNP-membership part.

We show coNP-hardness of this problem by a reduction to the coNP-complete prob-lem of deciding whether is the unique model of a positive DLP.

Let be a positive DLP over the alphabet,let be new atoms,and consider the program

.

We prove that is closed under here-intersection iff is the unique model of.

The if direction is straight forward:If is the unique model of then,by con-struction,is the unique model of,and since is positive–and hence constant under reduction–is trivially closed under here-intersection.

Eliminating Disjunction from Propositional Logic Programs163 For the only-if direction assume is closed under here-intersection and,towards a proof by contradiction,suppose there exists a model of.Then both,

and are models of and thus also both,

and are SE-models of.However,is not an SE-model of,since.This contradicts the assumption that is closed under here-intersection,hence is the unique model of.

We have shown coNP-hardness of deciding whether a DLP is closed under here-intersection which immediately implies coNP-hardness of checking whether there ex-ists a normal logic program strongly equivalent to by Theorem1.

Another interesting issue is the size of the rewriting of a given DLP into an equivalent NLP(if it exists).Using the constructive methods presented in this paper the rewriting is of exponential size,in general.However there is strong evidence that this is unavoidable.Let and denote the classes of all DLPs and NLPs, respectively(over atoms).

Theorem7.For each-hard family of DLPs such that there exist-equivalent NLPs,there does not exist a polynomial rewriting such that, ,for every DLP of,unless the Polynomial Hierarchy(PH)collapses. Proof.(Sketch)Towards a contradiction,assume that for a positive DLP,a polyno-mial rewriting exists and consider the-hard problem of checking whether,for some atom,is a cautious consequence of([9]).

Then,we could guess an NLP in nondeterministic polynomial time. Furthermore,checking is in coNP(even in case of uniform equivalence,since is positive and is normal,i.e.head-cycle free[7]).As well,checking whether is in coNP(since is normal).Thus,the-hard problem of deciding

would be in,a contradiction unless PH collapses.

Also for rewritings under ordinary equivalence,we cannot avoid an exponential blow up unless PH collapses.

Theorem8.There is no polynomial rewriting such that

for every,unless PH collapses.

Proof.(Sketch)This can be shown by using nonuniform complexity classes as in[3,2, 13],following closely the line of proof there that the existence of a polynomial model-preserving mapping from propositional circumscription to classical propositional logic would imply that coNP poly,i.e.,the class of problems decidable in polynomial time with polynomial advice.This inclusion implies a collapse of PH.The proof can be easily adapted,where the mapping,,and

are replaced with,(positive),and,respectively.

Clearly Theorem7implies Theorem8,but the proof of the latter does not refer to non-uniform complexity classes,and is thus more accessible.We remark that in terms of[13],is a stronger formalism than unless PH collapses.Furthermore, Theorem8remains true for generalized rewritings which admit projective extra vari-ables,i.e.,,where is on atoms and denotes the restriction of the stable models of to the original atoms.Indeed,such an would imply coNP NP poly,which again means that PH collapses.

164Thomas Eiter et al.

7Conclusion

In this work,we derived new results on the relationship between propositional dis-junctive and normal logic programs under the stable model semantics,by investigating

whether disjunctions in a given program can be replaced by normal rules,in such a way that this modi?cation does not change the set of SE-models(resp.UE-models,

stable models)of.In a bigger picture,such a rewriting technique allows to obtain a strongly(resp.uniformly,ordinary)equivalent normal logic program from a given disjunctive logic program.

Our results show that under ordinary and uniform equivalence,this rewriting is always possible.In the case of strong equivalence we identi?ed an appealing semantic criterion based on so-called here-intersection.The rewriting itself is based on the well-known(local)shifting of disjunctive rules,but extends this method by an addition of further rules,which take the semantic of the entire program into account.Hence,this rewriting is in general hard to obtain,and has to be exponential in the worst case under widely accepted complexity theoretic assumptions.

These results complement recent considerations on simpli?cation techniques un-

der different notions of equivalence,cf.[22,30,8],and thus add to the foundations of improving implementations of Answer Set Solvers.

Moreover,we showed that the problem of deciding whether a DLP is closed under

here-intersection,i.e.,deciding whether there exists a strongly equivalent NLP,is com-plete for coNP,answering a question left open in[8].As a by-product,we presented an encoding of this test via a normal logic program.Note,however,that this test may also be treated by SAT-solvers.Thus,we also contributed to a line of research in ASP,which relies on the application of classical propositional logic(or QBFs)to deal with certain problems in logic programming[15,5,24].

Further issues concern a closer investigation of adding extra variables,as well as of the newly derived class of DLPs closed under here-intersection.One the one hand,we are interested in the expressibility of such programs.On the other hand,it is an open question whether this class allows for optimizations of algorithms used in disjunctive logic programming engines such as DLV.

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