The complexity of some complementation problems

Information Processing Letters 71 (1999) 159–165

The complexity of some complementation problems

David A. Plaisteda,∗, Gregory Kucherovb,1a Department of Computer Science, University of North Carolina at Chapel Hill, Chapel Hill, NC 27599-3175, USA

b LORIA/INRIA-Lorraine, 615, rue du Jardin Botanique, B.P. 101, 54602 Villers-lès-Nancy, France

Received 18 August 1998Communicated by H. Ganzinger

Abstract

We study the computational complexity of the problem of computing a complement representation for a set of terms.Depending on the class of sets considered, some sets are shown to have an exponential complement representation in theworst case, and some are shown to have a polynomial one. The complexity of deciding if a set has an empty complement is alsostudied. 1999 Elsevier Science B.V. All rights reserved.

Keywords:Term; Ground term; Ground instance; Complement representation; Computational complexity

1. Introduction

In this paper we study the following natural prob-lem. Given a setS = {t1, . . . , tn} of terms (or literals),find another setc(S) with the set of instances consist-ing exactly of those ground terms which are not in-stances of terms ofS. c(S) is naturally called thecom-plement representationof S.

Probably the first study of how to compute a com-plement representation has been done by Lassez andMarriot [18] in the context of learning a set specifiedby counterexamples. It can be easily observed that ifterms ofS contain only linear terms (i.e., terms with-out repetitive occurrences of variables) then the finitecomplement representation always exists. Lassez andMarriot proved thatS has a finite complement repre-sentation if and only ifS is “equivalent” to a set oflinear terms, where “equivalence” means having thesame set of instances. Moreover, this equivalence is

∗ Corresponding author. Email: [email protected] Email: [email protected].

decidable, and such a set of linear terms can be alwaysobtained by instantiating nonlinear variables inS bysome ground terms.

The notion of complement representation was (ex-plicitly or implicitly) used in various applicationfields, such as functional and logic programming, au-tomated deduction, and machine learning. For exam-ple, in a logic program a clauseL :-L1,L2, . . . ,Lnwill only be invoked if no earlier clauses succeed.Thus, the set of goals for which this clause will be in-voked is the intersection of the instances ofL and thecomplement of the earlier instances. If we can com-pute a representation of the complement of the earlierinstances, then we can compute the set of goals thatwill cause this clause to be invoked. This permits us tochange the order of the clauses of the program with-out affecting its semantics. This could have applica-tions in concurrent execution of logic programs, sinceit would be possible to attempt execution of all clausesat the same time without concern about altering thesemantics of the program. Lassez et al. [20] discuss

0020-0190/99/$ – see front matter 1999 Elsevier Science B.V. All rights reserved.PII: S0020-0190(99)00102-7

160 D.A. Plaisted, G. Kucherov / Information Processing Letters 71 (1999) 159–165

some other application of complement construction inlogic programming and term rewriting. Fermuller andLeitsch discuss, as another example, the complementproblem and its application to model building, andgive an exponential algorithm. Computing a comple-ment representation can be also viewed from a moregeneral logic perspective as negation elimination insome class of equational formulae [4,20]. A recent pa-per by Gottlob and Pichler [7] studies the complexityof several problems related to “atomic representationsof Herbrand models”; some of them are closely relatedto the complement representation problem consideredhere.

The notion of complement has particularly oftenoccurred in the theory of term rewriting [5]. For ex-ample, in [17] an algorithm has been proposed forproving inductive theorems in an equational theory,which is explicitly based on computing a complement.A generalization of complementation for terms con-taining associative-commutative operations has alsobeen studied [10,19]. The notion of complement alsostands behind more general notions studied in theterm rewriting theory, such as sufficient completeness,ground reducibility, test set, and the set of normalforms. The latter, for example, can be viewed as a gen-eralization of the complement: given a set of termsS,a normal form is a ground term such that none of itssubterms (including the term itself) is an instance of aterm ofS.

There exists an extensive literature devoted to groundreducibility, the set of normal forms and related issues(some sample references are [12,9,1,4,8]). Relatedcomplexity questions have also been studied (see [21,16,11,14,15,3]). However, to the best of our knowl-edge, no work has been done on analyzing complexityproblems related to complement representations. Theonly result, trivially implied by the Lassez and Mar-riot’s construction, is the exponential upper bound onthe size of the complement representation. In this pa-per, we answer some further related complexity ques-tions. We show that the size of the complement repre-sentation is necessarily exponential, in the worst case,for a set of linear terms and, even more restrictively,for a set ofdisjoint(non-unifiable) linear terms. On theother hand, we show that in the case of ground terms,as well as in the special case ofhierarchical sets oflinear terms, a polynomial complement representationexists. We also study the complexity of deciding if a

set has an empty complement. While the latter prob-lem is co-NP-complete for general sets of linear terms,we show that it is polynomial (linear) for sets of dis-joint linear terms.

2. Basic definitions

T (F,X) is the set of terms formed from functionsymbols inF and variables inX. T (F ) is the setof ground terms formed from function symbols inF .In order forT (F ) to be non-empty,F must containfunction symbols of arity 0, calledconstant symbolsor constants. A term t ∈ T (F,X) is linear if novariable ofX occurs more than once int . We assumefamiliarity with basic notions of term rewriting [5](such as position in a term, substitution, matching,unification).

If t is a term, then‖t‖ is thesymbol sizeof t , whichis the number of occurrences of function and constantsymbols int . Thus‖f (a,f (a,f (b, x)))‖ = 6. If Tis a set of terms, then‖T ‖ is the sum of the symbolsizes of the terms inT . The depthof a symbol in aterm is its depth in the term tree. The depth of a termis the maximal depth of a symbol in it. For example,the depth of variablex in the term above is 3 which isthe depth of the term itself. We consider the order onpositions in a term corresponding to the pre-order inthe tree. For example, the leftmost variable in a termmeans the variable occurring at the smallest positionwith respect to the pre-order of positions.

Definition 2.1. If T is a set of terms,Gr(T ) is the setof ground terms inT (F ) that are instances of termsin T .

3. The complexity of some complement problems

Definition 3.1. If T andU are sets of terms, we saythat U represents the complement ofT if Gr(U) =T (F ) \Gr(T ).

We first consider the version of the complementrepresentation problem for sets of ground terms. In thiscase,Gr(U)= T (F ) \ T .

Definition 3.2. If t is a term and{x← f (x1, . . . , xn)}is a substitution and thexi are distinct fromx and

D.A. Plaisted, G. Kucherov / Information Processing Letters 71 (1999) 159–165 161

do not appear int , then {x ← f (x1, . . . , xn)} is anelementary substitution fort . t{x ← f (x1, . . . , xn)}denotes the result of applying{x ← f (x1, . . . , xn)}to t .

Elementary substitutions correspond to the ideaof step-wise expanding of terms, which has beenfrequently occurring in different constructions in termrewriting systems (see, e.g., [13,12]). We use them inthe following notion which will be very useful for ourconstructions.

Definition 3.3. SupposeT ⊆ T (F ) is a finite set ofground terms. LetT ↓, the set ofprefixesof T be thesetU ⊆ T (F,X) such that(i) x is inU for some variablex,(ii) if t is inU andy is the leftmost variable (with re-

spect to the pre-order of positions) int , and{y←f (x1, . . . , xn)} is an elementary substitution fort ,andt{y← f (x1, . . . , xn)} matches some term inT , thent{y← f (x1, . . . , xn)} is in U as well,

(iii) no other term is contained inU .

For example, ifT is {f (a, b, e), f (a, c, d)} thenT ↓is {x,f (x1, x2, x3), f (a, x2, x3), f (a, b, x3), f (a, c,x3), f (a, b, e), f (a, c, d)}. Note that for every posi-tion α in any t ∈ T , there is a terms ∈ T ↓ having avariable at positionα.

Lemma 3.4. For any finiteT ⊆ T (F ), T ↓ is finite and‖T ↓‖ =O(‖T ‖2).

Proof. The finiteness is obvious. The quadratic boundfollows from the fact that the function symbolfthat we add to a term inT ↓ according to (ii) ofDefinition 3.3 can be mapped to a distinct symbolf

of T . Thus, only a linear number of such symbolscan be added. Each symbol addition comes withcopying of an existing term. This implies the quadraticbound. 2

Using the set of prefixesT ↓ we now define anotherset which we will prove to be a complement represen-tation forT .

Definition 3.5. SupposeT ⊆ T (F ) is a finite set ofground terms. Letc(T ) be the set of termsu{x ←f (x1, . . . , xn)} such thatu ∈ T ↓, x is the leftmost

variable inu, {x← f (x1, . . . , xn)} is elementary foru, andu{x← f (x1, . . . , xn)} does not match any termof T .

Note that using the leftmost variable is the key fea-ture of our construction, which allows to obtain a com-pact complement representation, and which makes themain difference with the known constructions [13,12]based on term expansion.

Theorem 3.6. If T ⊆ T (F ) is a finite set of groundterms, thenc(T ) represents the complement ofT .Moreover, assuming a constant size ofF , c(T ) iscomputable in polynomial(quadratic) time on‖T ‖,and‖c(T )‖ is polynomial(quadratic) in ‖T ‖.

Proof. It is clear that if s ∈ T (F ) is an instance ofsome element ofc(T ), then s is not in Gr(T ) = T ,since elements ofc(T ) do not match any term ofT .Conversely, ifs ∈ T (F )\T , then consider the smallestpositionα of s (with respect to the pre-order) suchthat there exists a term inT which coincides withs atall positions strictly smaller thanα, but none of suchterms has the same symbol ass, sayf , at positionα.Note that positionα exists in some term ofT . Let tbe an element ofT ↓ which has its leftmost variablex at positionα. By the remark before Lemma 3.4,such a termt exists. Thent{x← f (x1, . . . , xn)} is anelement ofc(T ) which hass as an instance.

The facts thatc(T ) is computable in quadratic timeand of quadratic size follow from Lemma 3.4.2

Note that the quadratic upper bound of Theorem 3.6is tight, as any complement representation of the setT = {hn(a)} has at least quadratic symbol size, as itmust contain terms{a,h(a), . . . , hn−1(a)}.

We now consider the extension of this problem tolinear terms with variables. We first give a generalconstruction of a complement representation for a setof linear terms, and estimate its size.

Let T be a set of linear terms. Consider the set oftermsU defined as follows.(i) U consists of linear terms,(ii) each term ofU is non-unifiable with any term of

T ,(iii) in each term ofU , replacing any subterm of the

form f (x1, . . . , xn) (x1, . . . , xn are variables) by


a fresh variablex, leads to a term whichunifieswith some term inT ,

(iv) all terms verifying (i)–(iii) are inU , and no otherterm is inU .

Note that as a particular case of (iii),f (x1, . . . , xn)

may be a constanta. For example, if

T = {a,f (f (x, y), z), f (x, g(y))}over the symbolsF = {a,g,f } of arity 0,1,2, respec-tively, then

U = {g(x), f (a,f (x, y)), f (g(x), f (y, z)),f (a, a), f

(g(x), a

)}.

It is readily seen thatU is uniquely defined and finite.The finiteness follows from the fact that the depth ofterms ofU is at most one more than the maximaldepth of terms ofT because of condition (iii). Theuniqueness follows from condition (iv) asserting thatall terms verifying (i)–(iii) are taken.

We now show thatU is a complement representa-tion. Indeed, take a termt ∈ T (F ) \ Gr(T ). Startingfrom t , iterate the following non-deterministic pro-cedure: replace a subtermf (x1, . . . , xn) by a freshvariablex, provided that the resulting term is non-unifiable with any term ofT . We stop if no such trans-formation can be applied. The resulting term belongstoU and matchest .

Now observe that for any termt of U and anysubtermf (x1, . . . , xn) of t , say at positionα, thereexists a term inT with a non-variable symbol atpositionα. This is implied by condition (iii) above.This shows that there are at most|F |‖T ‖ terms inUand‖U‖ can be bounded by‖T ‖ · |F |‖T ‖.

We now show that this exponential bound is asymp-totically tight, i.e., there exist linear sets with the ex-ponential minimal size of complement representation.

Theorem 3.7. There is a setT of linear terms suchthat any setU of terms representing the complementof T is of exponential size.

Proof. Suppose the signature contains constant sym-bolsa, b andc, and ann-ary symbolf . Let T be theterms {a, b, c, f (c, x2, . . . , xn), f (x1, c, x3, . . . , xn),

. . . , f (x1, x2, . . . , c)}. SupposeU represents the com-plement ofT . Let u be a term inU . Thenu is of theform f (t1, . . . , tn) where theti are variables, constant

symbols, or terms of the formf (y1, . . . , yn). But notican be a variable, since thenU would not represent thecomplement ofT . (Such a termu is unifiable with aterm ofT havingc as theith argument.) Therefore, allthe ti must bea or b or of the formf (y1, . . . , yn). Allcombinations must be present in order to represent thecomplement ofT , and this requires at least 3n termsin U . 2

Note that the above proof can be trivially modifiedto the case of bounded-arity signature by represent-ing termsf (t1, t2, . . . , tn) by terms, say,f (t1, f (t2,f (. . . , f (tn, a), . . .))).

It is still possible that by placing more conditionson T , we could obtain sets of terms for which thecomplement problem is polynomially solvable. Onecan easily note that the proof above essentially usesthe property that terms ofT have common instances.Therefore, forbidding this would be a further naturalrestriction.

Definition 3.8. We say a setT of terms isdisjoint ifno two distinct terms inT are unifiable.

In other words,T is disjoint if for any t1, t2 ∈ T ,Gr({t1})∩Gr({t2})= ∅. It turns out that this conditionis not enough to guarantee that the complementproblem is polynomially solvable.

Theorem 3.9. There are disjoint setsT of linearterms such that any setU of terms representing thecomplement ofT is of exponential size on‖T ‖.

Proof. Let m = n(n − 1)/2 and letφ be a mappingfrom {(i, j): 16 i, j 6 n, i 6= j } to {1,2, . . . ,m} suchthatφ(i, j)= φ(k, l) iff {i, j } = {k, l}. Let T be a setof n linear terms{r1, . . . , rn} whereri is of the formf (t1, . . . , tm) and tφ(j,k) = ai if j = i or k = i, andtφ(j,k) is a distinct variable otherwise.

The idea is that the arguments off represent edgesof an undirected complete graph onn vertices, withtφ(j,k) representing the edge between verticesj andk,andri hasai at all the positions corresponding to edgesincident to the vertexi, and variables elsewhere.

Now, this is a disjoint set of terms, becauseri andrj haveai andaj , respectively, as theφ(i, j) argumentof f .


Suppose thatU represents the complement ofT .We claim that U has an exponential number ofelements.

First we observe that every termu in U which isnot a constant is of the formf (t1, . . . , tm), and atleastdn/2e of the termsti are not variables. Otherwisethere would existj such thattφ(j,k) is a variable forall 16 k 6 n, k 6= j , in which caseu would be aninstance ofrj .

Let us consider all terms of the formf (t1, . . . , tm)in which the ti are in the set{a1, a2, . . . , an}. Therearenm such terms altogether. By the above remark,each elementu of U matches at mostnm−dn/2e ofthem, since at leastdn/2e of the arguments off areconstants inu. How many terms are not instancesof terms of T ? It is nm − n · nm−(n−1) since T isdisjoint and each termri in T hasn − 1 argumentsnon-variable. IfU containsl termsf (t1, . . . , tm), weobtain the inequationnm−n ·nm−(n−1) > l ·nm−dn/2e,which implies the exponential lower bound�(nn/2)for l. 2

Finally, we consider more restrictive sets of linearterms in which the complement can be represented bya set of polynomial size.

Definition 3.10. We say that a setT of terms ishierarchical if T has at most one element, or thereis some positionα such that all termst in T have anon-variable symbol at positionα and coincide at allpositions ancestor toα, there are at least two termst1, t2 in T that have different symbols at positionα,and the subsets ofT having a given symbol at positionα are also hierarchical.

For example, the following set of terms is hierarchi-cal:{f (a, b, x), f (a, c, y), f (b, x, c),

f (b, y, a), g(x, y, z)}.

This set is hierarchical because all terms have a non-variable at the top position, and those terms having ag at this position form a set of one element, whichis hierarchical. Those terms having anf at the topposition form a set of four elements that is also hi-erarchical, since the first arguments of all terms inthis set {f (a, b, x), f (a, c, y), f (b, x, c), f (b, y, a)}are all non-variable symbols. Those having ana in the

first argument ({f (a, b, x), f (a, c, y)}) are also hier-archical, based on the second argument, and those hav-ing a b in the first argument ({f (b, x, c), f (b, y, a)})are hierarchical, based on the third argument. Notethat hierarchical sets are useful in automated deduc-tion (cf. [2]), as they allow a natural application of caseanalysis in deduction procedures.

Theorem 3.11.SupposeT is a hierarchical set oflinear terms. ThenT is disjoint, and moreover, thecomplement ofT has a polynomial representation.

Proof. The fact thatT is disjoint follows readily fromthe definition of hierarchical. The fact that it hasa polynomial size representation of the complementfollows by an algorithm similar to that for Theorem 3.6above, but instead of considering the pre-order ofpositions inT , we consider them in the order givenby the hierarchy. 2

However, even if a set is hierarchical, it maytake some work to detect this fact. Note that theunion of hierarchical sets is not hierarchical, sinceany singleton set is hierarchical, and any set can beexpressed as a union of singleton sets.

We finish with some results about how hard it isto test if the complement is empty. The followingtheorem is equivalent to a result proved in Section 3of [11] in the context of testing sufficient completenessof a rewriting system over free constructors.

Theorem 3.12 [11]. Given a setT of terms, it isco-NP-complete to determine if Gr(T )= T (F ).

Note that Theorem 3.12 was proved in [11] forany set of terms, containing possibly nonlinear terms.However, the co-NP-hardness was proved for sets oflinear terms, which implies that the problem is stillco-NP-complete for this restricted case.

However, this problem becomes easier if the setT

is required to be disjoint.

Theorem 3.13.Given a disjoint setT of linear terms,there is a polynomial(linear) time algorithm to test ifGr(T )= T (F ).

Proof. For any naturaln, defineQ(n) ⊆ T (F,X) asfollows.Q(n) consists of all those linear terms which


have each non-variable symbol at the depth at mostn,and each variable, if any, at the depth exactly(n+ 1).Note that for anyn, Gr(Q(n)) = T (F ). On the otherhand, it can be easily seen that any two distinct termsin Q(n) are disjoint.

Let d be the maximal depth of terms inT , andconsiderQ(d). For anyt1 ∈ T , t2 ∈ Q(d), either t1and t2 are not unifiable, ort2 is an instance oft1.This is implied by the linearity oft1 and by the factthat any variable int2 is deeper than the depth oft1.This observation further implies thatGr(T ) = T (F )if and only if every term ofQ(d) is matched by aterm fromT . SinceT is also disjoint, no term ofQ(d)can be matched by distinct terms ofT . Therefore, itis sufficient to count the number of terms ofQ(d)matched by each term ofT and to check if thesenumbers sum up to|Q(d)|.|Q(n)| can be computed by induction. Note that|Q(0)| = |F |. Let Fk , k > 0, be the symbols ofF ofarity k, and|Fk | their number. Then∣∣Q(n)∣∣= |F0| +

∑k>1

|Fk| ·(|Q(n− 1)|)k.

Assuming the signature of constant size and a constanttime of arithmetic operations, values|Q(n)| for n =1..d can be computed in time O(d).

Let t ∈ T , and assumet hasl variables occurrencesat the depthsd1, . . . , dl . Then the number of terms ofQ(d) matched byt is

M(t)=l∏i=1

∣∣Q(d − di)∣∣.To check thatGr(T )= T (F ) we have to check that∑t∈T

M(t)= ∣∣Q(d)∣∣.This test takes O(‖T ‖) operations altogether.2

4. Discussion

We have studied the complexity of some comple-ment problems involving finite sets of terms. It wouldbe interesting to extend this work in two ways. Onearea of interest is to find more applications where theseresults would be useful. Another possibility is to findnew classes of terms for which one can prove lower

or upper bounds on the complexity of the comple-mentation problem. It would also be possible to in-vestigate the complexity of the complementation prob-lem relative to various equational theories. In gen-eral, the study of disunification to date has emphasizedcompleteness results rather than results about the sizeof the representation of the complement set. Perhapsother disunification problems could be studied to de-termine bounds on the sizes of the sets of disunifiersreturned.

Acknowledgements

We are grateful to Miki Hermann and Laurent Jubanfor helpful discussions, to Michaël Rusinowitch andto anonymous referees for useful remarks. The firstauthor was supported by Hélène Kirchner’s group atINRIA-Lorraine which was invaluable for the comple-tion of this research.

References

[1] K. Bundgen, W. Küchlin, Computing ground reducibility andinductively complete positions, in: N. Dershowitz (Ed.), Pro-ceedings 3rd Conference on Rewriting Techniques and Appli-cations, Chapel Hill (NC, USA), Lecture Notes in ComputerSci., Springer, Berlin, 1989, pp. 59–75.

[2] A. Bouhoula, M. Rusinowitch, Automatic case analysis inproof by induction, in: R. Bajcsy (Ed.), Proceedings 13th In-ternational Joint Conference on Artificial Intelligence (IJCAI-93), Chambéry, France, Vol. 1, Morgan Kaufmann, San Mateo,CA, 1993, pp. 88–94.

[3] H. Comon, F. Jacquemard, Ground reducibility is exptime-complete, in: Proceedings of 12th IEEE Symposium on Logicin Computer Science, Warsaw, Poland, IEEE Computer Soci-ety Press, 1997.

[4] H. Comon, Disunification: A survey, in: J.-L. Lassez,G. Plotkin (Eds.), Computational Logic. Essays in Honor ofAlan Robinson, MIT Press, Cambridge, MA, 1991, Chapter 9,pp. 322–359.

[5] N. Dershowitz, J.-P. Jouannaud, Handbook of TheoreticalComputer Science, Vol. B, Chapter 6: Rewrite Systems,Elsevier Science Publishers, 1990, pp. 244–320.

[6] C. Fermuller, A. Leitsch, Hyperresolution and automatedmodel building, J. Logic Comput. 6 (2) (1996) 173–203.

[7] G. Gottlob, R. Pichler, Working with arms: Complexity re-sults on atomic representations of Herbrand models, in: Pro-ceedings of IEEE Symposium on Logic in Computer Sci-ence (LICS’99), 1999; available at http://www.dbai.tuwien.ac.at/staff/gottlob/arms.ps.

[8] D. Hofbauer, M. Huber, Linearizing term rewriting systemsusing test set, J. Symbolic Comput. 17 (1) (1994) 91–129.


[9] J.-P. Jouannaud, E. Kounalis, Automatic proofs by induction intheories without constructors, Inform. and Comput. 82 (1989)1–33.

[10] E. Kounalis, D. Lugiez, L. Pottier, A solution of the comple-ment problem in associative commutative theories, in: A. Tar-lecki (Ed.), 16th International Symposium Mathematical Foun-dation of Computer Sciences, Lecture Notes in Computer Sci.,Vol. 520, Springer, Berlin, 1991, pp. 287–297.

[11] D. Kapur, P. Narendran, D.J. Rosenkrantz, H. Zhang, Sufficientcompleteness, ground-reducibility and their complexity, ActaInform. 28 (1991) 311–350.

[12] D. Kapur, P. Narendran, H. Zhang, Proof by induction usingtest sets, in: Proceedings 8th International Conference on Au-tomated Deduction, Oxford (UK), Lecture Notes in ComputerSci., Vol. 230, Springer, Berlin, 1986, pp. 99–117.

[13] E. Kounalis, Testing for the ground (co-)reducibility propertyin term-rewriting systems, Theoret. Comput. Sci. 106 (1992)87–117.

[14] G. Kucherov, M. Rusinowitch, Complexity of testing groundreducibility for linear word rewriting systems with variables,in: Proceedings 4th International Workshop on Conditional andTyped Term Rewriting Systems, Jerusalem (Israel), LectureNotes in Computer Sci., Vol. 968, Springer, Berlin, 1995,pp. 262–275.

[15] G. Kucherov, M. Rusinowitch, Undecidability of ground re-ducibility for word rewriting systems with variables, Inform.Process. Lett. 53 (1995) 209–215.

[16] G. Kucherov, M. Tajine, Decidability of regularity and relatedproperties of ground normal form languages, Inform. andComput. 118 (1) (1995) 91–100.

[17] G. Kucherov, A new quasi-reducibilty testing algorithm and itsapplication to proofs by induction, in: J. Grabowski, P. Les-canne, W. Wechler (Eds.), Proceedings of the 1st Interna-tional Workshop on Algebraic and Logic Programming, Gaus-sig (GDR), Lecture Notes in Computer Sci., Vol. 343, Springer,Berlin, 1988, pp. 204–213.

[18] J.-L. Lassez, K. Marriot, Explicit representation of termsdefined by counter examples, J. Automated Reas. 3 (3) (1987)301–318.

[19] D. Lugiez, J.-L. Moysset, Complement problems and treeautomata in AC-like theories, in: P. Enjalbert, A. Finkel,K.W. Wagner (Eds.), Proceedings STACS 93, Lecture Notes inComputer Sci., Vol. 665, Springer, Berlin, 1993, pp. 515–524.

[20] J.-L. Lassez, M. Maher, K. Marriott, Elimination of negationin term algebras, in: Proceedings of MFCS’91, Vol. 520,Springer, Berlin, 1991, pp. 1–16.

[21] D. Plaisted, Semantic confluence and completion method,Inform. and Control 65 (2/3) (1985) 182–215.

Documents

The complexity of some complementation problems