Subclasses of Quantied Boolean Formulas. Andreas Flogel. University of Duisburg Duisburg 1. Marek Karpinski. University of Bonn.

Similar documents
Quantified Boolean Formulas: Complexity and Expressiveness

Part 1: Propositional Logic

A Collection of Problems in Propositional Logic

Propositional Logic: Models and Proofs

[3] R.M. Colomb and C.Y.C. Chung. Very fast decision table execution of propositional

2 Z. Lonc and M. Truszczynski investigations, we use the framework of the xed-parameter complexity introduced by Downey and Fellows [Downey and Fellow

Detecting Backdoor Sets with Respect to Horn and Binary Clauses

Clause/Term Resolution and Learning in the Evaluation of Quantified Boolean Formulas

Resolution on Quantified Generalized Clause-sets

Overview. I Review of natural deduction. I Soundness and completeness. I Semantics of propositional formulas. I Soundness proof. I Completeness proof.

Worst-Case Upper Bound for (1, 2)-QSAT

A New 3-CNF Transformation by Parallel-Serial Graphs 1

CSE 555 HW 5 SAMPLE SOLUTION. Question 1.

NP Complete Problems. COMP 215 Lecture 20

is a model, supported model or stable model, respectively, of P. The check can be implemented to run in linear time in the size of the program. Since

Propositional Calculus: Formula Simplification, Essential Laws, Normal Forms

Advanced Topics in LP and FP

Resolution and the Weak Pigeonhole Principle

On the Gap between the Complexity of SAT and Minimization for Certain Classes of Boolean Formulas

Mathematical Logic Propositional Logic - Tableaux*

Propositional and Predicate Logic - II

Propositional and Predicate Logic - V

Normal Forms of Propositional Logic

Knowledge base (KB) = set of sentences in a formal language Declarative approach to building an agent (or other system):

MATHEMATICAL ENGINEERING TECHNICAL REPORTS. Deductive Inference for the Interiors and Exteriors of Horn Theories

NP-Completeness of Refutability by Literal-Once Resolution

ECE473 Lecture 15: Propositional Logic

Chapter 2. Reductions and NP. 2.1 Reductions Continued The Satisfiability Problem (SAT) SAT 3SAT. CS 573: Algorithms, Fall 2013 August 29, 2013

Computational Logic. Davide Martinenghi. Spring Free University of Bozen-Bolzano. Computational Logic Davide Martinenghi (1/30)

Propositional Resolution Introduction

CS Lecture 29 P, NP, and NP-Completeness. k ) for all k. Fall The class P. The class NP

Language of Propositional Logic

Bounded-width QBF is PSPACE-complete

Lecture 9: The Splitting Method for SAT

NP-Completeness. f(n) \ n n sec sec sec. n sec 24.3 sec 5.2 mins. 2 n sec 17.9 mins 35.

Warm-Up Problem. Is the following true or false? 1/35

Lecture 11: Measuring the Complexity of Proofs

Propositional Resolution

An Improved Upper Bound for SAT

2.5.2 Basic CNF/DNF Transformation

Complexity of DNF and Isomorphism of Monotone Formulas

Computational Complexity and Intractability: An Introduction to the Theory of NP. Chapter 9

Guarded resolution for Answer Set Programming

Splitting a Default Theory. Hudson Turner. University of Texas at Austin.

Logic Part I: Classical Logic and Its Semantics

In a second part, we concentrate on interval models similar to the traditional ITL models presented in [, 5]. By making various assumptions about time

On the size of minimal unsatisfiable formulas

Halting and Equivalence of Program Schemes in Models of Arbitrary Theories

Boolean constraint satisfaction: complexity results for optimization problems with arbitrary weights

Regular Resolution Lower Bounds for the Weak Pigeonhole Principle

NP-Completeness Part II

Non-Deterministic Time

case in mathematics and Science, disposing of an auxiliary condition that is not well-understood (i.e., uniformity) may turn out fruitful. In particul

Propositional Resolution

Chapter 7 R&N ICS 271 Fall 2017 Kalev Kask

Lecture 7: Polynomial time hierarchy

CS173 Lecture B, September 10, 2015

Propositional Logic Basics Propositional Equivalences Normal forms Boolean functions and digital circuits. Propositional Logic.

Propositional Logic Language

Title: Logical Agents AIMA: Chapter 7 (Sections 7.4 and 7.5)

3 Propositional Logic

Inference in Propositional Logic

Inference Methods In Propositional Logic

Deductive Systems. Lecture - 3

INF3170 / INF4171 Notes on Resolution

Logical Agent & Propositional Logic


Spring Lecture 21 NP-Complete Problems

Resolution Lower Bounds for the Weak Pigeonhole Principle

Supplementary exercises in propositional logic

On the Complexity of the Reflected Logic of Proofs

Introduction to Kleene Algebra Lecture 15 CS786 Spring 2004 March 15 & 29, 2004

Propositional Logic Part 1

Show that the following problems are NP-complete

Tecniche di Verifica. Introduction to Propositional Logic

Vol. 2(1997): nr 4. Tight Lower Bounds on the. Approximability of Some. Peter Jonsson. Linkoping University

Rina Dechter and Irina Rish. Information and Computer Science. University of California, Irvine.

Propositional Logic. Logic. Propositional Logic Syntax. Propositional Logic

1 PSPACE-Completeness

COMP9414: Artificial Intelligence Propositional Logic: Automated Reasoning

Generalizations of Matched CNF Formulas

Tutorial 1: Modern SMT Solvers and Verification

Logical Agents (I) Instructor: Tsung-Che Chiang

Algebraic Dynamic Programming. Solving Satisfiability with ADP

Horn Clauses in Propositional Logic

PSPACE-completeness of LTL/CTL model checking

Theory of Computer Science. Theory of Computer Science. E4.1 Overview E4.2 3SAT. E4.3 Graph Problems. E4.4 Summary.

Inference Methods In Propositional Logic

Logic Part II: Intuitionistic Logic and Natural Deduction

Essential facts about NP-completeness:

Solutions to Sample Problems for Midterm

Part 1: Propositional Logic

2 PAVEL PUDLAK AND JIRI SGALL may lead to a monotone model of computation, which makes it possible to use available lower bounds for monotone models o

Methods of Partial Logic for Knowledge Representation and Deductive Reasoning in Incompletely Specified Domains

CS206 Lecture 03. Propositional Logic Proofs. Plan for Lecture 03. Axioms. Normal Forms

Integer Circuit Evaluation is PSPACE-complete. Ke Yang. Computer Science Department, Carnegie Mellon University, 5000 Forbes Ave.

Dependency Schemes in QBF Calculi: Semantics and Soundness

execution. Both are special cases of partially observable MDPs, in which the agent may receive incomplete (or noisy) information about the systems sta

Logic and Inferences

Matching Trace Patterns With Regular Policies

Transcription:

Subclasses of Quantied Boolean Formulas Andreas Flogel Department of Computer Science University of Duisburg 4100 Duisburg 1 Mare Karpinsi Department of Computer Science University of Bonn 5300 Bonn 1 and International Computer Science Institute Bereley, California Hans Kleine-Buning Department of Computer Science University of Duisburg 4100 Duisburg 1 Abstract Using the results of a former paper of two of the authors, for certain subclasses of quantied Boolean formulas it is shown, that the evalution problem is conp-complete. These subclasses can be seen as extensions of Horn and 2-CNF formulas. Further it is shown that the evalution problem for quantied Boolean formulas remains PSPACE-complete, even if at most one universal variable is allowed in each clause. Supported in part by the Leibniz Center for Research in Computer Science, by the DFG Grant KA 673/2-1 and by the SERC Grant GR-E 68297 1

1 Introduction It is well nown that the evaluation problem for Boolean formulas, that means to determine wether a formula Q 1 z 1 : : : Q n z n ( 1 ^ : : : ^ n ) is true, where Q is either 9 or 8 and i is a propositional clause, is PSPACE-complete [StM 73]. There are also subclasses decidable in polynominal time. In [APT 78] a linear time algorithm for quantied Boolean formulas with clauses of at most 2 variables only is given. In [KKBS 87] a cubic time algorithm for quantied Horn formulas was given. This result was improved in [KaKB 90]. It was shown that the evaluation problem for quantied Horn formulas can be decided in O(rn) time, where n is the length of the formula and r is the number of universal variables occuring positive in the formula. In the same paper a complete and sound resolution operation for quantied Boolean formulas, called Q-Resolution, was introduced. A Q-Resolution is an ordinary resolution, where only existential variables can be matched and additionally, each universal variable of a clause, for which no existential variable exists in that clause occuring after the universal variable in the prex, is omitted. A natural extension of quantied Horn formulas, where universal literals of a clause can be arbitrary, but the existential literals are in Horn form, was introduced. For this subclass, called extended quantied Horn formulas, it was shown by means of Q- Resolution, that the evaluation problem for such formulas with prex 8 : : : 8 : : :9 : : : 9 is conpcomplete. In this paper our interest is the complexity of extended quantied Horn formulas and extended quantied 2-CNF formulas, that are formulas with at most two existential variables and an arbitrary number of universal variables in each clause. It is shown that for any xed number of occurences of the pattern 8 : : : 8 : : :9 : : : 9 : : : in the prex the evalution problem for extended quantied Horn formulas remains conp-complete. A similar result is shown for extended quantied 2-CNF formulas. As mentioned above the evaluation problem for quantied Boolean formulas is PSPACEcomplete. In the last part of this paper it is shown, that this is true, even if at most one universal variable in each clause exists. So there is no improvement of the complexity possible by similar restrictions of the universal variables of a formula. 2 Preliminaries In what follows quantied Boolean formulas are of the form = 8x 1 9y 1 : : :8x 9y ( 1 ^ : : : ^ m ); where x i = x n i 1+1 : : :x n i and n 0 = 1; y i = y m i 1+1 : : :y m i and m 0 = 1: Further we assume that the variables in a clauses are given in a xed order. We say a literal L 1 is before a literal L 2, if the variable of L 1 occurs in the order of the prex before the variable of L 2. We write clauses in the form (L 1 _ : : : _ L t ), where L i 1 is before L i. An X-literal resp. Y-literal is a literal of the form x e or x e resp. y e or y e. A pure X-clause is a non-tautological clause consisting of X-literals only. In particular the empty clause is a pure X-clause. Finally a formula does not contain tautological clauses and each variable of the prex occurs in some clause. 2

Denition (1) A formula = 8x 1 9y 1 : : :8x 9y ( 1^: : :^ r ) is an extended quantied Horn Formula resp. 2-CNF formula, if for each i the Y-part is a Horn-clause resp. 2-CNF-clause, i.e. the clause i omitting all X-literals is a Horn-clause resp. 2-CNF-clause. In the following a prex 8 : : : 89 : : :9 is denoted as 1 and +1 is dened as 8 : : : 89 : : :9 Denition (2) For each 1 the class of extended quantied Horn resp. 2-CNF formulas with prex in is denoted as (extended Horn) resp. (extended 2-CNF). Now we introduce our generalized resolution operation, called Q-resolution: Denition (3) a: Let be a non-tautological X-clause, then we replace by the empty clause. In each other non-tautological clause we delete all X-literals, that are not before any Y-literal. b: Let 1 be a clause with Y-literal y l and 2 be a clause with Y-literal y l, then the resolvent of 1 and 2 is obtained as follows: 1. Remove all occurences of y l and y l in 1 _ 2. 2. Remove all occurences of X-literals, that are not before any Y-literal occuring in the disjunction. 3. If the resulting clause contains complementary literals, then no resolvent exists. Otherwise the resulting clause is the resolvent. Comparing ordinary resolution with Q-resolution, we see that only literals bounded by existential quantiers can be matched and universal variables not before an Y-literal will be eliminated. Example: 8x 1 x 2 9y 1 y 2 8x 3 ((x 1 _ x 2 _ y 1 _ y 2 _ x 3 ) ^ (x 1 _ y 1 ) ^ (x 2 _ y 2 ) ^ (x 1 _ y 2 )) The following Q-resolution steps can be performed: where j Q Res (x 1 _ x 2 _ y 1 _ y 2 _ x 3 ); (x 2 _ y 2 ) j Q Res (x 1 _ x 2 _ y 1 ) (x 1 _ y 2 ); (x 2 _ y 2 ) j Q Res t denotes the application of Q-resolution steps. It is well nown that resolution is complete and sound for propositional formulas in CNF, i.e. a formula 9y 1 : : :y n ( 1 ^ : : : ^ r ) is false i the empty clause can be derived by resolution applied to 1 ^ ::: ^ r. A similar result for Q-resolution and quantied Boolean formulas is shown in [KaKB 90]: Theorem 2.1 A quantied Boolean formula is false i j Q Res t Analogously to the case of unit-resolution for ordinary resolution we can dene a Q-unit resolution. A clause is called a Y-unit clause if contains exactly one Y-literal and arbitrarily many X-literals. A positive Y-unit clause is a unit clause with a positive Y-literal. 3

Denition (4) The Q-unit-resolution (Q-U-Res) is a Q-resolution, where one of the clauses is a positive Y-unit clause. The Q-unit-resolution is useful not only for quantied Horn formulas, but for extended quantied Horn formulas too (see again [KaKB 90]): Theorem 2.2 An extended quantied Horn formula is false i j Q U Res t. 3 The Evalution Problem for extended Formulas Theorem 3.1 The evaluation problem for 1 (extended Horn) and 1 (extended 2-CNF) is conp-complete. Proof: That the falsity of an extended quantied Horn formula and an extended quantied 2-CNF formula with prex 8 9 can be decided nondeterministically in polynominal time is obvious. Thus the evaluation problem belongs to conp. Now we associate to each formula = 1 ^ : : : ^ n in 3-CNF with variables x 1 ; : : : ; x m a quantied Boolean formula, which is in 1 (extended Horn) and in 1 (extended 2-CNF), such that the propositional formula is satisable i the extended quantied Boolean formula is false. Let y 0 ; : : : ; y n be new variables, then we associate to each clause i = (L i1 _ L i2 _ L i3 ) three clauses i1 = (L i1 _ y i 1 _ y i ) ; i2 = (L i2 _ y i 1 _ y i ) and i3 = (L i3 _ y i 1 _ y i ): Then we dene () := 8x 1 : : : x m 9y 0 : : :y n (y 0 ^ y n ^ 11 ^ 12 ^ : : : ^ n3 ): If () is false then it yields () j Q Res t. Since for each i(1 i n) one of the clauses i1 or i2 or i3 must be matched, there is some L i j i in i, such that fl 1j1 ; : : : ; L n g does not contain jn complementary literals. Then, we can dene a satisfying truth assignment J for choosing J (L i j i ) = 1. In the converse direction let be satisable. Then there is some truth assignment J, such that J () = 1. Hence, for each i there is some literal L i j i with J (L i j i ) = 1. Now we can apply the Q-resolution to () obtaining the empty clause as follows: (y 0 ); (L 1j1 _ y 0 _ y 1 ) j Q Res (L 1j1 _ y 1 ); (L 2j2 _ y 1 _ y 2 ) j Q Res (L 1j1 _ L 2j2 _ y 2 ); (L 3j3 _ y 2 _ y 3 ) j Q Res : : : j Q Res (L 1j1 _ : : : _ L n jn _ y n) ; y n j Q Res t. Since the satisability problem for 3-CNF is NP-complete and the transformation can be performed in polynominal time, we have proved our desired result. For technical reasons we need the following denition: A quantied Boolean formula = 8x 1 9y 1 : : : 8x 9y ( 1 ^ : : : ^ m ) is called reduced, if each variable in the prex occurs in 1 ^ : : : ^ m as literal and for each i the quantied Boolean formula f i g is true. In the case of the falsity of reduced means, that removing one clause we obtain a formula, which is true. 4

For each quantied Boolean formula = 8x 1 9y 1 : : : 8x 9y ( 1 ^ : : : ^ m ) a reduced formula r can be obtained by removing nondeterministically some clauses and removing all variables in the prex that do not occur in the remaining clauses. It is easy to see that if is false there is a selection of clauses so that r is false and reduced and that if is true, r remains true. Theorem 3.2 For each xed 1 and in (extended Horn): is false i j Q U Res in at most n (nondeterministic) Q-unit-resolution steps, where length() = n. t Proof: We now that for a formula, which is false, there is a derivation to the empty clause by means of Q-unit-resolution. We will prove by induction on that the number of Q-unit-resolution steps can be restricted to n. For = 1 the formula has the form 8x 1 9y 1 ( 1 ^ : : : ^ m ). Obviously at most length () resolution steps are sucient, because for a propositional Horn formula length() unitresolution steps lead to the empty clause. Now let be > 1. We assume = 8x 1 9y 1 : : : 8x 9y ( 1 ^ : : : ^ m ) is false and reduced. For a clause and a set S of variables (S) is the clause omitting all variables not in S. Since is false and reduced the set of clauses (x 1 ) = f 1 (x 1 ); : : : m (x 1 )g does not contain complementary literals, i.e. for each literal L in i (x 1 ) the complement L does not occur in (x 1 ). Otherwise the empty clause could not be derived by Q-unit-resolution. Then we can remove the x 1 -variables in without eect to the truth of obtaining the formula We dene 0 = 9y 1 8x 2 9y 2 ; : : : 8x 9y ( 0 1 ^ : : : ^ 0 m) Part(y 1 ) := f 0 ij1 i m and 0 i(y 1 ) is not emptyg Rest (y 1 ) := f 0 1 ; : : :; 0 mg Part(y 1 ): Since 0 is reduced for each 2 Part(y 1 ) the subclause (y 1 ) can be derived applying the Q-unit-resolution with Rest(y 1 ) [. Thus, Rest(y 1 ) [ f(x 2 ; y 2 ; : : : ; y )g j Q U Res t and the prex is in 1. Applying the induction hypothesis we now that (y 1 ) can be derived in length (Rest(y 1 ) [ fg) 1 Q-unitresolution steps. Altogether we obtain X 2 Part(y 1 ) length ((Rest(y 1 ) [ ) 1 Q-unit-resolution steps to generate 0 (y 1 ) and then length ( 0 (y 1 )) resolution steps to derive the empty clause. Since Part(y 1 ) must contain at least two clauses, otherwise 0 is not reduced and false, we obtain our desired upper bound length ( 0 ). Theorem 3.3 For each xed 1 the evalution problem for (extended Horn) is conpcomplete. 5

Proof: The evalution problem for (extended Horn) belongs to conp, because the falsity can be decided nondeterministically by choosing at most length () Q-unit-resolution steps (see theorem 3.2). Since 1 (extended Horn) is conp-complete we have proved our desired result. This result can not be improved by means of Q-Resolution only. The following lower bound for extended quantied Horn formulas is shown in [KaKB 90]: Theorem 3.4 There exist quantied extended Horn formulas t (t 1) of length 18t + 1, which are false and a refutation to the empty clause requires at least 2 t Q-resolution steps. The following denition is needed in the proof of the next theorem: A renaming is the replacing of all occurences of some literals by their complements. A renaming f is a one-to-one mapping Lit()! Lit(), such that for each L 2 Lit() : f(l) 2 fl; :Lg and f(l) = :f(:l): Lemma 3.5 : For each satisable formula in 2-CNF a renaming f can be found in O(length()) such that f() is a Horn formula. Proof: Since is satisable and in 2-CNF a satisfying truth assignment J for can be found in O(length()) [APT 78]. Now the renaming is dened as f(a) = ( A if J (A) = 1 A if J (A) = 0 for each variable A of : For at least one literal L i in each clause J (L i ) = 1, so we can analyse the following cases with the 5 possible structures of 2-CNF clauses 1) f(b _ C) = ( B _ f(c)) or f(b _ C) = ( C _ f(b)) 2) f( B _ C) = ( B _ f( C)) or f( B _ C) = ( C _ f( B)) 3) f(b _ C) = ( B _ f( C)) or f(b _ C) = ( C _ f(b)) 4) f(b) = ( B) 5) f( B) = ( B) In each case the result is a Horn-formula. Theorem 3.6 For each xed 1 the evalution problem for (extended 2-CNF) is conpcomplete. Proof: 1 (extended 2-CNF) is nown to be conp-complete, because of theorem 3.1. 6

Now we will show by induction on that the complement of the evalution problem for (extended 2-CNF) lies in NP. Let be given > 1. By induction hypothesis there is a nondeterministic algorithm NA 1 deciding whether an arbitrary formula 1 in 1 (extended 2-CNF) is false. We generate for each formula in (extended 2-CNF) nondeterministically in O(length( )) time two formulas 1;1 ; 1;2, such that 1;1 ; 1;2 2 1 (extended 2-CNF), so that both formulas are false if is false and length ( 1;i ) < length ( ) for i = 1; 2. Let be given = 8x 1 9y 1 : : : 8x 9y ( 1 ^ : : : ^ m ) Remember that (S) with a set S of variables is the clause omitting all variables not in S and (S) = f(s)j 2 g: Step 1: [Building a reduced formula] If is false we nondeterministically obtain a reduced and false formula 0. If is true the generated formula remains true. In the following whenever we observe that the resulting formula is true or not reduced we halt and nothing is nown. If a formula 0 is reduced and false then each clause is necessary to derive the empty clause. Then 0 (x 1) does not contain complementary literals. Step 2: [Deletion of x 1, renaming of y 1 ] If 0 (x 1) contains no complementary literals, then remove all x 1 -variables in 0 obtaining a formula 00 = 9y 1 8x 2 : : :8x 9y ( 1 ^ : : : ^ r ): In case of complementary literals in 0 (x 1) 0 is not reduced or true, so we halt. Let be 00 (y 1) i the set of clauses in 00 (y 1) with exactly i literals. For a non-empty set 00 (y 1) 2 the formula is false, if 00 (y 1) 2 is not satisable. In case of satisability there is some renaming, i.e. literals L may be replaced by L simultanously, such that the resulting formula 00 ;H (y 1) is a Horn formula. Since 00 (y 1) 2 contains 2-clauses only a renaming can be found in O(length( 00 (y 1) 2 )) time (see Lemma 3.5). Now in 00 we replace 00 (y 1) by the obtained Horn formula 00 ;H (y 1). The resulting formula is denoted as 000. Step 3: [Simplication of 000 (y 1)] If 000 (y 1) 2 is empty then only 1-clauses occur in 000 (y 1) and 000 (y 1) 1 must contain comple- is not reduced or true. (y 1) as follows: mentary literals, otherwise we halt, because 000 If 000 (y 1) 2 is not empty and assumed to be satisable we can simplify 000 All X-variables in the clauses of 000 (y 1) 2 are not before any Y-variable. Each clause in 000 (y 1) 2 is an implication of the form (y i1 y i2 ) or a negated clause of the form (y i1 _ y i2 ) for some y i1 ; y i2 y 1. Each clause s in 000 can be seen as a sort of unit-clause of the form ( ( ) ( ) with exactly one literal y i 2 y 1 y i _ s (x 2 : : :y )): A resolution of such a clause ( ( ) y i _ s (x 2 : : :y )) with a clause in 000 (y 1) 2 results in a clause ( ( ) y j _ s (x 2 : : : y )) with y j 2 y 1 : For all clauses s = ( ( ) y i _ s (x 2 : : :y )) 2 000 with exactly one y-variable y i 2 y 1 all 7

resulting clauses ( ( ) y j _ s (x 2 : : : y )) from resolution with clauses in 000 (y 1) 2 can be produced in linear time by means of unit resolution. If we add these clauses to 000, we can choose nondeterministically two clauses s = (y i _ s (x 2 : : : y )) and t = (y i _ t (x 2 : : :y )) with y i 2 y 1 : All other clauses containing variables from y 1 will be removed. If 000 is true 0v is false than there is some selection such that the resulting formula 0v remains true. The obtained formula 0v has the form 9y 1 8x 2 : : : 8x 9y ((y 1 _ 1 (x 2 : : :y )) ^ (y 1 _ 2 (x 2 : : : y )) ^ 3 ^ : : : ^ t ), where 3 ; : : : ; t does not contain the variable y 1. is false and if 000 We can divide the formula 0v into two formulas 0 and 1;1 0 with free variables as 1;2 follows: 0 1;i = 8x 2 : : : 9y : ((y i _ 1 i(x 2 : : : y )) ^ 3 ^ : : : ^ t ); where y 1 = y 1 1 and y 2 = y 1 1: If 0v is false and reduced then 0 resp. 1;1 0 must imply y 1;2 1 resp. y 1. Then it remains to decide whether 1;i = 8x 2 : : : 9y ( i (x 2 : : : y ) ^ 3 ^ : : : ^ t ) are false. Now we can apply our nondeterministic algorithm NA 1 with 1;i (i = 1; 2). Adding the above steps to NA 1 we obtain a nondeterministic algorithm NA, which decides in polynominal time, whether a formula in (extended 2-CNF) is false. Instead of restricting the existential part of the formulas to Horn-clauses resp. 2-clauses, one may also consider restrictions of the universal variables of the clauses. The following theorem shows, that from such a restriction, no gain in complexity will result. Theorem 3.7 The evalution problem for quantied Boolean formulas, for which each clause contains at most one universal variable is PSPACE-complete. Proof: Let be given any quantied Boolean formula = 8x 1 9y 1 : : :8x 9y ( 1 ^ : : : ^ m ), where x i = x n i 1+1 : : : x n i ; 1 i and n 0 = 1. Now we introduce new existential variables z 1 ; : : : ; z n and dene = 8x 1 9z 1 8x 2 9z 2 : : : 8x n1 9z n1 9y 1 : : : : : : 8x n 1+19z n 1+1 : : : 8x n 9z n 9y : ^ ^ 1jn j [x 1 =z 1 ; : : : ; x n =z n ]); 0 @ ^ 1in ((x i _ z i ) ^ (x i _ z i ) 1 A where j [x 1 =z 1 ; : : : ; x n =z n ] is the clause obtained by replacing in j x i resp. x i by the variable z i resp. z i. each occurence of 8

As easily can be seen is true i is true. contains clauses with at most one universal variable only. Thus we have proved our desired result, because the evalution problem for quantied Boolean formulas is PSPACE-complete and our transformation can be performed in polynominal time. 4 Conclusion We have shown that the evalution problem for the subclasses (extended Horn) and (extended 2-CNF) of quantied Boolean formulas for any xed 1 is conp-complete. We hope these results will help to solve the complexity of the evalution problem for extended quantied Horn resp. 2-CNF formulas, that still remains open. References [APT 78] B. Aspvall, M. F. Plass and R.E. Tarjan: A Linear-Time Algorithm for Testing the Truth of Certain Quantied Boolean Formulas, Information Processing Letters, Volume 8, number 3, 1978 [DoGa 84] W. F. Dowling and J. H. Gallier: Linear-Time Algorithms For Testing The Satis- ability Of Propositional Horn Formulae, J. of Logic Programming, 1984 [GJ 79] M. R. Garey and D. S: Johnson: Computers and Intractability, A Guide to the Theory of NP-Completeness, W. H. Freemann and Co., San Fransicao, 1979 [Ha 85] A. Haen: The intractability of resolution, Theoret. Comput. Sci. 39, 1985, 297{ 308 [KaKB 90] M.Karpinsi, H.Kleine Buning: Resolution for Quantied Boolean Formulas, submitted for publication [KKBS 87] M. Karpinsi, H. Kleine Buning and P.H. Schmitt: On the computational complexity of quantied Horn clauses, Lecture Notes in Computer Science 329, Springer- Verlag, 1987 [StM 73] L. J. Stocmeyer and A. R. Meyer: Word problems requiring exponential time, Proc. 5th Ann. ACM Symp. Theory Computer, 1973 9