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Generic Expression Hardness Results for Primitive Positive Formula Comparison Simone Bova Dept. of Mathematics Vanderbilt University Nashville, TN, USA [email protected] Hubie Chen Dept. de Tecnologies Universitat Pompeu Fabra Barcelona, Spain [email protected] Abstract We study the expression complexity of three basic problems involving the comparison of primitive positive formulas. We give two generic hardness results for the studied problems, and discuss evidence that they are optimal. 1 Introduction A primitive positive (pp) formula is a first-order formula defined from atomic formulas and equality of variables using conjunction and existential quantification. The class of primitive positive formulas includes, and is essentially equivalent to, the class of conjunctive queries, which is well-established in relational database theory as a pertinent and useful class of queries, and which has been studied complexity-theoretically from a number of perspectives (see for example [18, 16, 1]). In this paper, we study the complexity of the following fundamental problems, each of which involves the comparison of two pp-formulas π, πβ² having the same free variables, over a relational structure. β Equivalence: are the formulas π, πβ² equivalentβthat is, do they have the same satisfying assignmentsβover the structure? β Containment: are the satisfying assignments of π contained in those of πβ² , over the structure? β Isomorphism: does there exist a permutation of the free variables for one of the formulas under which the two formulas are equivalent? We study the complexity of these computational problems with respect to various fixed structures. That is, we parameterize each of these problems with respect to the structure to obtain a family of problems, containing one Matthew Valeriote Dept. of Mathematics & Statistics McMaster University Hamilton, Canada [email protected] member for each structure, and study the resulting families of problems. To employ the terminology of Vardi [20], we study the expression complexity of the presented comparison tasks. The suggestion here is that various relational structuresβwhich may represent databases or knowledge bases, according to useβmay possess structural characteristics that affect the complexity of the resulting problems, and our interest is in understanding this interplay. The present work focuses on relational structures that are finite (that is, have finite universe), and we assume that the structures under discussion are finite. Our study utilizes universal-algebraic tools that are currently being used to study computational problems related to primitive positive formulas, such as the constraint satisfaction problem (CSP), which can be viewed as the problem of deciding if a primitive positive formula is satisfiable over a given structure. It is known that, relative to a structure, the set of relations that are definable by a primitive positive formula forms a robust algebraic object known as a relational clone; a Galois correspondence associates, in a bijective manner, each such relational clone with a clone, a set of operations with certain closure properties. This correspondence provides a way to pass from a relational structure B to an algebra πΈB whose set of operations is the mentioned clone, in such a way that two structures having the same algebra have the same complexity (for each of the mentioned problems). In a companion paper [6] by the present authors, we developed this correspondence and presented some basic complexity results for the problems at hand, including a classification of the complexity of the problems on all twoelement structures. In this paper, we present two general expression hardness results on the problems of interest. In particular, each of our two main results provides a sufficient condition on a structure so that the problems are hard for certain complexity classes. Furthermore, we give evidence that our results are optimal, in that the conditions that they involve in fact describe dichotomies in the complexity of the studied problems. We now turn to describe each of our hardness results in greater detail. tractable CSP [6]. Our new result requires establishing a deeper understanding of the identified algebrasβ structure, some of which are CSP tractable, in order to obtain hardness. Our second hardness result, which concerns structures B whose associated algebra πΈB is idempotent, implies that for any such structure, if the variety π±(πΈB ) is not congruence modular, then the equivalence and containment problems are coNP-hard. Note that structures having all constants are well-known to have idempotent algebras, and so this result covers such structures.1 The question of whether or not this second hardness result can be extended to all structures is left as a tantalizing open question. Previous work identified one most general condition for the tractability of the equivalence and containment problems: if the algebra has few subpowersβa combinatorial condition [3, 14] involving the number of subalgebras of powers of an algebraβthen these problems are polynomialtime tractable. This second hardness result appears to perfectly complement this tractability result: there are no known examples of algebras πΈB (of structures B having finitely many relations) that are not covered by one of these results, and in fact the Edinburgh conjecture predicts that none exist, stating that every such algebra πΈB that generates a congruence modular variety also has few subpowers. Concerning this conjecture, it should be pointed out, on an optimistic note, that the resolution of the Zadori conjecture, a closely related conjecture of which the Edinburgh conjecture is a generalization, was recently announced by L. Barto. We also point out that this conjecture (as with the Zadori conjecture) is purely algebraic, making no references to notions of computation. This second hardness result similarly suggests a dichotomy for the isomorphism problem. All put together, the picture that emerges from this work is a trichotomy in the complexity of the studied problems. In the case of equivalence and containment, one appears to have Ξ π2 -completeness for an algebra with a variety admitting the unary type; coNP-completeness for an algebra with a non-congruence modular variety omitting the unary type, and polynomial-time tractability otherwise, with a picture for isomorphism that is similar but shifted slightly upwards. We certainly look forward to future work on these problems and the related challenging conjectures. Our first hardness result yields that for any structure B whose associated algebra πΈB gives rise to a variety π±(πΈB ) that admits the unary type, the equivalence and containment problems are Ξ π2 -complete. Note that this is the maximal complexity possible for these problems, as the problems are contained in the class Ξ π2 . The condition of admitting the unary type originates from tame congruence theory, a theory developed to understand the structure of finite algebras. We observe that this result implies a dichotomy in the complexity of the studied problems under the G-set conjecture for the CSP, a conjecture that predicts exactly where the tractability/intractability dichotomy lies for the CSP. In particular, under the G-set conjecture, the structures not obeying the described condition have equivalence and containment problems in coNP. The resolution of the G-set conjecture, on which there has been focused and steady progress over the past decade [10, 14, 11, 2], would thus, in combination with our hardness result, yield a coNP/Ξ π2 -complete dichotomy for the equivalence and containment problems. For the isomorphism problem, we also demonstrate that the G-set conjecture would yield a dichotomy between two modes of complexity behavior that cannot coincide, unless the polynomial hierarchy collapses. In fact, this hardness result already unconditionally implies dichotomies for our problems for all classes of structures where the G-set conjecture has already been established, including the class of three-element structures [10], the class of conservative structures [7], and the class of undirected graphs [8]. One formulation of the G-set conjecture is that, for a structure B whose associated algebra πΈB is idempotent, the absence of the unary type in the variety generated by πΈB implies that CSP(B) is polynomial-time tractable. The presence of the unary type is a known sufficient condition for intractability in the idempotent case [9, 11], and this conjecture predicts exactly where the tractability/intractability dichotomy lies for the CSP. It should be noted, however, that the boundary that is suggested by our hardness result for the equivalence, containment, and isomorphism problems is not the same as the boundary suggested by the G-set conjecture for the CSP. The G-set conjecture, which is typically phrased on idempotent algebras, yields a prediction on the CSP complexity of all structures via a theorem [9] showing that each structure B has the same CSP complexity as a structure Bβ² whose associated algebra is idempotent. The mapping from B to Bβ² does not preserve the complexity of our problems, and indeed, there are examples [6] of two-element structures B such that our hardness result applies to Bβthe equivalence and containment problems on B are Ξ π2 -completeβbut Bβ² does not admit the unary type and indeed has a polynomial-time 2 Preliminaries Here, a signature is a set of relation symbols, each having an associated arity; we assume that all signatures are of finite size. A relational structure over a signature π consists of a universe π΅ and, for each relation symbol π β π, a relation π B β π΅ π where π is the arity of π . We assume that 1 By 2 constants, here we mean singleton unary relations. PPISO to instances where the structure B is boolean (twoelement). Throughout the paper, the notion of reduction used is logspace many-one reducibility. all relational structures under discussion have universes of finite size. A primitive positive formula (pp-formula) on π is a first-order formula formed using equalities on variables (π₯ = π₯β² ), atomic formulas π (π₯1 , . . . , π₯π ) over π, conjunction (β§), and existential quantification (β). We now define the problems that will be studied. For the isomorphism problem, we will make use of the following notion. A sectioning for a formula with free variables π is a pair of mappings (π : π β π, π‘ : π β π ) such that the pair mapping (π , π‘) : π β π × π is a bijection. Let π, πβ² be formulas with free variables ππ , ππβ² and sectionings (π , π‘), (π β² , π‘β² ), respectively. We say that a bijection π : ππβ² β ππ respects the sectionings if for all π₯, π₯1 , π₯2 β ππβ² , (1) π β² (π₯1 ) = π β² (π₯2 ) if and only if π (π(π₯1 )) = π (π(π₯2 )), and (2) π‘β² (π₯) = π‘(π(π₯)). One can think of π (π₯) as the section of a variable π₯, and of π‘(π₯) as its type; the requirement on (π , π‘) requires that each section has exactly one variable of each type, and the given notion of respecting requires that sections are mapped to sections in a type-preserving way. Proposition 2.2 The problem PPISO reduces to the problem BOOL-PPISO. Proof. See Appendix A. β‘ In this paper, we will overload problems such as PPISO and use a problem to denote the set of all problems that reduce to it. This will allow us to talk about, for instance, PPISO-completeness. Proposition 2.3 For each structure B, the problem PPCON(B) reduces to the problem PPEQ(B). Proof. The reduction, given an instance (π, πβ² ) of PPCON(B), outputs the instance (π, π β§ πβ² ) of PPEQ(B). β‘ We now review the relevant algebraic concepts to be used. An algebra is a pair πΈ = (π΄, πΉ ) such that π΄ is a nonempty set, called the domain or universe of the algebra, and πΉ is a set of finitary operations on π΄. Let πΈ = (π΄, πΉ ) be an algebra; a term operation of πΈ is a finitary operation obtained by composition of (1) operations in πΉ and (2) projections on π΄, and a polynomial operation is a finitary operation obtained by composition of (1) operations in πΉ , (2) projections on π΄ and (3) constants from π΄. An operation π (π₯1 , . . . , π₯π ) on A is said to be idempotentif the equality π (π, π, . . . , π) = π holds for all π β π΄. An algebra πΈ is idempotent if all of its term operations are. Let π΅ be a nonempty set, let π be an π-ary operation on π΅, and let π be a π-ary relation on π΅. We say that π preserves π (or π is a polymorphism of π ), if for every length π sequence of tuples π‘1 , . . . , π‘π β π , denoting the tuple π‘π by (π‘π,1 , . . . , π‘π,π ), it holds that the tuple π (π‘1 , . . . , π‘π ) = (π (π‘1,1 , . . . , π‘π,1 ), . . . , π (π‘1,π , . . . , π‘π,π )) is in π . We extend this terminology to relational structures: an operation π is a polymorphism of a relational structure B if π is a polymorphism of every relation of B. We use Pol(B) to denote the set of all polymorphisms of a relational structure B, and use πΈB to denote the algebra (π΅, Pol(B)). We remark that it is well-known and straightforward to verify that a relational structure B having all constants (singleton unary relations) has an idempotent algebra πΈB . Dually, for an operation π , we use Inv(π ) to denote the set of all relations that are preserved β© by π , and for a set of operations πΉ , we define Inv(πΉ ) as π βπΉ Inv(π ). We will make use of the following result connecting the Pol(β ) and Inv(β ) operators to pp-definability. Definition 2.1 We define the following computational problems. For each of the first two, an instance consists of a relational structure B and a pair (π, πβ² ) of pp-formulas over the signature of B having the same set of free variables π; for the third problem (PPISO), an instance consists of these objects and, in addition, sectionings (π , π‘) and (π β² , π‘β² ) for the formulas. β PPEQ: decide if π and πβ² are equivalent, that is, whether for all π : π β π΅, it holds that B, π β£= π iff B, π β£= πβ² . β PPCON: decide if π is contained in πβ² , that is, whether for all π : π β π΅, it holds that B, π β£= π implies B, π β£= πβ² . β PPISO: decide if π and πβ² are isomorphic, relative to the given sectionings, that is, whether there exists a bijection π : π β π respecting the sectionings such that for all π : π β π΅, it holds that B, π β£= π if and only if B, π β π β£= πβ² . For every relational structure B, we define PPEQ(B) to be the problem PPEQ where the structure is fixed to be B; hence, an instance of PPEQ(B) is just a pair (π, πβ² ) of pp-formulas. We define the problems PPCON(B) and PPISO(B) similarly. β‘ It is straightforward to verify that the PPEQ and PPCON problems are contained in Ξ π2 , and that the PPISO problem is contained in Ξ£π3 . We use the described formulation of the PPISO problem as it is robust with respect to changes in representation, as shown, for example, by the following proposition. We use BOOL-PPISO to denote the restriction of the Theorem 2.4 (Geiger [12]/Bodcharnuk et al. [5]) Let B be a finite relational structure. The set of relations Inv(Pol(B)) is equal to the set of relations that are ppdefinable over B. 3 We associate to each algebra πΈ = (π΄, πΉ ) a set of problems PPEQ(πΈ), namely, the set containing all problems PPEQ(B) where B has universe π΄ and πΉ β Pol(B). We define PPCON(πΈ) and PPISO(πΈ) similarly. For a complexity class π, we say that the problem PPEQ(πΈ) is πhard if PPEQ(πΈ) contains a problem PPEQ(B) that is πhard. We define π-hardness similarly for PPCON(πΈ) and PPISO(πΈ). Theorem 3.1 If B is a finite relational structure such that the variety generated by πΈB admits the unary type, then β PPEQ(B) and PPCON(B) are Ξ π2 -complete, and β PPISO(B) is PPISO-complete. Prior to embarking on the proof of Theorem 3.1, we point out that from it we can deduce, modulo the G-Set conjecture, dichotomies for the class of pp-equivalence, ppcontainment, and pp-isomorphism problems. Theorem 2.5 Let B be a finite relational structure, and let π be a complexity class closed under logspace reduction. The problem PPEQ(B) is π-hard if and only if PPEQ(πΈB ) is π-hard. The same result holds for PPCON(β ) and PPISO(β ). Theorem 3.2 If the G-Set conjecture holds, then for all finite relational structures B, either PPEQ(B) and PPCON(B) are Ξ π2 -complete or they are in coNP. In addition, either PPISO(B) is PPISO-complete and Ξ π2 -hard or it is in Ξ£π2 . Proof. The proof of [6, Theorem 2] applies to all of the problems. β‘ It can be remarked that no Ξ π2 -hard problem is in Ξ£π2 unless Ξ π2 = Ξ£π2 and the polynomial hierarchy collapses. Proof. According to the G-Set conjecture, if B is a finite relational structure such that the variety generated by πΈB omits the unary type, then CSP(Bβ ) is in P, where Bβ is obtained from B by adding to it all relations of the form {π} for π β π΅. From this it follows that PPEQ(B) and PPCON(B) are in coNP and that PPISO(B) is in Ξ£π2 . On the other hand, if the variety generated by πΈB admits the unary type, then by Theorem 3.1 we conclude that PPEQ(B) and PPCON(B) are Ξ π2 -complete and PPISO(B) is PPISO-hard; the problem PPISO is straightforwardly verified to be Ξ π2 -hard. β‘ The notion of a variety is typically defined on indexed algebras; a variety is a class of similar algebras that is closed under the formation of homomorphic images, subalgebras, and products. For our purposes here, however, we may note that the variety generated by an algebra πΈ, denoted by π±(πΈ), is known to be equal to π»ππ ({πΈ}), where the operator π» (for instance) is the set of algebras derivable by taking homomorphic images of algebras in the given argument set. Theorem 2.6 Suppose that πΉ β π±(πΈ). Then, for every problem PPEQ(B) β PPEQ(πΉ), there exists a problem PPEQ(Bβ² ) β PPEQ(πΈ) such that PPEQ(B) reduces to PPEQ(Bβ² ), and likewise for PPCON(β ) and PPISO(β ). Lemma 3.3 There is a finite algebra πΈ in π± and some congruence πΌ on πΈ such that: Proof. See Appendix B. β‘ 3 β πΌ covers 0π΄ in Con (πΈ), β the type of the congruence pair β¨0π΄ , πΌβ© is unary, Unary Type β the β¨0π΄ , πΌβ©-traces are all polynomially equivalent to two-element sets. Throughout this section, let B be a finite relational structure and π± be the variety generated by πΈB . Further assume that π± admits the unary type . In [6] it is shown that if in addition πΈB is assumed to be idempotent, then PPEQ(B) is Ξ π2 -complete and PPISO(B) is BOOL-PPISO-hard. In this section we establish the same hardness results, but without assuming the idempotency of πΈB . Our proof of Theorem 3.1 makes use of the detailed information on tame congruence theory provided in [13] and [15]. This theory associates a typeset to a non-trivial finite algebra, which contains one or more of five types: (1) the unary type, (2) the affine type, (3) the boolean type, (4) the lattice type, and (5) the semilattice type. By extension, a typeset is associated to each variety, namely, the union of all typesets of finite algebras contained in the variety. A variety is said to admit a type if the type is contained in its typeset, and is otherwise said to omit the type. Proof. This lemma follows from Theorem 6.17 and Lemmma 6.18 of [13] along with our assumption that π± admits the unary type. β‘ Fix an algebra πΈ and congruence πΌ as in the lemma and choose some β¨0π΄ , πΌβ©-minimal set π , some β¨0π΄ , πΌβ©-trace π = {0, 1} contained in π and some unary polynomial π(π₯) of πΈ with π(π΄) = π and π(π₯) = π(π(π₯)) for all π₯. Definition 3.4 (see Definition 6.13 of [13]) For an element π β π΄ and π > 0, let π Λπ denote the π-tuple (π, π, . . . , π). We will drop the subscript when the arity of the tuple is clear. For an π-ary relation π over π that contains the constant tuples 0Μ and 1Μ, we define the π-ary relation πΈ(π ) over π΄ to be the universe of the subalgebra of πΈπ generated by π βͺ {Λ ππ : π β π΄}. 4 Lemma 3.5 If π is an π-ary relation over π that contains the constant tuples 0Μ and 1Μ, π = {π, π} is a β¨0π΄ , πΌβ©-trace and π(π₯) is a polynomial of πΈ with π(0) = π and π(1) = π then πΈ(π ) β© π π = π(π ). In particular πΈ(π ) β© π π = π . Furthermore, any member (π1 , . . . , ππ ) of πΈ(π ) is πΌconstant, i.e., (ππ , ππ ) β πΌ for all 1 β€ π β€ π β€ π and if π(π₯) is a unary polynomial of πΈ, then (π(π1 ), . . . , π(ππ )) β πΈ(π ). ππ then by the first part of this proposition, there is some multitrace π with (ππ1 , . . . , πππ ) β π π and thus π β π π . β‘ The following theorem records some relevant features of multitraces. Theorem 3.8 Let π be a multitrace, say π = π (π, π, . . . , π ) for some π-ary polynomial π of πΈ. There is a π-ary polynomial π β² (¯ π₯) of πΈ for some π β€ π and some unary polynomials (called coordinate maps) ππ (π₯) of πΈ, for 1 β€ π β€ π, such that Proof. The first part follows from Lemma 6.14 (2) of [13] and from the fact that any two β¨0π΄ , πΌβ©-traces are polynomially isomorphic in πΈ (see Corollary 5.2. (2) of [13]). The second holds since π is contained in a single πΌ-class of πΈ and πΈ(π ) contains all constant tuples and is a subuniverse of πΈπ . β‘ In the proof of the hardness results of this section, we will need detailed information about relations over π΄ of the form πΈ(π ), where π is a relation over π . We employ the theory of multitraces, developed in [15], to aid us. For a more general presentation of this notion, see Section 3 of that article. β π = π β² (π, π, . . . , π ) and ππ (π ) β π for all π, β for all π₯π β π , and all π, ππ (π β² (π₯1 , . . . , π₯π )) = π₯π , β for all π₯ β π , π₯ = π β² (π1 (π₯), . . . , ππ (π₯)), β the set π π is in bijective correspondence with π via the map that takes a π-tuple (π1 , . . . , ππ ) to π β² (π1 , . . . , ππ ). Proof. This follows from Theorem 3.10 of [15]. β‘ The following definition provides a way to translate primitive positive formulas over a set of Boolean relations to primitive positive formulas over relations compatible with πΈ. This translation will be used to establish our hardness results. Definition 3.6 A multitrace is a subset π of π΄ of the form π (π, π, . . . , π ) for some polynomial π (¯ π₯) of πΈ. We can use the notion of a multitrace to gain a clear picture of the relations πΈ(π ) in the special case where π > 0 and π = π π . For the rest of this section, let π = β£π΄β£. Definition 3.9 Let β be a set of finitary relations over {0, 1}. If π(π₯1 , . . . , π₯π ) is a pp-formula over the relations in β and if {π¦1 , . . . , π¦π } is its set of quantified variables, define πΈ(π)(π₯1 , . . . , π₯π ) to be the pp-formula over the relations {πΈ(π ) : π β β} βͺ {ππ : π β€ π} obtained from π by replacing each occurrence of a relation π β β by πΈ(π ) and by conjoining the formula ππ+π (π₯1 , . . . , π₯π , π¦1 , . . . , π¦π ). If π + π > π, we make use of the second part of Proposition 3.7 to express ππ+π in terms of ππ . π Proposition 3.7 For π > 0, let ππ = πΈ(π ). βͺ β ππ = {π π : π is a multitrace}. β For π β₯ π, we have ππ (π₯1 , . . . , π₯π ) iff β ππ (π₯π1 , . . . , π₯ππ ). {1β€π1 <β β β <ππ β€π} Proof. If π is a multitrace, then there is some πary polynomial π (π₯1 , . . . , π₯π ) of πΈ such that π = π (π, π, . . . , π ). Since ππ is the universe of the subalgebra of πΈπ generated by π π and all of the constant π-tuples, it follows that π π β ππ . Conversely, if π = (π1 , . . . , ππ ) is in ππ then there is some π > 0, some π-ary polynomial π‘(¯ π₯) of πΈ and π-tuples ππ β π π , for 1 β€ π β€ π, such that π = π‘(π1 , . . . , ππ ) (where π‘ is applied coordinatewise). But then π β π π , where π is the multitrace π‘(π, π, . . . , π ). For the second part of this proposition, suppose that π > π and π = (π1 , . . . , ππ ) is in ππ . Then there is some multitrace π with ππ β π for all 1 β€ π β€ π. In particular, if 1 β€ π1 < π2 < β β β < ππ β€ π then πππ β π for all 1 β€ π β€ π and from this it follows that (ππ1 , . . . , πππ ) β ππ . In the other direction, since π = β£π΄β£, we can choose some sequence 1 β€ ππ < π2 < β β β < ππ β€ π such that for all 1 β€ π β€ π, ππ = πππ for some π β€ π. If (ππ1 , . . . , πππ ) β Theorem 3.10 If π(π₯1 , . . . , π₯π ) is a pp-formula over the set of finitary relations β over {0, 1} and each π β β contains the constant tuples 0Μ and 1Μ then {π β {0, 1}π : π(π)} = {π β π΄π : πΈ(π)(π)} β© {0, 1}π and {π β π΄π : πΈ(π)(π)} = πΈ({π β {0, 1}π : π(π)}). Proof. Suppose that {π¦1 , . . . , π¦π } are the quantified variables of π. For the first equality, if π = (π1 , . . . , ππ ) β {0, 1}π and π(π) is witnessed by the elements ππ β {0, 1}, 1 β€ π β€ π, then all of these elements are in π and hence ππ+π (π1 , . . . , ππ , π1 , . . . , ππ ) holds. If some clause π (π’1 , . . . , π’π ) of π holds, 5 where π’π β {π1 , . . . , ππ , π1 , . . . , ππ } then by construction πΈ(π )(π’1 , . . . , π’π ) also holds. From this it follows that πΈ(π)(π1 , . . . , ππ ) holds, using the same witnesses ππ , 1 β€ π β€ π. Conversely, suppose that π = (π1 , . . . , ππ ) β {0, 1}π and πΈ(π)(π) holds, witnessed by the elements ππ β π΄, 1 β€ π β€ π. Since each relation πΈ(π ) that appears as a clause in πΈ(π) is closed under the unary operation π(π₯), applied coordinatewise (see Lemma 3.5), it follows that the elements π(ππ ), 1 β€ π β€ π also witness that πΈ(π)(π) holds. We use here that π is the identity map on its range, and in particular that π(0) = 0 and π(1) = 1. We claim that in fact, the elements π(ππ ) are all members of {0, 1}. Since the clause ππ+π (π1 , . . . , ππ , π1 , . . . , ππ ) holds then all of these elements lie in the same πΌ-class, and in particular belong to the πΌ-class that contains 0, since π1 β {0, 1}. We conclude that for each π, π(ππ ) β {0, 1} since π maps the entire πΌ-class that contains 0 into {0, 1}. Finally, Lemma 3.5 provides that πΈ(π ) β© {0, 1}π = π for all π β β and from this it follows that the elements π(ππ ), 1 β€ π β€ π, also witness that π(π) holds. Thus the first equality has been established. For the second equality, we can apply the πΈ(β ) operator to both sides of the first equality to obtain that πΈ({π β {0, 1}π : π(π)}) is equal to for any π. This is because for each coordinate map ππ , we have that ππ (π ) β {0, 1}. Extending this to our pp-formula πΈ(π), it follows that not only is (ππ (π1 ), . . . , ππ (ππ )) a solution, witnessed by ππ (ππ ), 1 β€ π β€ π, but it is also a member of {0, 1}π . So, each of these tuples belongs to the generating set of the relation πΈ({π β π΄π : πΈ(π)(π)} β© {0, 1}π ). Since for each π, ππ = π β² (π1 (ππ ), . . . , ππ (ππ )) and πΈ({π β π΄π : πΈ(π)(π)} β© {0, 1}π ) is closed under π β² , applied coordinatewise, we conclude that π = (π1 , . . . , ππ ) is also a member of this relation, as required. β‘ Corollary 3.11 Let β be a set of finitary relations over {0, 1} such that each π β β contains the constant tuples. Two pp-formulas π(π₯1 , . . . , π₯π ) and π(π₯1 , . . . , π₯π ) over the relations in β will be equivalent (contained, isomorphic with respect to sectionings) if and only if the ppformulas πΈ(π) and πΈ(π) are equivalent (contained, isomorphic with respect to sectionings) over {πΈ(π ) : π β β} βͺ {ππ : π β€ π}. Proof of Theorem 3.1. To establish that PPEQ(B) and PPCON(B) are Ξ π2 -complete, it will suffice to show that it is Ξ π2 -hard, since both problems are contained in Ξ π2 . By Theorem 4 and Lemma 4 from [6] it follows that there is some finite relational structure C = ({0, 1}, β) such that each π β β contains the constant tuples 0Μ and 1Μ and such that PPEQ(C) and PPCON(C) are Ξ π2 -hard. By those proofs, one also readily obtains that PPISO(C) is BOOL-PPISO-hard (for the definition of BOOL-PPISO given in this paper). From Corollary 3.11 there is a finite algebra πΈ in the variety generated by πΈB such that PPEQ(C), PPCON(C) and PPISO(C) reduce to PPEQ(πΈ), PPCON(πΈ), and PPISO(πΈ) respectively. Finally, using Theorems 2.6 and 2.5 we conclude that PPEQ(B) and PPCON(B) is Ξ π2 complete and PPISO(B) is BOOL-PPISO-hard; by Proposition 2.2, PPISO(B) is PPISO-hard and hence PPISOcomplete. β‘ πΈ({π β π΄π : πΈ(π)(π)} β© {0, 1}π ) and so it will suffice to prove that πΈ({π β π΄π : πΈ(π)(π)}β©{0, 1}π ) = {π β π΄π : πΈ(π)(π)} to complete the proof. The containment of the left hand side of this derived equality in the set {π β π΄π : πΈ(π)(π)} follows after observing that this set is a subuniverse of πΈπ and that it contains all constant π-tuples πΛ, for π β π΄. The latter follows from the fact that each relation πΈ(π ) contains all constant tuples and so every constant π-tuple over π΄ is a solution of πΈ(π). For the remaining containment, let π = (π1 , . . . , ππ ) be a solution of πΈ(π), witnessed by the elements ππ β π΄, 1 β€ π β€ π. Since ππ+π (π1 , . . . , ππ , π1 , . . . , ππ ) holds then there is some multitrace π of πΈ that contains all of these elements. By Theorem 3.8, it follows that for some π > 0, there is some π-ary polynomial π β² and coordinate maps ππ (π₯), 1 β€ π β€ π, that satisfy the properties stated in that theorem. In particular, π = π β² (π, π, . . . , π ) and for all π β π , π = π β² (π1 (π), . . . , ππ (π)). From Lemma 3.5 it follows that for any π-ary relation π β β, if ππ β π , for 1 β€ π β€ π, and (π1 , . . . , ππ ) β πΈ(π ), then 4 Non-Congruence Modularity A variety is idempotent if each algebra in the variety is idempotent. A variety is congruence modular if the congruence lattice of each algebra in the variety is modular. We let MFISO denote the problem of deciding whether two monotone Boolean formulas are isomorphic; recall that a monotone Boolean formula is one formed using just the connectives AND (β§) and OR (β¨). Theorem 4.1 Let B be a finite relational structure. If π±(πΈB ) is idempotent and not congruence modular, then: (ππ (π1 ), . . . , ππ (ππ )) β (πΈ(π ) β© {0, 1}π ) = π , (π) PPEQ(B) and PPCON(B) are coNP-hard, and 6 πΆ, equipped with the ternary sorted relation π β π΅×πΆ ×πΆ given by, (ππ) PPISO(B) is MFISO-hard. The Edinburgh conjecture, discussed in the introduction, predicts that for all structures B, either π±(πΈB ) is not congruence modular, or πΈB has few subpowers. This conjecture would imply a P/coNP-hard dichotomy for the PPEQ(β ) and PPCON(β ) problems for all such structures: by the conjecture, the structures not covered by Theorem 4.1 have that πΈB has few subpowers, in which case the problem PPEQ(B) is in P by [6, Theorem 7], and PPCON(B) is in P as well by an application of Proposition 2.3. Similarly, under this conjecture we have a dichotomy in the complexity of PPISO(B): either we have MFISO-hardness by the theoremβand hence coNP-hardness by [19, Theorem 19]βor, we have that PPISO(B) is in NP as a consequence of PPEQ(B) being in P. Note that this would give a dichotomy unless the polynomial hierarchy collapses, as no coNP-hard problem can be in NP unless NP = coNP. {(π, π, πβ² ) β£ π β π΅π β (π, πβ² ) β πΌπ , π = 0, 1}. A sorted pp-formula π on S is a conjunction of constraints of either the form π (π₯, π¦, π§), with π₯ of sort π΅ and π¦, π§ of sort πΆ, or the form π₯ = π¦, with π₯ and π¦ of the same sort, where some variables can be existentially quantified. A sorted assignment sends variables of sort π΅ to π΅, and variables of sort πΆ to πΆ. A sorted assignment π of the free variables of π satisfies π over S if there exists a sorted assignment of all the variables of π, extending π , that satisfies each constraint of π. Let π₯1 , . . . , π₯π and π¦1 , . . . , π¦π be variables of sort π΅ and πΆ respectively, and let π and π be sorted pp-formulas over S having free variables π₯1 , . . . , π₯π and π¦1 , . . . , π¦π . Then, π entails π over S, if and only if the satisfying assignments of π over S contain the satisfying assignments of π over S; the entailment problem on S is to decide, given two sorted pp-formulas π and π as above, whether or not π entails π over S. The proof proceeds by first reducing MBCON to the entailment problem on S, and then reducing the entailment problem on S to PPCON(A). The reduction from MBCON to the entailment problem on S works as follows. Let x = π₯1 , . . . , π₯π , and let π(x) be a monotone Boolean formula. By induction on the structure of π(x), we construct a sorted pp-formula πβ² (x, π¦1 , π¦2 ) on S, where the variables x are of sort π΅ and the variables π¦1 , π¦2 are of sort πΆ, as follows: for π β {1, . . . , π}, if π(x) is π₯π , then πβ² (x, π¦1 , π¦2 ) = π (π₯π , π¦1 , π¦2 ); if π(x) is π1 (x) β§ π2 (x), then πβ² (x, π¦1 , π¦2 ) = πβ²1 (x, π¦1 , π¦2 ) β§ πβ²2 (x, π¦1 , π¦2 ); if π(x) is π1 (x) β¨ π2 (x), then Proof. We exploit the fact, enlightened by [17, Lemma 1 and Lemma 2], that the failure of congruence modularity in π±(πΈB ) is nicely witnessed by a finite algebra πΈ β π±(πΈB ), and congruences πΌ, π½, and πΎ of πΈ with the following properties: there exist finite algebras πΉ and β in π±(πΈB ) such that πΈ = πΉ × β; π½ and πΎ are the kernels of the projections of πΈ onto πΉ and β respectively; there exist a partition of π΅ into two nonempty blocks π΅0 and π΅1 , and congruences πΌ0 < 1β and πΌ1 = 1β of β such that πΌ is {((π, π), (π, πβ² )) β£ π β π΅π β (π, πβ² ) β πΌπ , π = 0, 1}. (2) (1) Note that π½ and πΎ form a pair of factor congruences on πΈ and πΌ < π½; thus, 0πΈ , πΌ, π½, πΎ, and 1πΈ form a pentagon in the congruence lattice of πΈ, thus witnessing the failure of congruence modularity in π±(πΈB ). Let A be the relational structure with universe π΄ = π΅ × πΆ and relations πΌ, π½, and πΎ. πβ² = (βπ§) (πβ²1 (x, π¦1 , π§) β§ πβ²2 (x, π§, π¦2 )) , (3) where π§ is a fresh variable of sort πΆ. The reduction is now, given an instance (π, π) of MBCON, construct the instance (πβ² , π β² ) of the entailment problem on S. We now prove that the reduction is correct. The key step is to establish a correspondence between the satisfying assignments of π (over the two-element Boolean lattice) and πβ² over S. Let π be a Boolean assignment of variables π₯1 , . . . , π₯π , and let π be a sorted assignment of variables π₯1 , . . . , π₯π , π¦1 , π¦2 . Say that π and π match if: π (π₯π ) = 0 implies π(π₯π ) β π΅0 , and π (π₯π ) = 1 implies π(π₯π ) β π΅1 . (π) We now prove that PPCON(A) is coNP-hard. Note that PPCON(A) is in PPCON(πΈ), because πΌ, π½, and πΎ are congruences of πΈ, and hence are preserved by the fundamental operations of πΈ. Since πΈ β π±(πΈB ), by Theorem 2.6, PPCON(A) logspace reduces to some PPCON(Bβ² ) in PPCON(πΈB ), so that PPCON(πΈB ) is coNP-hard, which finally implies that PPCON(B) is coNPhard by Theorem 2.5. The coNP-hardness of PPEQ(B) follows immediately by Proposition 2.3. We describe a reduction from the coNP-complete problem of deciding entailment between two monotone Boolean formulas [4, Theorem 4.1], call it MBCON, to PPCON(A). The reduction has two stages, which we now prepare. We introduce the following intermediate problem. Let π΅ and πΆ be disjoint finite sets. Let S be a sorted structure, that is, a relational structure with universe π΅βͺπΆ and sorts π΅ and Claim 4.2 Let π be a sorted assignment of π₯1 , . . . , π₯π , π¦1 , π¦2 , matching the Boolean assignment π of π₯1 , . . . , π₯π . Then, π satisfies πβ² over S if and only if, π does not satisfy π implies (π(π¦1 ), π(π¦2 )) β πΌ0 . Proof. (β) Suppose that π satisfies πβ² over S. The proof is by induction on the structure of πβ² . If πβ² = π (π₯π , π¦1 , π¦2 ), then by (2) π(π₯π ) β π΅π implies (π(π¦1 ), π(π¦2 )) β πΌπ for 7 π = 0, 1. By construction, π = π₯π , and if π does not satisfy π, then π(π₯π ) β π΅0 (as π matches π), and then (π(π¦1 ), π(π¦2 )) β πΌ0 . If πβ² = πβ²1 β§ πβ²2 , then π satisfies both πβ²1 and πβ²2 over S. By construction, π = π1 β§ π2 . If π does not satisfy π, then either π does not satisfy π1 or π does not satisfy π2 ; say without loss of generality that π does not satisfy π1 . By the induction hypothesis, (π(π¦1 ), π(π¦2 )) β πΌ0 . If πβ² (x, π¦1 , π¦2 ) = (βπ§)(πβ²1 (x, π¦1 , π§) β§ πβ²2 (x, π§, π¦2 )), then there exists a point π β πΆ such that extending π by π(π§) = π, π satisfies both πβ²1 (x, π¦1 , π§) and πβ²2 (x, π§, π¦2 ) over S. By construction, π = π1 β¨ π2 . If π does not satisfy π, then π does not satisfy π1 and π does not satisfy π2 . Then, by the induction hypothesis, (π(π¦1 ), π(π§)) β πΌ0 and (π(π§), π(π¦2 )) β πΌ0 , so that by transitivity, (π(π¦1 ), π(π¦2 )) β πΌ0 . (β) Suppose that π does not satisfy πβ² over S. We show that π does not satisfy π but (π(π¦1 ), π(π¦2 )) β / πΌ0 . The proof is by induction on the structure of πβ² . If πβ² = π (π₯π , π¦1 , π¦2 ), then by (2) the only possibility for π to not satisfy πβ² is when π(π₯π ) β π΅0 and (π(π¦1 ), π(π¦2 )) β / πΌ0 (note that in fact, π(π₯π ) β π΅1 implies (π(π¦1 ), π(π¦2 )) β πΌ1 holds trivially because πΌ1 = 1β ). By construction, π = π₯π . As π matches π, by definition π (π₯π ) = 0, that is, π does not satisfy π, and we are done. If πβ² = πβ²1 β§ πβ²2 , then either π does not satisfy πβ²1 over S, or π does not satisfy πβ²2 over S; say without loss of generality that π does not satisfy πβ²1 over S. By construction, π = π1 β§ π2 , so by the induction hypothesis, π does not satisfy π1 but (π(π¦1 ), π(π¦2 )) β / πΌ0 . It follows that π does not satisfy π, and we are done. Finally, let πβ² (x, π¦1 , π¦2 ) = (βπ§)(πβ²1 (x, π¦1 , π§) β§ πβ²2 (x, π§, π¦2 )); recall that in this case, by construction, π = π1 β¨ π2 . If π does not satisfy πβ² over S, then there does not exist a point π β πΆ such that extending π by π(π§) = π, π satisfies both πβ²1 (x, π¦1 , π§) and πβ²2 (x, π§, π¦2 ) over S. In particular, the extension π(π§) = π(π¦1 ) of π satisfies πβ²1 (x, π¦1 , π§) over S, hence it does not satisfy πβ²2 (x, π§, π¦2 ) over S; here, the induction hypothesis gives that π does not satisfy π2 , but (π(π§), π(π¦2 )) = (π(π¦1 ), π(π¦2 )) β / πΌ0 . Similarly, the extension π(π§) = π(π¦2 ) of π satisfies πβ²2 (x, π§, π¦2 ) over S, hence it does not satisfy πβ²1 (x, π¦1 , π§) over S, and the induction hypothesis gives that π does not satisfy π1 , but (π(π¦1 ), π(π§)) = (π(π¦1 ), π(π¦2 )) β / πΌ0 . Hence, π does not satisfy π, but (π(π¦1 ), π(π§)) = (π(π¦1 ), π(π¦2 )) β / πΌ0 , and we are done. β‘ the Boolean assignments satisfying π contain the Boolean assignments satisfying π if and only if, the sorted assignments satisfying π β² over S contain the sorted assignments satisfying πβ² over S. This completes the first part of the proof. The second reduction, from the entailment problem on S to PPCON(A), works as follows. Let π be a sorted pp-formula on S over variables π₯1 , . . . , π₯π and π¦1 , . . . , π¦π of sort π΅ and πΆ respectively. We construct a pp-formula πβ² on A, as follows. For each variable π§ in π, introduce a fresh variable π§ β² = (π§1 , π§2 ); π§ β² is existentially quantified in πβ² if and only if π§ is existentially quantified in πβ² . If π contains the constraint π₯π = π₯π for some π, π β {1, . . . , π}, then πβ² contains the conjunct π½(π₯β²π , π₯β²π ); if π contains the constraint π¦π = π¦π for some π, π β {1, . . . , π}, then πβ² contains the conjunct πΎ(π¦πβ² , π¦πβ² ); if π contains the constraint π (π₯π , π¦π , π¦π ), then πβ² contains the conjunct (βπ€1β² )(βπ€2β² )(π½(π€1β² , π₯β²π ) β§ π½(π€2β² , π₯β²π )β§ πΎ(π€1β² , π¦πβ² ) β§ πΎ(π€2β² , π¦πβ² ) β§ πΌ(π€1β² , π€2β² )), (4) where π€1β² and π€2β² are fresh variables. The reduction is now, given an instance (π, π) of the entailment problem on S, construct the instance (πβ² , π β² ) of PPCON(A). The construction is a sequence of local substitutions, hence it is feasible in logspace. We now prove that the reduction is correct. The key step is to establish a correspondence between the satisfying assignments of π over S and πβ² over A. Let π be a sorted assignment of variables π₯1 , . . . , π₯π in π΅ and π¦1 , . . . , π¦π in πΆ, and let π be an assignment of variables β² in π΄ = π΅ × πΆ. Say that π and π π₯β²1 , . . . , π₯β²π , π¦1β² , . . . , π¦π match if: π (π₯π ) = π implies π(π₯β²π ) = (π, β ), and π (π¦π ) = π implies π(π¦πβ² ) = (β , π). For sake of notation, say that the free β² variables in π are π₯β²1 , . . . , π₯β²πβ² , π¦1β² , . . . , π¦π β². Claim 4.3 Let π be an assignment of β² π₯β²1 , . . . , π₯β²πβ² , π¦1β² , . . . , π¦π β² in π΅ × πΆ, matching the sorted assignment π of π₯1 , . . . , π₯πβ² , π¦1 , . . . , π¦πβ² . Then, π satisfies πβ² over A if and only if π satisfies π over S. Proof. (β) Suppose that π satisfies πβ² over A, and let π β² be an extension of π to the quantified variables of πβ² that satisfies each conjunct in πβ² . Let π β² be the sorted assignment of the variables π₯1 , . . . , π₯π , π¦1 , . . . , π¦π in π defined by π β² (π₯π ) = π if and only if π β² (π₯β²π ) = (π, β ), and π β² (π¦π ) = π if and only if π β² (π¦πβ² ) = (β , π); by definition, the restriction of π β² to the free variables of π, that is, π , matches π. We prove that π satisfies π over S, by checking that π β² satisfies each conjunct of π over S. If π β² satisfies the conjunct π½(π₯β²π , π₯β²π ) in πβ² , that is, β² β² π (π₯π ) = (π, β ) and π β² (π₯β²π ) = (π, β ) for some π β π΅, then π β² satisfies the counterpart π₯π = π₯π in π because π β² (π₯π ) = π β² (π₯π ) = π by definition of π β² . The case of conjuncts of the Let π and π be Boolean formulas, and let πβ² and π β² be the sorted pp-formulas given by the reduction. Note that Boolean assignments of π₯1 , . . . , π₯π determine a partition of sorted assignments of π₯1 , . . . , π₯π , π¦1 , π¦2 : for, sorted assignments π and π β² are in the same block of the partition if and only if they match the same Boolean assignment π . By Claim 4.2, a sorted assignment π satisfies πβ² (respectively, π β² ) if and only if its matching Boolean assignment satisfies π (respectively, π), or (π(π¦1 ), π(π¦2 )) β πΌ0 ; therefore, 8 form πΎ(π¦πβ² , π¦πβ² ) in πβ² is similar. If π β² satisfies a conjunct of the form in (4) in πβ² , then by direct inspection, the following holds: π β² (π₯β²π ) = (π, β ), π β² (π€1β² ) = (π, β ), and π β² (π€2β² ) = (π, β ) for some π β π΅; π β² (π¦πβ² ) = (β , π1 ) and π β² (π€1β² ) = (β , π1 ) for some π1 β πΆ; π β² (π¦πβ² ) = (β , π2 ) and π β² (π€2β² ) = (β , π2 ) for some π2 β πΆ; and, ((π, π1 ), (π, π2 )) β πΌ. By (1), this implies that, π β π΅π implies (π1 , π2 ) β πΌπ for π = 1, 2. But then, by (2), π β² satisfies the counterpart in π of the conjunct under consideration, π (π₯π , π¦π , π¦π ), because by definition, π β² (π₯π ) = π, π β² (π¦π ) = π1 , and π β² (π¦π ) = π2 . (β) Conversely, suppose that the sorted assignment π satisfies π over S. Let π β² be an extension of π to the quantified variables of π that satisfies each constraint in π. Let π β² β² be the assignment of the variables π₯β²1 , . . . , π₯β²π , π¦1β² , . . . , π¦π β² β² in π onto π΅ × πΆ defined by π (π₯π ) = (π, ππ ) if and only if π β² (π₯π ) = π and π β² (π¦πβ² ) = (ππ , π) if and only if π β² (π¦π ) = π, where ππ and ππ are arbitrary fixed points in π΅ and πΆ respectively. By definition, the restriction of π β² to the free variables of πβ² , call it π, matches π . We prove that π satisfies πβ² over A, by checking that π β² satisfies each conjunct of πβ² over A. If π β² satisfies the constraint π₯π = π₯π in π, that is, π β² (π₯π ) = π β² (π₯π ) = π for some π β π΅, then π β² satisfies the counterpart π½(π₯β²π , π₯β²π ) in πβ² , because π β² (π₯β²π ) = (π, β ) and π β² (π₯β²π ) = (π, β ) by definition of π β² . The case of constraints of the form π¦π = π¦π in π is similar. If π β² satisfies a constraint of the form π (π₯π , π¦π , π¦π ) in π, then by (2), π (π₯π ) = π β π΅π implies (π β² (π¦π ), π β² (π¦π )) = (π1 , π2 ) β πΌπ for π = 1, 2. In this case, a conjunct π of the form in (4) occurs in πβ² . We extend π β² to the existentially quantified variables π€1β² and π€2β² in (4) by letting π β² (π€1β² ) = (π, π1 ) and π β² (π€2β² ) = (π, π2 ). By direct inspection, this extension of π β² satisfies π. β‘ We describe a reduction from the problem of deciding whether two monotone Boolean formulas are isomorphic to the problem PPISO(A). Along the lines of part (π), the reduction proceeds in two stages. Say that two sorted ppformulas are isomorphic (on S) if and only if they have an isomorphism on S that fixes the sorts, that is, sends variables of sort π΅ (respectively, πΆ) to variables of sort π΅ (respectively, πΆ). The isomorphism problem on S is to decide, given two sorted pp-formulas π and π, whether or not π and π are isomorphic over S. First, we reduce from the monotone Boolean formulas isomorphism problem to PPISO(S). Let (π, π) be a pair of monotone Boolean formulas over variables π₯1 , . . . , π₯π , and let (πβ² , π β² ) be the pair of sorted pp-formulas on S, over variables π₯1 , . . . , π₯π of sort π΅ and variables π¦1 , π¦2 of sort πΆ, computed in part (π). The following claim shows that the reduction is correct. Claim 4.4 π and π are isomorphic if and only if πβ² and π β² are isomorphic. Proof. (β) Let π be an isomorphism of π and π, and let π β² (π₯π ) = π(π₯π ) and π β² (π¦π ) = π¦π for π = 1, . . . , π and π = 1, 2. We claim that π β² is an isomorphism of πβ² and π β² on S (clearly, π β² fixes the sorts). Let π be a sorted assignment of π₯1 , . . . , π₯π , π¦1 , π¦2 , and let π be the Boolean assignment matching π. First suppose that π satisfies πβ² on S. By Claim 4.2, either (π(π¦1 ), π(π¦2 )) β πΌ0 , or π satisfies π. In the former case, (π β π β² (π¦1 ), π β π β² (π¦2 )) = (π(π¦1 ), π(π¦2 )) β πΌ0 , and π β π β² satisfies π β² on S by Claim 4.2. In the latter case, by hypothesis, π β π satisfies π. Since π β π β² and π β π match, by Claim 4.2, π β π β² satisfies π β² on S. Conversely, suppose that π does not satisfy πβ² on S. By Claim 4.2, π does not satisfy π and (π(π¦1 ), π(π¦2 )) β / πΌ0 . By hypothesis, π β π does not satisfy π. As π β π β² and π β π match, by Claim 4.2, π β π β² does not satisfy π β² on S. (β) Let π β² witness that πβ² and π β² are isomorphic on S, and let π(π₯π ) = π β² (π₯π ) for π = 1, . . . , π. Let π be a Boolean assignment of π₯1 , . . . , π₯π , and let π be a sorted assignment of π₯1 , . . . , π₯π , π¦1 , π¦2 matching π . Without loss of generality, assume that (π(π¦1 ), π(π¦2 )) β / πΌ0 ; for otherwise, pick two points π1 and π2 such that (π1 , π2 ) β / πΌ0 (such points exist because πΌ0 < 1β ), settle π(π¦1 ) = π1 and π(π¦2 ) = π1 , and note that π still matches π . First suppose that π satisfies π. By Claim 4.2, π satisfies πβ² on S. By hypothesis, π β π β² satisfies πβ² on S. As (π(π¦1 ), π(π¦2 )) β / πΌ0 , and π βπ matches π β π β² , we conclude by Claim 4.2 that π β π satisfies π. Conversely, suppose that π does not satisfy π. By Claim 4.2, π does not satisfy πβ² on S, and by hypothesis, π β π β² does not satisfy π β² on S; similarly, π β π does not satisfy π. β‘ This completes the first part of the proof. Second, we reduce from PPISO(S) to PPISO(A). We import a technical lemma from [6]. Let π be a set, let {π1 , . . . , ππ } be a partition of π, and let π be a permuta- Now, let π and π be sorted pp-formulas on S, and let πβ² and π β² be the pp-formulas on A given by the reduction. Note that sorted assignments of π₯1 , . . . , π₯πβ² , π¦1 , . . . , π¦πβ² determine a partition of the assignβ² ments of π₯β²1 , . . . , π₯β²πβ² , π¦1β² , . . . , π¦π β² in π΅ × πΆ: for, assignments π and π β² are in the same block of the partition if and only if they match the same sorted assignment π . By Claim 4.3, the satisfying assignments of πβ² (respectively, π β² ) are exactly those in blocks that match satisfying assignments of π (respectively, π). Hence, the sorted assignments satisfying π over S are contained in the sorted assignments satisfying π over S, if and only if the assignments satisfying π over A are contained in the assignments satisfying π over A. This completes the second part of the proof. We conclude that PPCON(A) is coNP-hard, and the proof of statement (π) is complete. (ππ) We now prove that PPISO(A) is MFISO-hard. Along the lines of the first paragraph of part (π), noticing that PPISO(A) is in PPISO(πΈ), this implies that PPEQ(B) is MFISO-hard. 9 tion of π. We say that π fixes ππ if {π(π₯) β£ π₯ β ππ } = ππ . [6] S. Bova, H. Chen, and M. Valeriote. On the Expression Complexity of Equivalence and Isomorphism of Primitive Positive Formulas. In A. A. Bulatov, M. Grohe, P. G. Kolaitis, and A. Krokhin, editors, The Constraint Satisfaction Problem: Complexity and Approximability, number 09441 in Dagstuhl Seminar Proceedings, Dagstuhl, Germany, 2010. Schloss Dagstuhl - Leibniz-Zentrum fuer Informatik, Germany. [7] A. Bulatov. Tractable conservative constraint satisfaction problems. In Proceedings of 18th IEEE Symposium on Logic in Computer Science (LICS β03), pages 321β330, 2003. [8] A. Bulatov. H-Coloring dichotomy revisited. 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In Proceedings of the 18th Annual ACM Symposium on Theory of Computing (STOC), 1982. Lemma 4.5 [6] Let π be a signature, let B be a relational structure over π, and let π and π be pp-formulas on π. For each π β₯ 2, it is possible to construct in logspace, given a partition {π1 , . . . , ππ } of the set π of free variables of π and π, two pp-formulas πβ² and π β² on π satisfying: πβ² and π β² are isomorphic if and only if π and π have an isomorphism that fixes π1 , . . . , ππ . Let (π, π) be a pair of sorted pp-formulas on S over variables π₯1 , . . . , π₯π of sort π΅, and π¦1 , . . . , π¦π of sort πΆ, let (πβ² , π β² ) be the pair of pp-formulas on A over variβ² ables π₯β²1 , . . . , π₯β²π , π¦1β² , . . . , π¦π computed (in logspace) in part β²β² β²β² (π), and let (π , π ) be the pair of pp-formulas on A computed (in logspace) by Lemma 4.5 given the partition β² {{π₯β²1 , . . . , π₯β²π }, {π¦1β² , . . . , π¦π }}. The following claim shows that the reduction is correct. Say that a sectioning for a formula is simple if it has exactly as many sections as the free variables of the formula, and one type. Over any relational structure, two sorted ppformulas are isomorphic with respect to the simple sectionings, if and only if they have an isomorphism; in fact, any bijection respects the simple sectionings. Claim 4.6 π and π are isomorphic on S if and only if πβ²β² and π β²β² are isomorphic on A with respect to the simple sectionings. Proof. See Appendix C. β‘ This completes the second part of the proof. We conclude that PPISO(A) is MFISO-hard, and the proof of statement (ππ) is complete. β‘ References [1] A. Atserias. Conjunctive Query Evaluation by Search-Tree Revisited. Theoretical Computer Science, 371(3):155β168, 2007. [2] L. Barto and M. Kozik. Constraint satisfaction problems of bounded width. In Proceedings of the 50th Annual IEEE Symposium on Foundations of Computer Science, FOCSβ09, pages 595β603, 2009. [3] J. Berman, P. Idziak, P. MarkovicΜ, R. McKenzie, M. Valeriote, and R. Willard. Varieties with Few Subalgebras of Powers. Transactions of the American Mathematical Society, 362(3):1445β1473, 2010. [4] O. Beyersdorff, A. Meiera, M. Thomas, and H. Vollmer. The Complexity of Propositional Implication. Information Processing Letters, 109(18):1071β1077, 2009. [5] V. Bodnarchuk, L. Kaluzhnin, V. Kotov, and B. Romov. Galois Theory for Post Algebras. I, II. Cybernetics, 5:243β252, 531β539, 1969. 10 Appendix A apply the same reduction to the original pair of formulas (π, πβ² ) to obtain a new pair (π, π β² ), and associate the sectioning of π with π, and that of πβ² with π β² . β‘ Proof of Proposition 2.2 Proof. Let π, πβ² , B, (π , π‘), (π β² , π‘β² ) be an instance of PPISO with signature π. Fix a π β₯ 1 such that there is an injective mapping π : π΅ β {0, 1}π . Define π Λ to be the signature with the same relation symbols as π, but where the arity of π β π Λ is ππ, where π is the arity of π β π. We create a boolean relational structure BΜ by defining π BΜ = {(π(π1 ), . . . , π(ππ )) β£ (π1 , . . . , ππ ) β π B } for all relation symbols π . The new instance is created as follows. The new pair of formulas (π, π β² ) is created from the old pair (π, πβ² ) by replacing each variable π£ with a sequence of variables π£ 1 , . . . , π£ π . The free variables of π and π β² is thus π β² = {π₯π β£ π₯ β π, π β [π]} where π denotes the free variables of π and πβ² . The sectioning of π is given by (Λ π , π‘Λ) : π β² β π × (π × [π]), where π Λ(π₯π ) = π (π₯) and π π‘Λ(π₯ ) = (π‘(π₯), π), and the sectioning of π β² is given from that of πβ² in an analogous way. It is straightforward to verify that this reduction is correct. β‘ B C Proof of Claim 4.6 Proof. (β) Let π be an isomorphism of π and π, and let π β² (π₯β²π ) = π₯β²π if and only if π(π₯π ) = π₯π for π, π β {1, . . . , π}, and π β² (π¦πβ² ) = π¦πβ² if and only if π(π¦π ) = π¦π for π, π β {1, . . . , π}. We check that π β² is an isomorphism of πβ² and β² π β² . Let π be an assignment of π₯β²1 , . . . , π₯β²π , π¦1β² , . . . , π¦π , and let π be the sorted assignment of π₯1 , . . . , π₯π , π¦1 , . . . , π¦π matching π. Now, π satisfies πβ² on A if and only if π satisfies π on S (by Claim 4.3), if and only if π β π satisfies π on S (π is an isomorphism), if and only if π β π β² satisfies πβ² on A (by Claim 4.3, since π β π and π β π β² match). Therefore, π β² is an isomorphism of πβ² and π β² . Moreover, π β² fixes β² }. Hence, by Lemma 4.5, πβ²β² {π₯β²1 , . . . , π₯β²π } and {π¦1β² , . . . , π¦π β²β² and π are isomorphic on A. (β) Let π β²β² be an isomorphism of πβ²β² and π β²β² on A. By Lemma 4.5, there exists an isomorphism of πβ² and π β² on β² A that fixes {π₯β²1 , . . . , π₯β²π } and {π¦1β² , . . . , π¦π }. Let π(π₯π ) = β² β² β² π₯π if and only if π (π₯π ) = π₯π for π, π β {1, . . . , π}, and π(π¦π ) = π¦π if and only if π β² (π¦πβ² ) = π¦πβ² for π, π β {1, . . . , π}. Let π be a sorted assignment of π₯1 , . . . , π₯π , π¦1 , . . . , π¦π and β² matching let π be an assignment of π₯β²1 , . . . , π₯β²π , π¦1β² , . . . , π¦π π . It is easy to check that, by Claim 4.3, π satisfies π on S if and only if π β π satisfies π on S. Hence, π and π are isomorphic on S. β‘ Proof of Theorem 2.6 Proof. We first treat powers; suppose πΉ = πΈπ . Consider a problem PPEQ(B) β PPEQ(πΈB ), and let π denote the signature of B. Let π β² be the signature that has the same symbols as π, but where the arity of a symbol of π β π β² is ππ, where π is the arity of π β π. Deβ² fine Bβ² to be the structure whose relation π B contains the tuple (π11 , . . . , ππ1 , . . . , π1π , . . . , πππ ) if and only if the tuple ((π11 , . . . , ππ1 ), . . . , (π1π , . . . , πππ )) belongs to the relation π B . Clearly, we have PPEQ(Bβ² ) β PPEQ(πΈ). To reduce an instance (π, πβ² ) of PPEQ(B) to PPEQ(Bβ² ), we simply replace, in each of π, πβ² , each variable π£ with a sequence of π variables π£ 1 , . . . , π£ π . It is straightforward to verify that the original instance (π, πβ² ) was a yes instance if and only if the new formulas are. The same reduction applies to PPCON(β ). For the problem PPISO(β ), we apply the same transformation to the formulas, and define new sectionings as follows. Let (π , π‘), (π β² , π‘β² ) : π β π × π be the original sectionings. The new sectionings are denoted by (π 1 , π‘1 ), (π β²1 , π‘β²1 ) : π β π × π1 where π1 = π × [π] and the mappings are defined by π 1 (π£ π ) = π (π£), π β²1 (π£ π ) = π β² (π£), π‘1 (π£ π ) = (π‘(π£), π), and π‘β²1 (π£ π ) = (π‘(π£), π). It is straightforward to verify that the original formulas are isomorphic with respect to (π , π‘), (π β² , π‘β² ) if and only if the new formulas are isomorphic with respect to (π 1 , π‘1 ), (π β²1 , π‘β²1 ). In the case that πΉ is a subalgebra or homomorphic image of πΈ, the result is proved in [6, Proposition 4] for PPEQ(β ), and from the argumentation there it is clear that exactly the same reduction works for PPCON(β ). For PPISO(β ), we 11