$\Pi_{2}^{P}$ vs PSpace Dichotomy for the Quantified Constraint Satisfaction Problem

Dmitriy Zhuk The author is funded by the European Union (ERC, POCOCOP, 101071674). Views and opinions expressed are however those of the author(s) only and do not necessarily reflect those of the European Union or the European Research Council Executive Agency. Neither the European Union nor the granting authority can be held responsible for them.

Abstract

The Quantified Constraint Satisfaction Problem is the problem of evaluating a sentence with both quantifiers, over relations from some constraint language, with conjunction as the only connective. We show that for any constraint language on a finite domain the Quantified Constraint Satisfaction Problem is either in $\Pi_{2}^{P}$ , or PSpace-complete. Additionally, we build a constraint language on a 6-element domain such that the Quantified Constraint Satisfaction Problem over this language is $\Pi_{2}^{P}$ -complete.

1 Introduction

The Quantified Constraint Satisfaction Problem $\operatorname{QCSP}(\Gamma)$ is the generalization of the Constraint Satisfaction Problem $\operatorname{CSP}(\Gamma)$ which, given the latter in its logical form, augments its native existential quantification with universal quantification. That is, $\operatorname{QCSP}(\Gamma)$ is the problem to evaluate a sentence of the form $\forall x_{1}\exists y_{1}\ldots\forall x_{n}\exists y_{n}\ \Phi$ , where $\Phi$ is a conjunction of relations from the constraint language $\Gamma$ , all over the same finite domain $A$ . Since the resolution of the Feder-Vardi “Dichotomy” Conjecture, classifying the complexity of $\operatorname{CSP}(\Gamma)$ , for all finite $\Gamma$ , between P and NP-complete [7, 8, 17, 19], a desire has been building for a classification for $\operatorname{QCSP}(\Gamma)$ . Indeed, since the classification of the Valued CSPs was reduced to that for CSPs [14], the QCSP remains the last of the older variants of the CSP to have been systematically studied but not classified. More recently, other interesting open classification questions have appeared such as that for Promise CSPs [5] and finitely-bounded, homogeneous infinite-domain CSPs [3].

1.1 Complexity of the QCSP

While $\operatorname{CSP}(\Gamma)$ remains in NP for any finite $\Gamma$ , $\operatorname{QCSP}(\Gamma)$ can be PSpace-complete, as witnessed by Quantified 3-Satisfiability or Quantified Graph 3-Colouring (see [4]). It is well-known that the complexity classification for QCSPs embeds the classification for CSPs: if $\Gamma+1$ is $\Gamma$ with the addition of a new isolated element not appearing in any relations, then $\operatorname{CSP}(\Gamma)$ and $\operatorname{QCSP}(\Gamma+1)$ are polynomially equivalent. Thus, and similarly to the Valued CSPs, the CSP classification will play a part in the QCSP classification. For a long time the complexities P, NP-complete, and PSpace-complete were the only complexity classes that could be achieved by $\operatorname{QCSP}(\Gamma)$ [4, 10, 9, 12, 15]. Nevertheless, in [22, 23] a constraint language $\Gamma$ on a 3-element domain was discovered such that $\operatorname{QCSP}(\Gamma)$ is coNP-complete. Combining this language with an NP-complete language the authors also built a DP-complete constraint language on a 4-element domain and a $\Theta_{2}^{P}$ -complete language on a 10-element domain [22, 23]. Discovering these exotic complexity classes ruined hope to obtain a simple and complete classification of the complexity of the QCSP for all constraint languages on a finite domain. On the other hand, the possibility to express those complexity classes by fixing a constraint language makes the QCSP a powerful tool for studying complexity classes between P and PSpace. Finding a concrete border between complexity classes in terms of constraint languages may shed some light on the fundamental differences between them, and may bring us closer to understanding why P and PSpace are different (if they are).

The exotic complexity classes appeared only on domains of size at least 4, while on a domain of size 2 we have a complete classification between P and PSpace-complete, and on a domain of size 3 we have a partial classification between P, NP-complete, coNP-complete, and PSpace-complete.

Theorem 1 ([16]).

Suppose $\Gamma$ is a constraint language on $\{0,1\}$ . Then $\operatorname{QCSP}(\Gamma)$ is in P if $\operatorname{CSP}(\Gamma\cup\{x=0,x=1\})$ is in P, $\operatorname{QCSP}(\Gamma)$ is PSpace-complete otherwise.

Theorem 2 ([22, 23]).

Suppose $\Gamma\supseteq\{x=a\mid a\in A\}$ is a constraint language on $\{0,1,2\}$ . Then $\operatorname{QCSP}(\Gamma)$ is either in P, or NP-complete, or coNP-complete, or PSpace-complete.

The statement proved in [22, 23] is stronger than Theorem 2 as the authors provide necessary and sufficient conditions for the $\operatorname{QCSP}(\Gamma)$ to be in each of these classes. Notice that for the QCSP we do not know a simple trick that allows us to find an equivalent constraint language with all constant relations $\{x=a\mid a\in A\}$ for a constraint language without. Recall that for the usual CSP we first consider the core of the language and then safely add all the constant relations to it [13, 6]. For the QCSP reducing the domain is not an option as the universal quantifier lives on the whole domain. That is why, Theorem 2, has been proved only for constraint languages with all constant relations, and a complete classification for all constraint languages on a 3-element domain is wide open.

1.2 Reduction to CSP

It is natural to try to reduce the QCSP to its older brother CSP. In fact, any QCSP instance $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\;\Psi$ can be viewed as a CSP instance of an exponential size. If a QCSP-sentence holds, then there exists a winning strategy for the Existential Player (EP) defined by Skolem functions, i.e., $y_{i}=f_{i}(x_{1},\dots,x_{i})$ . We encode every value of $f_{i}(a_{1},\dots,a_{i})$ by a new variable $y_{i}^{a_{1},\dots,a_{i}}$ , and for any play of the Universal Player (UP) we list all the constraints that have to be satisfied (see Section 5.3 for more details).

Clearly, this procedure gives us nothing algorithmically, because the obtained CSP instance is of exponential size. Nevertheless, we might ask whether it is necessary to look at the whole instance to learn that it does not hold, which can be formulated as follows. We say that the UP wins on $S\subseteq A^{n}$ in $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\;\Psi$ if the instance

\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}((x_{1},\dots,x_{n})\in S\rightarrow\Psi)

does not hold.

Question 1.

For a No-instance of $\operatorname{QCSP}(\Gamma)$ with $n$ universal variables, what is the minimal $S\subseteq A^{n}$ such that the UP wins on $S$ ?

In this paper we answer this fundamental question by showing that unless the problem is PSpace-hard, the set $S$ can be chosen of polynomial size. Notice that for the PSpace-hard case we should not expect $S$ to be of non-exponential size, as it would send our problem to some class below PSpace.

It would be even better if the set $S$ , on which the UP wins, could be fixed for all No-instances or could be calculated efficiently. We can ask the following question.

Question 2.

What is the minimal $S\subseteq A^{n}$ such that for any No-instance of $\operatorname{QCSP}(\Gamma)$ with $n$ universal variables the UP wins on $S$ ?

If $S$ can always be chosen of polynomial-size and can be computed efficiently, then $\operatorname{QCSP}(\Gamma)$ immediately goes to the complexity class NP, as it is reduced to a polynomial-size CSP instance that can be efficiently computed. Surprisingly, all the problems $\operatorname{QCSP}(\Gamma)$ known to be in NP by 2018 satisfy the above property [11, 12, 15]. In fact, as it is shown in [20], for all constraint languages whose polymorphisms satisfy the Polynomially Generated Powers (PGP) Property, the set $S$ can be chosen to be very simple: there exists $k$ such that the UP wins in any No-instance on the set of all tuples having at most $k$ switches, where a switch in a tuple $(a_{1},\dots,a_{n})$ is a pair $(a_{i},a_{i+1})$ such that $a_{i}\neq a_{i+1}$ . Moreover, as it was shown in [18], if polymorphisms do not satisfy the PGP property, they satisfy the Exponential Generated Powers (EGP) Property, which automatically implies that such a polynomial-size $S$ cannot exist (at least if the number of existential variables is not limited).

Surprisingly, in [22, 23] two constraint languages on a 3-element domain were discovered such that the QCSP over these languages is solvable in polynomial time, but they do not satisfy the PGP property and, therefore, we cannot fix a polynomial-size $S$ . Nevertheless, for every instance we can efficiently calculate a polynomial-size $S$ such that if the UP can win, the UP wins on $S$ . We can formulate the following open question.

Question 3.

Suppose $\operatorname{QCSP}(\Gamma)$ is in NP. Is it true that for any instance of $\operatorname{QCSP}(\Gamma)$ with $n$ universal variables there exists a polynomial-time computable set $S\subseteq A^{n}$ such that the UP can win if and only if the UP wins on $S$ ?

2 Main Results

2.1 $\Pi_{2}^{P}$ vs PSpace Dichotomy

The main result of this paper comes from Question 1 from the introduction. We show that if $\operatorname{QCSP}(\Gamma)$ is not PSpace-complete and the UP has a winning strategy in a concrete QCSP instance, then this winning strategy can be chosen to be rather simple. We cannot expect the winning strategy for the UP to be polynomial-time computable because this would imply that $\operatorname{QCSP}(\Gamma)$ is in NP, and we know that $\operatorname{QCSP}(\Gamma)$ can be coNP-complete [23]. Nevertheless, as we show in the next theorem, the UP wins in any No-instance on a set $S$ of polynomial size, that is, we can restrict the UP to polynomially many possible moves and he still wins.

Theorem 3.

Suppose $\Gamma$ is a constraint language on a finite set $A$ , $\operatorname{QCSP}(\Gamma)$ is not PSpace-hard. Then for any No-instance $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\Psi$ of $\operatorname{QCSP}(\Gamma)$ there exists $S\subseteq A^{n}$ with $|S|\leqslant|A|^{2}\cdot(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ such that

\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}((x_{1},\dots,x_{n})\in S\rightarrow\Psi)

does not hold.

In other words, the above theorem states that unless $\operatorname{QCSP}(\Gamma)$ is PSpace-hard, for any No-instance the UP wins on a set $S$ of polynomial-size (notice that the domain $A$ is fixed). If the polynomial-size set $S$ is fixed, then to confirm that the instance does not hold we need to check all the strategies of the EP defined on prefixes of the words (tuples) from $S$ , which is also a polynomial-size set. Thus, if $\operatorname{QCSP}(\Gamma)$ is not PSpace-hard, then to solve an instance $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\Psi$ we need to check that for all $S\subseteq A^{n}$ with $|S|\leqslant|A|^{2}\cdot(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ there exists a winning strategy for the EP for the restricted problem, which sends the problem to the complexity class $\Pi_{2}^{P}$ . In fact, $\Pi_{2}^{P}$ is the class of problems $\mathcal{U}$ that can be defined as

\mathcal{U}(Z)=\forall X^{\color[rgb]{0.6,0.6,0.6}\definecolor[named]{pgfstrokecolor}{rgb}{0.6,0.6,0.6}\pgfsys@color@gray@stroke{0.6}\pgfsys@color@gray@fill{0.6}|X|<p(|Z|)}\exists Y^{\color[rgb]{0.6,0.6,0.6}\definecolor[named]{pgfstrokecolor}{rgb}{0.6,0.6,0.6}\pgfsys@color@gray@stroke{0.6}\pgfsys@color@gray@fill{0.6}|Y|<q(|Z|)}\mathcal{V}(X,Y,Z),

for some $\mathcal{V}\in\mathrm{P}$ and some polynomials $p$ and $q$ . In our case the set $S$ plays the role of $X$ and the restricted Skolem functions play the role of $Y$ . Then, in $\mathcal{V}$ we need to check for every tuple from $S$ (play of the UP) that the corresponding strategy of the EP works, which is obviously computable in polynomial time. Thus, we have the following Dichotomy Theorem.

Theorem 4.

Suppose $\Gamma$ is a constraint language on a finite set. Then $\operatorname{QCSP}(\Gamma)$ is

•

PSpace-complete or
•

in $\Pi_{2}^{P}$ .

2.2 What is inside $\Pi_{2}^{P}$ ?

We show that the gap between Pspace and $\Pi_{2}^{P}$ cannot be enlarged, and there is a constraint language whose QCSP is $\Pi_{2}^{P}$ -complete.

Theorem 5.

There exists $\Gamma$ on a 6-element domain such that $\operatorname{QCSP}(\Gamma)$ is $\Pi_{2}^{P}$ -complete.

Thus, we already have 7 complexity classes that can be expressed as the QCSP for some constraint language: P, NP, coNP, DP, $\Theta_{2}^{P}$ , $\Pi_{2}^{P}$ , and PSpace. In Figure 1 we show all the complexity classes known to be expressible as the QCSP and inclusions between them, where the edge is solid if we know that there are no classes between them, and dotted otherwise.

Question 4.

Are there any other complexity classes up to polynomial reduction that can be expressed as $\operatorname{QCSP}(\Gamma)$ for some $\Gamma$ on a finite set?

In fact, we want to prove or disprove the following dichotomy claims:

Question 5.

Suppose $\Gamma$ is a constraint language on a finite set. Is it true that

1.

$\operatorname{QCSP}(\Gamma)$ is either $\Pi_{2}^{P}$ -hard, or in $\Theta_{2}^{P}$ ?
2.

$\operatorname{QCSP}(\Gamma)$ is either $\Theta_{2}^{P}$ -hard, or in $\mathrm{DP}$ ?
3.

$\operatorname{QCSP}(\Gamma)$ is either $\mathrm{DP}$ -hard, or in $\mathrm{NP}\cup\mathrm{coNP}$ ?
4.

$\operatorname{QCSP}(\Gamma)$ is either $\mathrm{NP}$ -hard, or in $\mathrm{coNP}$ ?
5.

$\operatorname{QCSP}(\Gamma)$ is either $\mathrm{coNP}$ -hard, or in $\mathrm{NP}$ ?
6.

$\operatorname{QCSP}(\Gamma)$ is either in $\mathrm{P}$ , or $\mathrm{NP}$ -hard, or $\mathrm{coNP}$ -hard?

Figure 1: Complexity classes expressible as

\operatorname{QCSP}(\Gamma)

for some

\Gamma

It is not hard to build an example showing that we cannot just move all universal quantifiers left to reduce $\operatorname{QCSP}(\Gamma)$ to a $\Pi_{2}^{P}$ -sentence even if $\operatorname{QCSP}(\Gamma)$ is in $\Pi_{2}^{P}$ . Nevertheless, it is still not clear whether a smarter polynomial reduction to a $\Pi_{2}$ -sentence over the same language exists. We denote the modification of $\operatorname{QCSP}(\Gamma)$ in which only $\Pi_{2}$ -sentences are allowed by $\Pi_{2}$ - $\operatorname{QCSP}(\Gamma)$ . Then this question can be formulated as follows.

Question 6.

Suppose $\Gamma$ is a constraint language on a finite set and $\operatorname{QCSP}(\Gamma)$ is $\Pi_{2}^{P}$ -complete. Is it true that $\Pi_{2}$ - $\operatorname{QCSP}(\Gamma)$ is $\Pi_{2}^{P}$ -complete?

One may also ask whether it is sufficient to consider only $\Pi_{2}$ -sentences for all complexity classes but PSpace.

Question 7.

Suppose $\Gamma$ is a constraint language on a finite domain and $\operatorname{QCSP}(\Gamma)$ is in $\Pi_{2}^{P}$ . Is it true that $\Pi_{2}$ - $\operatorname{QCSP}(\Gamma)$ is polynomially equivalent to $\operatorname{QCSP}(\Gamma)$ ?

A positive answer to this question would make a complete classification of the complexity of $\operatorname{QCSP}(\Gamma)$ for each $\Gamma$ much closer. Checking a $\Pi_{2}$ -sentence is equivalent to solving a Constraint Satisfaction Problem for every evaluation of universal variables, but if we need to check exponentially many of them, it does not give us an efficient algorithm. It is very similar to Question 1 from the introduction on whether the UP can win only playing strategies from a polynomial-size subset, but for the $\Pi_{2}$ -sentence the situation is much easier as the UP plays first and the EP just reacts.

Earlier Hubie Chen noticed [11] that in some cases it is sufficient to check only polynomially many evaluations to guarantee that the $\Pi_{2}$ -sentence holds, which implies that the problem is equivalent to the CSP and belongs to NP. Precisely, this reduction works for constraint languages satisfying the Polynomially Generated Powers (PGP) Property already mentioned in the introduction. These are languages such that all the tuples of $A^{n}$ can be generated from polynomially many tuples by applying polymorphisms of $\Gamma$ coordinate-wise. Notice that in the PGP case this polynomial set of evaluations can be chosen independently of the instance and can be calculated efficiently, as it is just the set of all tuples with at most $k$ switches [18]. This gives us a very simple polynomial reduction to CSP [11].

Sometimes a similar strategy works even if the polymorphisms of the constraint language do not satisfy the PGP property: two such constraint languages were presented in [23]. The polynomial algorithm for them works as follows. First, by solving many CSP instances it calculates the polynomial set of tuples (evaluations of the universal variables). Then, again by solving CSP instances, it checks that the quantifier-free part of the instance is satisfiable for every tuple (evaluation) it found. This gives us a Turing reduction to the CSP, and if the CSP is solvable in polynomial time, it gives us a polynomial algorithm. This idea completed the classification of the complexity of the QCSP for all constraint languages on a 3-element set containing all constant relations [23], and we hope that a generalization of this idea will lead to a complete classification of the complexity inside $\Pi_{2}^{P}$ .

2.3 PSpace-complete languages

The complexity of the CSP for a (finite) constraint language $\Gamma$ has a very simple characterization in terms of polymorphisms. Precisely, $\operatorname{CSP}(\Gamma)$ is solvable in polynomial time if $\Gamma$ admits a cyclic polymorphism, and it is NP-complete otherwise [7, 8, 17, 19]. It is also known that the complexity of $\operatorname{QCSP}(\Gamma)$ is determined by surjective polymorphisms of $\Gamma$ [4], but we are not aware of a nice characterization of $\Pi_{2}^{P}$ -membership in terms of polymorphisms, moreover polymorphisms do not play any role in this paper. Nevertheless, we have a nice characterization in terms of relations. It turned out all the PSpace-hard cases are similar in the sense that they can express certain relations giving us PSpace-hardness. We say that a constraint language $\Gamma$ q-defines a relation $R$ if there exists a quantified conjunctive formula over $\Gamma$ that defines the relation $R$ . Similarly, we say that $\Gamma$ q-defines a set $S$ of relations if it q-defines each relation from $S$ . In this case we also say that $S$ is q-definable over $\Gamma$ . It is an easy observation that $\operatorname{QCSP}(\Gamma_{1})$ can be (LOGSPACE) reduced to $\operatorname{QCSP}(\Gamma_{2})$ if $\Gamma_{2}$ q-defines $\Gamma_{1}$ [4].

To formulate the classification of all PSpace-complete languages, we introduce the notion of a mighty tuple. Suppose $k\geqslant 0$ , $m\geqslant 1$ , $Q\subseteq A^{|A|+k+m+2}$ , $B,C,D\subseteq A^{|A|+k+m+1}$ , $\Delta\subseteq A^{|A|+k}$ . The relation $Q$ can be viewed as a binary relation having three additional parameters $\mathbf{z}\in A^{|A|}$ , $\delta\in A^{k}$ , and $\alpha\in A^{m}$ . Similarly, $\Delta$ is a $k$ -ary relation with an additional parameter $\mathbf{z}\in A^{|A|}$ , $B,C,D$ are unary relations with additional parameters $\mathbf{z}\in A^{|A|}$ and $\delta\in A^{k}$ . By $\prescript{\mathbf{z}}{}{\Delta}$ , $\prescript{\mathbf{z}}{\delta}{Q}^{\alpha}$ , $\prescript{\mathbf{z}}{\delta}{D}$ , $\prescript{\mathbf{z}}{\delta}{B}$ , $\prescript{\mathbf{z}}{\delta}{C}$ we denote the respective $k$ -ary, binary, and unary relations where these parameters are fixed. For instance, we say that $(a,b)\in\prescript{\mathbf{z}}{\delta}{Q}^{\alpha}$ if $(\mathbf{z},\delta,\alpha,a,b)\in Q$ . Denote

	$\displaystyle\prescript{\mathbf{z}}{\delta}{Q}^{\forall}(y_{1},y_{2})$	$\displaystyle=\forall x\;\prescript{\mathbf{z}}{\delta}{Q}^{x,x,\dots,x}(y_{1},y_{2}),$		(1)
	$\displaystyle\prescript{\mathbf{z}}{\delta}{Q}^{\forall\forall}(y_{1},y_{2})$	$\displaystyle=\forall x_{1}\dots\forall x_{m}\;\prescript{\mathbf{z}}{\delta}{Q}^{x_{1},x_{2},\dots,x_{m}}(y_{1},y_{2}).$		(2)

A tuple $(Q,D,B,C,\Delta)$ is called a mighty tuple I if

1.

$\prescript{\mathbf{z}}{}{\Delta}\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ ;
2.

$\prescript{\mathbf{z}}{\delta}{B}$ , $\prescript{\mathbf{z}}{\delta}{C}$ , and $\prescript{\mathbf{z}}{\delta}{D}$ are nonempty for every $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ ;
3.

$\prescript{\mathbf{z}}{\delta}{Q}^{\alpha}$ is an equivalence relation on $\prescript{\mathbf{z}}{\delta}{D}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\alpha\in A^{m}$ ;
4.

$\prescript{\mathbf{z}}{\delta}{Q}^{\forall}=\prescript{\mathbf{z}}{\delta}{D}\times\prescript{\mathbf{z}}{\delta}{D}$ for every $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ ;
5.

$\prescript{\mathbf{z}}{\delta}{B}$ and $\prescript{\mathbf{z}}{\delta}{C}$ are equivalence classes of $\prescript{\mathbf{z}}{\delta}{Q}^{\forall\forall}$ ;
6.

there exists $\mathbf{z}\in A^{|A|}$ such that $\prescript{\mathbf{z}}{\delta}{B}\neq\prescript{\mathbf{z}}{\delta}{C}$ for every $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ .

The idea behind this definition is as follows. For fixed $\mathbf{z}$ and $\delta$ we have a parameterized binary relation $\prescript{\mathbf{z}}{\delta}{Q}$ , which is the full equivalence relation if $\alpha$ is a constant tuple and some equivalence relation otherwise. Relations $B$ and $C$ are just two equivalence classes that the EP has to connect by a complicated formula over $Q$ and the UP is trying to prevent this by choosing the parameters $\alpha$ .

In the next two theorems and later in the paper we assume that $\mathrm{PSpace}\neq\Pi_{2}^{P}$ .

Theorem 6.

Suppose $\Gamma$ is a constraint language on a finite set $A$ . Then the following conditions are equivalent:

1.

$\operatorname{QCSP}(\Gamma)$ is PSpace-complete;
2.

there exists a mighty tuple I q-definable over $\Gamma$ .

For constraint languages containing all constant relations we get an easier characterization.

Theorem 7.

Suppose $\Gamma\supseteq\{x=a\mid a\in A\}$ is a constraint language on a finite set $A$ . Then the following conditions are equivalent:

1.

$\operatorname{QCSP}(\Gamma)$ is PSpace-complete;
2.

there exist an equivalence relation $\sigma$ on $D\subseteq A$ and $B,C\subsetneq A$ such that $B\cup C=A$ and $\Gamma$ q-defines the relations $(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in B))$ and $(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in C))$ .

The above theorems show that all the hardness cases have the same nature. In the next section we provide a sketch of a proof of the PSpace-hardness in Theorem 7 for the case when $A=\{+,-,0,1\}$ , $D=\{+,-\}$ , $\sigma$ is the equality on $D$ , $B=\{+,-,1\}$ and $C=\{+,-,0\}$ , but one may check that the same proof works word for word for the arbitrary $A$ , $B$ , $C$ , $D$ , and $\sigma$ . Moreover, as we show in Section 7.2 a very similar reduction works for the general case in Theorem 6.

2.4 Idea of the proof

The proof of Theorem 3, which is the main result of the paper, comes from the exponential-size CSP instance we discussed in Section 1.2. Even though we cannot actually run any algorithm on it, one may ask whether it is solvable by local consistency methods. Surprisingly, unless $\operatorname{QCSP}(\Gamma)$ is PSpace-hard, a slight modification of the exponential-size CSP instance can be solved even by arc-consistency, which is the biggest discovery of this paper (see Theorem 22). The reader should not think that the trick is hidden in the modification, as we just replace the constraint language $\Gamma$ by the relation that is defined by the quantifier-free part of the instance (see Section 5.3).

This result does not give immediate consequences on the complexity of the QCSP as the instance is still of exponential size. Nevertheless, we prove that unless the $\operatorname{QCSP}(\Gamma)$ is PSpace-hard, the arc-consistency can show that the instance has no solutions only by looking at the polynomial part of it, and this is the second main discovery of the paper (see Corollary 33 and Theorem 18).

Notice that most of the previous results on the complexity of the QCSP were proved for constraint languages with all constant relations $x=a$ [11, 15, 22]. Here, we obtain results for the general case replacing constant relations by $|A|$ new variables that are universally quantified at the very beginning, and therefore can be viewed as external parameters. The price we pay for the general case is that all the relations and instances are parameterized by two additional parameters $\mathbf{z}$ and $\delta$ , which you already saw in the classification of all PSpace-complete languages.

2.5 Are there other complexity classes?

As we now know, $\operatorname{QCSP}(\Gamma)$ can be solvable in polynomial time, NP-complete, coNP-complete, DP-complete, $\Theta_{2}^{P}$ -complete, $\Pi_{2}^{P}$ -complete, and PSpace-complete. Knowing this, most of the readers probably expect infinitely many other complexity classes up to polynomial equivalence that can be expressed via the QCSP by fixing the constraint language. In our opinion it is highly possible that these 7 complexity classes are everything we can attain, as we mostly expected new classes between $\Pi_{2}^{P}$ and PSpace, and now we know that there are none. In this section we share our speculations on the question.

First, let us formulate what each of the classes means from the game theoretic point of view. A QCSP instance is a game between the Universal Player (he) and the Existential Player (she): he tries to make the quantifier-free part false, and she tries to make it true [1]. We say that a move of a player is trivial if the optimal move can be calculated in polynomial time. Then, those complexity classes just show how much they can interact with each other.

$\mathbf{P}$ : the play of both players is trivial;

$\mathbf{NP}$ : only the EP plays, the play of the UP is trivial;

$\mathbf{coNP}$ : only the UP plays, the play of the EP is trivial;

$\mathbf{DP}=\mathbf{NP}\wedge\mathbf{coNP}$ : each player plays their own game. Yes-instance: EP wins, UP loses;

$\mathbf{\Theta_{2}^{P}}=(\mathbf{NP}\vee\mathbf{coNP})\wedge\dots\wedge(\mathbf{NP}\vee\mathbf{coNP})$ : each player plays many games (no interaction), the result is a boolean combination of the results of those games;

$\mathbf{\Pi_{2}^{P}}$ : the UP plays first, then the EP plays;

$\mathbf{PSpace}$ : they play against each other (no restrictions).

We do not expect anything new between P and DP because we do not have anything between P and NP for the CSP. As the conjunction is given for free in the QCSP, whenever we can combine NP and coNP by anything else than conjunction, we expect to obtain a disjunction and, therefore, get $\Theta_{2}^{P}$ . Thus, the only place where we expect new classes is between $\Theta_{2}^{P}$ and $\Pi_{2}^{P}$ . Nevertheless, even here we cannot imagine an interaction which is weaker than in $\Pi_{2}^{P}$ , and we must have some interaction as the class $\Theta_{2}^{P}$ is the strongest class where we just combine the results of the independent games.

Notice that everything we wrote in this section is only speculation, and we need a real proof of all of the dichotomies formulated in Section 2.2.

2.6 Structure of the paper

The rest of the paper is organized as follows. In Section 3 we show a concrete constraint language on a 4-element domain whose complexity of the QCSP is PSpace-complete. Then, in Section 4 we present a constraint language on a 6-element domain whose QCSP is $\Pi_{2}^{P}$ -complete.

In Section 5 we provide necessary definitions and derive all the main results of the paper from the statements that are proved later. Here, we define 5 tuples of relations, called mighty tuples, such that the QCSP over any of them is PSpace-hard. In Section 6 we prove all the necessary statements under the assumption that a mighty tuple is not q-definable. Finally, in Section 7 we prove PSpace-hardness for a mighty tuple I, show the equivalence and reductions between mighty tuples.

3 The most general PSpace-hard constraint language

In this section for a concrete constraint language $\Gamma$ on a 4-element domain we show how to reduce the complement of Quantified-3-CNF to $\operatorname{QCSP}(\Gamma)$ and therefore prove PSpace-hardness. This constraint language is important because a similar reduction works for all the PSpace-hard cases.

Lemma 8.

Suppose $A=\{+,-,0,1\}$ , $R_{0}(y_{1},y_{2},x)=(y_{1},y_{2}\in\{+,-\})\wedge(x=0\rightarrow y_{1}=y_{2})$ , $R_{1}(y_{1},y_{2},x)=(y_{1},y_{2}\in\{+,-\})\wedge(x=1\rightarrow y_{1}=y_{2})$ , $\Gamma=\{R_{0},R_{1},\{+\},\{-\}\}$ . Then $\operatorname{QCSP}(\Gamma)$ is PSpace-hard.

This example may be viewed as the weakest constraint language whose QCSP is PSpace-hard. It can be seen from the definition that the UP should only play $x$ -variables (the last coordinates of $R_{0}$ and $R_{1}$ ) and the EP should only play $y$ -variables (the first two coordinates), and originally we did not see how they can interact with each other and, therefore, did not expect the QCSP to be PSpace-hard. Below we demonstrate on a concrete example how the EP can control the moves of the UP and therefore, interact in the area of the UP.

Sketch of the proof:.

We build a reduction from the complement of the Quantified-3-CNF. Let the sentence be

\neg(\exists x_{1}\forall x_{2}\exists x_{3}\;\;((x_{1}\vee\overline{x}_{2}\vee x_{3})\wedge(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3})).

Instead of formulas we draw graphs whose vertices are variables and edges are relations. $R_{0}(y_{1},y_{2},x)$ is drawn as a red edge from $y_{1}$ to $y_{2}$ labeled with $x$ . Similarly $R_{1}(y_{1},y_{2},x)$ is drawn as a blue edge from $y_{1}$ to $y_{2}$ labeled with $x$ . If a vertex has no name, then we assume that the variable is existentially quantified after all other variables are quantified. If the vertex is marked with $+$ or $-$ , then we assume that the corresponding variable is equal to $+$ or $-$ respectively. In Figure 2 you can see an example of a graph and the corresponding formula.

Figure 2: A graph for

\exists u_{1}\exists u_{2}\exists u_{3}R_{1}(y_{1},u_{1},x_{1})\wedge R_{0}(u_{1},u_{2},x_{2})\wedge R_{1}(u_{2},u_{3},x_{3})\wedge(u_{3}=+)

Thus, these graphs can be viewed as electrical circuits where the ends of an edge are connected (equal) whenever the variable written on it has the corresponding value. Then the encoding of the quantifier-free part

((x_{1}\vee\overline{x}_{2}\vee x_{3})\wedge(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3}))

is shown in Figure 3. If we assume that all the $x$ -variables are from $\{0,1\}$ , which will be the case, then the formula in Figure 3 holds if and only if the 3-CNF does not hold. In fact, if the 3-CNF holds then $+$ is connected (equal) to $-$ through three edges, which gives a contradiction. If the formula in Figure 3 holds, then at some point we go from $+$ to $-$ , which means that the corresponding clause does not hold.

Figure 3: A graph expressing

\neg((x_{1}\vee\overline{x}_{2}\vee x_{3})\wedge(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3})).

If we add universal quantifiers to the formula in Figure 3 we get

	$\displaystyle\forall x_{1}\forall x_{2}\forall x_{3}\;\;\neg((x_{1}\vee\overline{x}_{2}\vee x_{3})\wedge$	$\displaystyle(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3}))=$
		$\displaystyle\neg(\exists x_{1}\exists x_{2}\exists x_{3}\;((x_{1}\vee\overline{x}_{2}\vee x_{3})\wedge(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3})))$

Notice that it does not make sense for the UP to play values $+$ and $-$ because the relations $R_{0}$ and $R_{1}$ hold whenever the last coordinate is from $\{+,-\}$ . Thus, we already encoded the complement to 3-CNF-Satisfability, which means that $\operatorname{QCSP}(\Gamma)$ is coNP-hard.

To show PSpace-hardness we need to add existential quantifiers. We cannot just add $\exists x_{2}$ because the obvious choice for the EP would be $+$ or $-$ . As shown in Figure 4, whenever we want to add $\exists x_{2}$ , we add a new existential variable $y_{2}$ and universally quantify $x_{2}$ . The goal of the UP is to connect $+$ and $-$ . Hence, if the EP plays $y_{2}=+$ , then the only reasonable choice for the UP is to play $x_{2}=1$ ; and if the EP plays $y_{2}=-$ , then the UP must play $x_{2}=0$ . Thus, the EP controls the moves of the UP, which is equivalent to the EP playing on the set $\{0,1\}$ .

Figure 4: A graph expressing

\forall x_{1}\exists x_{2}\forall x_{3}\;\;\neg((x_{1}\vee\overline{x}_{2}\vee x_{3})\wedge(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3}))

Thus, we encoded

	$\displaystyle\forall x_{1}\exists x_{2}\forall x_{3}\;\;\neg((x_{1}\vee\overline{x}_{2}\vee x_{3})$	$\displaystyle\wedge(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3}))=$
		$\displaystyle\neg(\exists x_{1}\forall x_{2}\exists x_{3}\;\;((x_{1}\vee\overline{x}_{2}\vee x_{3})\wedge(\overline{x}_{1}\vee x_{2}\vee\overline{x}_{3})\wedge(x_{1}\vee\overline{x}_{2}\vee\overline{x}_{3})),$

which is a complement to the Quantified-3-CNF. ∎

As we saw in Theorem 7 two relations equivalent to $R_{0}$ and $R_{1}$ can be q-defined from $\Gamma$ whenever $\Gamma$ contains all constant relations and $\operatorname{QCSP}(\Gamma)$ is not in $\Pi_{2}^{P}$ . The criterion for the general case is formulated using mighty tuples, but as you can see in the proof of Theorem 14 in Section 7.2 we still build two relations $R_{0}$ and $R_{1}$ and use almost the same construction to prove PSpace-hardness.

4 $\Pi_{2}^{P}$ -complete constraint language

In this section we define a concrete constraint language $\Gamma$ on a 6-element domain $A=\{0,1,2,0^{\prime},1^{\prime},2^{\prime}\}$ such that $\operatorname{QCSP}(\Gamma)$ is $\Pi_{2}^{P}$ -complete.

First, we define two ternary relations $\mathrm{AND}_{2}$ and $\mathrm{OR}_{2}$ corresponding to the operations $\wedge$ and $\vee$ on $\{0,1\}$ . If one of the first two coordinates is from $\{2,0^{\prime},1^{\prime},2^{\prime}\}$ , then the remaining elements can be chosen arbitrary, i.e.,

	$\displaystyle\{2,0^{\prime},1^{\prime},2^{\prime}\}\times A\times A\subseteq$	$\displaystyle\mathrm{AND}_{2},$	$\displaystyle\{2,0^{\prime},1^{\prime},2^{\prime}\}\times A\times A\subseteq\mathrm{OR}_{2},$
	$\displaystyle A\times\{2,0^{\prime},1^{\prime},2^{\prime}\}\times A\subseteq$	$\displaystyle\mathrm{AND}_{2},$	$\displaystyle A\times\{2,0^{\prime},1^{\prime},2^{\prime}\}\times A\subseteq\mathrm{OR}_{2}.$

If $a,b\in\{0,1\}$ , then $(a,b,c)\in\mathrm{AND}_{2}\Rightarrow(a\wedge b=c)$ and $(a,b,c)\in\mathrm{OR}_{2}\Rightarrow(a\vee b=c)$ . In other words

\mathrm{AND}_{2}\cap(\{0,1\}\times\{0,1\}\times A)=\begin{pmatrix}0&0&1&1\\ 0&1&0&1\\ 0&0&0&1\end{pmatrix},\mathrm{OR}_{2}\cap(\{0,1\}\times\{0,1\}\times A)=\begin{pmatrix}0&0&1&1\\ 0&1&0&1\\ 0&1&1&1\end{pmatrix},

where each matrix should be understood as the set of tuples written as columns. The remaining four relations are defined by

	$\displaystyle\operatorname{1IN3^{\prime}}$	$\displaystyle=\{(2^{\prime},2^{\prime},2^{\prime}),(1^{\prime},0^{\prime},0^{\prime}),(0^{\prime},1^{\prime},0^{\prime}),(0^{\prime},0^{\prime},1^{\prime})\},$
	$\displaystyle\delta_{0}$	$\displaystyle=\{1\}\times\{0^{\prime},2^{\prime}\}\cup(A\setminus\{1\})\times\{0^{\prime},1^{\prime},2^{\prime}\},$
	$\displaystyle\delta_{1}$	$\displaystyle=\{1\}\times\{1^{\prime},2^{\prime}\}\cup(A\setminus\{1\})\times\{0^{\prime},1^{\prime},2^{\prime}\},$
	$\displaystyle\epsilon$	$\displaystyle=\{0\}\times\{0^{\prime},1^{\prime}\}\cup(A\setminus\{0\})\times\{0^{\prime},1^{\prime},2^{\prime}\}.$

The relation $\operatorname{1IN3^{\prime}}$ is the usual relation $\mathrm{1IN3}$ on $\{0^{\prime},1^{\prime}\}$ with an additional tuple $(2^{\prime},2^{\prime},2^{\prime})$ . The relations $\delta_{0}$ , $\delta_{1}$ and $\epsilon$ can also be viewed as

	$\displaystyle\delta_{0}(x,y)$	$\displaystyle=(y\in\{0^{\prime},1^{\prime},2^{\prime}\})\wedge(x=1\Rightarrow y\neq 1^{\prime}),$
	$\displaystyle\delta_{1}(x,y)$	$\displaystyle=(y\in\{0^{\prime},1^{\prime},2^{\prime}\})\wedge(x=1\Rightarrow y\neq 0^{\prime}),$
	$\displaystyle\epsilon(x,y)$	$\displaystyle=(y\in\{0^{\prime},1^{\prime},2^{\prime}\})\wedge(x=0\Rightarrow y\neq 2^{\prime}).$

Note that $\delta_{1}$ can be derived from $\delta_{0}$ and $\operatorname{1IN3^{\prime}}$ by the formula

\delta_{1}(x,y)=\exists u_{1}\exists u_{2}\exists u_{3}\;\delta_{0}(x,u_{1})\wedge\operatorname{1IN3^{\prime}}(y,u_{1},u_{2})\wedge\operatorname{1IN3^{\prime}}(u_{2},u_{2},u_{3}).

Let $\Gamma=\{\mathrm{AND}_{2},\mathrm{OR}_{2},\operatorname{1IN3^{\prime}},\delta_{0},\delta_{1},\epsilon\}$ .

Lemma 9.

$\operatorname{QCSP}(\Gamma)$ is $\Pi_{2}^{P}$ -hard.

Proof.

First, we derive the $\mathrm{OR}$ relation of larger arity by

\mathrm{OR}_{n+1}(x_{1},\dots,x_{n+1},y)=\exists y^{\prime}\;\mathrm{OR}_{n}(x_{1},\dots,x_{n},y^{\prime})\wedge\mathrm{OR}_{2}(y^{\prime},x_{n+1},y).

Then we assume that $\mathrm{OR}_{n}$ is in our language. To prove the $\Pi_{2}^{P}$ -hardness we build a reduction from $\Pi_{2}$ - $\operatorname{QCSP}(\mathrm{1IN3})$ , where $\mathrm{1IN3}=\{(1,0,0),(0,1,0),(0,0,1)\}$ . Let

\Phi=\forall x_{1}\dots\forall x_{m}\exists x_{m+1}\dots\exists x_{n}\;\mathrm{1IN3}(x_{i_{1}},x_{j_{1}},x_{k_{1}})\wedge\dots\wedge\mathrm{1IN3}(x_{i_{s}},x_{j_{s}},x_{k_{s}}).

Since we can always add dummy variables, we assume that $\Phi$ has at least two universally quantified variables for the general construction to make sense. The problem of checking whether $\Phi$ holds is $\Pi_{2}^{P}$ -complete [1]. We will encode $\Phi$ by the following instance of $\operatorname{QCSP}(\Gamma)$ .

	$\displaystyle\Psi=\forall x_{1}^{0}\forall x_{1}^{1}\dots\forall x_{m}^{0}\forall x_{m}^{1}\;\;\exists x_{1}\exists x_{2}\dots\exists x_{n}\;\;\exists z_{1}\dots\exists z_{n}\;\;\exists z\;\;\bigwedge\limits_{i=1}^{m}\left(\delta_{0}(x_{i}^{0},x_{i})\wedge\delta_{1}(x_{i}^{1},x_{i})\right)\wedge$
	$\displaystyle\;\bigwedge\limits_{i=1}^{m}\mathrm{AND}_{2}(x_{i}^{0},x_{i}^{1},z_{i})\wedge\mathrm{OR}_{m}(z_{1}\dots,z_{n},z)\wedge\bigwedge\limits_{i=1}^{n}\epsilon(z,x_{i})\wedge\bigwedge\limits_{\ell=1}^{s}\operatorname{1IN3^{\prime}}(x_{i_{\ell}},x_{j_{\ell}}$	$\displaystyle,x_{k_{\ell}})$

Figure 5: Reduction from

\Pi_{2}

\operatorname{QCSP}(\mathrm{1IN3})

\operatorname{QCSP}(\Gamma)

The quantifier free part of $\Psi$ is shown in Figure 5, where triangle elements are $\mathrm{AND}_{2}$ and $\mathrm{OR}_{m}$ , rectangular elements are $\delta_{0}$ , $\delta_{1}$ , and $\epsilon$ , and the big block at the bottom is just the conjunction of the corresponding $\operatorname{1IN3^{\prime}}$ -relations. The variable $x_{i}$ in $\Psi$ will take values from $\{0^{\prime},1^{\prime},2^{\prime}\}$ , which makes the use of universal quantifiers directly impossible. For the universal variables to be applicable we introduce two new variables $x_{i}^{0}$ and $x_{i}^{1}$ for each $x_{i}$ where $i\in\{1,2,\dots,m\}$ . We expect exactly one of the two values $x_{i}^{0}$ and $x_{i}^{1}$ to be equal to 1. $x_{i}^{0}=1$ means that $x_{i}=0^{\prime}$ , $x_{i}^{1}=1$ means that $x_{i}=1^{\prime}$ . Using the relations $\delta_{0}$ and $\delta_{1}$ we make $x_{i}$ equal to the value we need. Notice that $x_{i}$ can also be equal to $2^{\prime}$ and this value should be forbidden by $\epsilon$ . Let us prove that $\Phi$ and $\Psi$ are equivalent.

$\Phi\Rightarrow\Psi$ . Suppose we have a winning strategy for the Existential Player (EP) in $\Phi$ , let us define a winning strategy for the EP in $\Psi$ . If the Universal Player (UP) in $\Psi$ plays $x_{i}^{0}=x_{i}^{1}=1$ , or $x_{i}^{0}\notin\{0,1\}$ , or $x_{i}^{1}\notin\{0,1\}$ for some $i$ , then the winning strategy for the EP is to choose $x_{1}=\dots=x_{n}=2^{\prime}$ . Then the $\operatorname{1IN3^{\prime}}$ -block of $\Psi$ is satisfied as $(2^{\prime},2^{\prime},2^{\prime},\dots,2^{\prime})$ is its trivial solution. Only value 1 restricts the second coordinate of the relations $\delta_{0}$ and $\delta_{1}$ , hence the best choice for the UP is to make $(x_{i}^{0},x_{i}^{1})\in\{(0,1),(1,0)\}$ . We interpret $(x_{i}^{0},x_{i}^{1})=(0,1)$ as $x_{i}=1$ and $(x_{i}^{0},x_{i}^{1})=(1,0)$ as $x_{i}=0$ . Then the EP in $\Psi$ plays $x_{1},\dots,x_{m}$ according to $(x_{i}^{0},x_{i}^{1})$ and plays $x_{m+1},\dots,x_{n}$ just copying the moves of the EP in $\Phi$ but $0^{\prime}$ instead of $0$ and $1^{\prime}$ instead of $1$ . Since the quantifier-free part of $\Phi$ is satisfied, the $\operatorname{1IN3^{\prime}}$ -block of $\Psi$ is also satisfied.

$\Psi\Rightarrow\Phi$ . Suppose the UP in $\Phi$ plays $x_{1},\dots,x_{m}$ . Let the UP in $\Psi$ play $x_{i}^{0}=1$ , $x_{i}^{1}=0$ if $x_{i}=0$ and $x_{i}^{0}=0$ , $x_{i}^{1}=1$ if $x_{i}=1$ . Then the EP in $\Psi$ should play only values from $\{0^{\prime},1^{\prime}\}$ for $x_{1},\dots,x_{n}$ . The EP in $\Phi$ just copies the moves of the EP in $\Psi$ playing $0$ instead of $0^{\prime}$ , and $1$ instead of $1^{\prime}$ . The satisfiability of the $\operatorname{1IN3^{\prime}}$ -block of $\Psi$ implies the satisfiability of the quantifier-free part of $\Phi$ . ∎

Lemma 10.

$\operatorname{QCSP}(\Gamma)$ is in $\Pi_{2}^{P}$ .

Proof.

One of the definitions of the class $\Pi_{2}^{P}$ is $\mathrm{coNP}^{\mathrm{NP}}$ , that is the class of problem solvable by a nondetermenistic Turing machine augmented by an oracle for some NP-complete problem [1]. Thus, to prove the membership in $\Pi_{2}^{P}$ , it is sufficient to show that an optimal strategy for the EP can be calculated in polynomial time using the NP-oracle.

Suppose we have an instance $\forall x_{1}\exists y_{1}\forall x_{2}\exists y_{2}\dots\forall x_{n}\exists y_{n}\Phi$ . Suppose the variables $x_{1},\dots,x_{i}$ and $y_{1},\dots,y_{i-1}$ are already evaluated with $a_{1},\dots,a_{i}$ and $b_{1},\dots,b_{i-1}$ , respectively. We need to calculate an optimal value $b_{i}$ for $y_{i}$ , i.e., suppose the EP can still win in this position then she should be able to win after making the move $y_{i}=b_{i}$ .

We will explain the algorithm first and then we argue why it returns an optimal move. Using the NP-oracle for every $d\in A$ we check the satisfiability of the instance

\displaystyle\Phi\wedge\bigwedge_{j=1}^{j=i}x_{j}=a_{j}\wedge\bigwedge_{j=i+1}^{j=n}x_{j}=1\wedge\bigwedge_{j=1}^{j=i-1}y_{j}=b_{j}\wedge y_{i}=d.

(3)

Thus, we just send the previous variables to their values, evaluate all further universal variables to 1 and the variable $y_{i}$ to $d$ . Let $D$ be the set of $d$ such that (3) has a solution. Then we choose an optimal value as follows:

1.

If $D$ is empty, then the EP cannot win and an optimal move does not exist.
2.

If $D=\{b\}$ for some $b$ then $y_{i}=b$ is the optimal move.
3.

If $c\in D\cap\{0^{\prime},1^{\prime},2\}$ then $y_{i}=c$ is an optimal move.

The algorithm is obviously polynomial. Hence it is sufficient to prove that cases 1-3 cover all the possible cases and the move chosen by 3 is optimal.

The binary operation $g(x,y)=\begin{cases}x,&\text{if $y=1$}\\ x,&\text{if $x=y$}\\ y,&\text{if $y\in\{0^{\prime},1^{\prime}\}$}\\ x,&\text{if $x\in\{0^{\prime},1^{\prime}\}$ and $y=2^{\prime}$}\\ 2,&\text{otherwise}\end{cases}$ preserves all the relations from $\Gamma$ and all constant relations, which follows from the following properties of $g$ and manual checking of some cases:

•

$g$ either returns the first variable or an element of $\{0^{\prime},1^{\prime},2\}$ .
•

$g$ preserves $\{0^{\prime},1^{\prime},2^{\prime}\}$ and $g$ restricted to $\{0^{\prime},1^{\prime},2^{\prime}\}$ returns the last non- $2^{\prime}$ value if it exists.
•

$g$ returns 1 only on the tuple $(1,1)$ and it returns $2^{\prime}$ only on the tuple $(2^{\prime},2^{\prime})$ .

This implies (see [2]) that $g$ preserves the solution set of any instance of $\operatorname{CSP}(\Gamma)$ .

The only cases that are not covered by cases 1-3 are $\{0,1\}\subseteq D$ , $\{0,2^{\prime}\}\subseteq D$ , or $\{1,2^{\prime}\}\subseteq D$ . Since $g(1,0)=g(0,2^{\prime})=g(1,2^{\prime})=2$ , we have $2\in D$ , which means that $D$ satisfies case 3.

It remains to show that any value $c\in D\cap\{0^{\prime},1^{\prime},2\}$ is an optimal move for the EP. Let $y_{j}:=f_{j}(x_{i+1},\dots,x_{j})$ , where $j=i,i+1,\dots,n$ , be the winning strategy for the EP. Then the tuple

\displaystyle(a_{1},b_{1},\dots,a_{i-1},b_{i-1},a_{i},f_{i}(),a_{i+1},f_{i+1}(a_{i+1}),a_{i+2},f_{i+2}(a_{i+1},a_{i+2}),\dots,a_{n},f_{n}(a_{i+1},\dots,a_{n}))

(4)

is a solution of the quantifier-free part $\Phi$ for any $a_{i+1},\dots,a_{n}\in A$ . By the definition of $D$ the tuple

\displaystyle(a_{1},b_{1},\dots,a_{i-1},b_{i-1},a_{i},c,1,c_{i+1},1,c_{i+2},\dots,1,c_{n})

(5)

is a solution of $\Phi$ for some $c_{i+1},\dots,c_{n}\in A$ . Applying $g$ to the tuples (4) and (5) coordinate-wise we derive that the tuple

	$\displaystyle(a_{1},b_{1},\dots,a_{i-1},b_{i-1},a_{i},c,$
	$\displaystyle a_{i+1},g(f_{i+1}(a_{i+1}),$	$\displaystyle c_{i+1}),a_{i+2},g(f_{i+2}(a_{i+1},a_{i+2}),c_{i+2}),\dots,a_{n},g(f_{n}(a_{i+1},\dots,a_{n}),c_{n}))$

is a solution of $\Phi$ for any $a_{i+1},\dots,a_{n}\in A$ . Hence $y_{i}=c$ is an optimal move for the EP, which completes the proof. ∎

Thus, we proved Theorem 5 from Section 2.

5 Proof of the main result

5.1 Necessary definitions and notations

In the paper we assume that the overall domain $A$ is finite and fixed. To simplify notations we even assume that $A=\{1,2,\dots,|A|\}$ . For a positive integer $m$ by $[m]$ we denote the set $\{1,2,\dots,m\}$ .

For two binary relations $S_{1}$ and $S_{2}$ by $S_{1}+S_{2}$ we denote the composition of two relations, that is $S_{1}+S_{2}=S$ , where $S(x,y)=\exists z\;S_{1}(x,z)\wedge S_{2}(z,y)$ . By $S_{1}-S_{2}$ we denote the binary relation $S(x,y)=\exists z\;S_{1}(x,z)\wedge S_{2}(y,z)$ , that is, $S_{1}-S_{2}=S_{1}+S_{2}^{\prime}$ , where $S_{2}^{\prime}$ is obtained from $S_{2}$ by switching the coordinates. Similarly, we can write $U_{1}+S_{2}$ if $U_{1}$ is unary, that is $U_{1}+S_{2}=U$ , where $U(x)=\exists z\;U_{1}(z)\wedge S_{2}(z,x)$ .

For a formula $\Phi$ and some free variables $u_{1},\dots,u_{s}$ of this formula by $\Phi\downarrow\begin{smallmatrix}u_{1}&\dots&u_{s}\\ v_{1}&\dots&v_{s}\end{smallmatrix}$ we denote the formula obtained from $\Phi$ by substituting each $u_{i}$ by $v_{i}$ .

For a relation $R\subseteq A^{n}$ and $i\in[n]$ by $\operatorname{pr}_{i}(R)$ we denote the projection of $R$ onto the $i$ -th coordinate. Similarly for a constraint $C=R(u_{1},\dots,u_{s})$ by $\operatorname{pr}_{u_{i}}(C)$ we denote $\operatorname{pr}_{i}(R)$ .

Sometimes it will be convenient to assume that some of the variables of a relation are external parameters. Thus, a relation of arity $|A|+2$ is called a $\mathbf{z}$ -parameterized binary relation, where $\mathbf{z}\in A^{|A|}$ . Some relations have two or even three parameters. Thus, we may consider $(\mathbf{z},\delta,\alpha)$ -parameterized binary relation $Q$ , where $\mathbf{z}\in A^{|A|}$ , $\delta\in A^{k}$ , and $\alpha\in A^{m}$ , which is a relation of arity $|A|+k+m+2$ . To refer to the binary relation for the fixed parameters $\mathbf{z},\delta$ , and $\alpha$ we write $\prescript{\mathbf{z}}{\delta}{Q^{\alpha}}$ . Sometimes we replace the $\alpha$ -parameter with $\forall$ or $\forall\forall$ meaning that we universally quantify this parameter in two different ways (see equations (1) and (2)).

A parameterized unary relation is called nonempty if it is nonempty for every choice of the parameters. For an instance $\mathcal{I}$ by $\operatorname{Var}(\mathcal{I})$ we denote the set of all variables appearing in this instance.

For a CSP instance $\mathcal{I}$ (conjunctive formula) and some variables $u_{1},\dots,u_{k}\in\operatorname{Var}(\mathcal{I})$ by $\mathcal{I}(u_{1},\dots,u_{k})$ we denote the set of all tuples $(a_{1},\dots,a_{k})$ such that $\mathcal{I}$ has a solution with $u_{i}=a_{i}$ for every $i$ . Thus, $\mathcal{I}(x_{1},\dots,x_{k})$ defines a $k$ -ary relation. Since this relation is defined from $\mathcal{I}$ by adding existential quantifiers, the relation $\mathcal{I}(x_{1},\dots,x_{k})$ is q-definable from relations in $\mathcal{I}$ .

5.2 Universal subset

Suppose $S\subseteq W\subseteq A^{t}$ and $\Sigma$ is a set of relations on $A$ . We say that $S$ is a universal subset of $W$ over $\Sigma$ , denote $S\mathrel{\trianglelefteq}^{\Sigma}W$ , if there exist $s$ and a relation $R\subseteq A^{t+s}$ q-definable from $\Sigma$ such that

	$\displaystyle S(y_{1},\dots,y_{t})$	$\displaystyle=\forall x_{1}\dots\forall x_{s}\;R(y_{1},\dots,y_{t},x_{1},\dots,x_{s}),$
	$\displaystyle W(y_{1},\dots,y_{t})$	$\displaystyle=\forall x\;R(y_{1},\dots,y_{t},x,\dots,x).$

Notice that $S\mathrel{\trianglelefteq}^{\Sigma}W$ also implies that $\Sigma$ q-defines both $S$ and $W$ . To emphasize that $S$ and $W$ are different we write $S\mathrel{\lhd}^{\Sigma}W$ instead of $S\mathrel{\trianglelefteq}^{\Sigma}W$ .

Lemma 11 (proved in Section 6.2).

Suppose

	$\displaystyle W(y_{1},\dots,y_{t})$	$\displaystyle=\exists u_{1}\exists u_{2}\dots\exists u_{\ell}\;\;\;W_{1}(z_{1,1},\dots,z_{1,n_{1}})\wedge\dots\wedge W_{m}(z_{m,1},\dots,z_{m,n_{m}}),$
	$\displaystyle S(y_{1},\dots,y_{t})$	$\displaystyle=\exists u_{1}\exists u_{2}\dots\exists u_{\ell}\;\;\;S_{1}(z_{1,1},\dots,z_{1,n_{1}})\wedge\dots\wedge S_{m}(z_{m,1},\dots,z_{m,n_{m}}),$

where each $z_{i,j}\in\{y_{1},\dots,y_{t},u_{1},\dots,u_{\ell}\}$ , and $S_{i}\mathrel{\trianglelefteq}^{\Sigma}W_{i}$ for every $i$ . Then $S\mathrel{\trianglelefteq}^{\Sigma}W$ .

For $k\geqslant 0$ we write $S\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}_{k}^{\Sigma}W$ if $S\mathrel{\trianglelefteq}^{\Sigma}C_{1}\mathrel{\trianglelefteq}^{\Sigma}C_{2}\mathrel{\trianglelefteq}^{\Sigma}\dots\mathrel{\trianglelefteq}^{\Sigma}C_{k}\mathrel{\trianglelefteq}^{\Sigma}W$ for some relations $C_{1},\dots,C_{k}$ q-definable from $\Sigma$ . We often omit $k$ , if we do not want to specify the length of the sequence. In Section 6 we usually omit $\Sigma$ and write just $\mathrel{\trianglelefteq}$ , $\mathrel{\lhd}$ , or $\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}$ meaning that $\Sigma=\{R\}$ .

It follows from the definition that for any $\alpha$ -parameterized relation $Q$ we have $Q^{\forall\forall}\mathrel{\trianglelefteq}^{\{Q\}}Q^{\forall}$ .

5.3 Induced CSP Instances

Suppose $\Psi=\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\Phi$ is an instance of $\operatorname{QCSP}(\Gamma)$ , where $\Phi$ a quantifier-free conjunctive formula. Let us show how to build an equivalent CSP instance of an exponential size. If the sentence $\Psi$ holds, then there exist Skolem functions $f_{0},\dots,f_{n}$ defining a winning strategy for the existential player (EP), that is, she can play $y_{i}=f_{i}(x_{1},\dots,x_{i})$ . Since it is a winning strategy, if the universal player (UP) plays $(x_{1},x_{2},\dots,x_{n})=(a_{1},a_{2},\dots,a_{n})$ and the EP plays $(y_{0},y_{1},\dots,y_{n})=(f_{0}(),f_{1}(a_{1}),f_{2}(a_{1},a_{2}),\dots,f_{n}(a_{1},\dots,a_{n}))$ , then the obtained evaluation should satisfy the quantifier-free part $\Phi$ . We introduce an existential variable $y_{i}^{a_{1},\dots,a_{i}}$ for every $i$ and $a_{1},\dots,a_{i}\in A$ . Then $\Psi$ is equivalent to the satisfiability of the CSP instance $\bigwedge\limits_{a_{1},\dots,a_{n}\in A}(\Phi\downarrow\begin{smallmatrix}x_{1}&\dots&x_{n}&y_{1}&y_{2}&\dots&y_{n}\\ a_{1}&\dots&a_{n}&y_{1}^{a_{1}}&y_{2}^{a_{1},a_{2}}&\dots&y_{n}^{a_{1},\dots,a_{n}}\end{smallmatrix})$ . Notice that this instance is of exponential size, that is why we cannot really use it in the algorithm. Since usually we do not have constant relations in our constraint language $\Gamma$ we replace the constants $1,2,\dots,|A|\in A$ by the respective universally quantified variables $z_{1},\dots,z_{|A|}$ . Since in the paper we do not care about a concrete conjunctive formula $\Phi$ , we usually define the relation $R\subseteq A^{2n+1}$ by $R(y_{0},y_{1},\dots,y_{n},x_{1},\dots,x_{n})=\Phi$ and work with it instead of constraints of $\Phi$ .

In Section 6 the crucial idea is to show that the induced exponential-size CSP instance can be solved by arc-consistency, and to make it work we replace the relations $R$ by stronger relations defined below. For $R\subseteq A^{2n+1}$ and $m\in\{0,1,\dots,n-1\}$ put

	$\displaystyle\mathcal{W}_{R}^{m}(y_{0},\dots,y_{m},x_{1},\dots,x_{m})=$	$\displaystyle\forall x\exists y_{m+1}\exists y_{m+2}\dots\exists y_{n}R(y_{0},\dots,y_{n},x_{1},\dots,x_{m},x,x,\dots,x),$
	$\displaystyle\mathcal{S}_{R}^{m}(y_{0},\dots,y_{m},x_{1},\dots,x_{m})=$	$\displaystyle\forall x\exists y_{m+1}\forall x^{\prime}\exists y_{m+2}\dots\exists y_{n}R(y_{0},\dots,y_{n},x_{1},\dots,x_{m},x,x^{\prime},\dots,x^{\prime}).$

Notice that $\mathcal{S}_{R}^{n-1}=\mathcal{W}_{R}^{n-1}$ . We set by definition that $\mathcal{S}_{R}^{n}=\mathcal{W}_{R}^{n}=R$ . The crucial property of $\mathcal{W}_{R}^{m}$ and $\mathcal{S}_{R}^{m}$ is formulated in the following lemma.

Lemma 12.

Suppose $R\subseteq A^{2n+1}$ and $m\in\{0,1,\dots,n\}$ , then $\mathcal{S}_{R}^{m}\mathrel{\trianglelefteq}^{\{R\}}\mathcal{W}_{R}^{m}$ .

Proof.

We define a relation $Q$ by

	$\displaystyle Q(y_{0},\dots,y_{m},$	$\displaystyle x_{1},\dots,x_{m},x,x^{1},\dots,x^{\|A\|})=$
	$\displaystyle\exists y_{m+1}\bigwedge\limits_{a\in A}$	$\displaystyle\left(\exists y_{m+2}\exists y_{m+3}\dots\exists y_{n}\;R(y_{0},\dots,y_{m},y_{m+1},y_{m+2},\dots,y_{n},x_{1},\dots,x_{m},x,x^{a},\dots,x^{a})\right).$

Then the relation $Q$ witnesses that $\mathcal{S}_{R}^{m}\mathrel{\trianglelefteq}\mathcal{W}_{R}^{m}$ :

	$\displaystyle\mathcal{W}_{R}^{m}(y_{0},\dots,y_{m},x_{1},\dots,x_{m})$	$\displaystyle=\forall x\;Q(y_{0},\dots,y_{m},x_{1},\dots,x_{m},x,x,\dots,x),$
	$\displaystyle\mathcal{S}_{R}^{m}(y_{0},\dots,y_{m},x_{1},\dots,x_{m})$	$\displaystyle=\forall x\forall x^{1}\forall x^{2}\dots\forall x^{\|A\|}\;Q(y_{0},\dots,y_{m},x_{1},\dots,x_{m},x,x^{1},x^{2},\dots,x^{\|A\|}).$

∎

Thus, we can define a bunch of CSP instances equivalent to the original QCSP instance.

Lemma 13.

Suppose $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\Phi$ is an instance of $\operatorname{QCSP}(\Gamma)$ and $R\subseteq A^{2n+1}$ is defined by $R(y_{0},y_{1},\dots,y_{n},x_{1},\dots,x_{n})=\Phi$ . Then the following conditions are equivalent:

1.

$\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\Phi$ holds;
2.

$\bigwedge\limits_{a_{1},\dots,a_{n}\in A}(\Phi\downarrow\begin{smallmatrix}x_{1}&\dots&x_{n}&y_{1}&y_{2}&\dots&y_{n}\\ a_{1}&\dots&a_{n}&y_{1}^{a_{1}}&y_{2}^{a_{1},a_{2}}&\dots&y_{n}^{a_{1},\dots,a_{n}}\end{smallmatrix})$ has a solution;
3.

$\bigwedge\limits_{a_{1},\dots,a_{n}\in A}R(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{n}^{a_{1},\dots,a_{n}},a_{1},\dots,a_{n})$ has a solution;
4.

$\bigwedge\limits_{a_{1},\dots,a_{n}\in A}(\Phi\downarrow\begin{smallmatrix}x_{1}&\dots&x_{n}&y_{1}&y_{2}&\dots&y_{n}\\ z_{a_{1}}&\dots&z_{a_{n}}&y_{1}^{a_{1}}&y_{2}^{a_{1},a_{2}}&\dots&y_{n}^{a_{1},\dots,a_{n}}\end{smallmatrix})$ has a solution for every $z_{1},\dots,z_{|A|}\in A$ ;
5.

$\bigwedge\limits_{a_{1},\dots,a_{n}\in A}R(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{n}^{a_{1},\dots,a_{n}},z_{a_{1}},\dots,z_{a_{n}})$ has a solution for every $z_{1},\dots,z_{|A|}\in A$ ;
6.

$\bigwedge\limits_{\begin{subarray}{c}m\in\{0,1,\dots,n\}\\ a_{1},\dots,a_{m}\in A\end{subarray}}\mathcal{S}_{R}^{m}(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}})$ has a solution for every $z_{1},\dots,z_{|A|}\in A$ .

Proof.

Trivially, we have $1\leftrightarrow 2\leftrightarrow 3$ and $6\rightarrow 5\leftrightarrow 4\rightarrow 2$ . To complete the proof let us show that $3\rightarrow 6$ . Let us define a solution to 6 for a concrete $z_{1},\dots,z_{|A|}\in A$ . Let $y_{i}^{a_{1},\dots,a_{i}}=b_{i}^{a_{1},\dots,a_{i}}$ be a solution of 3. Then a solution to 6 can be defined by $y_{i}^{a_{1},\dots,a_{i}}=b_{i}^{z_{a_{1}},\dots,z_{a_{i}}}$ . ∎

Denote $\mathcal{I}_{R}=\bigwedge\limits_{\begin{subarray}{c}m\in\{0,1,\dots,n\}\\ a_{1},\dots,a_{m}\in A\end{subarray}}\mathcal{S}_{R}^{m}(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}})$ , that is the equivalent CSP instance from item 6. Notice that the variables $z_{1},\dots,z_{|A|}$ are viewed as external parameters of the instance $\mathcal{I}_{R}$ and we call such instances $\mathbf{z}$ -parameterized. That is why, we assume that $z_{1},\dots,z_{|A|}$ are not in $\operatorname{Var}(\mathcal{I}_{R})$ . Also, when we refer to the constraint

\mathcal{S}_{R}^{m}(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}})

we usually omit these variables and write $\mathcal{S}_{R}^{m}(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{m}^{a_{1},\dots,a_{m}})$ instead.

5.4 Mighty tuples

In this subsection, we formulate five sufficient conditions for the QCSP over a constraint language $\Gamma$ to be PSpace-hard. One of them, already defined in Section 2, is also a necessary condition.

Mighty tuple I. A tuple $(Q,D,B,C,\Delta)$ is called a mighty tuple I if $\Delta$ is a $\mathbf{z}$ -parameterized $k$ -ary relation, $Q$ is a $(\mathbf{z},\delta,\alpha)$ -parameterized binary relations, $D$ , $B$ , and $C$ are $(\mathbf{z},\delta)$ -parameterized unary relations, and they satisfy the following conditions:

1.

$\prescript{\mathbf{z}}{}{\Delta}\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ ;
2.

$\prescript{\mathbf{z}}{\delta}{B}$ , $\prescript{\mathbf{z}}{\delta}{C}$ , and $\prescript{\mathbf{z}}{\delta}{D}$ are nonempty for every $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ ;
3.

$\prescript{\mathbf{z}}{\delta}{Q}^{\alpha}$ is an equivalence relation on $\prescript{\mathbf{z}}{\delta}{D}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ and $\alpha\in A^{m}$ ;
4.

$\prescript{\mathbf{z}}{\delta}{Q}^{\forall}=\prescript{\mathbf{z}}{\delta}{D}\times\prescript{\mathbf{z}}{\delta}{D}$ for every $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ ;
5.

$\prescript{\mathbf{z}}{\delta}{B}$ and $\prescript{\mathbf{z}}{\delta}{C}$ are equivalence classes of $\prescript{\mathbf{z}}{\delta}{Q}^{\forall\forall}$ ;
6.

there exists $\mathbf{z}\in A^{|A|}$ such that $\prescript{\mathbf{z}}{\delta}{B}\neq\prescript{\mathbf{z}}{\delta}{C}$ for every $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ .

In Section 7 we prove the following theorem.

Theorem 14.

Suppose $(Q,D,B,C,\Delta)$ is a mighty tuple I. Then $\operatorname{QCSP}(\{Q,D,B,C,\Delta\})$ is PSpace-hard.

In the paper instead of deriving a mighty tuple I, which is rather complicated, we derive one of the easier tuples that we call a mighty tuple II, a mighty tuple III, a mighty tuple IV, and a mighty tuple V. They are defined below.

Mighty tuple II. A tuple $(Q,D,B,C)$ is called a mighty tuple II if $Q$ is a $(\mathbf{z},\alpha)$ -parameterized binary relation, $D$ , $B$ , and $C$ are $\mathbf{z}$ -parameterized unary relations, and they satisfy the following conditions:

1.

$\prescript{\mathbf{z}}{}{B}\neq\varnothing$ and $\prescript{\mathbf{z}}{}{C}\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ ;
2.

$\prescript{\mathbf{z}}{}{Q^{\alpha}}$ is an equivalence relations on $\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}$ and $\alpha$ ;
3.

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{Q^{\forall\forall}}=\prescript{\mathbf{z}}{}{B}$ and $\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{Q^{\forall\forall}}=\prescript{\mathbf{z}}{}{C}$ for every $\mathbf{z}\in A^{|A|}$ ;
4.

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{Q^{\forall}}=\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{Q^{\forall}}=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
5.

$\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}=\varnothing$ for some $\mathbf{z}\in A^{|A|}$ .

Mighty tuple III. A tuple $(Q,B,C)$ is called a mighty tuple III if $Q$ is a $(\mathbf{z},\alpha)$ -parameterized binary relation, $B$ and $C$ are $\mathbf{z}$ -parameterized unary relations, and they satisfy the following conditions:

1.

$\prescript{\mathbf{z}}{}{B}\neq\varnothing$ and $\prescript{\mathbf{z}}{}{C}\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ ;
2.

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{Q^{\forall\forall}}=\prescript{\mathbf{z}}{}{B}$ for every $\mathbf{z}\in A^{|A|}$ ;
3.

$\prescript{\mathbf{z}}{}{Q^{\forall\forall}}+\prescript{\mathbf{z}}{}{C}=\prescript{\mathbf{z}}{}{C}$ for every $\mathbf{z}\in A^{|A|}$ ;
4.

$\prescript{\mathbf{z}}{}{Q}^{\forall}\cap(\prescript{\mathbf{z}}{}{B}\times\prescript{\mathbf{z}}{}{C})\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ ;
5.

$\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}=\varnothing$ for some $\mathbf{z}\in A^{|A|}$ .

Mighty tuple IV. A tuple $(Q,D,B,C)$ is called a mighty tuple IV if $Q$ is a $(\mathbf{z},\alpha)$ -parameterized binary relation, $D$ , $B$ , and $C$ are $\mathbf{z}$ -parameterized unary relations, and they satisfy the following conditions:

1.

$\varnothing\neq\prescript{\mathbf{z}}{}{B}\subseteq\prescript{\mathbf{z}}{}{D}$ and $\varnothing\neq\prescript{\mathbf{z}}{}{C}\subseteq\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
2.

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{Q^{\forall\forall}}=\prescript{\mathbf{z}}{}{B}$ for every $\mathbf{z}\in A^{|A|}$ ;
3.

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{Q^{\forall}}=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
4.

$\prescript{\mathbf{z}}{}{D}+\prescript{\mathbf{z}}{}{Q^{\forall\forall}}=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
5.

$\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}=\varnothing$ for some $\mathbf{z}\in A^{|A|}$ .

As we prove in Section 7.3 the existence of a mighty tuple II, a mighty tuple III, and a mighty tuple IV are equivalent:

Lemma 15.

Suppose $\Sigma$ is a set of relations on $A$ . Then the following conditions are equivalent:

1.

$\Sigma$ q-defines a mighty tuple II;
2.

$\Sigma$ q-defines a mighty tuple III;
3.

$\Sigma$ q-defines a mighty tuple IV.

Moreover, each of them implies a mighty tuple I and therefore guarantees PSpace-hardness.

Lemma 16.

Suppose $T$ is a mighty tuple of type II, III, or IV. Then relations of $T$ q-define a mighty tuple I.

Mighty tuple V. A tuple $(Q,D)$ is called a mighty tuple V if $Q$ is a $(\mathbf{z},\alpha)$ -parameterized binary relation, $D$ is a nonempty $\mathbf{z}$ -parameterized unary relation, and they satisfy the following conditions:

1.

$\{(d,d)\mid d\in\prescript{\mathbf{z}}{}{D}\}\subseteq\prescript{\mathbf{z}}{}{Q}^{\forall}$ for every $\mathbf{z}\in A^{|A|}$ ; ( $\prescript{\mathbf{z}}{}{Q}^{\forall}$ is reflexive)
2.

$\operatorname{pr}_{1}(\prescript{\mathbf{z}}{}{Q}^{\forall\forall})=\operatorname{pr}_{2}(\prescript{\mathbf{z}}{}{Q}^{\forall\forall})=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
3.

$\prescript{\mathbf{z}}{}{Q}^{\forall\forall}\cap\{(d,d)\mid d\in A\}=\varnothing$ for some $\mathbf{z}\in A^{|A|}$ . ( $\prescript{\mathbf{z}}{}{Q}^{\forall\forall}$ is irreflexive)

In Section 7.4 we prove that a mighty tuple V implies a mighty tuple I.

Lemma 17.

Suppose $(Q,D)$ is a mighty tuple V. Then there exists a mighty tuple I q-definable from $\{Q,D\}$ .

5.5 Reductions

A $\mathbf{z}$ -parameterized reduction $D^{(\top)}$ for a $\mathbf{z}$ -parameterized CSP instance $\mathcal{I}$ is a mapping that assigns a $\mathbf{z}$ -parameterized unary relation $D_{u}^{(\top)}$ to every variable $u$ of $\mathcal{I}$ . Then for any constraint $C=R(u_{1},\dots,u_{s})$ by $C^{(\top)}$ we denote the constraint $R^{(\top)}(u_{1},\dots,u_{s})$ , where $\prescript{\mathbf{z}}{}{R}^{(\top)}(u_{1},\dots,u_{s})=\prescript{\mathbf{z}}{}{R}(u_{1},\dots,u_{s})\wedge\bigwedge_{i=1}^{s}(u_{i}\in\prescript{\mathbf{z}}{}{D}_{u_{i}}^{(\top)})$ . A reduction $D^{(\top)}$ of $\mathcal{I}$ is called 1-consistent if for any constraint $C=R(u_{1},\dots,u_{s})$ in $\mathcal{I}$ and any $i\in\{1,2,\dots,s\}$ we have $\operatorname{pr}_{u_{i}}(C^{(\top)})=D_{u_{i}}^{(\top)}$ .

We say that a $\mathbf{z}$ -parameterized reduction $D^{(\bot)}$ is smaller than a $\mathbf{z}$ -parameterized reduction $D^{(\top)}$ if $\prescript{\mathbf{z}}{}{D}_{u}^{(\bot)}\subseteq\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}$ for every $u$ and $\mathbf{z}$ , and $D^{(\bot)}\neq D^{(\top)}$ . In this case we write $D^{(\bot)}\subsetneq D^{(\top)}$ . A reduction $D^{(\top)}$ is called nonempty if $\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}$ is nonempty for every $u$ and $\mathbf{z}$ .

For a $\mathbf{z}$ -parameterized reduction $D^{(\top)}$ of $\mathcal{I}_{R}$ by $D^{(\top,0)}_{y_{m}^{a_{1},\dots,a_{m}}}$ we denote the $\mathbf{z}$ -parameterized unary relation defined by

\left(\mathcal{W}_{R}^{m}(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{m}^{a_{1},\dots,a_{m}})\wedge\bigwedge\limits_{i=0}^{m-1}(y_{i}^{a_{1},\dots,a_{i}}\in D_{y_{i}^{a_{1},\dots,a_{i}}}^{(\top)})\right)(y_{m}^{a_{1},\dots,a_{m}}),

which is the set of all possible values of $y_{m}^{a_{1},\dots,a_{m}}$ in solutions of the conjunction in the brackets. In other words, $D^{(\top,0)}_{u}$ is the restriction on $u$ we get by restricting the variables $u_{0},u_{1},\dots,u_{m-1}$ in $\mathcal{W}_{R}^{m}(u_{0},u_{1},\dots,u_{m-1},u)$ to $D^{(\top)}$ . Notice that if $D^{(\top)}$ is 1-consistent then $D^{(\top)}_{u}\subseteq D^{(\top,0)}_{u}$ for every $u$ . A reduction $D^{(\top)}$ of $\mathcal{I}_{R}$ is called a universal reduction if $D^{(\top)}_{u}\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}D^{(\top,0)}_{u}$ for every $u\in\operatorname{Var}(\mathcal{I}_{R})$ .

5.6 Proof of Theorems 3, 6, and 7

As we said before for a given instance of the QCSP we define an equivalent $\mathbf{z}$ -parameterized exponential CSP $\mathcal{I}_{R}$ whose only relations are $\mathcal{S}_{R}^{m}$ . The next theorem shows that either we can find a 1-consistent reduction of $\mathcal{I}_{R}$ , or we can find a small subinstance without a solution for some $\mathbf{z}$ , or the QCSP over this language is PSpace-hard.

Theorem 18.

Suppose $R\subseteq A^{2n+1}$ . Then one of the following conditions holds:

1.

there exists a $\mathbf{z}$ -parameterized nonempty 1-consistent reduction for $\mathcal{I}_{R}$ ;
2.

there exists a subinstance $\mathcal{J}\subseteq\mathcal{I}_{R}$ with at most $(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ variables not having a solution for some $\mathbf{z}\in A^{|A|}$ ;
3.

there exists a mighty tuple III q-definable from $R$ .

The next theorem as well as Theorem 18 is proved in Section 6.3.

Theorem 19.

Suppose $R\subseteq A^{2n+1}$ , $D^{(\top)}$ is an inclusion-maximal $\mathbf{z}$ -parameterized 1-consistent nonempty reduction for $\mathcal{I}_{R}$ . Then $D^{(\top)}$ is a universal reduction.

Then we consider the case when there exists a 1-consistent universal reduction for $\mathcal{I}_{R}$ . We will show that if the instance has no solutions, then we can find a smaller 1-consistent reduction. We do this in two steps. First, in Section 6.4 we prove the following theorem that states that we can find a universal subset on some $D_{u}^{(\top)}$ .

Theorem 20.

Suppose $R\subseteq A^{2n+1}$ , $\mathcal{I}_{R}$ has no solutions for some $\mathbf{z}$ , $D^{(\top)}$ is a $\mathbf{z}$ -parameterized universal 1-consistent reduction for $\mathcal{I}_{R}$ . Then one of the following conditions holds:

1.

there exists a variable $u$ of $\mathcal{I}_{R}$ and a $\mathbf{z}$ -parameterized unary relation $B$ such that $B\mathrel{\lhd}D_{u}^{(\top)}$ ;
2.

there exists a mighty tuple V q-definable from $R$ .

Then we show (in Section 6.5) that this universal subset can be extended to a smaller 1-consistent reduction.

Theorem 21.

Suppose $R\subseteq A^{2n+1}$ , $D^{(\top)}$ is a $\mathbf{z}$ -parameterized universal 1-consistent reduction for $\mathcal{I}_{R}$ , $u\in\operatorname{Var}(\mathcal{I}_{R})$ , $B\mathrel{\lhd}D_{u}^{(\top)}$ is a $\mathbf{z}$ -parameterized nonempty unary relation. Then one of the following conditions holds:

1.

there exists a $\mathbf{z}$ -parameterized universal 1-consistent reduction $D^{(\bot)}$ for $\mathcal{I}_{R}$ that is smaller than $D^{(\top)}$ ;
2.

there exists a mighty tuple IV q-definable from $R$ .

In both cases, there is an option that it cannot be done, but this implies that the QCSP is PSpace-hard. Combining above theorems we obtain the fundamental fact that it is sufficient to run arc-consistency algorithm on $\mathcal{I}_{R}$ to be sure that it has a solution.

Theorem 22.

Suppose $R\subseteq A^{2n+1}$ , there exists a $\mathbf{z}$ -parameterized nonempty 1-consistent reduction for $\mathcal{I}_{R}$ . Then one of the following conditions holds:

1.

$\mathcal{I}_{R}$ has a solution for every $\mathbf{z}\in A^{|A|}$ ;
2.

there exists a mighty tuple IV or V q-definable from $R$ .

Proof.

Assume the converse. Suppose there does not exist a mighty tuple IV or V q-definable from $R$ and $\mathcal{I}_{R}$ has no solutions for some $\mathbf{z}$ . By Theorem 19 there exists a $\mathbf{z}$ -parameterized universal 1-consistent reduction $D^{(1)}$ for $\mathcal{I}_{R}$ . By Theorem 20 there exists a variable $u$ of $\mathcal{I}_{R}$ and a $\mathbf{z}$ -parameterized unary relation $B$ such that $B\mathrel{\lhd}D_{u}^{(1)}$ . By Theorem 21 there exists a smaller $\mathbf{z}$ -parameterized 1-consistent universal reduction $D^{(2)}$ . Then iteratively applying Theorems 20 and 21 we build reductions $D^{(1)},D^{(2)},\dots,D^{(s)}$ . Since we never stop and we cannot reduce domains forever, we get a contradiction. ∎

Summarizing above theorems we obtain the following theorem.

Theorem 23.

Suppose $\Gamma$ is a constraint language on a finite set $A$ . Then one of the following conditions holds:

1.

for any No-instance $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}\;\Psi$ of $\operatorname{QCSP}(\Gamma)$ there exists $S\subseteq A^{n}$ with $|S|\leqslant|A|^{2}\cdot(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ such that $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}((x_{1},\dots,x_{n})\in S\rightarrow\Psi)$ does not hold;
2.

there exists a mighty tuple III, IV, or V q-definable over $\Gamma$ .

Proof.

Below we assume that the second condition does not hold, that is, there does not exist a mighty tuple III, IV, or V q-definable over $\Gamma$ .

Let us prove condition 1. Put $R(y_{0},y_{1},\dots,y_{n},x_{1},\dots,x_{n})=\Psi$ . By Lemma 13, $\mathcal{I}_{R}$ has no solutions for some $\mathbf{z}\in A^{|A|}$ . Then by Theorem 22 there does not exist a $\mathbf{z}$ -parameterized nonempty 1-consistent reduction for $\mathcal{I}_{R}$ . Applying Theorem 18 to $\mathcal{I}_{R}$ we obtain that only case 2 is possible.

Thus, there exists a subinstance $\mathcal{J}\subseteq\mathcal{I}_{R}$ with at most $(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ variables not having a solution for some $\mathbf{z}=(b_{1},\dots,b_{|A|})$ . Let us define an appropriate $S\subseteq A^{n}$ . For a tuple $(a_{1},\dots,a_{i})$ and $i\in\{0,1,\dots,n-2\}$ by $E(a_{1},\dots,a_{i})$ we denote the set

\{(b_{a_{1}},b_{a_{2}},\dots,b_{a_{i}},\underbrace{c,d,d,\dots,d}_{n-i})\mid c,d\in A\}.

Put $E(a_{1},\dots,a_{n})=\{(b_{a_{1}},\dots,b_{a_{n}})\}$ and $E(a_{1},\dots,a_{n-1})=\{(b_{a_{1}},\dots,b_{a_{n-1}},c)\mid c\in A\}$ . Then put $S=\bigcup\limits_{y_{i}^{a_{1},\dots,a_{i}}\in\operatorname{Var}(\mathcal{J})}E(a_{1},\dots,a_{i})$ . We have

|S|\leqslant|A|^{2}\cdot|\operatorname{Var}(\mathcal{J})|\leqslant|A|^{2}\cdot(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}.

Observe that tuples from $S$ cover all the constraints of $\mathcal{J}$ for $\mathbf{z}=(b_{1},\dots,b_{|A|})$ . Therefore the CSP instance $\bigwedge\limits_{(a_{1},\dots,a_{n})\in S}R(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{n}^{a_{1},\dots,a_{n}},a_{1},\dots,a_{n})$ cannot be satisfied. Notice that the existence of a winning strategy for the EP in

\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}((x_{1},\dots,x_{n})\in S\rightarrow\Psi)

is equivalent to the satisfiability of $\bigwedge\limits_{(a_{1},\dots,a_{n})\in S}R(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{n}^{a_{1},\dots,a_{n}},a_{1},\dots,a_{n})$ . Hence, the sentence $\exists y_{0}\forall x_{1}\exists y_{1}\dots\forall x_{n}\exists y_{n}((x_{1},\dots,x_{n})\in S\rightarrow\Psi)$ does not hold, which completes the proof. ∎

Combining the above theorem with Theorem 14 and Lemmas 16 and 17, we derive Theorem 3. Assuming that $\mathrm{PSpace}\neq\Pi_{2}^{P}$ , we obtain the following classification of PSpace-complete languages that is a bit stronger than the classification in Theorem 6:

Theorem 24.

Suppose $\Gamma$ is a constraint language on a finite set $A$ . Then the following conditions are equivalent:

1.

$\operatorname{QCSP}(\Gamma)$ is PSpace-complete;
2.

$\Gamma$ q-defines a mighty tuple I;
3.

$\Gamma$ q-defines a mighty tuple II or V.

Proof.

2 implies 1 by Theorem 14. 3 implies 2 by Lemmas 16 and 17. Let us prove that 1 implies 3. As we assumed that $\mathrm{PSpace}\neq\Pi_{2}^{P}$ , and the first case of Theorem 23 implies $\Pi_{2}^{P}$ -membership, we derive the second case of Theorem 23, that is, there exists a mighty tuple III, IV, or V q-definable from $\Gamma$ . Using lemma 15 there exists a mighty tuple II or V q-definable from $\Gamma$ , which compltes the proof. ∎

For constraint languages containing all constant relations an easier classification of PSpace-complete languages is proved in Section 7.5.

Lemma 25.

Suppose $\Gamma\supseteq\{x=a\mid a\in A\}$ is a set of relations on $A$ . Then the following conditions are equivalent:

1.

$\Gamma$ q-defines a mighty tuple I;
2.

$\Gamma$ q-defines a mighty tuple II;
3.

there exist an equivalence relation $\sigma$ on $D\subseteq A$ and $B,C\subsetneq A$ such that $B\cup C=A$ and $\Gamma$ q-defines the relations $(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in B))$ and $(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in C))$ .

The above lemma together with Theorem 14 implies Theorem 7.

As we prove in Lemmas 15 and 16, mighty tuples II, III, and IV are equivalent, and each of them implies a mighty tuple I. Nevertheless, it is still not clear whether a mighty tuple I implies a mighty tuple II. Moreover, we do not know how to q-define a mighty tuple II or V from a mighty tuple I even though we know it is possible by Theorem 24. Considering constraint languages with all constant relations from Lemma 25, we can observe that a mighty tuple V cannot be derived from the relations $\sigma(y_{1},y_{2})\vee(x\in B)$ , $\sigma(y_{1},y_{2})\vee(x\in C)$ and constant relations. Hence, it is not always true that a mighty tuple I implies a mighty tuple V, but we can formulate the following conjecture.

Conjecture 1.

$\operatorname{QCSP}(\Gamma)$ is PSpace-complete if and only if $\Gamma$ q-defines a mighty tuple II.

6 Finding a solution of $\mathcal{I}_{R}$

6.1 Definitions

We say that $u_{0}-C_{1}-u_{1}-\dots-C_{\ell}-u_{\ell}$ is a path in a CSP instance $\mathcal{I}$ if $C_{i}$ is a constraint of $\mathcal{I}$ and $u_{i},u_{i+1}\in\operatorname{Var}(C_{i})$ for every $i$ . The number $\ell$ is called the length of the path. We say that an instance is connected if any two variables are connected by a path. We say that an instance is a tree-instance if it is connected and there is no path $u_{0}-C_{1}-u_{1}-\dots-u_{\ell-1}-C_{\ell}-u_{\ell}$ such that $\ell\geqslant 2$ , $u_{0}=u_{l}$ , and all the constraints $C_{1},\ldots,C_{\ell}$ are different.

An instance $\mathcal{I}^{\prime}$ is called a universal weakening of $\mathcal{I}$ if $\mathcal{I}$ can be obtained from $\mathcal{I}^{\prime}$ by replacing some constraint relations by their universal subsets over $\{R\}$ . We also denote it by $\mathcal{I}\mathrel{\trianglelefteq}\mathcal{I}^{\prime}$ .

A ( $\mathbf{z}$ -parameterized) CSP instance $\mathcal{I}^{\prime}$ is called a covering of a ( $\mathbf{z}$ -parameterized) CSP instance $\mathcal{I}$ if there exists a mapping $\phi:\operatorname{Var}(\mathcal{I}^{\prime})\to\operatorname{Var}(\mathcal{I})$ such that for every constraint $S(u_{1},\dots,u_{t})$ of $\mathcal{I}^{\prime}$ the constraint $S(\phi(u_{1}),\dots,\phi(u_{t}))$ is in $\mathcal{I}$ . We say that $\phi(u)$ is the parent of $u$ and $u$ is a child of $\phi(u)$ . The same child/parent terminology will also be applied to constraints. An instance is called a tree-covering if it is a covering and also a tree-instance. Notice that reductions for an instance can be naturally extended to their coverings.

For a ( $\mathbf{z}$ -parameterized) instance $\mathcal{I}$ and a ( $\mathbf{z}$ -parameterized) reduction $D^{(\top)}$ by $\mathcal{I}^{(\top)}$ we denote the instance whose constraints are reduced to $D^{(\top)}$ .

All the variables of $\mathcal{I}_{R}$ can be drawn as a tree with a root $y_{0}$ and leaves $y_{n}^{a_{1},\dots,a_{n}}$ . We assume that the root is at the top and the leaves are at the bottom. Thus, whenever we refer to a lowest/highest variable we mean $y_{i}^{a_{1},\dots,a_{i}}$ with the maximal/minimal $i$ . Also we say that a variable $y_{i}^{a_{1},\dots,a_{i}}$ is from the $i$ -th level.

For a ( $\mathbf{z}$ -parameterized) instance $\mathcal{I}$ and some variables $u_{1},\dots,u_{m}\in\operatorname{Var}(\mathcal{I})$ by $\mathcal{I}(u_{1},\dots,u_{m})$ we denote the set of all tuples $(a_{1},\dots,a_{m})$ such that $\mathcal{I}$ has a solution with $u_{i}=a_{i}$ for every $i$ . Thus, $\mathcal{I}(u_{1},\dots,u_{m})$ defines an $m$ -ary ( $\mathbf{z}$ -parameterized) relation. Notice that $\mathcal{I}(u_{1},\dots,u_{m})$ is q-definable over the relations appearing in $\mathcal{I}$ .

Suppose $D^{(\top)}$ is a $\mathbf{z}$ -parameterized universal reduction of $\mathcal{I}_{R}$ , that is, for every variable $u$ we have $D_{u}^{(\top)}\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}D_{u}^{(\top,0)}$ . As we can repeat elements in the sequence $\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}$ and any sequence longer than $|A|$ has repetitions, we have $B\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}_{|A|}C\Leftrightarrow B\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}C$ for any $B,C\subseteq A$ . The sequence witnessing that $D_{u}^{(\top)}\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}_{|A|}D_{u}^{(\top,0)}$ we denote by

D_{u}^{(\top,|A|)},D_{u}^{(\top,|A|-1)},\dots,D_{u}^{(\top,1)},D_{u}^{(\top,0)},

where $D_{u}^{(\top,|A|)}=D_{u}^{(\top)}$ . Also, notice that the reduction $D^{(\top,i)}$ is defined independently on different variables, that is why we should not expect it to be 1-consistent.

To simplify the explanation we give names to some constraints we will need later.

	$\displaystyle C_{S,\top}^{a_{1},\dots,a_{m}}:=\;\;\;\;\;$	$\displaystyle\left(\mathcal{S}_{R}^{m}(y_{0},y_{1}^{a_{1}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}})\wedge\bigwedge\limits_{i=0}^{m}y_{i}\in D_{y_{i}^{a_{1},\dots,a_{i}}}^{(\top)}\right)$
	$\displaystyle C_{W,\top}^{a_{1},\dots,a_{m}}:=\;\;\;\;\;$	$\displaystyle\left(\mathcal{W}_{R}^{m}(y_{0},y_{1}^{a_{1}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}})\wedge\bigwedge\limits_{i=0}^{m}y_{i}\in D_{y_{i}^{a_{1},\dots,a_{i}}}^{(\top)}\right)$
	$\displaystyle C_{W,\top,j}^{a_{1},\dots,a_{m}}:=\;\;\;\;\;$	$\displaystyle\left(\exists y_{m}^{a_{1},\dots,a_{m}}\;\mathcal{W}_{R}^{m}(y_{0},y_{1}^{a_{1}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}})\wedge\right.$
		$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left.\bigwedge\limits_{i=0}^{m-1}y_{i}\in D_{y_{i}^{a_{1},\dots,a_{i}}}^{(\top)}\wedge y_{m}^{a_{1},\dots,a_{m}}\in D_{y_{m}^{a_{1},\dots,a_{m}}}^{(\top,j)}\right)$

Notice that $C_{S,\top}^{a_{1},\dots,a_{m}}$ and $C_{W,\top}^{a_{1},\dots,a_{m}}$ have $m+1$ $y$ -variables, but $C_{W,\top,j}^{a_{1},\dots,a_{m}}$ has only $m$ $y$ -variables. Then $\mathcal{I}_{R}^{(\top)}$ is the instance consisting of the constraints $C_{S,\top}^{a_{1},\dots,a_{m}}$ , where $m\in\{0,1,\dots,n\}$ and $a_{1},\dots,a_{m}\in A$ .

By Lemma 11, $C_{S,\top}^{a_{1},\dots,a_{m}}\mathrel{\trianglelefteq}C_{W,\top}^{a_{1},\dots,a_{m}}$ and $C_{W,\top,j+1}^{a_{1},\dots,a_{m}}\mathrel{\trianglelefteq}C_{W,\top,j}^{a_{1},\dots,a_{m}}$ .

6.2 Auxiliary statements

Lemma 11.

Suppose

	$\displaystyle W(y_{1},\dots,y_{t})$	$\displaystyle=\exists u_{1}\exists u_{2}\dots\exists u_{\ell}\;\;\;W_{1}(z_{1,1},\dots,z_{1,n_{1}})\wedge\dots\wedge W_{m}(z_{m,1},\dots,z_{m,n_{m}}),$
	$\displaystyle S(y_{1},\dots,y_{t})$	$\displaystyle=\exists u_{1}\exists u_{2}\dots\exists u_{\ell}\;\;\;S_{1}(z_{1,1},\dots,z_{1,n_{1}})\wedge\dots\wedge S_{m}(z_{m,1},\dots,z_{m,n_{m}}),$

where each $z_{i,j}\in\{y_{1},\dots,y_{t},u_{1},\dots,u_{\ell}\}$ , and $S_{i}\mathrel{\trianglelefteq}^{\Sigma}W_{i}$ for every $i$ . Then $S\mathrel{\trianglelefteq}^{\Sigma}W$ .

Proof.

Let $S_{i}\mathrel{\trianglelefteq}^{\Sigma}W_{i}$ be witnessed by a relation $R_{i}\subseteq A^{n_{i}+k_{i}}$ . Let $k=|A|$ . Define the relation $R$ witnessing that $S\mathrel{\trianglelefteq}^{\Sigma}W$ by

	$\displaystyle R(y_{1},\dots,y_{t},x_{1},\dots,$	$\displaystyle x_{k})=$
	$\displaystyle\exists u_{1}\exists u_{2}\dots\exists u_{\ell}$	$\displaystyle\left(\bigwedge\limits_{i=1}^{m}\;\bigwedge\limits_{\phi\colon[k_{i}]\to[k]}R_{i}(z_{i,1},\dots,z_{i,n_{i}},x_{\phi(1)},\dots,x_{\phi(k_{i})})\wedge\bigwedge\limits_{i=1}^{m}W_{i}(z_{i,1},\dots,z_{i,n_{i}})\right)$

Notice that $W_{i}$ is q-definable from $R_{i}$ , hence $R$ is q-definable over $\Sigma$ . ∎

Lemma 26 ([21], Lemma 5.6).

Suppose $D^{(\top)}$ is a reduction for an instance $\mathcal{I}$ , $D^{(\bot)}$ is an inclusion maximal 1-consistent reduction for $\mathcal{I}$ such that $D_{u}^{(\bot)}\subseteq D_{u}^{(\top)}$ for every $u$ . Then for every variable $y\in\operatorname{Var}(\mathcal{I})$ there exists a tree-covering $\Upsilon_{y}$ of $\mathcal{I}$ such that $\Upsilon_{y}^{(\top)}(y)$ defines $D_{y}^{(\bot)}$ .

The above lemma can be generalized for $\mathbf{z}$ -parameterized reductions as follows:

Lemma 27.

Suppose $D^{(\top)}$ is a $\mathbf{z}$ -parameterized reduction for a $\mathbf{z}$ -parameterized instance $\mathcal{I}$ , $D^{(\bot)}$ is an inclusion maximal $\mathbf{z}$ -parameterized 1-consistent reduction for $\mathcal{I}$ such that $D^{(\bot)}_{u}\subseteq D^{(\top)}_{u}$ for every $u$ . Then for every variable $y\in\operatorname{Var}(\mathcal{I})$ there exists a tree-covering $\Upsilon_{y}$ of $\mathcal{I}$ such that $\prescript{\mathbf{z}}{}{\Upsilon}_{y}^{(\top)}(y)$ defines $\prescript{\mathbf{z}}{}{D}_{y}^{(\bot)}$ for every $\mathbf{z}$ .

Proof.

By Lemma 26 for every $\mathbf{z}\in A^{|A|}$ and $v$ there exists a tree-covering $\Upsilon_{v,\mathbf{z}}$ such that $\prescript{\mathbf{z}}{}{\Upsilon_{v,\mathbf{z}}^{(\top)}}(v)$ defines $\prescript{\mathbf{z}}{}{D_{v}^{(\bot)}}$ . Let $\Upsilon_{v}$ be $\bigwedge_{\mathbf{z}\in A^{|A|}}\Upsilon_{v,\mathbf{z}}$ , where we assume that the only common variable of $\Upsilon_{v,\mathbf{z}_{1}}$ and $\Upsilon_{v,\mathbf{z}_{2}}$ , if $\mathbf{z}_{1}\neq\mathbf{z}_{2}$ , is $v$ . Then $\Upsilon_{v}$ is a tree-covering of $\mathcal{I}$ and $\prescript{\mathbf{z}}{}{\Upsilon_{v}^{(\top)}}(v)$ defines $\prescript{\mathbf{z}}{}{D_{v}^{(\bot)}}$ for every $\mathbf{z}$ . ∎

We can always take a trivial reduction $\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}=A$ for every $\mathbf{z}$ and $u$ , and derive the following lemma.

Lemma 28.

Suppose $D^{(\bot)}$ is an inclusion maximal $\mathbf{z}$ -parameterized 1-consistent reduction for a $\mathbf{z}$ -parameterized instance $\mathcal{I}$ . Then for every variable $u\in\operatorname{Var}(\mathcal{I})$ there exists a tree-covering $\Upsilon_{u}$ of $\mathcal{I}$ such that $\prescript{\mathbf{z}}{}{\Upsilon}_{u}(u)$ defines $\prescript{\mathbf{z}}{}{D}_{u}^{(\bot)}$ for every $\mathbf{z}$ .

6.3 Finding a 1-consistent reduction

In this section we prove that either there exists a 1-consistent reduction for $\mathcal{I}_{R}$ , or there exists a polynomial size subinstance of $\mathcal{I}_{R}$ without a solution, or we can build a mighty tuple III that guarantees PSpace-hardness. To be able to simplify our instance to a polynomial size we will need even stronger relations. Put

	$\displaystyle\mathcal{\widetilde{W}}_{R}^{m}(y_{0},\dots,y_{m},x_{1},\dots,x_{m})$	$\displaystyle=\bigwedge\limits_{i=0}^{m}\mathcal{W}_{R}^{i}(y_{0},\dots,y_{i},x_{1},\dots,x_{i}),$
	$\displaystyle\mathcal{\widetilde{S}}_{R}^{m}(y_{0},\dots,y_{m},x_{1},\dots,x_{m})$	$\displaystyle=\mathcal{S}_{R}^{m}(y_{0},\dots,y_{m},x_{1},\dots,x_{m})\wedge\bigwedge\limits_{i=0}^{m-1}\mathcal{W}_{R}^{i}(y_{0},\dots,y_{i},x_{1},\dots,x_{i}).$

The following lemma follows immediately from the definition and Lemmas 12 and 11.

Lemma 29.

Suppose $R\subseteq A^{2n+1}$ , then $\mathcal{\widetilde{S}}_{R}^{m}\mathrel{\trianglelefteq}\mathcal{{\widetilde{W}}}_{R}^{m}$ .

Denote

	$\displaystyle\mathcal{\widetilde{I}}_{R}=\bigwedge\limits_{\begin{subarray}{c}m\in\{0,1,\dots,n\}\\ a_{1},\dots,a_{m}\in A\end{subarray}}$	$\displaystyle\mathcal{\widetilde{S}}_{R}^{m}(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}})\wedge$
		$\displaystyle\bigwedge\limits_{\begin{subarray}{c}m\in\{0,1,\dots,n\}\\ a_{1},\dots,a_{m}\in A\end{subarray}}\mathcal{\widetilde{W}}_{R}^{m}(y_{0},y_{1}^{a_{1}},y_{2}^{a_{1},a_{2}},\dots,y_{m}^{a_{1},\dots,a_{m}},z_{a_{1}},\dots,z_{a_{m}}).$

Notice that $\mathcal{\widetilde{I}}_{R}$ is obtained from $\mathcal{I}_{R}$ by adding constraints that are satisfied by any solution of $\mathcal{I}_{R}$ . Hence $\mathcal{\widetilde{I}}_{R}$ is satisfiable if and only if $\mathcal{I}_{R}$ is satisfiable.

First, we prove some technical lemmas showing the connection of the length of a path and the size of a tree-covering.

Lemma 30.

Suppose $\mathcal{T}$ is a $\mathbf{z}$ -parameterized tree-instance such that $\prescript{\mathbf{z}}{}{\mathcal{T}}$ has no solutions for some $\mathbf{z}$ , but if we remove any constraint from $\mathcal{T}$ we get an instance with a solution for every $\mathbf{z}$ . Then each variable appears in $\mathcal{T}$ at most $|A|$ times.

Proof.

Let $u$ be some variable of $\mathcal{T}$ . Since $\mathcal{T}$ is a tree-instance, we can split it into tree-subinstances in $u$ , that is, for any constraint $C$ containing $u$ we take the (maximal) tree-subinstance containing $C$ but not the other constraints containing $u$ . Let $\Phi_{1},\dots,\Phi_{s}$ be the subinstances we obtain if we split the instance $\mathcal{T}$ in $u$ . By $B_{i}$ we denote the $\mathbf{z}$ -parameterized unary relation defined by $\Phi_{i}(u)$ . Then there exists $\mathbf{z}$ such that $\bigcap\limits_{i\in[s]}\prescript{\mathbf{z}}{}{B}_{i}=\varnothing$ . Since removing any constraint from $\mathcal{T}$ gives an instance with a solution, $\bigcap\limits_{i\in[s]\setminus\{j\}}\prescript{\mathbf{z}}{}{B}_{i}\neq\varnothing$ for every $j\in[s]$ . Therefore $s\leqslant|A|$ , which completes the proof. ∎

Lemma 31.

Suppose $\mathcal{T}$ is a tree-instance having $N\geqslant 2$ variables, the arity of every constraint of $\mathcal{T}$ is at most $n$ , and every variable appears at most $|A|$ times. Then there exists a path in $\mathcal{T}$ of length at least $\lceil{\log_{k}(N\cdot(k-1)+1)}\rceil$ , where $k=(n-1)\cdot|A|$ .

Proof.

We prove even a stronger claim by induction on $N$ . We prove that there exists a path of length $\lceil{\log_{k}(N\cdot(k-1)+1)}\rceil$ starting with any variable $u$ .

Suppose $u$ appears in constraints $C_{1},C_{2},\dots,C_{s}$ , where $s\leqslant|A|$ . Let $V$ be the set of all variables appearing in $C_{1},\dots,C_{s}$ except for $u$ . Notice that every variable $v\in V$ appears in exactly one constraint $C_{i}$ . By $\Phi_{v}$ we denote the (maximal) tree-subinstance of $\mathcal{T}$ containing all the constraints with $v$ but constraints from $\{C_{1},\dots,C_{s}\}$ . Then $\mathcal{T}=\bigwedge_{v\in V}\Phi_{v}\wedge\bigwedge_{i\in[s]}C_{i}$ , and $\Phi_{v_{1}}$ and $\Phi_{v_{2}}$ do not share any variables if $v_{1}\neq v_{2}$ . Since $|V|\leqslant s\cdot(n-1)\leqslant k$ , there exists $v\in V$ such that $\Phi_{v}$ contains at least $(N-1)/k$ variables. By the inductive assumption $\Phi_{v}$ contains a path starting with $v$ of length at least $\lceil\log_{k}(((N-1)/k)\cdot(k-1)+1)\rceil$ . Then $\mathcal{T}$ has a path of length at least

1+\lceil\log_{k}(((N-1)/k)\cdot(k-1)+1)\rceil=\lceil\log_{k}((N-1)\cdot(k-1)+k)\rceil=\lceil\log_{k}(N\cdot(k-1)+1)\rceil.

∎

Suppose $\mathcal{T}$ is a tree-covering of $\mathcal{\widetilde{I}}_{R}$ . We define several transformations of $\mathcal{T}$ , which we will apply to make it easier.

(w)

replace a constraint ${\mathcal{\widetilde{S}}}_{R}^{i}(u_{0},u_{1},\dots,u_{i})$ by ${\mathcal{\widetilde{W}}}_{R}^{i}(u_{0},u_{1},\dots,u_{i})$ ;
(s)

if $u_{i}$ appears only once in $\mathcal{T}$ in a constraint ${\mathcal{\widetilde{W}}}_{R}^{i}(u_{0},u_{1},\dots,u_{i-1},u_{i})$ then replace this constraint by ${\mathcal{\widetilde{S}}}_{R}^{i}(u_{0},u_{1},\dots,u_{i-1})$ ;
(r)

remove some constraint;
(j)

suppose $u_{i}\in\operatorname{Var}(\mathcal{T})$ appears in constraints $Q(u_{0},u_{1},\dots,u_{i},\dots,u_{j})$ and ${\mathcal{\widetilde{W}}}_{R}^{i}(v_{0},v_{1},\dots,v_{i-1},u_{i})$ , where $Q\in\{{\mathcal{\widetilde{W}}}_{R}^{j},{\mathcal{\widetilde{S}}}_{R}^{j}\}$ and $j\geqslant i$ . Then we identify the variables $v_{k}=u_{k}$ for every $k\in\{0,1,\dots,i-1\}$ and remove the constraint ${\mathcal{\widetilde{W}}}_{R}^{i}(v_{0},v_{1},\dots,v_{i-1},u_{i})$ .

Notice that transformations (w) and (r) make the instance weaker (more solutions) but (s) and (j) make the instance stronger (less solutions). The next lemma shows that if a tree-covering without a solution cannot be simplified using the transformations and it is large enough then we can cut it into three pieces satisfying nice properties.

Lemma 32.

Suppose

1.

$R\subseteq A^{2n+1}$ , where $n>0$ ;
2.

$\mathcal{T}$ is a tree-covering of $\mathcal{\widetilde{I}}_{R}$ having at least $(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ variables;
3.

$\prescript{\mathbf{z}_{0}}{}{\mathcal{T}}$ has no solutions for some $\mathbf{z}_{0}\in A^{|A|}$ ;
4.

applying transformations (w) and (r) to $\mathcal{T}$ gives an instance with a solution for every $\mathbf{z}\in A^{|A|}$ ;
5.

transformations (s) and (j) are not applicable.

Then $\mathcal{T}$ can be divided into 3 nonempty connected parts $\mathcal{I}_{1}$ , $\mathcal{I}_{2}$ , and $\mathcal{I}_{3}$ such that

(l1)

the only common variable of $\mathcal{I}_{1}$ and $\mathcal{I}_{2}$ is a variable $u$ ;
(r1)

the only common variable of $\mathcal{I}_{2}$ and $\mathcal{I}_{3}$ is a variable $v$ ;
(l2)

$\prescript{\mathbf{z}}{}{\mathcal{I}_{1}(u)}=(\prescript{\mathbf{z}}{}{\mathcal{I}_{1}}\wedge\prescript{\mathbf{z}}{}{\mathcal{I}_{2}})(v)$ for every $\mathbf{z}\in A^{|A|}$ ;
(r2)

$\prescript{\mathbf{z}}{}{\mathcal{I}_{3}(v)}=(\prescript{\mathbf{z}}{}{\mathcal{I}_{2}}\wedge\prescript{\mathbf{z}}{}{\mathcal{I}_{3}})(u)$ for every $\mathbf{z}\in A^{|A|}$ ;
(m)

$\mathcal{I}_{2}$ contains a constraint $\mathcal{\widetilde{S}}_{R}^{i}(v_{0},\dots,v_{i})$ with $i<n-1$ .

Proof.

Let $N=|\operatorname{Var}(\mathcal{T})|$ , $k=n\cdot|A|$ . Notice that the arity of constraints in $\mathcal{\widetilde{I}}_{R}$ is at most $n+1$ if we ignore $\mathbf{z}$ -variables. Since we cannot remove any constraints (property 4), Lemmas 30 and 31 imply that there exists a path $u_{0}-C_{1}-u_{1}-C_{2}-\dots-C_{s}-u_{s}$ with

s\geqslant\lceil{\log_{k}(N\cdot(k-1)+1)}\rceil\geqslant\log_{k}N\geqslant 2^{{2|A|}^{|A|+1}}={(2^{|A|})}^{2|A|^{|A|}}>1+{(2^{|A|}-1)}^{2|A|^{|A|}}.

For every $i\in[s-1]$ we split $\mathcal{T}$ into two tree-coverings $\Phi_{i}$ and $\Psi_{i}$ as follows. If we split $\mathcal{T}$ in the variable $u_{i}$ , then the part containing $C_{i+1}$ goes to $\Psi_{i}$ and all the remaining parts go to $\Phi_{i}$ . Thus, the only common variable of $\Phi_{i}$ and $\Psi_{i}$ is $u_{i}$ . Let $B_{i}$ and $C_{i}$ be the $\mathbf{z}$ -parameterized unary relations defined by $\Phi_{i}(u_{i})$ and $\Psi_{i}(u_{i})$ , respectively.

Since the transformation (r) cannot be applied, $\prescript{\mathbf{z}}{}{B}_{i}$ and $\prescript{\mathbf{z}}{}{C}_{i}$ are nonempty for every $i\in[s-1]$ and $\mathbf{z}\in A^{|A|}$ . Since $\prescript{\mathbf{z}_{0}}{}{\mathcal{T}}$ has no solutions, $\prescript{\mathbf{z}_{0}}{}{B}_{i}\cap\prescript{\mathbf{z}_{0}}{}{C}_{i}=\varnothing$ for every $i\in[s-1]$ .

There are exactly ${(2^{|A|}-1)}^{|A|^{|A|}}$ distinct nonempty $\mathbf{z}$ -parameterized unary relations. Since $s>1+{(2^{|A|}-1)}^{2|A|^{|A|}}$ , there should be $1\leqslant i<i^{\prime}\leqslant s-1$ such that $\prescript{\mathbf{z}}{}{B}_{i}=\prescript{\mathbf{z}}{}{B}_{i^{\prime}}$ and $\prescript{\mathbf{z}}{}{C}_{i}=\prescript{\mathbf{z}}{}{C}_{i^{\prime}}$ for every $\mathbf{z}\in A^{|A|}$ . By $\mathcal{I}_{1}$ we denote $\Phi_{i}$ , by $\mathcal{I}_{3}$ we denote $\Psi_{i^{\prime}}$ , and by $\mathcal{I}_{2}$ we denote the intersection of $\Psi_{i}$ and $\Phi_{i^{\prime}}$ . Then conditions (l1), (l2), (r1), and (r2) are satisfied for $u=u_{i}$ and $v=u_{i^{\prime}}$ . It remains to prove property (m).

We say that a variable is a leaf if it occurs only in one constraint. We say that a constraint is a leaf-constraint if only one of its variables is not a leaf. Suppose a leaf-constraint is ${\mathcal{\widetilde{W}}}_{R}^{j}(u_{0},u_{1},\dots,u_{j})$ for some $j$ . If its nonleaf variable is $u_{j^{\prime}}$ , where $j^{\prime}<j$ , then we can use the transformation (s). If its nonleaf variable is $u_{j}$ , then from the definition of ${\mathcal{\widetilde{S}}}_{R}^{i}$ and ${\mathcal{\widetilde{W}}}_{R}^{i}$ we derive that this leaf-constraint can be removed from the instance not changing the property that it has no solutions for some $\mathbf{z}$ . Both situations contradict our assumptions that (r) and (s) are not applicable. Hence, any leaf-constraint of $\mathcal{T}$ is of the form ${\mathcal{\widetilde{S}}}_{R}^{j}(v_{0},v_{1},\dots,v_{j})$ . Notice that $j$ must be smaller than $n-1$ because otherwise ${\mathcal{\widetilde{S}}}_{R}^{j}={\mathcal{\widetilde{W}}}_{R}^{j}$ and the transformation (w) does not really change the instance and can always be applied. Consider several cases:

Case 1. The constraint $C_{i+1}$ is ${\mathcal{\widetilde{S}}}_{R}^{j}(v_{0},v_{1},\dots,v_{j})$ for some $j$ . Then $C_{i+1}$ is the required constraint to satisfy (m).

Case 2. The constraint $C_{i+1}$ is ${\mathcal{\widetilde{W}}}_{R}^{j}(v_{0},v_{1},\dots,v_{j})$ where $v_{j}\neq u_{i}$ and $v_{j}\neq u_{i+1}$ . If $v_{j}$ is a leaf, we can apply the transformation (s). Otherwise, consider a part of $\mathcal{I}_{2}$ containing $v_{j}$ . This part must contain a leaf-constraint, which implies property (m).

Case 3. The constraint $C_{i+1}$ is ${\mathcal{\widetilde{W}}}_{R}^{j}(v_{0},v_{1},\dots,v_{j-1},u_{i})$ . Then we apply transformation (j) to $C_{i}$ and $C_{i+1}$ , which contradicts our assumptions.

Case 4. The constraint $C_{i+1}$ is ${\mathcal{\widetilde{W}}}_{R}^{j}(v_{0},v_{1},\dots,v_{j-1},u_{i+1})$ . Then we apply transformation (j) to $C_{i+1}$ and $C_{i+2}$ , which again contradicts our assumptions. ∎

Corollary 33.

Suppose

1.

$R\subseteq A^{2n+1}$ , where $n>0$ ;
2.

$\mathcal{T}$ is a tree-covering of $\mathcal{\widetilde{I}}_{R}$ with the minimal number of variables such that $\prescript{\mathbf{z}}{}{\mathcal{T}}$ has no solutions for some $\mathbf{z}$ ;
3.

$|\operatorname{Var}(\mathcal{T})|\geqslant(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ .

Then $R$ q-defines a mighty tuple III.

Proof.

We apply transformations (w) and (r) to $\mathcal{T}$ while we can maintain the condition that $\prescript{\mathbf{z}}{}{\mathcal{T}}$ has no solutions for some $\mathbf{z}$ . Also, we apply transformations (s) and (j) when applicable. Notice that we cannot apply these transformations forever, and we never increase the number of variables. Thus, the obtained tree-instance $\mathcal{T}^{\prime}$ still contains the minimal number of variables and satisfies the same conditions.

By Lemma 32 we can split $\mathcal{T}^{\prime}$ into 3 parts $\mathcal{I}_{1}$ , $\mathcal{I}_{2}$ , and $\mathcal{I}_{3}$ satisfying conditions (l1), (r1), (l2), (r2), and (m). Let $\mathcal{I}_{2}^{\prime}$ be obtained from $\mathcal{I}_{2}$ by replacing the constraint $\mathcal{\widetilde{S}}_{R}^{i}(v_{0},\dots,v_{i})$ coming from condition (m) by $\mathcal{\widetilde{W}}_{R}^{i}(v_{0},\dots,v_{i})$ . Notice that $\mathcal{I}_{1}\wedge\mathcal{I}_{2}^{\prime}\wedge\mathcal{I}_{3}$ has a solution for every $\mathbf{z}$ . Let $\mathcal{I}_{1}(u)$ define $B$ , $\mathcal{I}_{3}(v)$ define $C$ , $\mathcal{I}_{2}(u,v)$ define $S$ , and $\mathcal{I}_{2}^{\prime}(u,v)$ define $W$ . By Lemma 29, $\mathcal{I}_{2}\mathrel{\trianglelefteq}\mathcal{I}_{2}^{\prime}$ , hence by Lemma 11 we have $S\mathrel{\trianglelefteq}W$ . Let $Q$ be a $(\mathbf{z},\alpha)$ -parameterized relation q-definable from $R$ such that $Q^{\forall}=W$ and $Q^{\forall\forall}=S$ . Let us show that $(Q,B,C)$ is a mighty tuple III. Condition 1 follows from the fact that the transformation (r) gives an instance with a solution for every $\mathbf{z}$ . Condition 2 follows from (l2), condition 3 follows from (r2), condition 4 follows from the existence of a solution of $\mathcal{I}_{1}\wedge\mathcal{I}_{2}^{\prime}\wedge\mathcal{I}_{3}$ for every $\mathbf{z}$ . Condition 5 follows from the fact that $\prescript{\mathbf{z}_{0}}{}{\mathcal{T}^{\prime}}=\prescript{\mathbf{z}_{0}}{}{\mathcal{I}_{1}}\wedge\prescript{\mathbf{z}_{0}}{}{\mathcal{I}_{2}}\wedge\prescript{\mathbf{z}_{0}}{}{\mathcal{I}_{3}}$ has no solutions for some $\mathbf{z}_{0}$ . ∎

Now we are ready to prove two theorems from Section 5.

Theorem 18.

Suppose $R\subseteq A^{2n+1}$ . Then one of the following conditions holds:

1.

there exists a $\mathbf{z}$ -parameterized nonempty 1-consistent reduction for $\mathcal{I}_{R}$ ;
2.

there exists a subinstance $\mathcal{J}\subseteq\mathcal{I}_{R}$ with at most $(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ variables not having a solution for some $\mathbf{z}\in A^{|A|}$ ;
3.

there exists a mighty tuple III q-definable from $R$ .

Proof.

Let us consider a maximal $\mathbf{z}$ -parameterized 1-consistent reduction for $\mathcal{\widetilde{I}_{R}}$ . By Lemma 28 either this reduction is nonempty, or there exists a $\mathbf{z}$ -parameterized tree-covering $\mathcal{T}$ of $\mathcal{\widetilde{I}_{R}}$ such that the instance $\prescript{\mathbf{z}}{}{\mathcal{T}}$ has no solutions for some $\mathbf{z}$ . In the first case the same reduction is also a nonempty 1-consistent reduction for $\mathcal{I}_{R}$ , and we satisfied condition 1.

In the second case we consider a tree-covering $\mathcal{T}$ with the minimal number of variables. If $|\operatorname{Var}(\mathcal{T})|\geqslant(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ , then Corollary 33 implies that a mighty tuple III is q-definable from $R$ . If $|\operatorname{Var}(\mathcal{T})|<(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ , then let $\mathcal{J}$ be the subinstance of $\mathcal{I}_{R}$ containing all the constraints $C$ of $\mathcal{I}_{R}$ such that a child of each variable of $C$ appears in $\mathcal{T}$ . Notice that if $\prescript{\mathbf{z}}{}{\mathcal{J}}$ has a solution then $\prescript{\mathbf{z}}{}{\mathcal{T}}$ has a solution. Thus, $\prescript{\mathbf{z}}{}{\mathcal{J}}$ has no solutions for some $\mathbf{z}$ . Hence $|\operatorname{Var}(\mathcal{J})|\leqslant|\operatorname{Var}(\mathcal{T})|<(n\cdot|A|)^{2^{{2|A|}^{|A|+1}}}$ . ∎

Theorem 19.

Suppose $R\subseteq A^{2n+1}$ , $D^{(\top)}$ is an inclusion-maximal $\mathbf{z}$ -parameterized 1-consistent nonempty reduction for $\mathcal{I}_{R}$ . Then $D^{(\top)}$ is a universal reduction.

Proof.

Choose some variable $u\in\operatorname{Var}(\mathcal{I}_{R})$ and prove that $D^{(\top)}_{u}\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}D^{(\top,0)}_{u}$ .

By Lemma 28, there exists a tree-covering $\Upsilon_{0}$ of $\mathcal{I}_{R}$ such that $\prescript{\mathbf{z}}{}{\Upsilon}_{0}(u)$ defines $\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}$ for every $\mathbf{z}$ . We apply the following transformations to $\Upsilon_{0}$ similar to the transformations we used before:

(w)

replace a constraint ${\mathcal{S}}_{R}^{i}(u_{0},u_{1},\dots,u_{i})$ by ${\mathcal{W}}_{R}^{i}(u_{0},u_{1},\dots,u_{i})$ ;
(s)

if $u_{i}$ appears only once in the instance in a constraint ${\mathcal{W}}_{R}^{i}(u_{0},u_{1},\dots,u_{i-1},u_{i})$ , and $u_{i}\neq u$ ; then replace the constraint by ${\mathcal{S}}_{R}^{i}(u_{0},u_{1},\dots,u_{i-1})$ ;
(j)

suppose $u_{i}\in\operatorname{Var}(\mathcal{T})$ appears in constraints ${\mathcal{W}}_{R}^{i}(u_{0},u_{1},\dots,u_{i-1},u_{i})$ and ${\mathcal{W}}_{R}^{i}(v_{0},v_{1},\dots,v_{i-1},u_{i})$ , and $u\notin\{v_{1},\dots,v_{i-1}\}$ ; then we identify the variables $v_{k}=u_{k}$ for every $k\in\{0,1,\dots,i-1\}$ and remove the constraint ${\mathcal{W}}_{R}^{i}(v_{0},v_{1},\dots,v_{i-1},u_{i})$ .

Notice that transformation (w) makes the instance weaker (more solutions) but (s) and (j) make the instance stronger (less solutions).

We apply transformations (w), (s), and (j) in any order and obtain a sequence $\Upsilon_{0},\Upsilon_{1},\dots,\Upsilon_{t}$ of tree-coverings of $\mathcal{I}_{R}$ . Notice that at least one transformation is applicable unless the lowest variable of $\Upsilon_{t}$ is $u$ and $u$ appears exactly once. Let the constraint containing $u$ be ${\mathcal{W}}_{R}^{i}(u_{0},u_{1},\dots,u_{i-1},u)$ . Since $D^{(\top)}$ is a 1-consistent reduction and $\Upsilon_{t}$ is a tree-covering, $D_{u_{k}}^{(\top)}\subseteq\Upsilon_{t}(u_{k})$ for every $k\in\{0,1,\dots,i-1\}$ . Hence, $\Upsilon_{t}(u)\supseteq D_{u}^{(\top,0)}$ . Let the $\mathbf{z}$ -parameterized unary relation $C_{j}$ be defined by $\Upsilon_{j}(u)$ for $j=0,1,2,\dots,t$ . By the construction and Lemma 11 we have $C_{j}\supseteq C_{j+1}$ or $C_{j}\mathrel{\trianglelefteq}C_{j+1}$ for every $j$ . Put $E_{j}=C_{j}\cap C_{j+1}\cap\dots\cap C_{t}\cap D_{u}^{(\top,0)}$ for $j=0,1,2,\dots,t$ . Since the reduction $D^{(\top)}$ is 1-consistent and each $\Upsilon_{j}$ is a tree-covering, we have $C_{j}\supseteq D_{u}^{(\top)}$ . Hence, $E_{0}=C_{0}=\Upsilon_{0}(u)=D_{u}^{(\top)}$ and $E_{t}=D_{u}^{(\top,0)}$ . By Lemma 11 $E_{j}\mathrel{\trianglelefteq}E_{j+1}$ for every $j\in\{0,1,\dots,t\}$ . Thus, $D_{u}^{(\top)}\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}D_{u}^{(\top,0)}$ and $D^{(\top)}$ is a universal reduction. ∎

6.4 The existence of a universal subset

In this section we prove that for any 1-consistent $\mathbf{z}$ -parameterized universal reduction $D^{(\top)}$ of $\mathcal{I}_{R}$ there exists a $\mathbf{z}$ -parameterized unary relation $B$ and a variable $y_{i}^{a_{1},\dots,a_{i}}$ such that $B\mathrel{\lhd}D_{y_{i}^{a_{1},\dots,a_{i}}}^{(\top)}$ .

For a sequence $a_{1},\dots,a_{m}$ , where $m\in\{0,1,\dots,n\}$ and it can be empty, we put

{\mathcal{I}^{a_{1},\dots,a_{m}}_{R}}^{(\top)}=\bigwedge\limits_{i=m}^{n}\;\;\bigwedge\limits_{a_{m+1},\dots,a_{i}\in A}C_{S,\top}^{a_{1},\dots,a_{i}}.

Thus, ${\mathcal{I}^{a_{1},\dots,a_{m}}_{R}}^{(\top)}$ is the part of $\mathcal{I}_{R}^{(\top)}$ containing the variable $y_{m}^{a_{1},\dots,a_{m}}$ .

Theorem 20.

1.

there exists a variable $u$ of $\mathcal{I}_{R}$ and a $\mathbf{z}$ -parameterized nonempty unary relation $B$ such that $B\mathrel{\lhd}D_{u}^{(\top)}$ ;
2.

there exists a mighty tuple V q-definable from $R$ .

Proof.

Since $\mathcal{I}_{R}$ has no solutions for some $\mathbf{z}$ , ${\mathcal{I}_{R}}^{(\top)}$ also does not have solutions for some $\mathbf{z}$ . Consider maximal $m$ such that ${\mathcal{I}_{R}^{c_{1},\dots,c_{m}}}^{(\top)}$ has no solutions for some $c_{1},\dots,c_{m}\in A$ and $\mathbf{z}$ . We fix $c_{1},\dots,c_{m}$ and denote $\mathcal{I}_{0}={\mathcal{I}_{R}^{c_{1},\dots,c_{m}}}^{(\top)}$ . Then we apply the following transformations to the instance $\mathcal{I}_{0}$ while possible to obtain a sequence of instances $\mathcal{I}_{0},\mathcal{I}_{1},\dots,\mathcal{I}_{T}$ .

(w)

replace the constraint $C_{S,\top}^{a_{1},\dots,a_{i}}$ by $C_{W,\top}^{a_{1},\dots,a_{i}}$ ;
(e)

if a variable $y_{i}^{a_{1},\dots,a_{i}}$ , where $i>m$ , appears only once in the instance in a constraint $C_{W,\top}^{a_{1},\dots,a_{i}}$ , then replace $C_{W,\top}^{a_{1},\dots,a_{i}}$ by $C_{W,\top,|A|}^{a_{1},\dots,a_{i}}$ ;
(z)

replace $C_{W,\top,j}^{a_{1},\dots,a_{i}}$ by $C_{W,\top,j-1}^{a_{1},\dots,a_{i}}$ ;
(s)

replace $C_{W,\top,0}^{a_{1},\dots,a_{i}}$ by $C_{S,\top}^{a_{1},\dots,a_{i-1}}$ .

Notice that (w) replaces an instance by its universal weakening, (s) makes the instance stronger, (e) just existentially quantifies a variable that appears only once, (z) replaces a constraint by its universal weakening.

It follows from the definition that we cannot apply these transformations forever, and we can never remove the constraint $C_{W,\top}^{c_{1},\dots,c_{m}}$ . Let us show that the final instance $\mathcal{I}_{T}$ consists of just one constraint $C_{W,\top}^{c_{1},\dots,c_{m}}$ . If $\mathcal{I}_{T}$ contains some $C_{S,\top}^{a_{1},\dots,a_{i}}$ or $C_{W,\top,j}^{a_{1},\dots,a_{i}}$ , then we can apply the transformations (w) , (z), or (s), which contradicts the assumption that $\mathcal{I}_{T}$ is final. Otherwise, let $y_{i}^{a_{1},\dots,a_{i}}$ be the lowest variable of $\mathcal{I}_{T}$ . If $i>m$ , then $y_{i}^{a_{1},\dots,a_{i}}$ appears only in the constraint $C_{W,\top}^{a_{1},\dots,a_{i}}$ and we can apply transformation (e), which again contradicts the assumption. Thus, $i=m$ and the only constraint in $\mathcal{I}_{T}$ is $C_{W,\top}^{c_{1},\dots,c_{m}}$ .

Consider the last instance $\mathcal{I}_{t}$ in the sequence $\mathcal{I}_{0},\mathcal{I}_{1},\dots,\mathcal{I}_{T}$ not satisfying the following property: $\mathcal{I}_{t}$ has a solution with $y_{i}^{c_{1},\dots,c_{i}}=d$ for any $\mathbf{z}\in A^{|A|}$ , any $i\in\{0,\dots,m\}$ , and any $d\in\prescript{\mathbf{z}}{}{D}_{y_{i}^{c_{1},\dots,c_{i}}}^{(\top)}$ . We refer to this property as the subdirectness property. Since $\mathcal{I}_{0}$ does not satisfy the subdirectness property, such $t$ always exists. Notice that $\mathcal{I}_{T}$ and even $\mathcal{I}_{T-1}$ satisfy the subdirectness property, hence $t<T-1$ . By the definition of $t$ , the instance $\mathcal{I}_{t+1}$ cannot be stronger than $\mathcal{I}_{t}$ . Hence, $\mathcal{I}_{t}\mathrel{\trianglelefteq}\mathcal{I}_{t+1}$ . Consider two cases:

Case 1. $\mathcal{I}_{t}$ has a solution for every $\mathbf{z}\in A^{|A|}$ . Then consider some variable $y_{i}^{c_{1},\dots,c_{i}}$ wintessing that $\mathcal{I}_{t}$ does not have the subdirectness property. Since $\mathcal{I}_{t+1}$ has the subdirectness property, $\mathcal{I}_{t+1}(y_{i}^{c_{1},\dots,c_{i}})$ defines $D_{y_{i}^{c_{1},\dots,c_{i}}}^{(\top)}$ . Let $\mathcal{I}_{t}(y_{i}^{c_{1},\dots,c_{i}})$ define a $\mathbf{z}$ -parameterized unary relation $B$ . Since $\mathcal{I}_{t}\mathrel{\trianglelefteq}\mathcal{I}_{t+1}$ , Lemma 11 implies $B\mathrel{\lhd}D_{y_{i}^{c_{1},\dots,c_{i}}}^{(\top)}$ , which satisfies condition 1 and completes this case.

Case 2. $\mathcal{I}_{t}$ does not have a solution for some $\mathbf{z}\in A^{|A|}$ . Put $\mathcal{J}_{0}=\mathcal{I}_{t}$ . Then we apply another transformation to $\mathcal{I}_{t}$ and obtain a sequence of instances $\mathcal{J}_{0},\mathcal{J}_{1},\dots,\mathcal{J}_{s}$ , where $\mathcal{J}_{0}=\mathcal{I}_{t}$ . If a variable $u$ is a child of $y_{i}^{c_{1},\dots,c_{i}}$ , where $i\in\{0,1,\dots,m\}$ , and $u$ appears several times in $\mathcal{J}_{k}$ , then we rename some (but not all) of the variables $u$ into $u^{\prime}$ and obtain a covering $\mathcal{J}_{k+1}$ of $\mathcal{J}_{k}$ . If $\mathcal{J}_{k+1}$ has a solution for every $\mathbf{z}$ , we finish the sequence. Notice that if we split all the children of each $y_{i}^{c_{1},\dots,c_{i}}$ so that each of them appears exactly once, then the obtained instance has a solution for every $\mathbf{z}$ by the maximality of $m$ . Thus, we get a sequence $\mathcal{J}_{0},\mathcal{J}_{1},\dots,\mathcal{J}_{s}$ of coverings of $\mathcal{I}_{t}$ such that $\mathcal{J}_{s}$ has a solution for every $\mathbf{z}$ .

Since $\mathcal{I}_{t+1}$ is a universal weakening of $\mathcal{I}_{t}$ , this universal weakening can be transferred to $\mathcal{J}_{s}$ , where we replace the child of every constraint of $\mathcal{I}_{t}$ by the corresponding weakened version. As a result we get a universal weakening $\mathcal{J}_{s}^{\prime}$ of $\mathcal{J}_{s}$ . Notice that $\mathcal{J}_{s}^{\prime}$ is a covering of $\mathcal{I}_{t+1}$ , which implies that $\mathcal{J}_{s}^{\prime}$ satisfies the modification of the subdirectness property for coverings. That is, $\mathcal{J}_{s}^{\prime}$ has a solution with $v=d$ , if $v$ is a child of $y_{i}^{c_{1},\dots,c_{i}}$ , for any $\mathbf{z}\in A^{|A|}$ , any $i\in\{0,\dots,m\}$ , and any $d\in\prescript{\mathbf{z}}{}{D}_{y_{i}^{c_{1},\dots,c_{i}}}^{(\top)}$ . Let $u$ be the variable we split while defining $\mathcal{J}_{s}$ and $u^{\prime}$ be the new variable we added. Consider two subcases:

Subcase 2A. There exist $\mathbf{z}$ and $d\in\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}$ such that $\prescript{\mathbf{z}}{}{\mathcal{J}_{s}}$ has no solution with $u=d$ or has no solution with $u^{\prime}=d$ . Without loss of generality let it be $u$ . By the subdirectness property for $\mathcal{J}_{s}^{\prime}$ , the formula $\mathcal{J}_{s}(u)$ defines the $\mathbf{z}$ -parameterized unary relation $D_{u}^{(\top)}$ . Suppose $\mathcal{J}_{s}(u)$ defines a $\mathbf{z}$ -parameterized unary relation $B$ . Since $\mathcal{J}_{s}\mathrel{\trianglelefteq}\mathcal{J}_{s}^{\prime}$ , Lemma 11 implies that $B\mathrel{\lhd}D_{u}^{(\top)}$ and $B$ satisfies condition 1, which completes this case.

Subcase 2B. For every $\mathbf{z}$ and every $d\in D_{u}^{(\top)}$ the instance $\mathcal{J}_{s}$ has a solution with $u=d$ and a solution with $u^{\prime}=d$ . Let $S$ be the binary $\mathbf{z}$ -parameterized relation defined by $\mathcal{J}_{s}(u,u^{\prime})$ and $W$ be the binary $\mathbf{z}$ -parameterized relation defined by $\mathcal{J}_{s}^{\prime}(u,u^{\prime})$ . Since $\mathcal{J}_{s}\mathrel{\trianglelefteq}\mathcal{J}_{s}^{\prime}$ , Lemma 11 implies that $S\mathrel{\trianglelefteq}W$ . Let $Q$ be a $(\mathbf{z},\alpha)$ -parameterized relation q-definable from $R$ witnessing $S\mathrel{\trianglelefteq}W$ , that is, $Q^{\forall}=W$ and $Q^{\forall\forall}=S$ . Let us show that $(Q,D)$ forms a mighty tuple V. Property 1 follows from the subdirectness of $\mathcal{J}_{s}^{\prime}$ . Property 2 follows from the definition of subcase 2B. Since $\mathcal{J}_{s-1}$ has no solutions for some $\mathbf{z}$ , $\prescript{\mathbf{z}}{}{Q}^{\forall\forall}$ is irreflexive for this $\mathbf{z}$ , and we get property 3. ∎

6.5 Finding a smaller reduction

In this section we will show that for any 1-consistent universal reduction $D^{(\top)}$ for $\mathcal{I}_{R}$ and a unary $\mathbf{z}$ -parameterized relation $B\mathrel{\lhd}D_{y_{i}^{a_{1},\dots,a_{i}}}^{(\top)}$ we can build a smaller 1-consistent universal reduction $D^{(\bot)}\subsetneq D^{(\top)}$ .

By $\mathcal{I}_{R}^{\prime}$ we denote the instance $\mathcal{I}_{R}^{(\top)}$ with additional constraints $C_{W,\top}^{a_{1},\dots,a_{i}}$ , $C_{W,\top,j}^{a_{1},\dots,a_{i}}$ , for all $i\in\{0,1,\dots,n\}$ , $a_{1},\dots,a_{i}\in A$ , and $j\in\{0,1,\dots,|A|\}$ . Notice that all the constraints we added to $\mathcal{I}_{R}$ to get $\mathcal{I}_{R}^{\prime}$ are weaker than the constraints that are already there. Hence, $\mathcal{I}_{R}^{\prime}$ has a solution if and only if $\mathcal{I}_{R}$ has a solution, and a reduction is 1-consistent for $\mathcal{I}_{R}^{\prime}$ if and only if it is 1-consistent for $\mathcal{I}_{R}$ .

To simplify presentation we fix a highest variable $y_{m}^{c_{1},\dots,c_{m}}$ such that there exists a $\mathbf{z}$ -parameterized unary relation $B$ satisfying $B\mathrel{\lhd}D_{y_{m}^{c_{1},\dots,c_{m}}}^{(\top)}$ . Denote this variable by $u=y_{m}^{c_{1},\dots,c_{m}}$ . By $\mathcal{B}$ we denote the set of all $\mathbf{z}$ -parameterized nonempty unary relations $B$ satisfying $B\mathrel{\lhd}D_{u}^{(\top)}$ .

By $\mathcal{T}$ we denote the set of all tree-coverings of $\mathcal{I}_{R}^{\prime}$ such that some of the children of $u$ are marked as leaves and exactly one child of $u$ is marked as the root. Elements of $\mathcal{T}$ are called trees. Notice that a vertex can be simultaneously a leaf and the root. Any path in an instance $\mathcal{I}_{R}^{\prime}$ can be viewed as a tree-covering. Marking the first element of the path as a leaf and the last element as the root we can make a tree from any path. By $\mathcal{P}$ we denote the set of all paths in $\mathcal{T}$ . Notice that by choosing one leaf in a tree we can always make a path from a tree.

Suppose $t\in\mathcal{T}$ with leaves $u_{1},\dots,u_{s}$ and the root $u_{0}$ . For a $\mathbf{z}$ -parameterized unary relation $B$ by $B+t$ we denote the $\mathbf{z}$ -parameterized unary relation defined by $(t\wedge u_{1}\in B\wedge\dots\wedge u_{s}\in B)(u_{0})$ . Informally, $B+t$ is the restriction we get on the root if restrict we all the leaves to $B$ . Notice that, since the reduction $D^{(\top)}$ is 1-consistent, we have $D_{u}^{(\top)}+t=D_{u}^{(\top)}$ for any $t\in\mathcal{T}$ .

To prove the existence of a smaller reduction, it is sufficient to satisfy the following Lemma.

Lemma 34.

Suppose $B\in\mathcal{B}$ and $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}\neq\varnothing$ for every $t\in\mathcal{T}$ and every $\mathbf{z}\in A^{|A|}$ . Then there exists a $\mathbf{z}$ -parameterized 1-consistent universal reduction $D^{(\bot)}$ for $\mathcal{I}_{R}$ such that $D^{(\bot)}\subsetneq D^{(\top)}$ .

Proof.

Let $D^{(\triangle)}$ be the $\mathbf{z}$ -parameterized reduction for $\mathcal{I}_{R}$ such that $D_{u}^{(\triangle)}=B$ and $D_{v}^{(\triangle)}=D_{v}^{(\top)}$ for every $v\neq u$ . For every $\mathbf{z}\in A^{|A|}$ let $\prescript{\mathbf{z}}{}{D}^{(\bot)}$ be the inclusion maximal 1-consistent reduction such that $\prescript{\mathbf{z}}{}{D}^{(\bot)}\subseteq\prescript{\mathbf{z}}{}{D}^{(\triangle)}$ . Consider two cases:

Case 1. Assume that $\prescript{\mathbf{z}_{0}}{}{D_{v}^{(\bot)}}$ is empty for some $\mathbf{z}_{0}\in A^{|A|}$ and some variable $v$ . Then by Lemma 26 there exists a tree-covering $\Upsilon$ of $\mathcal{I}_{R}^{\prime}$ such that $\prescript{\mathbf{z}_{0}}{}{\Upsilon}$ has no solutions if all the children of $u$ are restricted to $\prescript{\mathbf{z}_{0}}{}{B}$ . Let $t\in\mathcal{T}$ be the tree obtained from $\Upsilon^{(\top)}$ by marking all the children of $u$ as leaves and marking one of the children as the root. Then $\prescript{\mathbf{z}_{0}}{}{B}+\prescript{\mathbf{z}_{0}}{}{t}=\varnothing$ , which contradicts our assumptions.

Case 2. $\prescript{\mathbf{z}}{}{D_{v}^{(\bot)}}$ is not empty for every $\mathbf{z}\in A^{|A|}$ and every $v$ . By Lemma 27 for every $v$ there exists a tree-covering $\Upsilon_{v}$ of $\mathcal{I}_{R}$ such that $\prescript{\mathbf{z}}{}{\Upsilon_{v}^{(\triangle)}}(v)$ defines $\prescript{\mathbf{z}}{}{D_{v}^{(\bot)}}$ for every $\mathbf{z}$ . Since $D^{(\top)}$ is 1-consistent and $\Upsilon_{v}$ is a tree-formula, $\prescript{\mathbf{z}}{}{\Upsilon_{v}^{(\top)}}(v)$ defines $\prescript{\mathbf{z}}{}{D_{v}^{(\top)}}$ . Since $B\mathrel{\lhd}D^{(\top)}_{u}$ , Lemma 11 implies that $D_{v}^{(\bot)}\mathrel{\trianglelefteq}D_{v}^{(\top)}$ for every $v$ . Again, by Lemma 11

D_{v}^{(\bot)}=D_{v}^{(\bot)}\cap D_{v}^{(\bot,0)}\mathrel{\trianglelefteq}D_{v}^{(\top)}\cap D_{v}^{(\bot,0)}\mathrel{\mathrel{\lhd}\!\!\mathrel{\lhd}\!\!\mathrel{\lhd}}D_{v}^{(\top,0)}\cap D_{v}^{(\bot,0)}=D_{v}^{(\bot,0)}.

Hence, $D^{(\bot)}$ is a $\mathbf{z}$ -parameterized 1-consistent universal reduction for $\mathcal{I}_{R}$ that is smaller than $D^{(\top)}$ . ∎

We define two directed graphs $G_{\mathcal{P}}$ and $G_{\mathcal{T}}$ whose vertices are elements of $\mathcal{B}$ . There is an edge $B_{1}\to B_{2}$ in $G_{\mathcal{P}}$ if there exists $p\in\mathcal{P}$ such that $B_{1}+p=B_{2}$ . Similarly, the edge $B_{1}\to B_{2}$ is in $G_{\mathcal{T}}$ if there exists $t\in\mathcal{T}$ such that $B_{1}+t=B_{2}$ . Since we consider trivial paths/trees, both graphs are reflexive (have all the loops). Since we can compose paths and trees, $B_{1}\to B_{2}$ and $B_{2}\to B_{3}$ implies $B_{1}\to B_{3}$ , that is both graphs are transitive. Let $\mathcal{B}_{\mathcal{T}}$ be a strongly connected component of $G_{\mathcal{T}}$ not having edges going outside of the component. Let $\mathcal{B}_{\mathcal{P}}$ be a strongly connected component of $G_{\mathcal{P}}$ inside $\mathcal{B}_{\mathcal{T}}$ not having edges going outside of the component. Thus, we have $\mathcal{B}_{\mathcal{P}}\subseteq\mathcal{B}_{\mathcal{T}}\subseteq\mathcal{B}$ . Put $\mathcal{B}_{\mathcal{P}}^{*}=B_{\mathcal{P}}\cup\{D_{u}^{(\top)}\}$ . Then for every $B\in\mathcal{B}_{\mathcal{P}}^{*}$ , and $p\in\mathcal{P}$ we have $B+p\in\mathcal{B}_{\mathcal{P}}^{*}$ .

In this section we prove that there exists a smaller 1-consistent reduction for $\mathcal{I}_{R}$ or $R$ q-defines a mighty tuple IV. As we show in Lemma 35, to prove this, it is sufficient to satisfy the following property: there exist $B\in\mathcal{B}_{\mathcal{P}}$ and $\mathbf{z}$ -parameterized binary relations $S$ and $W$ q-definable from $R$ such that

1.

$S\triangleleft W$ ;
2.

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{S}=\prescript{\mathbf{z}}{}{B}$ for every $\mathbf{z}\in A^{|A|}$ ;
3.

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{W}=\prescript{\mathbf{z}}{}{D_{u}^{(\top)}}$ for every $\mathbf{z}\in A^{|A|}$ ;
4.

$\prescript{\mathbf{z}}{}{D_{u}^{(\top)}}+\prescript{\mathbf{z}}{}{S}=\prescript{\mathbf{z}}{}{D_{u}^{(\top)}}$ for every $\mathbf{z}\in A^{|A|}$ .

In this case we say that the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Lemma 35.

Suppose $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is a strong tuple. Then one of the following conditions holds:

1.

there exists a $\mathbf{z}$ -parameterized 1-consistent universal reduction $D^{(\bot)}$ for $\mathcal{I}_{R}$ such that $D^{(\bot)}\subsetneq D^{(\top)}$ ;
2.

there exists a mighty tuple IV q-definable from $R$ .

Proof.

By the definition, there exist $B\in\mathcal{B}_{\mathcal{P}}$ and $\mathbf{z}$ -parameterized binary relations $S$ and $W$ q-definable from $R$ satisfying the required 4 conditions. If $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}\neq\varnothing$ for any $t\in\mathcal{T}$ and $\mathbf{z}$ then Lemma 34 implies that condition 1 holds. Otherwise, let $t$ be the tree with the minimal number of leaves such that $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}=\varnothing$ . Define a new tree $t^{\prime}$ by moving the root of $t$ to one of the leaves and removing its leaf mark. By the minimality of the number of leaves $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}^{\prime}\neq\varnothing$ for any $\mathbf{z}$ . Denote $C=B+t^{\prime}$ . Since $S\mathrel{\lhd}W$ , there exists a q-definable relation $Q$ such that $Q^{\forall\forall}=S$ and $Q^{\forall}=W$ . Let us check that $(Q,D_{u}^{(\top)},B,C)$ forms a mighty tuple IV. All the conditions but 5 follow from the definition of a strong tuple. Condition 5 follows from the fact that $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}=\varnothing$ for some $\mathbf{z}$ and therefore $\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}=\varnothing$ for this $\mathbf{z}$ . ∎

In the first case of the above lemma we obtain a smaller reduction, and in the second case we can build a mighty tuple and therefore prove PSpace-hardness. Thus, whenever the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong, we can achieve the required result. That is why, in many further lemmas we have an assumption that it is not strong.

We say that $B_{1}$ is a supervised universal subset of $B_{2}$ if there exist $B_{0}\in\mathcal{B}_{\mathcal{P}}$ and $p_{1},p_{2}\in\mathcal{P}$ such that $p_{1}\mathrel{\trianglelefteq}p_{2}$ , $B_{0}+p_{1}=B_{1}$ , and $B_{0}+p_{2}=B_{2}$ . We write it as $B_{1}\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}B_{2}$ .

The following lemma follows immediately from the definition and the fact that we can compose paths.

Lemma 36.

Suppose $B_{1}\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}B_{2}$ and $p\in\mathcal{P}$ . Then $B_{1}+p\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}B_{2}+p$ .

The following lemma is the crucial fact in the whole proof. We show that we are done whenever $B\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}D_{u}^{(\top)}$ , and in the next lemmas we just try to achieve the condition $B\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}D_{u}^{(\top)}$ .

Lemma 37.

Suppose $B\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}D_{u}^{(\top)}$ . Then $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is a strong tuple.

Proof.

Consider $B_{0}\in\mathcal{B}_{\mathcal{P}}$ and two paths $p_{1},p_{2}\in\mathcal{P}$ such that $B_{0}+p_{1}=B$ , $B_{0}+p_{2}=D_{u}^{(\top)}$ , and $p_{1}\mathrel{\trianglelefteq}p_{2}$ . Consider a path $p_{0}\in\mathcal{P}$ such that $B+p_{0}=B_{0}$ , and define two new paths by $p_{1}^{\prime}=p_{0}+p_{1}$ and $p_{2}^{\prime}=p_{0}+p_{2}$ . Then $B+p_{1}^{\prime}=B$ , $B+p_{2}^{\prime}=D_{u}^{(\top)}$ , and $p_{1}^{\prime}\mathrel{\trianglelefteq}p_{2}^{\prime}$ . Let $u_{1}$ and $u_{2}$ be the two ends of the paths $p_{1}^{\prime}$ and $p_{2}^{\prime}$ . Let $p_{1}^{\prime}(u_{1},u_{2})$ define $S$ , $p_{2}^{\prime}(u_{1},u_{2})$ define $W$ . Then $W$ , $S$ , and $B$ witness that $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is a strong tuple. ∎

Lemma 38.

Suppose $B_{1},B_{2}\in\mathcal{B}_{\mathcal{P}}$ . Then for every $\mathbf{z}$ either both $\prescript{\mathbf{z}}{}{B}_{1}$ and $\prescript{\mathbf{z}}{}{B}_{2}$ are different from $\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}$ , or both are equal to $\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}$ .

Proof.

Since $B_{1},B_{2}\in\mathcal{B}_{\mathcal{P}}$ , there exist paths $p_{1},p_{2}\in\mathcal{P}$ such that $B_{1}+p_{1}=B_{2}$ and $B_{2}+p_{2}=B_{1}$ . Since the reduction $D^{(\top)}$ is 1-consistent, $\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}+p=\prescript{\mathbf{z}}{}{D}_{u}^{(\top)}$ for any $p\in\mathcal{P}$ and $\mathbf{z}\in A^{|A|}$ . This implies the required property. ∎

A supervised zig-zag from $B_{1}$ to $B_{2}$ is a sequence $C_{0},C_{1},\dots,C_{k}\in\mathcal{B}_{\mathcal{P}}^{*}$ such that

•

$C_{0}=B_{1}$ , $C_{k}=B_{2}$ ;
•

$C_{i-1}\supseteq C_{i}$ or $C_{i-1}\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}C_{i}$ for every $i\in[k]$ .

If there exists a supervised zig-zag of length $k$ from $B_{1}$ to $B_{2}$ , then we write $B_{1}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}_{k}B_{2}$ or just $B_{1}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}B_{2}$ if we do not want to specify the length.

Lemma 39.

Suppose $B_{1},B_{2}\in\mathcal{B}_{\mathcal{P}}$ . Then $B_{1}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}B_{2}$ .

Proof.

Consider a path $p_{0}\in\mathcal{P}$ such that $B_{2}+p_{0}=B_{1}$ . We will build a sequence of paths $p_{0},p_{1},\dots,p_{k}$ such that $C_{i}=B_{2}+p_{i}$ . We have $C_{0}=B_{2}+p_{0}=B_{1}$ .

By the choice of the variable $u$ there is no $\mathbf{z}$ -parameterized unary relation $B\triangleleft D_{v}^{(\top)}$ for any variable $v$ that is above $u$ in $\mathcal{I}_{R}$ . Therefore, by Lemma 11, these variables cannot appear in the path from the leaf to the root but can appear in some constraints.

We apply the following transformations to $p_{0}$ and define a sequence $p_{0},p_{1},\dots,p_{k}\in\mathcal{P}$ .

(w)

replace a child of the constraint $C_{S,\top}^{a_{1},\dots,a_{i}}$ by the corresponding child of $C_{W,\top}^{a_{1},\dots,a_{i}}$ ;
(e)

if the lowest variable of a child of $C_{W,\top}^{a_{1},\dots,a_{i}}$ appears only once, then replace it by the corresponding child of $C_{W,\top,|A|}^{a_{1},\dots,a_{i}}$ ;
(j)

if a variable from the $i$ -th level appears in two children of $C_{W,\top}^{a_{1},\dots,a_{i}}$ , then we replace these children by one child of $C_{W,\top}^{a_{1},\dots,a_{i}}$ identifying the corresponding variables of the children;
(z)

replace a child of $C_{W,\top,j}^{a_{1},\dots,a_{i}}$ by the corresponding child of $C_{W,\top,j-1}^{a_{1},\dots,a_{i}}$ ;
(s)

replace a child of $C_{W,\top,0}^{a_{1},\dots,a_{i}}$ by the corresponding child of $C_{S,\top}^{a_{1},\dots,a_{i-1}}$ .

Notice that (w) replaces an instance by its universal weakening, (s) makes the instance stronger, (e) just existentially quantifies a variable that appears only once, (j) joins several constraints together and makes the instance stronger, (z) replaces the instance by its universal weakening. Notice that we do not apply (e) if the lowest variable is a child of $u$ because this would mean removing a root.

Let us show that we can apply these transformation till the moment when we have only one constraint and this constraint is a child of $C_{W}^{c_{1},\dots,c_{m}}$ . Suppose we already have $p_{0},\dots,p_{\ell}$ . If $p_{\ell}$ has a child of $C_{W,j}^{a_{1},\dots,a_{i}}$ or a child of $C_{S}^{a_{1},\dots,a_{i}}$ then we can apply (w), (z), or (s). Otherwise, let $v$ be the lowest variable of $p_{\ell}$ . Notice that $v$ has to be from a level below $u$ , since otherwise $p_{\ell}$ already consists of just one constraint. If $v$ appears only once, then we can apply (e). Otherwise, we can apply (j).

Since we always reduce the tree and reduce the arity of a constraint, we cannot apply transformations forever. Thus, we have the sequence $p_{0},p_{1},\dots,p_{k}\in\mathcal{P}$ such that for every $i$ either $p_{i+1}$ is stronger than $p_{i}$ , or $p_{i}\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}p_{i+1}$ . Since the last path in the sequence consists of a child of $C_{W}^{c_{1},\dots,c_{m}}$ , we have $B_{2}+p_{k}=B_{2}$ and the sequence $C_{0},C_{1},C_{2},\dots,C_{k}$ witnesses that $B_{1}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}B_{2}$ . ∎

Lemma 40.

Suppose $B_{1},B_{2}\in\mathcal{B}_{\mathcal{P}}$ , $p\in\mathcal{P}$ , $B_{1}+p\neq D_{u}^{(\top)}$ . Then $B_{2}+p\neq D_{u}^{(\top)}$ or the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Proof.

Assume that $B_{2}+p=D_{u}^{(\top)}$ . By Lemma 39 there is a supervised zig-zag $C_{0},C_{1},\dots,C_{k}$ from $B_{1}$ to $B_{2}$ . Consider the last element in the sequence $C_{0}+p,C_{1}+p,\dots,C_{k}+p$ that is different from $D_{u}^{(\top)}$ . Let it be $C_{i}+p$ . Then by Lemma 36, $C_{i}+p\mathrel{\mathchoice{\ooalign{\cr$\displaystyle\leq$\cr\raisebox{0.86108pt}{$\displaystyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\textstyle\leq$\cr\raisebox{0.86108pt}{$\textstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}{\ooalign{\cr$\scriptscriptstyle\leq$\cr\raisebox{0.86108pt}{$\scriptscriptstyle\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft$}}}}C_{i+1}+p=D_{u}^{(\top)}$ , which by Lemma 37 implies that the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong. ∎

Lemma 41.

Suppose $B_{1},B_{2}\in\mathcal{B}_{\mathcal{P}}$ , $B_{1}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}_{k}B_{2}$ , and $\prescript{\mathbf{z}}{}{B_{1}}\not\supseteq\prescript{\mathbf{z}}{}{B_{2}}$ for some $\mathbf{z}\in A^{|A|}$ . Then there exists $p\in\mathcal{P}$ such that $B_{1}+p\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}_{k-1}B_{2}+p\neq D_{u}^{(\top)}$ or the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Proof.

Let $C_{0},\dots,C_{k}$ be a supervised zig-zag from $B_{1}$ to $B_{2}$ . Since $\prescript{\mathbf{z}}{}{B_{1}}\not\supseteq\prescript{\mathbf{z}}{}{B_{2}}$ , $C_{i}\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}C_{i+1}$ for some $i$ . Choose an inclusion maximal $B\in\mathcal{B}_{\mathcal{P}}$ and a path $p\in\mathcal{P}$ such that $C_{i}+p=B$ . Unless the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong, Lemma 40 implies that $C_{j}+p\neq D_{u}^{(\top)}$ for every $j$ . Since $C_{i}\subseteq C_{i+1}$ and $B$ is inclusion maximal, we have $C_{i}+p=C_{i+1}+p$ . Then by Lemma 36, $C_{0}+p,C_{1}+p,\dots,C_{i}+p,C_{i+2}+p,\dots,C_{k}+p$ is a supervised zig-zag from $B_{1}+p$ to $B_{2}+p$ of length $k-1$ . ∎

Lemma 42.

Suppose $B_{1},B_{2}\in\mathcal{B}_{\mathcal{P}}$ . Then there exists $p\in\mathcal{P}$ such that $B_{1}+p\subseteq B_{2}+p\neq D_{u}^{(\top)}$ or the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Proof.

By Lemma 39, $B_{2}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}_{k}B_{1}$ for some $k$ . Applying Lemma 41 we obtain $B_{2}+p_{1}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}_{k-1}B_{1}+p_{1}$ . Applying Lemma 41 again we obtain $B_{2}+p_{1}+p_{2}\mathrel{\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}\!\!\supset\!\!\!\mathrel{\color[rgb]{0.45,0.45,0.45}\definecolor[named]{pgfstrokecolor}{rgb}{0.45,0.45,0.45}\pgfsys@color@gray@stroke{0.45}\pgfsys@color@gray@fill{0.45}\blacktriangleleft}}_{k-2}B_{1}+p_{1}+p_{2}$ . We can do this till the moment when $B_{2}+p_{1}+p_{2}+\dots+p_{s}\supseteq B_{1}+p_{1}+p_{2}+\dots+p_{s}$ . It remains to put $p=p_{1}+p_{2}+\dots+p_{s}$ . ∎

Corollary 43.

Suppose $B_{1},B_{2}\in\mathcal{B}_{\mathcal{P}}$ . Then there exists $p\in\mathcal{P}$ such that $B_{1}+p=B_{2}+p\neq D_{u}^{(\top)}$ or the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Proof.

By Lemma 42 there exists $p$ such that $B_{1}+p\supseteq B_{2}+p$ . Again by Lemma 42 there exists $p^{\prime}$ such that $B_{1}+p+p^{\prime}\subseteq B_{2}+p+p^{\prime}$ . Combining this with $B_{1}+p\supseteq B_{2}+p$ , we obtain $B_{1}+p+p^{\prime}=B_{2}+p+p^{\prime}$ . ∎

Lemma 44.

Suppose $B\in\mathcal{B}_{\mathcal{P}}$ . Then there exists $p\in\mathcal{P}$ such that $B^{\prime}+p=B$ for every $B^{\prime}\in\mathcal{B}_{\mathcal{P}}$ or the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Proof.

First, let us show that there exists a path sending all $B^{\prime}\in\mathcal{B}_{\mathcal{P}}$ to the same element of $\mathcal{B}_{\mathcal{P}}$ . Put $\mathcal{B}_{0}=\mathcal{B}_{\mathcal{P}}$ . If $|\mathcal{B}_{0}|=1$ then we can take a trivial path $p$ . Otherwise, consider different $B_{1},B_{2}\in\mathcal{B}_{0}$ . By Corollary 43, there exists a path $p_{1}$ such that $B_{1}+p_{1}=B_{2}+p_{1}\neq D_{u}^{(\top)}$ . Put $\mathcal{B}_{1}=\{B^{\prime}+p_{1}\mid B^{\prime}\in\mathcal{B}_{0}\}$ . If $|\mathcal{B}_{1}|=1$ then we finish. Otherwise, choose different $B_{1},B_{2}\in\mathcal{B}_{1}$ and make them equal using a path $p_{2}$ . Then define $\mathcal{B}_{2}=\{B^{\prime}+p_{2}\mid B^{\prime}\in\mathcal{B}_{1}\}$ . Proceeding this way we get paths $p_{1},\dots,p_{s}$ and sets $\mathcal{B}_{1},\dots,\mathcal{B}_{s}$ such that $\mathcal{B}_{i}=\{B^{\prime}+p_{i}\mid B^{\prime}\in\mathcal{B}_{i-1}\}$ and $|\mathcal{B}_{i}|<|\mathcal{B}_{i-1}|$ for $i=1,2,\dots,s$ . Notice that by Lemma 40 $D_{u}^{(\top)}\notin\mathcal{B}_{i}$ for any $i$ . We finish when $|\mathcal{B}_{s}|=1$ , so let $\mathcal{B}_{s}=\{B_{0}\}$ .

Thus, for any $B^{\prime}\in\mathcal{B}_{\mathcal{P}}$ we have $B^{\prime}+p_{1}+p_{2}+\dots+p_{s}=B_{0}\in\mathcal{B}_{\mathcal{P}}$ . By the definition of $\mathcal{B}_{\mathcal{P}}$ there exists a path $p_{s+1}\in\mathcal{P}$ such that $B_{0}+p_{s+1}=B$ . It remains to put $p=p_{1}+p_{2}+\dots+p_{s}+p_{s+1}$ . ∎

Lemma 45.

Suppose $B\in\mathcal{B}_{\mathcal{T}}$ . Then there exists $B^{\prime}\in\mathcal{B}_{\mathcal{P}}$ such that $B^{\prime}\supseteq B$ , or the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Proof.

By the definition of $\mathcal{B}_{\mathcal{T}}$ there exist $B_{0}\in\mathcal{B}_{\mathcal{P}}$ and a tree $t_{1}\in\mathcal{T}$ such that $B_{0}+t_{1}=B$ . Let $M\in\mathcal{B}_{\mathcal{T}}$ be chosen inclusion maximal. Choose $t_{0},t_{2}\in\mathcal{T}$ such that $M+t_{0}=B_{0}$ and $B+t_{2}=M$ . Put $t=t_{0}+t_{1}+t_{2}$ . Then $M+t=M$ .

Consider the minimal set $L$ of leaves of $t$ we need to restrict to $M$ to obtain a $\mathbf{z}$ -parameterized unary relation in the root that is different from $D_{u}^{(\top)}$ . Since $M$ is inclusion maximal, this unary relation must be $M$ . Let $L=\{u_{1},\dots,u_{\ell}\}$ and $u_{0}$ be the root of $t$ . We consider two cases:

Case 1. $\ell>1$ . Let $(t\wedge u_{2}\in C\wedge\dots\wedge u_{\ell}\in C)(u_{1},u_{0})$ define a $\mathbf{z}$ -parameterized binary relation $S$ and $t(u_{1},u_{0})$ define a $\mathbf{z}$ -parameterized binary relation $W$ . By the minimality of $\ell$ we have $\prescript{\mathbf{z}}{}{D_{u}^{(\top)}}+\prescript{\mathbf{z}}{}{S}=\prescript{\mathbf{z}}{}{D_{u}^{(\top)}}$ for any $\mathbf{z}$ . Then $W$ , $S$ , and $B$ witness that $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is a strong tuple.

Case 2. $\ell=1$ . Let $p$ be the path in $t$ from the leaf $u_{1}$ to the root $u_{0}$ . Then $p=p_{0}+p_{1}+p_{2}$ , where $p_{0}$ , $p_{1}$ , and $p_{2}$ are parts of $p$ coming from $t_{0}$ , $t_{1}$ , and $t_{2}$ , respectively. Notice that $M+p=M$ , $M+p_{0}\supseteq B_{0}$ , $M+p_{0}+p_{1}\supseteq B$ , and $M+p_{0}+p_{1}\neq D_{u}^{(\top)}$ . Hence, $B_{0}+p_{1}\subseteq M+p_{0}+p_{1}\neq D_{u}^{(\top)}$ and $B_{0}+p_{1}\supseteq B$ . It remains to put $B^{\prime}=B_{0}+p_{1}$ . ∎

Lemma 46.

Suppose $B_{1}\in\mathcal{B}_{\mathcal{P}}$ and $B_{2}\in\mathcal{B}_{\mathcal{T}}$ . Then $B_{1}\cap B_{2}\neq\varnothing$ , or the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong.

Proof.

By Lemma 44 there exists a path $p\in\mathcal{P}$ such that $B+p=B_{1}$ for every $B\in\mathcal{B}_{\mathcal{P}}$ . Let $B_{0}=B_{2}+p^{\prime}$ , where $p^{\prime}$ is obtained from $p$ by switching ends (we could also write $B_{0}=B_{2}-p$ ). If $B_{0}=D_{u}^{(\top)}$ then $B_{1}+p=B_{1}$ implies $B_{1}\cap B_{2}\neq\varnothing$ . Otherwise, $B_{0}\in\mathcal{B}_{\mathcal{T}}$ and by Lemma 45 there exists $B_{0}^{\prime}\in\mathcal{B}_{\mathcal{P}}$ such that $B_{0}^{\prime}\supseteq B_{0}$ . By the definition of $p$ we must have $B_{0}^{\prime}+p=B_{1}$ . Therefore, $B_{2}\subseteq B_{2}-p+p=B_{0}+p\subseteq B_{0}^{\prime}+p=B_{1}$ , which completes the proof. ∎

We are ready to prove the main theorem of this subsection.

Theorem 21.

1.

there exists a $\mathbf{z}$ -parameterized universal 1-consistent reduction $D^{(\bot)}$ for $\mathcal{I}_{R}$ that is smaller than $D^{(\top)}$ ;
2.

there exists a mighty tuple IV q-definable from $R$ .

Proof.

First, we repeat assumptions from the beginning of this section. We choose the highest variable $u$ with the same property, then we define $\mathcal{B}$ and choose $\mathcal{B}_{\mathcal{T}}$ and $\mathcal{B}_{\mathcal{P}}$ .

Choose some $B\in\mathcal{B}_{\mathcal{P}}$ . If $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}\neq\varnothing$ for every $\mathbf{z}$ and $t\in\mathcal{T}$ , then by Lemma 34 there exists a required smaller reduction. Otherwise, choose a tree $t\in\mathcal{T}$ with the minimal number of leaves such that $\prescript{\mathbf{z}_{0}}{}{B}+\prescript{\mathbf{z}_{0}}{}{t}=\varnothing$ for some $\mathbf{z}_{0}$ . Moving the root to one of the leaves and removing its leaf mark we get another tree $t^{\prime}$ such that $\prescript{\mathbf{z}_{0}}{}{B}+\prescript{\mathbf{z}_{0}}{}{t}^{\prime}\cap\prescript{\mathbf{z}_{0}}{}{B}=\varnothing$ . Since $t$ has the minimal number of leaves, $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}^{\prime}\neq\varnothing$ for any $\mathbf{z}$ and $\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{t}^{\prime}\in\mathcal{B}_{\mathcal{T}}$ . Then Lemma 46 implies that the tuple $(R,D^{(\top)},u,\mathcal{B}_{\mathcal{P}})$ is strong. It remains to apply Lemma 35. ∎

7 Hardness Claims

7.1 Definitions

Binary relations in this section are often viewed as directed graphs, and we use terminology from the graph theory such as paths and cycles. A binary relation $R$ is called transitive if $R+R=R$ . Even though the general domain of any relation $R$ is $A$ , we often define a subset $D$ such that $R\subseteq D^{2}$ . Then we call the relation $R$ reflexive if $\{(d,d)\mid d\in D\}\subseteq R$ . Suppose $R$ is a reflexive relation on $D$ . Then the transitive symmetric closure of $R$ is the minimal transitive symmetric relation $R^{\prime}\supseteq R$ . Notice that $R^{\prime}=R-R+R-R+\dots+R-R+R$ , for sufficiently many pluses and minuses, hence $R^{\prime}$ is q-definable over $R$ .

For a positive integer $m$ and a binary relation $S$ denote $m\cdot S=\underbrace{S+S+\dots+S}_{m}$ .

Lemma 47.

Suppose $R\subseteq A\times A$ , $S=(|A|!\cdot|A|^{2})\cdot R$ . Then $S+S=S$ .

Proof.

First, put $S_{1}=(|A|!)\cdot R$ . Notice that if $(a,b)\in S_{1}$ , then there is a path from $a$ to $b$ in $R$ of length $|A|!$ . Since the domain is of size $|A|$ , there must be a cycle of length $m\leqslant|A|$ in the path. Repeating this cycle $|A|!/m$ times we make a path of length $2|A|!$ , which implies $S_{1}+S_{1}\supseteq S_{1}$ . Put $S_{n}=n\cdot S_{1}$ . Since $S_{1}+S_{1}\supseteq S_{1}$ , we have $S_{i}\subseteq S_{i+1}$ . Moreover, if $S_{i}=S_{i+1}$ , then $S_{j}=S_{i}$ for any $j>i$ . Thus, the sequence $S_{1},S_{2},\dots,$ stabilises at some $S_{i}$ , where $i\leqslant|A|^{2}$ . Hence, $S_{|A|^{2}}+S_{|A|^{2}}=S_{|A|^{2}}$ , which completes the proof. ∎

For two equivalence relations $R_{1}$ and $R_{2}$ on some set $D$ by $R_{1}\ovee R_{2}$ we denote the minimal equivalence relation on $D$ containing $R_{1}$ and $R_{2}$ . Thus, it is the usual join of two equivalence relations, but we prefer to use this symbol to distinguish it from the disjunction. Notice that $R_{1}\ovee R_{2}$ is q-definable from $R_{1}$ and $R_{2}$ as we can always write a quantified formula defining $R_{1}+R_{2}+R_{1}+R_{2}+\dots+R_{1}+R_{2}$ .

Recall that we agreed that $A=\{1,\dots,|A|\}$ . Then we put $\kappa=(1,\dots,|A|)$ , that is, $\kappa$ is a concrete tuple of length $|A|$ with all the elements of $A$ .

7.2 PSpace-hardness for a mighty tuple I

In this section we show that the QCSP over a mighty tuple I is PSpace-hard.

For technical reasons we will need mighty tuples with an additional property:

( $\kappa$ )

$\prescript{\mathbf{z}}{\delta}{R}^{\kappa}\subseteq\prescript{\mathbf{z}}{\delta}{R}^{\alpha}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\alpha$ .

A mighty tuple I satisfying property ( $\kappa$ ) is called a mighty tuple I’.

Lemma 48.

Suppose $(Q,D,B,C,\Delta)$ is a mighty tuple I. Then $\{Q,D,B,C,\Delta\}$ q-defines a mighty tuple I’.

Proof.

Define a mighty tuple I’ as follows. Suppose the $\alpha$ -parameter is from $A^{k}$ . Put

\prescript{\mathbf{z}}{\delta}{R}^{x_{1},\dots,x_{|A|}}(y_{1},y_{2})=\bigwedge\limits_{i_{1},\dots,i_{k}\in\{1,2,\dots,|A|\}.}\prescript{\mathbf{z}}{\delta}{Q}^{x_{i_{1}},\dots,x_{i_{k}}}(y_{1},y_{2}).

Then $R^{\kappa}=R^{\forall\forall}=Q^{\forall\forall}$ and $R^{\forall}=Q^{\forall}$ . Hence $(R,D,B,C,\Delta)$ is a mighty tuple I’. ∎

As we mentioned in Section 3 we have one reduction covering all the PSpace-hard cases of the QCSP. Precisely, we will show that using a mighty tuple we can build relations very similar to the relations from Section 3. Then we use the same reduction from the Quantified-3-DNF, which is the complement of the Quantified-3-CNF.

First, for any instance of the Quantified-3-DNF we define a sentence corresponding to this reduction. Suppose we have two relational symbols $\Upsilon_{0}$ and $\Upsilon_{1}$ of arity $m+2$ for some $m\geqslant 1$ . Then $\Upsilon_{0}$ and $\Upsilon_{1}$ can be viewed as $\mathbf{x}$ -parameterized binary relations, where $\mathbf{x}\in A^{m}$ . Let $\Phi$ be an instance of the Quantified-3-DNF of the form

Q_{1}x_{1}Q_{2}x_{2}\dots Q_{n}x_{n}\;(x_{a_{1}}=a_{1}^{\prime}\wedge x_{b_{1}}=b_{1}^{\prime}\wedge x_{c_{1}}=c_{1}^{\prime})\vee\dots\vee(x_{a_{s}}=a_{s}^{\prime}\wedge x_{b_{s}}=b_{s}^{\prime}\wedge x_{c_{s}}=c_{s}^{\prime}),

where $Q_{1},Q_{2},\dots,Q_{n}\in\{\forall,\exists\}$ , $a_{i},b_{i},c_{i}\in[n]$ and $a_{i}^{\prime},b_{i}^{\prime},c_{i}^{\prime}\in\{0,1\}$ for every $i\in[n]$ .

We define recursively formulas $\Psi_{n},\Psi_{n-1},\dots,\Psi_{1},\Psi_{0}$ . Put

\displaystyle\Psi_{n}=\exists y_{1}\dots\exists y_{s-1}\bigwedge_{1\leqslant i\leqslant s}(\Upsilon_{\overline{a_{i}^{\prime}}}^{\mathbf{x}_{a_{i}}}(y_{i-1},y_{i})\wedge\Upsilon_{\overline{b_{i}^{\prime}}}^{\mathbf{x}_{b_{i}}}(y_{i-1},y_{i})\wedge\Upsilon_{\overline{c_{i}^{\prime}}}^{\mathbf{x}_{c_{i}}}(y_{i-1},y_{i})),

where $\overline{a}$ is a negation of $a$ for any $a\in\{0,1\}$ . Notice that for any variable $x_{i}$ of the original instance we introduce a variable $\mathbf{x}_{i}$ , which takes values from $A^{m}$ .

For every $i$ by $l_{i}$ and $r_{i}$ we denote the minimal and the maximal indices of $y$ -variables appearing in the formula $\Psi_{i}$ . Thus, we have $l_{n}=0$ and $r_{n}=s$ . Let us show how to define $\Psi_{k-1}$ from $\Psi_{k}$ . If $Q_{k}=\forall$ , then we put $\Psi_{k-1}=\forall\mathbf{x}_{k}\;\Psi_{k}$ .

If $Q_{k}=\exists$ , then we put

\Psi_{k-1}=\exists y_{r_{k}}\forall\mathbf{x}_{k}\exists y_{l_{k}}\;\Psi_{k}\wedge\Upsilon_{0}^{\mathbf{x}_{k}}(y_{l_{k}-1},y_{l_{k}})\wedge\Upsilon_{1}^{\mathbf{x}_{k}}(y_{r_{k}+1},y_{l_{k}}).

Notice that in the formula $\Psi_{0}$ all the variables except for $y_{l_{0}}$ and $y_{r_{0}}$ are quantified. By $\mathcal{Q}^{\Phi}$ we denote the formula $\Psi_{0}$ whose variables $y_{l_{0}}$ and $y_{r_{0}}$ are replaced by $y$ and $y^{\prime}$ respectively.

For $\mathbf{x}$ -parameterized relations $R_{0}$ , $R_{1}$ , and an instance $\Phi$ of the Quantified-3-DNF by $\mathcal{Q}^{\Phi}(R_{0},R_{1})$ we denote the formula obtained from $\mathcal{Q}^{\Phi}$ by substituting $R_{0}$ for $\Upsilon_{0}$ and $R_{1}$ for $\Upsilon_{1}$ . By $\mathcal{T}^{\Phi}(R_{0},R_{1})$ we denote the transitive symmetric closure of $\sigma$ , where $\sigma(y,y^{\prime})=\mathcal{Q}^{\Phi}(R_{0},R_{1})$ .

Arguing as in Section 3 we can prove the following lemma.

Lemma 49.

Suppose $\Phi$ is an instance of the Quantified-3-DNF, $A=\{+,-,0,1\}$ , $V_{0}^{x}(y_{1},y_{2})=(y_{1},y_{2}\in\{+,-\})\wedge(x=0\rightarrow y_{1}=y_{2})$ , $V_{1}^{x}(y_{1},y_{2})=(y_{1},y_{2}\in\{+,-\})\wedge(x=1\rightarrow y_{1}=y_{2})$ . Then $\mathcal{T}^{\Phi}(V_{0},V_{1})=\{(+,+),(-,-)\}$ if $\Phi$ does not hold; $\mathcal{T}^{\Phi}(V_{0},V_{1})=\{+,-\}^{2}$ if $\Phi$ holds.

The next lemmas describe important properties of the operator $\mathcal{T}^{\Phi}$ .

Lemma 50.

Suppose

1.

$\Phi$ is a No-instance of the Quantified-3-DNF;
2.

$R_{0}$ and $R_{1}$ are $\mathbf{x}$ -parameterized equivalence relations on $D$ ;
3.

$(B\times C)\cap(R_{0}^{\beta_{0}}\ovee R_{1}^{\beta_{1}})=\varnothing$ for some $\beta_{0}$ and $\beta_{1}$ .

Then $(B\times C)\cap\mathcal{T}^{\Phi}(R_{0},R_{1})=\varnothing$ .

Proof.

Let $\delta=R_{0}^{\beta_{0}}\ovee R_{1}^{\beta_{1}}$ . Let $B^{\prime}$ be the union of all classes of $\delta$ having a nonempty intersection with $B$ . Let $C^{\prime}=D\setminus B^{\prime}$ , $\delta^{\prime}=B^{\prime 2}\cup C^{\prime 2}$ , $L_{0}^{\mathbf{x}}=\begin{cases}\delta^{\prime}&\text{ if $\mathbf{x}=\beta_{0}$}\\ D\times D&\text{ otherwise}\end{cases}$ , $L_{1}^{\mathbf{x}}=\begin{cases}\delta^{\prime}&\text{ if $\mathbf{x}=\beta_{1}$}\\ D\times D&\text{ otherwise}\end{cases}$ .

Notice that $L_{0}\supseteq R_{0}$ and $L_{1}\supseteq R_{1}$ , hence replacement of $R_{0}$ by $L_{0}$ and $R_{1}$ by $L_{1}$ would make it even harder for the UP to win. Also, we may assume that the UP only plays $\beta_{0}$ and $\beta_{1}$ . Interpreting $\beta_{0}$ as $0$ and $\beta_{1}$ as 1, and interpreting the domain $D/\delta^{\prime}$ as $\{+,-\}$ we derive that $\{(+,-)\}\cap\mathcal{T}^{\Phi}(V_{0},V_{1})=\varnothing$ if and only if $(B\times C)\cap\mathcal{T}^{\Phi}(L_{0},L_{1})=\varnothing$ , where $V_{0}$ and $V_{1}$ are the canonical relations from Lemma 49. This implies $(B\times C)\cap\mathcal{T}^{\Phi}(R_{0},R_{1})=\varnothing$ . ∎

Lemma 51.

Suppose

1.

$\Phi$ is a Yes-instance of the Quantified-3-DNF;
2.

$R_{0}$ and $R_{1}$ are $\mathbf{x}$ -parameterized equivalence relations on $D$ ;
3.

$(b,c)\in R_{0}^{\alpha}$ or $(b,c)\in R_{1}^{\alpha}$ for every $\alpha$ .

Then $(b,c)\in\mathcal{T}^{\Phi}(R_{0},R_{1})$ .

Proof.

Notice that if $b=c$ , then there is an obvious winning strategy for the EP where she always plays the element $b$ . Thus, we assume that $b\neq c$ . To make it harder for the EP to win we let her play only elements $b$ and $c$ . That is, we replace the relation $R_{0}^{\mathbf{x}}$ by the relation $L_{0}^{\mathbf{x}}=R_{0}^{\mathbf{x}}\cap\{b,c\}^{2}$ and the relation $R_{1}^{\mathbf{x}}$ by $L_{1}^{\mathbf{x}}=R_{1}^{\mathbf{x}}\cap\{b,c\}^{2}$ . By condition 3 for any $\mathbf{x}$ one of the two relations $L_{0}^{\mathbf{x}}$ and $L_{1}^{\mathbf{x}}$ is equal to $\{b,c\}^{2}$ and another is either $\{(b,b),(c,c)\}$ , or $\{b,c\}^{2}$ .

Since $\Phi$ is a Yes-instance, $(+,-)\in\mathcal{T}^{\Phi}(V_{0},V_{1})$ for the canonical relations $V_{0}$ and $V_{1}$ in Lemma 49. Interpreting $b$ and $c$ as $-$ and $+$ we can derive that $(b,c)\in\mathcal{T}^{\Phi}(L_{0},L_{1})$ . In fact, for any choice of $\mathbf{x}$ either $L_{0}^{\mathbf{x}}$ connects $b$ and $c$ , or $L_{1}^{\mathbf{x}}$ connects $b$ and $c$ , or both connect. Hence, if the UP cannot win in $\mathcal{T}^{\Phi}(V_{0},V_{1})$ , then he cannot win in $\mathcal{T}^{\Phi}(L_{0},L_{1})$ . This implies that $(b,c)\in\mathcal{T}^{\Phi}(R_{0},R_{1})$ . ∎

We will need parameterized relations having arbitrary many parameters. Formally, we say that $S$ is a multi-parameter equivalence relation if it assigns an equivalence relation $S^{\alpha_{1},\dots,\alpha_{n}}$ to every sequence $\alpha_{1},\dots,\alpha_{n}\in A^{m}$ and satisfies the following properties:

(s)

$S^{\alpha_{1},\dots,\alpha_{n_{1}}}=S^{\beta_{1},\dots,\beta_{n_{2}}}$ whenever $\{\alpha_{1},\dots,\alpha_{n_{1}}\}=\{\beta_{1},\dots,\beta_{n_{2}}\}$
(m)

$S^{\alpha_{1},\dots,\alpha_{n}}\subseteq S^{\alpha_{1},\dots,\alpha_{n},\alpha_{n+1}}$ for any $\alpha_{1},\dots,\alpha_{n},\alpha_{n+1}\in A^{m}$

Since the set $A^{m}$ is finite, (s) implies that we may think of $S$ as a relation of an arity $N:=m\cdot|A|^{m}+2$ such that $S^{\alpha_{1},\dots,\alpha_{N}}$ depends only on the set $\{\alpha_{1},\dots,\alpha_{N}\}$ . Thus, $S$ is still a finite relation of a fixed arity, but it will be convenient for us to assume that it can have arbitrary many parameters. We say that a multi-parameter equivalence relation $S_{1}$ is larger than a multi-parameter equivalence relation $S_{2}$ if $S_{1}^{\alpha_{1},\dots,\alpha_{n}}\supseteq S_{2}^{\alpha_{1},\dots,\alpha_{n}}$ for every $\alpha_{1},\dots,\alpha_{n}\in A^{m}$ . If additionally $S_{1}\neq S_{2}$ , we say that $S_{1}$ is strictly larger than $S_{2}$ .

We extend $\mathcal{T}^{\Phi}$ to multi-parameter equivalence relations. For a multi-parameter equivalence relation $S$ and a parameterized equivalence relation $R$ by $\mathcal{T}^{\Phi}(S,R)$ we denote the multi-parameter equivalence relation $S_{0}$ defined as follows. To define $S_{0}^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}$ we take the formula $\mathcal{Q}^{\Phi}$ , replace each $\Upsilon_{0}^{\mathbf{x}_{i}}$ by $S^{\mathbf{u}_{1},\dots,\mathbf{u}_{n},\mathbf{x}_{i}}$ , replace each $\Upsilon_{1}^{\mathbf{x}_{i}}$ by $S^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}\ovee R^{\mathbf{x}_{i}}$ . For fixed $\mathbf{u}_{1},\dots,\mathbf{u}_{n}$ the obtained formula has only two free variables $y$ and $y^{\prime}$ and defines a binary relation $\sigma$ . Then $S_{0}^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}$ is the transitive symmetric closure of $\sigma$ .

Lemma 52.

Suppose

1.

$S$ is a multi-parameter relation on a set $D$ ;
2.

$R$ is an $\mathbf{x}$ -parameterized equivalence relation on $D$ ;
3.

$\Phi$ is an instance of the Quantified-3-DNF.

Then $\mathcal{T}^{\Phi}(S,R)$ is a multi-parameter equivalence relation that is larger than $S$ .

Proof.

Suppose $\mathcal{T}^{\Phi}(S,R)=S_{0}$ . Since $S$ satisfies properties (s) and (m), it immediately follows from the definition that $S_{0}$ also satisfies properties (s) and (m). Let us show that $S_{0}$ is larger than $S$ . For any $\mathbf{u}_{1},\dots,\mathbf{u}_{n}$ we have $S^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}\subseteq S^{\mathbf{u}_{1},\dots,\mathbf{u}_{n},\mathbf{x}_{i}}$ and $S^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}\subseteq S^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}\ovee R^{\mathbf{x}_{i}}$ . Hence, the interpretations of both $\Upsilon_{0}^{\mathbf{x}_{i}}$ and $\Upsilon_{1}^{\mathbf{x}_{i}}$ contain every pair $(b,c)$ from $S^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}$ . Thus, for any play of the UP the EP can always play $b$ to confirm that $(b,c)\in S_{0}^{\mathbf{u}_{1},\dots,\mathbf{u}_{n}}$ . ∎

Lemma 53.

Suppose

(1)

$S$ is a multi-parameter equivalence relation on a set $D$ ;
(2)

$R$ is an $\mathbf{x}$ -parameterized equivalence relation on $D$ ;
(3)

$(b,c)\in S^{\alpha}\ovee R^{\alpha}$ for every $\alpha\in A^{m}$ ;
(4)

$(b,c)\notin S^{\alpha}$ for some $\alpha\in A^{m}$ ;
(5)

$\Phi$ is a Yes-instance of the Quantitied-3-DNF.

Then $\mathcal{T}^{\Phi}(S,R)$ is strictly larger than $S$ .

Proof.

Suppose $\mathcal{T}^{\Phi}(S,R)=S_{0}$ . Consider a maximal set of tuples $\alpha_{1},\dots,\alpha_{n}$ such that $(b,c)\notin S^{\alpha_{1},\dots,\alpha_{n}}$ . By condition (4) such a set exists. Note that this set may contain all tuples. Let us show that $(b,c)\in S_{0}^{\alpha_{1},\dots,\alpha_{n}}$ , which together with Lemma 52 would mean that $\mathcal{T}^{\Phi}(S,R)$ is strictly larger than $S$ .

Let us consider the interpretations of $\Upsilon_{0}^{\mathbf{x}_{i}}$ and $\Upsilon_{1}^{\mathbf{x}_{i}}$ in the definition of $\mathcal{T}^{\Phi}(S,R)$ . If $\mathbf{x}_{i}$ is not from the set $\{\alpha_{1},\dots,\alpha_{n}\}$ , then by the maximality of the set the relation $\Upsilon_{0}^{\mathbf{x}_{i}}=S^{\alpha_{1},\dots,\alpha_{n},\mathbf{x}_{i}}$ contains $(b,c)$ . If $\mathbf{x}_{i}$ is from the set $\{\alpha_{1},\dots,\alpha_{n}\}$ , then $\Upsilon_{1}^{\mathbf{x}_{i}}=S^{\alpha_{1},\dots,\alpha_{n}}\ovee R^{\mathbf{x}_{i}}\supseteq S^{\mathbf{x}_{i}}\ovee R^{\mathbf{x}_{i}}$ , which contains $(b,c)$ by condition (3). Hence, by Lemma 51 $(b,c)\in S_{0}^{\alpha_{1},\dots,\alpha_{n}}$ . ∎

Lemma 54.

Suppose

(1)

$S$ is a multi-parameter equivalence relation on a set $D$ ;
(2)

$R$ is an $\mathbf{x}$ -parameterized equivalence relation on $D$ ;
(3)

$(B\times C)\cap S^{\beta}=\varnothing$ for some $\beta\in A^{m}$ ;
(4)

there exists $\alpha$ such that $R^{\alpha}\subseteq S^{\mathbf{x}}$ for every $\mathbf{x}$ ;
(5)

$\Phi$ is a No-instance of the Quantitied-3-DNF.

Then $(B\times C)\cap S_{0}^{\beta}=\varnothing$ , where $S_{0}=\mathcal{T}^{\Phi}(S,R)$ .

Proof.

Recall that $S_{0}^{\beta}$ is defined using the formula $\mathcal{Q}^{\Phi}$ , where we substitute $S^{\beta,\mathbf{x}_{i}}$ for each $\Upsilon_{0}^{\mathbf{x}_{i}}$ and $S^{\beta}\ovee R^{\mathbf{x}_{i}}$ for each $\Upsilon_{1}^{\mathbf{x}_{i}}$ . We derive from (3) and (4) that $(B\times C)\cap(S^{\beta,\beta}\ovee(S^{\beta}\ovee R^{\alpha}))=\varnothing$ . Then Lemma 50 implies that $(B\times C)\cap S_{0}^{\beta}=\varnothing$ . ∎

Lemma 55.

Suppose $(R,D,B,C,\Delta)$ is a mighty tuple I’. Then there exist $(\mathbf{z},\delta,\mathbf{x})$ -parameterized equivalence relations $R_{0}$ and $R_{1}$ on $D$ q-definable from $\{R,D,B,C,\Delta\}$ and satisfying the following conditions:

(1)

$\forall\mathbf{z}\in A^{|A|}\exists\delta\in\prescript{\mathbf{z}}{}{\Delta}\;\forall\mathbf{x}\;(\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C}\subseteq\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}}\ovee\prescript{\mathbf{z}}{\delta}{R}_{1}^{\mathbf{x}})$ ;
(2)

$\exists\mathbf{z}\in A^{|A|}\forall\delta\in\prescript{\mathbf{z}}{}{\Delta}\;\exists\mathbf{x}((\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C})\cap\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}}=\varnothing)$ ;
(3)

there exists $\alpha$ such that $\prescript{\mathbf{z}}{\delta}{R}_{1}^{\alpha}\subseteq\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\mathbf{x}$ .

Proof.

Let $\sigma_{1},\dots,\sigma_{N}$ be the set of all injective mappings from $\{1,2,\dots,|A|\}$ to $\{1,2,\dots,|A|^{2}\}$ . Let

\prescript{\mathbf{z}}{\delta}{U}_{n}^{x_{1},\dots,x_{|A|^{2}}}=\prescript{\mathbf{z}}{\delta}{R}^{x_{\sigma_{1}(1)},\dots,x_{\sigma_{1}(|A|)}}\ovee\prescript{\mathbf{z}}{\delta}{R}^{x_{\sigma_{2}(1)},\dots,x_{\sigma_{2}(|A|)}}\ovee\dots\ovee\prescript{\mathbf{z}}{\delta}{R}^{x_{\sigma_{n}(1)},\dots,x_{\sigma_{n}(|A|)}}

Since at least $|A|$ elements in the set $x_{1},\dots,x_{|A|^{2}}$ are equal, there exists $i\in\{1,2,\dots,N\}$ such that $x_{\sigma_{i}(1)}=x_{\sigma_{i}(2)}=\dots=x_{\sigma_{i}(|A|)}$ . Since $\prescript{\mathbf{z}}{\delta}{R}^{\forall}=\prescript{\mathbf{z}}{\delta}{D}\times\prescript{\mathbf{z}}{\delta}{D}$ for every $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , the relation $\prescript{\mathbf{z}}{\delta}{U}_{N}^{x_{1},\dots,x_{|A|^{2}}}$ is equal to $\prescript{\mathbf{z}}{\delta}{D}\times\prescript{\mathbf{z}}{\delta}{D}$ .

Consider the maximal $n$ such that the following condition holds

\exists\mathbf{z}\in A^{|A|}\forall\delta\in\prescript{\mathbf{z}}{}{\Delta}\;\exists\mathbf{x}_{0}((\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C})\cap(\prescript{\mathbf{z}}{\delta}{U}_{n}^{\mathbf{x}_{0}})=\varnothing).

Put $\prescript{\mathbf{z}}{\delta}{R}_{0}^{x_{1},\dots,x_{|A|^{2}}}=\prescript{\mathbf{z}}{\delta}{U}_{n}^{x_{1},\dots,x_{|A|^{2}}}$ and $\prescript{\mathbf{z}}{\delta}{R}_{1}^{x_{1},\dots,x_{|A|^{2}}}=\prescript{\mathbf{z}}{\delta}{R}^{x_{\sigma_{n+1}(1)},\dots,x_{\sigma_{n+1}(|A|)}}$ and show that they satisfy the required properties.

Property (1) follows from the fact that $n$ was chosen maximal and the corresponding condition for $\prescript{\mathbf{z}}{\delta}{U}_{n+1}^{\mathbf{x}}=\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}}\ovee\prescript{\mathbf{z}}{\delta}{R}_{1}^{\mathbf{x}}$ does not hold. Property (2) again follows from the choice of $n$ . To prove property (3) consider a tuple $\alpha=(a_{1},\dots,a_{|A|^{2}})$ such that $(a_{\sigma_{n+1}(1)},\dots,a_{\sigma_{n+1}(|A|)})=\kappa$ . Then $\prescript{\mathbf{z}}{\delta}{R}_{1}^{\alpha}=\prescript{\mathbf{z}}{\delta}{R}^{\kappa}\subseteq\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}_{0}}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\mathbf{x}_{0}$ . ∎

Theorem 14.

Suppose $(Q,D,B,C,\Delta)$ is a mighty tuple I. Then $\operatorname{QCSP}(\{Q,D,B,C,\Delta\})$ is PSpace-hard.

Proof.

By Lemma 48 there exists a mighty tuple I’ $(R,D,B,C,\Delta)$ q-definable from the set $\{Q,D,B,C,\Delta\}$ . By Lemma 55 there exist $R_{0}$ and $R_{1}$ satisfying the corresponding conditions (1)-(3). For every $\mathbf{z}$ and $\delta$ we define a multi-parameter equivalence relation $\prescript{\mathbf{z}}{\delta}{S}_{0}$ by

\prescript{\mathbf{z}}{\delta}{S}_{0}^{\mathbf{x}_{1},\dots,\mathbf{x}_{k}}=\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}_{1}}\ovee\dots\ovee\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}_{k}}.

Using the operator $\mathcal{T}^{\Phi}$ we will build a sequence of multi-parameter equivalence relations $\prescript{\mathbf{z}}{\delta}{S}_{0},\prescript{\mathbf{z}}{\delta}{S}_{1},\dots,\prescript{\mathbf{z}}{\delta}{S}_{N}$ . The idea is to reduce an instance $\Phi$ of the Quantified-3-DNF to $\operatorname{QCSP}(\Gamma)$ by substituting $S_{N}$ into the formula $\mathcal{Q}^{\Phi}$ . If this reduction works, then we proved the PSpace-hardness. If it does not work, we define a new bigger multi-parameter equivalence relation $S_{N+1}$ and continue. Thus, we want to build a sequence $S_{0},\dots,S_{N}$ maintaining the following properties:

(s1)

$S_{i}$ is q-definable over $\{R,D,B,C,\Delta\}$ ;
(s2)

for every $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ the universal relation $\prescript{\mathbf{z}}{\delta}{S}_{i+1}$ is larger than $\prescript{\mathbf{z}}{\delta}{S}_{i}$ ;
(s3)

there exist $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ such that $\prescript{\mathbf{z}}{\delta}{S}_{i+1}$ is strictly larger than $\prescript{\mathbf{z}}{\delta}{S}_{i}$ ;
(s4)

$\forall\mathbf{z}\in A^{|A|}\exists\delta\in\prescript{\mathbf{z}}{}{\Delta}\;\forall\mathbf{x}\;(\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C}\subseteq\prescript{\mathbf{z}}{\delta}{S}_{i}^{\mathbf{x}}\ovee\prescript{\mathbf{z}}{\delta}{R}_{1}^{\mathbf{x}})$ ;
(s5)

$\exists\mathbf{z}\in A^{|A|}\forall\delta\in\prescript{\mathbf{z}}{}{\Delta}\exists\mathbf{x}((\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C})\cap\prescript{\mathbf{z}}{\delta}{S}_{i}^{\mathbf{x}}=\varnothing)$ .

Let us check that $S_{0}$ satisfies conditions (s1), (s4), and (s5). Condition (s1) follows from the definition. Condition (s4) and (s5) come from (1) and (2) in Lemma 55.

Properties (s2) and (s3) guarantee that the sequence will not be infinite. Assume that we have a sequence $S_{0},S_{1},\dots,S_{N}$ . Let us build $S_{N+1}$ satisfying the above properties or prove the PSpace-hardness using $S_{N}$ . For every instance $\Phi$ of the Quantified-3-DNF by $\prescript{\mathbf{z}}{\delta}{S}_{N+1,\Phi}$ we denote $\mathcal{T}^{\Phi}(\prescript{\mathbf{z}}{\delta}{S}_{N},\prescript{\mathbf{z}}{\delta}{R}_{1})$ . Consider two cases:

Case 1. There exists a Yes-instance $\Phi$ of the Quantified-3-DNF such that $S_{N+1,\Phi}$ satisfies condition (s5). Put $S_{N+1}=S_{N+1,\Phi}$ and check that each of the properties (s1)-(s5) holds. Property (s1) follows from the definition. Property (s2) follows from Lemma 52. To prove property (s3) consider $\mathbf{z}$ from condition (s5) for $S_{N}$ , and the corresponding $\delta$ from condition (s4) for $S_{N}$ . Then $\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C}\subseteq\prescript{\mathbf{z}}{\delta}{S}_{N}^{\mathbf{x}}\ovee\prescript{\mathbf{z}}{\delta}{R}_{1}^{\mathbf{x}}$ for every $\mathbf{x}$ and $(\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C})\cap\prescript{\mathbf{z}}{\delta}{S}_{N}^{\mathbf{x}}=\varnothing$ for some $\mathbf{x}$ . Then Lemma 53 implies that $\prescript{\mathbf{z}}{\delta}{S}_{N+1}$ is strictly larger than $\prescript{\mathbf{z}}{\delta}{S}_{N}$ which proves condition (s3). Property (s4) follows from the fact that $S_{N+1}$ is larger that $S_{N}$ and $S_{N}$ satisfies (s4). Property (s5) is just the definition of Case 1. Thus, we defined $S_{N+1}$ satisfying the required properties (s1)-(s5).

Case 2. $S_{N+1,\Phi}$ does not satisfy property (s5) for any Yes-instance $\Phi$ of the Quantified-3-DNF. Thus, for every Yes-instance $\Phi$ we have

\displaystyle\forall\mathbf{z}\in A^{|A|}\exists\delta\in\prescript{\mathbf{z}}{}{\Delta}\;\forall\mathbf{x}\;(\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C}\subseteq\prescript{\mathbf{z}}{\delta}{S}_{N+1,\Phi}^{\mathbf{x}}).

(6)

Let us show that any (Yes- or No-) instance $\Phi$ of the Quantified-3-DNF is equivalent to

\displaystyle\forall\mathbf{z}\in A^{|A|}\exists\delta\in\prescript{\mathbf{z}}{}{\Delta}\;\forall\mathbf{u}\;(S_{N+1,\Phi}^{\mathbf{u}}(y,y^{\prime})\wedge y\in\prescript{\mathbf{z}}{\delta}{B}\wedge y^{\prime}\in\prescript{\mathbf{z}}{\delta}{C}).

(7)

Notice that the above formula can be efficiently built from the instance $\Phi$ , which gives us a polynomial reduction from the Quantified-3-DNF. If $\Phi$ is a Yes-instance, then it follows from (6). Suppose $\Phi$ is a No-instance. Recall that (see condition (3) in Lemma 55) there exists $\alpha$ such that $\prescript{\mathbf{z}}{\delta}{R}_{1}^{\alpha}\subseteq\prescript{\mathbf{z}}{\delta}{R}_{0}^{\mathbf{x}}\subseteq\prescript{\mathbf{z}}{\delta}{S}_{N}^{\mathbf{x}}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\mathbf{x}$ . Combining this with property (s5) for $S_{N}$ and using Lemma 54 we obtain that

\exists\mathbf{z}\in A^{|A|}\forall\delta\in\prescript{\mathbf{z}}{}{\Delta}\;\exists\mathbf{x}\;((\prescript{\mathbf{z}}{\delta}{B}\times\prescript{\mathbf{z}}{\delta}{C})\cap\prescript{\mathbf{z}}{\delta}{S}_{N+1}^{\mathbf{x}}=\varnothing).

Hence, (7) does not hold and the instance $\Phi$ is equivalent to (7). Thus we built a reduction from the Quantified-3-DNF and proved PSpace-hardness of $\operatorname{QCSP}(\{Q,D,B,C,\Delta\})$ . ∎

7.3 Mighty tuples II, III, and IV

It this section we show that mighty tuples II, III, and IV are equivalent in the sense that any of them q-defines any other.

Below, $R$ is always a $(\mathbf{z},\alpha)$ -parameterized binary relation, $D,B$ and $C$ are $\mathbf{z}$ -parameterized unary relation, where $\mathbf{z}\in A^{|A|}$ and $\alpha\in A^{k}$ . The tuple $(R,D,B,C)$ is called a quadruple in this section. We will need the following properties of a quadruple:

( $\kappa$ )

$k=|A|$ and $\prescript{\mathbf{z}}{}{R}^{\kappa}\subseteq\prescript{\mathbf{z}}{}{R}^{\alpha}$ for every $\mathbf{z}\in A^{|A|}$ and $\alpha\in A^{k}$ ;
( $d+$ )

$\prescript{\mathbf{z}}{}{D}+\prescript{\mathbf{z}}{}{R}^{\forall\forall}=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
(un)

$\prescript{\mathbf{z}}{}{B}\neq\varnothing$ , $\prescript{\mathbf{z}}{}{C}\neq\varnothing$ , $\prescript{\mathbf{z}}{}{B}\subseteq\prescript{\mathbf{z}}{}{D}$ , $\prescript{\mathbf{z}}{}{C}\subseteq\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
( $bc$ )

$\prescript{\mathbf{z}}{}{R}^{\forall}\cap(\prescript{\mathbf{z}}{}{B}\times\prescript{\mathbf{z}}{}{C})\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ ;
( $\varnothing$ )

$\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}=\varnothing$ for some $\mathbf{z}\in A^{|A|}$ ;
( $b+$ )

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{R}^{\forall\forall}=\prescript{\mathbf{z}}{}{B}$ for every $\mathbf{z}\in A^{|A|}$ ;
( $+c$ )

$\prescript{\mathbf{z}}{}{R}^{\forall\forall}+\prescript{\mathbf{z}}{}{C}=\prescript{\mathbf{z}}{}{C}$ for every $\mathbf{z}\in A^{|A|}$ ;
(t)

$\prescript{\mathbf{z}}{}{R}^{\alpha}+\prescript{\mathbf{z}}{}{R}^{\alpha}=\prescript{\mathbf{z}}{}{R}^{\alpha}$ for every $\mathbf{z}\in A^{|A|}$ and $\alpha\in A^{k}$ ;
(sd)

$\operatorname{pr}_{1}(\prescript{\mathbf{z}}{}{R}^{\alpha})=\operatorname{pr}_{2}(\prescript{\mathbf{z}}{}{R}^{\alpha})=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ and $\alpha\in A^{k}$ ;
(r)

$\{(d,d)\mid d\in\prescript{\mathbf{z}}{}{D}\}\subseteq\prescript{\mathbf{z}}{}{R}^{\alpha}$ for every $\mathbf{z}\in A^{|A|}$ and $\alpha\in A^{k}$ ;
( $bd$ )

$\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{R}^{\forall}=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
( $cd$ )

$\prescript{\mathbf{z}}{}{R}^{\forall}+\prescript{\mathbf{z}}{}{C}=\prescript{\mathbf{z}}{}{D}$ for every $\mathbf{z}\in A^{|A|}$ ;
( $c+$ )

$\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\forall\forall}=\prescript{\mathbf{z}}{}{C}$ for some $\mathbf{z}\in A^{|A|}$ ;
(s)

$\prescript{\mathbf{z}}{}{R}^{\alpha}$ is symmetric for every $\mathbf{z}\in A^{|A|}$ and $\alpha\in A^{k}$ .

Notice that a mighty tuple II is just a quadruple satisfying all the above properties except for ( $\kappa$ ), a mighty tuple III $(R,B,C)$ forms a quadruple $(R,A,B,C)$ , where $D=A$ , satisfying properties $\{\text{un},bc,\varnothing,b+,+c\}$ , a mighty tuple IV is a quadruple satisfying properties $\{d+,\text{un},bc,\varnothing,b+,bd\}$ . Let $\mathrm{II}$ be the set of all the above properties except for ( $\kappa$ ), $\mathrm{III}=\{\text{un},bc,\varnothing,b+,+c\}$ , and $\mathrm{IV}=\{d+,\text{un},bc,\varnothing,b+,bd\}$ .

Below we prove many claims that allow us to moderate the quadruple to satisfy more properties from the above list. Usually the claims are of the form $P_{1}\vdash P_{2}$ , where $P_{1}$ and $P_{2}$ are some sets of properties of a quadruple, and should be understood as follows. Suppose a quadruple satisfies properties $P_{1}$ , then there exists a quadruple q-definable from the first one and satisfying properties $P_{2}$ . Also sometimes we add “ $+\text{reduce }\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ ” meaning that the sum $\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ calculated for the new quadruple is smaller than the sum calculated for the old one. We write increase or keep instead of reduce if the sum is increased or stays the same, respectively. Most of the properties are from the above list but some of them are given by a quantified formula.

First, we want to be able to add the additional property ( $\kappa$ ) to existing properties from III or IV.

Claim 7.3.1.

Suppose $P\subseteq\mathrm{III}\cup\mathrm{IV}$ . Then $P\vdash P\cup\{\kappa\}$ .

Proof.

We change only the $(\mathbf{z},\alpha)$ -parameterized relation $R$ . The new relation $R_{0}$ is defined by

\prescript{\mathbf{z}}{}{R}_{0}^{x_{1},\dots,x_{|A|}}(y_{1},y_{2})=\bigwedge\limits_{i_{1},\dots,i_{k}\in\{1,2,\dots,|A|\}.}\prescript{\mathbf{z}}{}{R}^{x_{i_{1}},\dots,x_{i_{k}}}(y_{1},y_{2}).

Then $R_{0}^{\kappa}=R_{0}^{\forall\forall}=R^{\forall\forall}$ and $R_{0}^{\forall}=R^{\forall}$ . It is straightforward to check that the quadruple $(R_{0},D,B,C)$ satisfies all the properties satisfied by $(R,D,B,C)$ , which completes the proof. ∎

Notice that property ( $\kappa$ ) implies that $R^{\kappa}=R^{\forall\forall}$ and in the following claims we usually write $R^{\kappa}$ instead of $R^{\forall\forall}$ .

Claim 7.3.2.

$\mathrm{III}\cup\{\kappa\}\vdash\mathrm{III}\cup\{\kappa,t\}$ .

Proof.

Put $\prescript{\mathbf{z}}{}{R}_{0}^{\beta}=N\cdot\prescript{\mathbf{z}}{}{R}^{\beta}$ , where $N=|A|!\cdot|A|^{2}$ . Note that we have $\prescript{\mathbf{z}}{}{R}_{0}^{\forall}\supseteq N\cdot\prescript{\mathbf{z}}{}{R}^{\forall}$ . We claim that $R_{0}$ , $B$ , $C$ , and $D$ satisfy properties $\{\kappa,\text{un},bc,\varnothing,b+,+c,\text{t}\}$ . For all the properties but ( $bc$ ) and (t) it follows immediately from the same properties for $R$ . To prove ( $bc$ ) we choose some $(\prescript{\mathbf{z}}{}{b}_{1},\prescript{\mathbf{z}}{}{c})\in\prescript{\mathbf{z}}{}{R}^{\forall}\cap(\prescript{\mathbf{z}}{}{B}\times\prescript{\mathbf{z}}{}{C})$ . Since ( $b+$ ), we can find a sequence $\prescript{\mathbf{z}}{}{b}_{N}-\prescript{\mathbf{z}}{}{b}_{N-1}-\dots-\prescript{\mathbf{z}}{}{b}_{2}-\prescript{\mathbf{z}}{}{b}_{1}$ such that each $\prescript{\mathbf{z}}{}{b}_{i}$ is from $\prescript{\mathbf{z}}{}{B}$ and $(\prescript{\mathbf{z}}{}{b}_{i+1},\prescript{\mathbf{z}}{}{b}_{i})\in\prescript{\mathbf{z}}{}{R}^{\kappa}\subseteq\prescript{\mathbf{z}}{}{R}^{\forall}$ . Then $(b_{N},c)\in\prescript{\mathbf{z}}{}{R}_{0}^{\forall}$ , which implies ( $bc$ ). Lemma 47 implies property (t). ∎

Claim 7.3.3.

$\mathrm{IV}\cup\{\kappa\}\vdash\mathrm{IV}\cup\{\kappa,t\}$ .

Proof.

The proof repeats the proof of the previous claim word for word. Additional properties ( $bd$ ) and ( $d+$ ) follow from ( $bd$ ) and ( $d+$ ) for $R$ . ∎

Claim 7.3.4.

$\mathrm{IV}\cup\{\kappa,\text{t}\}\vdash\mathrm{IV}\cup\{\kappa,t,+c\}$ .

Proof.

Put $\prescript{\mathbf{z}}{}{C}_{0}=(\prescript{\mathbf{z}}{}{R}^{\kappa}+\prescript{\mathbf{z}}{}{C})\cap\prescript{\mathbf{z}}{}{D}$ and $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}=\prescript{\mathbf{z}}{}{R}^{\alpha}\cap(\prescript{\mathbf{z}}{}{D}\times\prescript{\mathbf{z}}{}{D})$ . We claim that $R_{0}$ , $B$ , $C_{0}$ , and $D$ satisfy the required properties. Restriction of $R$ does not affect any properties as we have property (un). Changing $C$ could affect only properties (un) and ( $\varnothing$ ). (un) follows from ( $d+$ ) for $R$ and property (un) for $C$ . By property (t) we have

\prescript{\mathbf{z}}{}{C}_{0}=\prescript{\mathbf{z}}{}{R}_{0}^{\kappa}+\prescript{\mathbf{z}}{}{C}=\prescript{\mathbf{z}}{}{R}_{0}^{\kappa}+\prescript{\mathbf{z}}{}{R}_{0}^{\kappa}+\prescript{\mathbf{z}}{}{C}=\prescript{\mathbf{z}}{}{R}_{0}^{\kappa}+\prescript{\mathbf{z}}{}{C}_{0}

hence we have ( $+c$ ). To prove property ( $\varnothing$ ) we use this property for $C$ and consider $\mathbf{z}\in A^{|A|}$ such that $\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}=\varnothing$ . Then using property ( $b+$ ) for $R$ we derive

\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}=\varnothing\Rightarrow(\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{R}^{\kappa})\cap\prescript{\mathbf{z}}{}{C}=\varnothing\Rightarrow\prescript{\mathbf{z}}{}{B}\cap(\prescript{\mathbf{z}}{}{R}^{\kappa}+\prescript{\mathbf{z}}{}{C})=\varnothing\Rightarrow\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{C}_{0}=\varnothing.

∎

Notice that $\mathrm{IV}\cup\{\kappa,\text{t},+c\}=\mathrm{III}\cup\{\kappa,\text{t},d+,bd\}$ .

Claim 7.3.5.

$\mathrm{III}\cup\{\kappa,\text{t}\}\vdash\mathrm{III}\cup\{\kappa,\text{t},d+,\text{r},\text{sd}\}$ .

Proof.

Put $\prescript{\mathbf{z}}{}{D}_{0}(x)=\prescript{\mathbf{z}}{}{R}^{\kappa}(x,x)$ , $\prescript{\mathbf{z}}{}{B}_{0}=\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{D}_{0}$ , $\prescript{\mathbf{z}}{}{C}_{0}=\prescript{\mathbf{z}}{}{C}\cap\prescript{\mathbf{z}}{}{D}_{0}$ , $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}=\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}\cap(\prescript{\mathbf{z}}{}{D}_{0}\times\prescript{\mathbf{z}}{}{D}_{0})$ .

Let us prove that $R_{0}$ , $B_{0}$ , $C_{0}$ , and $D_{0}$ satisfy the required properties. Properties ( $\kappa$ ) and $(\varnothing)$ follow from the corresponding properties for $R,B,C$ , and $D$ . Properties (r), ( $d+$ ), and (sd) follow immediately from the definition.

Consider a tuple $(b_{0},c_{0})\in\prescript{\mathbf{z}}{}{R}^{\forall}$ , which exists by property ( $bc$ ). By property ( $b+$ ) we can find a path $b_{N}-b_{N-1}-\dots-b_{1}-b_{0}$ of any length $N$ such that each $b_{i}$ is from $\prescript{\mathbf{z}}{}{B}$ and $(b_{i+1},b_{i})\in\prescript{\mathbf{z}}{}{R}^{\kappa}$ . Similarly, by property ( $+c$ ) we can find $c_{0}-c_{1}-\dots-c_{N}$ of length $N$ such that each $c_{i}$ is from $\prescript{\mathbf{z}}{}{C}$ and $(c_{i},c_{i+1})\in\prescript{\mathbf{z}}{}{R}^{\kappa}$ . If $N$ is large enough then both sequences will have repetitive elements. Let these elements be $b_{i}$ and $c_{j}$ . By property (t), these repetitive elements $b_{i}$ and $c_{j}$ should be from $\prescript{\mathbf{z}}{}{B}_{0}$ and $\prescript{\mathbf{z}}{}{C}_{0}$ , respectively, which implies (un). Again (t) for $R$ implies that $(b_{i},c_{j})\in\prescript{\mathbf{z}}{}{R}^{\forall}$ and therefore $(b_{i},c_{j})\in\prescript{\mathbf{z}}{}{R}_{0}^{\forall}$ , which confirms ( $bc$ ). Properties ( $b+$ ) and ( $+c$ ) for $R_{0}$ follows from ( $b+$ ) and ( $+c$ ) for $R$ and (r) for $R_{0}$ .

By reflexivity (property (r)) of $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}$ we have $\prescript{\mathbf{z}}{}{R}^{\alpha}+\prescript{\mathbf{z}}{}{R}^{\alpha}\supseteq\prescript{\mathbf{z}}{}{R}^{\alpha}$ and by property (t) for $R$ we have $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}+\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}\subseteq\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}$ . Thus, we have property (t) for $R_{0}$ . ∎

Denote $J=\{\kappa,d+,\text{un},bc,\varnothing,b+,+c,\text{t},\text{r},\text{sd}\}$ . Then $J=\mathrm{III}\cup\{\kappa,d+,\text{t},\text{r},\text{sd}\}=\mathrm{IV}\cup\{\kappa,\text{t},+c,\text{r},\text{sd}\}\setminus\{bd\}=\mathrm{II}\cup\{\kappa\}\setminus\{bd,cd,c+,\text{s}\}$ .

Claim 7.3.6.

$J\cup\{\neg bd\}\vdash J\cup\{bd\}+\text{reduce }\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ .

Proof.

Put $\prescript{\mathbf{z}}{}{D}_{0}=\prescript{\mathbf{z}}{}{B}+\prescript{\mathbf{z}}{}{R}^{\forall}$ , $\prescript{\mathbf{z}}{}{B}_{0}=\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{D}_{0}=\prescript{\mathbf{z}}{}{B}$ , $\prescript{\mathbf{z}}{}{C}_{0}=\prescript{\mathbf{z}}{}{C}\cap\prescript{\mathbf{z}}{}{D}_{0}$ , $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}=\prescript{\mathbf{z}}{}{R}^{\alpha}\cap(\prescript{\mathbf{z}}{}{D}_{0}\times\prescript{\mathbf{z}}{}{D}_{0})$ . Let us prove that $R_{0}$ , $B_{0}$ , $C_{0}$ , $D_{0}$ satisfy the required properties. Properties ( $\kappa$ ), ( $d+$ ), ( $\varnothing$ ), ( $b+$ ), ( $+c$ ), and (r) follow immediately from the definition and the corresponding properties for $R$ . Property (t) follows from (t) and (r) for $R$ . By property (r) for $R$ we have $\prescript{\mathbf{z}}{}{B}_{0}=\prescript{\mathbf{z}}{}{B}$ . Hence, properties ( $bd$ ) and ( $bc$ ) follow from the definition. Property (un) follows from (un) and ( $bc$ ) for $R$ . ∎

Claim 7.3.7.

$J\cup\{\neg cd\}\vdash J\cup\{cd\}+\text{reduce }\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ .

Proof.

Put $\prescript{\mathbf{z}}{}{D}_{0}=\prescript{\mathbf{z}}{}{R}^{\forall}+\prescript{\mathbf{z}}{}{C}$ , $\prescript{\mathbf{z}}{}{B}_{0}=\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{D}_{0}$ , $\prescript{\mathbf{z}}{}{C}_{0}=\prescript{\mathbf{z}}{}{C}\cap\prescript{\mathbf{z}}{}{D}_{0}=\prescript{\mathbf{z}}{}{C}$ , $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}=\prescript{\mathbf{z}}{}{R}^{\alpha}\cap(\prescript{\mathbf{z}}{}{D}_{0}\times\prescript{\mathbf{z}}{}{D}_{0})$ and repeat the proof of the previous claim switching $B$ and $C$ . ∎

Claim 7.3.8.

$J\cup\{bd,cd,\forall\mathbf{z}(\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa})\cap\prescript{\mathbf{z}}{}{B}\neq\varnothing,\exists\mathbf{z}\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa}\neq\prescript{\mathbf{z}}{}{D}\}\vdash J+\text{reduce }\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ .

Proof.

Put $\prescript{\mathbf{z}}{}{D}_{0}=\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa}$ , $\prescript{\mathbf{z}}{}{B}_{0}=\prescript{\mathbf{z}}{}{B}\cap\prescript{\mathbf{z}}{}{D}_{0}$ , $\prescript{\mathbf{z}}{}{C}_{0}=\prescript{\mathbf{z}}{}{C}\cap\prescript{\mathbf{z}}{}{D}_{0}=\prescript{\mathbf{z}}{}{C}$ , $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}=\prescript{\mathbf{z}}{}{R}^{\alpha}\cap(\prescript{\mathbf{z}}{}{D}_{0}\times\prescript{\mathbf{z}}{}{D}_{0})$ . Notice that by (r) we have $\prescript{\mathbf{z}}{}{D}_{0}\supseteq\prescript{\mathbf{z}}{}{C}$ , and by the property $\forall\mathbf{z}(\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa})\cap\prescript{\mathbf{z}}{}{B}\neq\varnothing$ we have $\prescript{\mathbf{z}}{}{B}_{0}\neq\varnothing$ . Then properties ( $\kappa$ ), ( $d+$ ), (un), ( $\varnothing$ ), ( $b+$ ), ( $+c$ ), (t), and (r) follow from the corresponding properties for $R$ . Property ( $bc$ ) follows from ( $cd$ ) for $R$ . ∎

Claim 7.3.9.

$J\cup\{\neg c+,\exists\mathbf{z}(\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa})\cap\prescript{\mathbf{z}}{}{B}=\varnothing\}\vdash J+\text{keep }\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|+\text{increase }\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{C}|$ .

Proof.

We put $\prescript{\mathbf{z}}{}{C}_{0}=\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa}$ , $\prescript{\mathbf{z}}{}{C}_{1}=\prescript{\mathbf{z}}{}{R}^{\kappa}+\prescript{\mathbf{z}}{}{C}_{0}$ , and claim that $R$ , $D$ , $B$ , and $C_{1}$ satisfy the required properties. Since we only increased $C$ the only properties we need to check are ( $\varnothing$ ) and ( $+c$ ). Property ( $+c$ ) follows from property (r) and (t) for $R$ . To prove property ( $\varnothing$ ) choose $\mathbf{z}\in A^{|A|}$ from the property $\exists\mathbf{z}(\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa})\cap\prescript{\mathbf{z}}{}{B}=\varnothing$ . Then $\prescript{\mathbf{z}}{}{C}_{0}\cap\prescript{\mathbf{z}}{}{B}=\varnothing$ . By property ( $b+$ ) $\prescript{\mathbf{z}}{}{C}_{1}\cap\prescript{\mathbf{z}}{}{B}=\varnothing$ , which gives us property ( $\varnothing$ ). ∎

Claim 7.3.10.

$J\cup\{bd,cd,\forall\mathbf{z}\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa}=\prescript{\mathbf{z}}{}{D}\}\vdash J\cup\{c+,\text{s}\}+\text{keep }\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ .

Proof.

Put $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}(x,y)=\prescript{\mathbf{z}}{}{R}^{\alpha}(x,y)\wedge\prescript{\mathbf{z}}{}{R}^{\alpha}(y,x)$ and prove the required properties for $R_{0}$ , $B$ , $C$ , and $D$ . Properties ( $\kappa$ ), ( $d+$ ), (un), ( $\varnothing$ ), ( $b+$ ), ( $+c$ ), ( $c+$ ), (t), (r), and (s) follow from the definition and the respective properties for $R$ . It remains to prove property ( $bc$ ).

Notice that it follows from (t) for $R$ that $\prescript{\mathbf{z}}{}{R}^{\forall}$ is transitive. Let us build an infinite path $d_{0}-d_{1}-d_{2}-d_{3}-d_{4}-d_{5}\dots$ such that each $d_{2i}$ is from $\prescript{\mathbf{z}}{}{B}$ , each $d_{2i+1}$ is from $\prescript{\mathbf{z}}{}{C}$ , each $(d_{i+1},d_{i})$ is from $\prescript{\mathbf{z}}{}{R}^{\forall}$ . Choose some $d_{0}\in\prescript{\mathbf{z}}{}{B}$ . By the property $\forall\mathbf{z}\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa}=\prescript{\mathbf{z}}{}{D}$ , there exists $d_{1}\in\prescript{\mathbf{z}}{}{C}$ such that $(d_{1},d_{0})\in\prescript{\mathbf{z}}{}{R}^{\kappa}\subseteq\prescript{\mathbf{z}}{}{R}^{\forall}$ . By property ( $bd$ ) there exists $d_{2}\in\prescript{\mathbf{z}}{}{B}$ such that $(d_{2},d_{1})\in\prescript{\mathbf{z}}{}{R}^{\forall}$ . Proceeding this way we can make an infinite sequence. Since $\prescript{\mathbf{z}}{}{B}$ is finite, we have $d_{2i}=d_{2j}$ for some $i<j$ . By transitivity of $\prescript{\mathbf{z}}{}{R}^{\forall}$ we have $(d_{2i},d_{2i+1}),(d_{2i+1},d_{2i})\in\prescript{\mathbf{z}}{}{R}^{\forall}$ , which gives us property ( $bc$ ). ∎

Claim 7.3.11.

$J\cup\{bd,cd,c+\}\vdash J\cup\{bd,cd,c+,\text{s}\}$ .

Proof.

Put $B_{0}=B$ and $R_{0}=R$ . Define sequences $\prescript{\mathbf{z}}{}{B}_{i+1}=\prescript{\mathbf{z}}{}{B}_{i}-\prescript{\mathbf{z}}{}{R}^{\kappa}+\prescript{\mathbf{z}}{}{R}^{\kappa}$ and $\prescript{\mathbf{z}}{}{R}_{i+1}^{\alpha}=\prescript{\mathbf{z}}{}{R}_{i}^{\alpha}-\prescript{\mathbf{z}}{}{R}^{\alpha}+\prescript{\mathbf{z}}{}{R}^{\alpha}$ .

Since $\prescript{\mathbf{z}}{}{R}^{\kappa}$ is reflexive, these sequences of relations are growing. From the finiteness we conclude that these sequences will stabilize at some $N$ .

Let us prove that $R_{N}$ , $B_{N}$ , $C$ , and $D$ satisfy the required properties. Properties ( $\kappa$ ), ( $d+$ ), (un), (bc), ( $+c$ ), (r), (bd), (cd), and ( $c+$ ) easily follow from the corresponding properties for $R$ . Properties ( $b+$ ) and (t) follow from the fact that sequences stabilized. By ( $c+$ ) and ( $+c$ ) for $R$ we derive that we can never escape from $\prescript{\mathbf{z}}{}{C}$ and therefore, by property ( $\varnothing$ ) for $R$ , we can never come to $\prescript{\mathbf{z}}{}{B}$ . Hence we have ( $\varnothing$ ). Property (s) follows from properties (r) and (t) for $R$ and from the fact that the sequences stabilized. ∎

Claim 7.3.12.

$\mathrm{III}\vdash\mathrm{II}\cup\{\kappa\}$ .

Proof.

By Claim 7.3.1 we get property ( $\kappa$ ). By Claim 7.3.2 we additionally get property (t). By Claim 7.3.5 we additionally get ( $d+$ ), (r), and (sd). Thus, we get all the properties from $J$ .

Iteratively applying Claims 7.3.6, 7.3.7, 7.3.8, and 7.3.9 whenever possible we either achieve additional properties ( $bd$ ), ( $cd$ ), and ( $c+$ ), or we get additional properties ( $bd$ ), ( $cd$ ) and $\forall\mathbf{z}\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa}=\prescript{\mathbf{z}}{}{D}$ . Notice that the process cannot last forever because at every step we either reduce $\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ or increase $\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{C}|$ . If we get additional properties ( $bd$ ), ( $cd$ ), and ( $c+$ ) then the statement follows from Claim 7.3.11. If we get additional properties ( $bd$ ), ( $cd$ ) and $\forall\mathbf{z}\prescript{\mathbf{z}}{}{C}+\prescript{\mathbf{z}}{}{R}^{\kappa}=\prescript{\mathbf{z}}{}{D}$ then we apply Claim 7.3.10. If the obtained quadruple satisfies ( $bd$ ) and ( $cd$ ) then we satisfied all the required properties. Otherwise we apply Claims 7.3.6 and 7.3.7, and reduce the sum $\sum_{\mathbf{z}\in A^{|A|}}|\prescript{\mathbf{z}}{}{D}|$ . Then we again apply Claims 7.3.6, 7.3.7, 7.3.8, and 7.3.9 whenever possible and so on. ∎

Note that properties (r),(t),(s), and (sd) imply that $\prescript{\mathbf{z}}{}{R}^{\alpha}$ is an equivalence relation on $\prescript{\mathbf{z}}{}{D}$ , and properties ( $b+$ ) and $(+c)$ imply that $\prescript{\mathbf{z}}{}{B}$ and $\prescript{\mathbf{z}}{}{C}$ are unions of some equivalence classes of $\prescript{\mathbf{z}}{}{R}^{\kappa}$ .

Lemma 56.

Suppose a quadruple $(R,D,B,C)$ satisfies all the properties from $\mathrm{II}\cup\{\kappa\}$ . Then there exists a mighty tuple I q-definable from $R,D,B,C$ .

Proof.

Define a mighty tuple $(R_{1},D_{1},B_{1},C_{1},\Delta_{1})$ by

	$\displaystyle\prescript{\mathbf{z}}{}{\Delta}_{1}(u,v)=$	$\displaystyle\exists x(\prescript{\mathbf{z}}{}{B}(u)\wedge\prescript{\mathbf{z}}{}{C}(v)\wedge\prescript{\mathbf{z}}{}{R}^{\forall}(u,x)\wedge\prescript{\mathbf{z}}{}{R}^{\forall}(v,x))$
	$\displaystyle\prescript{\mathbf{z}}{uv}{D}_{1}(x)=$	$\displaystyle\prescript{\mathbf{z}}{}{B}(u)\wedge\prescript{\mathbf{z}}{}{C}(v)\wedge\prescript{\mathbf{z}}{}{R}^{\forall}(u,x)\wedge\prescript{\mathbf{z}}{}{R}^{\forall}(v,x)$
	$\displaystyle\prescript{\mathbf{z}}{uv}{R}_{1}^{\alpha}=$	$\displaystyle\prescript{\mathbf{z}}{}{R}^{\alpha}\cap(\prescript{\mathbf{z}}{uv}{D}_{1}\times\prescript{\mathbf{z}}{uv}{D}_{1})$
	$\displaystyle\prescript{\mathbf{z}}{uv}{B}_{1}(x)=$	$\displaystyle\prescript{\mathbf{z}}{}{B}(u)\wedge\prescript{\mathbf{z}}{}{C}(v)\wedge\prescript{\mathbf{z}}{}{R}^{\kappa}(u,x)\wedge\prescript{\mathbf{z}}{}{R}^{\forall}(v,x)$
	$\displaystyle\prescript{\mathbf{z}}{uv}{C}_{1}(x)=$	$\displaystyle\prescript{\mathbf{z}}{}{B}(u)\wedge\prescript{\mathbf{z}}{}{C}(v)\wedge\prescript{\mathbf{z}}{}{R}^{\forall}(u,x)\wedge\prescript{\mathbf{z}}{}{R}^{\kappa}(v,x)$

By property ( $bc$ ) for the quadruple we derive $\prescript{\mathbf{z}}{}{\Delta}_{1}\neq\varnothing$ , $\prescript{\mathbf{z}}{}{D}_{1}\neq\varnothing$ , $\prescript{\mathbf{z}}{}{B}_{1}\neq\varnothing$ , and $\prescript{\mathbf{z}}{}{C}_{1}\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ . Thus, we already satisfied first two properties of a mighty tuple. Property 3 from (r), (t), (s). Property 4 follows from the definition and the fact that $\prescript{\mathbf{z}}{}{R}^{\forall}$ is an equivalence relation. Property 5 follows from the definition, and property 6 follows from $(\varnothing$ ) for $R$ . ∎

We will prove the following two lemmas from Section 5.4 simultaneously.

Lemma 15.

Suppose $\Sigma$ is a set of relations on $A$ . Then the following conditions are equivalent:

1.

$\Sigma$ q-defines a mighty tuple II;
2.

$\Sigma$ q-defines a mighty tuple III;
3.

$\Sigma$ q-defines a mighty tuple IV.

Lemma 16.

Suppose $T$ is a mighty tuple of type II, III, or IV. Then relations of $T$ q-define a mighty tuple I.

Proof.

We want to prove that the existence of a mighty tuple II, III, or IV implies the existence of I, II, III, and IV. First, let us show that we can derive a mighty tuple III. A mighty tuple II is also a mighty tuple III, hence for II and III it is obvious. If $(Q,D,B,C)$ is a mighty tuple IV, then $(Q,D,B,C)$ satisfies conditions from IV. By Claims 7.3.1, 7.3.3 and 7.3.4 we derive a quadruple satisfying all the properties of III, which gives us a mighty tuple III.

Let $(Q,B,C)$ be a mighty tuple III. Put $\prescript{\mathbf{z}}{}{D}=A$ for every $\mathbf{z}$ . Then $(Q,D,B,C)$ satisfies all properties of III. Claim 7.3.12 implies the existence of a quadruple satisfying properties $\mathrm{II}\cup\{\kappa\}$ . Notice that this quadruple is simultaneously a mighty tuple II, III, and IV. Additionally, Lemma 56 implies that this quadruple q-defines a mighty tuple I. ∎

7.4 Mighty tuple V

First, let us define a modification of a mighty tuple V, which we call a mighty tuple V’.

A tuple $(Q,D,\Delta)$ , where $\Delta$ is a $\mathbf{z}$ -parameterized $m$ -ary relation, $Q$ is a $(\mathbf{z},\delta,\alpha)$ -parameterized binary relation, and $D$ is a $(\mathbf{z},\delta)$ -parameterized unary relation, is called a mighty tuple V’ if

1.

$\prescript{\mathbf{z}}{}{\Delta}\neq\varnothing$ for every $\mathbf{z}\in A^{|A|}$ ;
2.

$\prescript{\mathbf{z}}{\delta}{Q}^{\kappa}\subseteq\prescript{\mathbf{z}}{\delta}{Q}^{\alpha}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\alpha\in A^{|A|}$ ;
3.

$\{(d,d)\mid d\in\prescript{\mathbf{z}}{\delta}{D}\}\subseteq\prescript{\mathbf{z}}{\delta}{Q}^{\forall}$ for every $\mathbf{z}\in A^{|A|}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ ; ( $\prescript{\mathbf{z}}{\delta}{Q}^{\forall}$ is reflexive)
4.

$\operatorname{pr}_{1}(\prescript{\mathbf{z}}{\delta}{Q}^{\alpha})=\operatorname{pr}_{2}(\prescript{\mathbf{z}}{\delta}{Q}^{\alpha})=\prescript{\mathbf{z}}{\delta}{D}$ for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\alpha\in A^{|A|}$ .
5.

$\prescript{\mathbf{z}}{\delta}{Q}^{\forall\forall}\cap\{(d,d)\mid d\in A\}=\varnothing$ for some $\mathbf{z}\in A^{|A|}$ and every $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ . ( $\prescript{\mathbf{z}}{\delta}{Q}^{\forall\forall}$ has no loops)

Notice that we allow $\Delta$ to be of arity $0$ . Then condition 1 means that $\Delta=\{\Lambda\}$ , where $\Lambda$ is an empty tuple/word. In this case we can omit a parameter $\delta$ in relations.

Lemma 57.

Suppose $(Q,D)$ is a mighty tuple V. Then $\{Q,D\}$ q-defines a mighty tuple V’ $(R,D,\{\Lambda\})$ .

Proof.

Define a mighty tuple V’ as follows. Let $\Delta$ be the relation of arity $0$ containing the empty tuple. The relation $R$ is defined by

\prescript{\mathbf{z}}{}{R}^{x_{1},\dots,x_{|A|}}(y_{1},y_{2})=\bigwedge\limits_{i_{1},\dots,i_{k}\in\{1,2,\dots,|A|\}.}\prescript{\mathbf{z}}{}{Q}^{x_{i_{1}},\dots,x_{i_{k}}}(y_{1},y_{2})\wedge(y_{1}\in\prescript{\mathbf{z}}{}{D})\wedge(y_{2}\in\prescript{\mathbf{z}}{}{D}).

Then $R^{\kappa}=R^{\forall\forall}=Q^{\forall\forall}$ and $R^{\forall}=Q^{\forall}$ . It is straightforward to check that $(R,D,\Delta)$ is a mighty tuple V’. ∎

Lemma 58.

Suppose $(R,D,\Delta)$ is a mighty tuple V’, $\prescript{\mathbf{z}}{\delta}{R}^{\alpha}$ is symmetric for every $\mathbf{z}\in A^{|A|}$ , $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , and $\alpha$ . Then there exists a mighty tuple I q-definable from $R$ , $D$ , and $\Delta$ .

Proof.

First, we assign a pair to every mighty tuple V’ and evaluations of $\mathbf{z}$ and $\delta$ . Put $\phi_{R,D,\Delta}(\mathbf{z},\delta)=(m,|\prescript{\mathbf{z}}{\delta}{D}|)$ , where $m$ is the minimal odd positive integer such that $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ has cycles of length $m$ . By $\phi^{1}_{R,D,\Delta}(\mathbf{z},\delta)$ we denote the first element of the pair, that is $m$ . Notice that $m$ can be $\infty$ if $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ does not have odd cycles. Then we define a linear order on pairs by $(m_{1},s_{1})\leqslant(m_{2},s_{2})\Leftrightarrow(m_{1}<m_{2})\vee(m_{1}=m_{2}\wedge s_{1}\geqslant s_{2})$ . We put $\phi_{R,D,\Delta}=\max\limits_{\mathbf{z}\in A^{|A|}}\min\limits_{\delta\in\prescript{\mathbf{z}}{}{\Delta}}\phi_{R,D,\Delta}(\mathbf{z},\delta)$ .

We prove the lemma by induction on $\phi_{R,D,\Delta}$ . Assume that it does not hold. Choose a mighty tuple V’ $(R,D,\Delta)$ such that we cannot q-define a mighty tuple I from it and the pair $\phi_{R,D,\Delta}$ is maximal. Thus, to complete the proof it is sufficent to q-define a mighty-tuple V’ with larger pair or q-define a mighty tuple I. Suppose $(m,s)=\phi_{R,D,\Delta}$ . By property 5 of a mighty tuple V’ we have $m\geqslant 3$ . We consider two cases.

Case 1 (base of the induction). $m=\infty$ , i.e., for some $\mathbf{z}_{0}\in Z$ and every $\delta\in\prescript{\mathbf{z}_{0}}{}{\Delta}$ the relation $\prescript{\mathbf{z}_{0}}{\delta}{R}^{\kappa}$ has only cycles of an even length. Then define $\prescript{\mathbf{z}}{\delta}{R}^{\alpha}_{0}=N\cdot\prescript{\mathbf{z}}{\delta}{R}^{\alpha}$ , where $N=|A|!\cdot|A|^{2}$ . Since $\prescript{\mathbf{z}}{\delta}{R}^{\alpha}$ is symmetric, the relation $\prescript{\mathbf{z}}{\delta}{R}^{\alpha}_{0}$ is reflexive and symmetric. By Lemma 47 $\prescript{\mathbf{z}}{\delta}{R}^{\alpha}_{0}$ is transitive, that is $\prescript{\mathbf{z}}{\delta}{R}^{\alpha}_{0}+\prescript{\mathbf{z}}{\delta}{R}^{\alpha}_{0}=\prescript{\mathbf{z}}{\delta}{R}^{\alpha}_{0}$ . Therefore, $\prescript{\mathbf{z}}{\delta}{R}^{\alpha}_{0}$ is an equivalence relation on $\prescript{\mathbf{z}}{\delta}{D}$ for every $\mathbf{z}$ , $\delta$ , and $\alpha$ . This implies that $\prescript{\mathbf{z}}{\delta}{R}^{\forall}_{0}$ is also an equivalence relation. Since $\prescript{\mathbf{z}_{0}}{\delta}{R}^{\kappa}$ has no cycles of an odd length, we have $\prescript{\mathbf{z}_{0}}{\delta}{R}^{\kappa}_{0}\cap\prescript{\mathbf{z}_{0}}{\delta}{R}^{\kappa}=\varnothing$ . By the reflexivity of $\prescript{\mathbf{z}}{\delta}{R}^{\forall}$ we have $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}\subseteq\prescript{\mathbf{z}}{\delta}{R}_{0}^{\forall}$ . Then a mighty tuple I $(R_{1},D_{1},B_{1},C_{1},\Delta_{1})$ can be defined as follows:

	$\displaystyle\prescript{\mathbf{z}}{}{\Delta}_{1}(\delta,u,v)=$	$\displaystyle\prescript{\mathbf{z}}{\delta}{R}^{\kappa}(u,v)\wedge\prescript{\mathbf{z}}{}{\Delta}(\delta)$
	$\displaystyle\prescript{\mathbf{z}}{\delta uv}{D}_{1}(x)=$	$\displaystyle\prescript{\mathbf{z}}{\delta}{R}^{\kappa}(u,v)\wedge\prescript{\mathbf{z}}{\delta}{R}_{0}^{\forall}(u,x)$
	$\displaystyle\prescript{\mathbf{z}}{\delta uv}{R}_{1}^{\alpha}=$	$\displaystyle\prescript{\mathbf{z}}{\delta}{R}_{0}^{\alpha}\cap(\prescript{\mathbf{z}}{\delta uv}{D}_{1}\times\prescript{\mathbf{z}}{\delta uv}{D}_{1})$
	$\displaystyle\prescript{\mathbf{z}}{\delta uv}{B}_{1}(x)=$	$\displaystyle\prescript{\mathbf{z}}{\delta}{R}^{\kappa}(u,v)\wedge\prescript{\mathbf{z}}{\delta}{R}_{0}^{\kappa}(u,x)$
	$\displaystyle\prescript{\mathbf{z}}{\delta uv}{C}_{1}(x)=$	$\displaystyle\prescript{\mathbf{z}}{\delta}{R}^{\kappa}(u,v)\wedge\prescript{\mathbf{z}}{\delta}{R}_{0}^{\kappa}(v,x)$

Since we can choose any tuple $(a,b)\in\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ as $(u,v)$ , we obtain $\prescript{\mathbf{z}}{}{\Delta_{1}}\neq\varnothing$ , $a\in\prescript{\mathbf{z}}{\delta ab}{B}_{1}\subseteq\prescript{\mathbf{z}}{\delta ab}{D}_{1}$ , and $b\in\prescript{\mathbf{z}}{\delta ab}{C}_{1}\subseteq\prescript{\mathbf{z}}{\delta ab}{D}_{1}$ . The relation $\prescript{\mathbf{z}_{0}}{\delta uv}{R}_{1}^{\alpha}$ is an equivalence relation because it is just a restriction of the equivalence relation $\prescript{\mathbf{z}_{0}}{\delta}{R}_{0}^{\alpha}$ to $\prescript{\mathbf{z}}{\delta uv}{D}_{1}$ . It follows from the definition of $\prescript{\mathbf{z}}{\delta}{D}_{1}$ that $\prescript{\mathbf{z}}{\delta}{R}_{1}^{\forall}=\prescript{\mathbf{z}}{\delta}{D}_{1}\times\prescript{\mathbf{z}}{\delta}{D}_{1}$ . The relations $\prescript{\mathbf{z}}{\delta uv}{B}_{1}$ and $\prescript{\mathbf{z}}{\delta uv}{C}_{1}$ are the equivalence classes of $\prescript{\mathbf{z}}{\delta uv}{R}_{1}^{\kappa}$ containing $u$ and $v$ , respectively. Moreover, we have $\prescript{\mathbf{z}_{0}}{\delta uv}{B}_{1}\cap\prescript{\mathbf{z}_{0}}{\delta uv}{C}_{1}=\varnothing$ because otherwise we would get a path of an even length from $u$ to $v$ in $\prescript{\mathbf{z}_{0}}{\delta}{R}^{\kappa}$ , which together with the edge $(u,v)$ would give us a cycle of an odd length from $u$ to $u$ and contradicts our assumption about $\mathbf{z}_{0}$ .

Case 2 (inductive step). $m<\infty$ . Then there exists $\mathbf{z}$ such that for any $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ the relation $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ has no cycles of length smaller than $m$ but for some $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ the relation $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ has cycles of length $m$ . Let $\prescript{\mathbf{z}}{\delta}{R}_{0}^{\alpha}=\lfloor\frac{m}{2}\rfloor\cdot\prescript{\mathbf{z}}{\delta}{R}^{\alpha}$ . Define new relations by

	$\displaystyle\prescript{\mathbf{z}}{}{\Delta}_{1}(\delta,y)$	$\displaystyle=\exists x\exists x^{\prime}\prescript{\mathbf{z}}{\delta}{R}_{0}^{\kappa}(y,x)\wedge\prescript{\mathbf{z}}{\delta}{R}_{0}^{\kappa}(y,x^{\prime})\wedge\prescript{\mathbf{z}}{\delta}{R}^{\kappa}(x,x^{\prime})\wedge\Delta(\delta),$
	$\displaystyle\prescript{\mathbf{z}}{\delta y}{D}_{1}(x)$	$\displaystyle=\exists x^{\prime}\prescript{\mathbf{z}}{\delta}{R}_{0}^{\kappa}(y,x)\wedge\prescript{\mathbf{z}}{\delta}{R}_{0}^{\kappa}(y,x^{\prime})\wedge\prescript{\mathbf{z}}{\delta}{R}^{\kappa}(x,x^{\prime}),$
	$\displaystyle\prescript{\mathbf{z}}{\delta y}{R}_{1}^{\alpha}$	$\displaystyle=\prescript{\mathbf{z}}{\delta}{R}^{\alpha}\cap(\prescript{\mathbf{z}}{\delta y}{D}_{1}\times\prescript{\mathbf{z}}{\delta y}{D}_{1}).$

That is, $\prescript{\mathbf{z}}{}{\Delta_{1}}(\delta,y)$ holds if $\Delta(\delta)$ holds and $y$ is on some cycle of $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ of length $m$ . By the definition of maximality of $m$ over $\mathbf{z}$ the relation $\prescript{\mathbf{z}}{}{\Delta_{1}}$ is not empty for any $\mathbf{z}$ . Also, $\prescript{\mathbf{z}}{\delta y}{D}_{1}$ is the set of all elements such that there exist paths from $y$ to it of lengths $\lfloor\frac{m}{2}\rfloor$ and $\lceil\frac{m}{2}\rceil$ . Hence, $\prescript{\mathbf{z}}{\delta}{D}_{1}$ is not empty for any $\mathbf{z}$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta}_{1}$ . Let us show that $\phi_{R_{1},D_{1},\Delta_{1}}>\phi_{R,D,\Delta}$ . Notice that $\prescript{\mathbf{z}}{\delta y}{R}_{1}^{\kappa}$ is just a restriction of $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ . Also if $\phi^{1}_{R,D,\Delta}(\mathbf{z},\delta)=m$ then $a\notin\prescript{\mathbf{z}}{\delta a}{D}_{1}$ for any $\delta a\in\prescript{\mathbf{z}}{}{\Delta}_{1}$ . Hence, $\prescript{\mathbf{z}}{\delta}{D}\supsetneq\prescript{\mathbf{z}}{\delta a}{D}_{1}$ in this case and we have

\displaystyle\forall\mathbf{z}\;\forall\delta a\in\prescript{\mathbf{z}}{}{\Delta}_{1}\;\left(\phi^{1}_{R,D,\Delta}(\mathbf{z},\delta)=m\rightarrow\phi_{R,D,\Delta}(\mathbf{z},\delta)<\phi_{R_{1},D_{1},\Delta_{1}}(\mathbf{z},\delta a)\right)

(8)

Let $Z$ be the set of all $\mathbf{z}$ such that $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ has no cycle of length smaller than $m$ for every $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ , i.e. $\min\limits_{\delta\in\prescript{\mathbf{z}}{}{\Delta}}\phi^{1}_{R,D,\Delta}(\mathbf{z},\delta)=m$ . Let $\prescript{\mathbf{z}}{}{\Delta}^{\prime}$ be the set of all $\delta\in\prescript{\mathbf{z}}{}{\Delta}$ such that $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ has a cycle of length $m$ . Notice that $\prescript{\mathbf{z}}{}{\Delta}^{\prime}$ is a projection of $\prescript{\mathbf{z}}{}{\Delta}_{1}$ onto all the coordinates but the last one. Then $\phi^{1}_{R,D,\Delta}(\mathbf{z},\delta)=m$ for any $\mathbf{z}\in Z$ and $\delta\in\prescript{\mathbf{z}}{}{\Delta^{\prime}}$ . By (8) we have

	$\displaystyle\phi_{R,D,\Delta}=\max\limits_{\mathbf{z}\in A^{\|A\|}}\min\limits_{\delta\in\prescript{\mathbf{z}}{}{\Delta}}\phi_{R,D,\Delta}(\mathbf{z},\delta)=\max\limits_{\mathbf{z}\in Z}\min\limits_{\delta\in\prescript{\mathbf{z}}{}{\Delta}}\phi_{R,D,\Delta}(\mathbf{z},\delta)=\max\limits_{\mathbf{z}\in Z}\min\limits_{\delta\in\prescript{\mathbf{z}}{}{\Delta}^{\prime}}\phi_{R,D,\Delta}(\mathbf{z},\delta)<$
	$\displaystyle\max\limits_{\mathbf{z}\in Z}\min\limits_{\delta\in\prescript{\mathbf{z}}{}{\Delta}_{1}}\phi_{R_{1},D_{1},\Delta_{1}}(\mathbf{z},\delta)\leqslant\max\limits_{\mathbf{z}\in A^{\|A\|}}\min\limits_{\delta\in\prescript{\mathbf{z}}{}{\Delta}_{1}}\phi_{R_{1},D_{1},\Delta_{1}}(\mathbf{z},\delta)=$	$\displaystyle\phi_{R_{1},D_{1},\Delta_{1}}.$

It remains to check that $(R_{1},D_{1},\Delta_{1})$ is a mighty tuple V’. Property 1 was already mentioned. Properties 2 and 3 follow from the respective properties for $(R,D,\Delta)$ . Property 5 follows from the fact that we only restrict the relation $R$ . To prove property 4 notice that by the definition of $D_{1}$ for every element $x$ there is an element $x^{\prime}$ connected to $x$ in $\prescript{\mathbf{z}}{\delta}{R}^{\kappa}$ and both $x$ and $x^{\prime}$ are in $\prescript{\mathbf{z}}{\delta}{D}_{1}$ . Hence, by the inductive assumption, $(R_{1},D_{1},\Delta_{1})$ q-defines a mighty tuple I, which completes the proof. ∎

Lemma 17.

Suppose $(R,D)$ is a mighty tuple V. Then there exists a mighty tuple I q-definable from $\{R,D\}$ .

Proof.

By Lemma 57 there exists a mighty tuple V’ $(R,D,\{\Lambda\})$ q-definable from $Q$ and $D$ . Put $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}=N\cdot\prescript{\mathbf{z}}{}{R}^{\alpha}$ , where $N=|A|!\cdot|A|^{2}$ . By Lemma 47, $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}$ is transitive for any $\mathbf{z}$ and $\alpha$ .

Put $\prescript{\mathbf{z}}{}{D}_{1}(x)=\prescript{\mathbf{z}}{}{R}_{0}^{\kappa}(x,x)$ . Since any element from a cycle of $\prescript{\mathbf{z}}{}{R}^{\kappa}$ is in $\prescript{\mathbf{z}}{}{D}_{1}$ and $\prescript{\mathbf{z}}{}{R}^{\kappa}$ has cycles by property 4 of a mighty tuple V’, the set $\prescript{\mathbf{z}}{}{D}_{1}$ is not empty.

Let $\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}(x,y)=\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}(x,y)\wedge\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}(y,x)\wedge\prescript{\mathbf{z}}{}{D}_{1}(x)\wedge\prescript{\mathbf{z}}{}{D}_{1}(y)$ . Notice that the relation $\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}$ is reflexive on $\prescript{\mathbf{z}}{}{D}_{1}$ . Then $\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}+\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}\supseteq\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}$ and by transitivity of $\prescript{\mathbf{z}}{}{R}_{0}^{\alpha}$ we get the transitivity of $\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}$ . Thus, $\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}$ is an equivalence relation for every $\mathbf{z}$ , $\delta$ , and $\alpha$ . Consider two cases.

Case 1. There exists $\mathbf{z}_{0}$ such that we have $\prescript{\mathbf{z}_{0}}{}{R}_{1}^{\kappa}\cap\prescript{\mathbf{z}_{0}}{}{R}^{\kappa}=\varnothing$ . Then we define a mighty tuple I $(R_{2},D_{2},B_{2},C_{2},\Delta_{2})$ as follows:

	$\displaystyle\prescript{\mathbf{z}}{}{\Delta}_{2}(u,v)=$	$\displaystyle\prescript{\mathbf{z}}{}{R}^{\kappa}(u,v)$
	$\displaystyle\prescript{\mathbf{z}}{uv}{D}_{2}(x)=$	$\displaystyle\prescript{\mathbf{z}}{}{R}^{\kappa}(u,v)\wedge\prescript{\mathbf{z}}{}{R}_{1}^{\forall}(x,u)$
	$\displaystyle\prescript{\mathbf{z}}{uv}{R}_{2}^{\alpha}=$	$\displaystyle\prescript{\mathbf{z}}{}{R}_{1}^{\alpha}\cap(\prescript{\mathbf{z}}{uv}{D}_{2}\times\prescript{\mathbf{z}}{uv}{D}_{2})$
	$\displaystyle\prescript{\mathbf{z}}{uv}{B}_{2}(x)=$	$\displaystyle\prescript{\mathbf{z}}{}{R}^{\kappa}(u,v)\wedge\prescript{\mathbf{z}}{}{R}_{1}^{\kappa}(u,x)$
	$\displaystyle\prescript{\mathbf{z}}{uv}{C}_{2}(x)=$	$\displaystyle\prescript{\mathbf{z}}{}{R}^{\kappa}(u,v)\wedge\prescript{\mathbf{z}}{}{R}_{1}^{\kappa}(v,x)$

Let us check that all the properties of a mighty tuple I are satisfied. Since we can take any $(u,v)$ on a cycle (of length at most $|A|$ ) in $\prescript{\mathbf{z}}{}{R}^{\kappa}$ , we have $\prescript{\mathbf{z}}{}{\Delta_{2}}\neq\varnothing$ , $u\in\prescript{\mathbf{z}}{uv}{B}_{2}\subseteq\prescript{\mathbf{z}}{uv}{D}_{2}$ , and $v\in\prescript{\mathbf{z}}{uv}{C}_{2}\subseteq\prescript{\mathbf{z}}{uv}{D}_{2}$ . Property $\prescript{\mathbf{z}_{0}}{uv}{B}_{2}\cap\prescript{\mathbf{z}_{0}}{uv}{C}_{2}=\varnothing$ follows from the definition of case 1. Other properties are straightforward.

Case 2. For every $\mathbf{z}$ we have $\prescript{\mathbf{z}}{}{R}_{1}^{\kappa}\cap\prescript{\mathbf{z}}{}{R}^{\kappa}\neq\varnothing$ . This means that we have $(b,c)\in\prescript{\mathbf{z}}{}{R}^{\kappa}$ such that there exists a path from $c$ to $b$ of length $N$ . Hence, $b$ is on a cycle of length $N+1$ . Since a minimal cycle going through $b$ is of size at most $|A|$ , by repeating this cycle we can get a cycle of length $|A|!$ . Combining cycles of lengths $|A|!$ and $|A|!+1$ we can build a cycle of any sufficiently large length. Let $k\geqslant 1$ be the minimal number such that for every $\mathbf{z}$ the graph $\prescript{\mathbf{z}}{}{R}^{\kappa}$ has cycles of length $2^{k}$ . Since $\prescript{\mathbf{z}}{}{R}^{\kappa}$ has no loops for some $\mathbf{z}$ and has all sufficiently large cycles, $k$ is well-defined. Put $\prescript{\mathbf{z}}{}{R}_{3}^{\alpha}=2^{k-1}\cdot\prescript{\mathbf{z}}{}{R}^{\alpha}$ , $\prescript{\mathbf{z}}{}{D}_{4}(x)=\exists y\prescript{\mathbf{z}}{}{R}_{3}^{\kappa}(x,y)\wedge\prescript{\mathbf{z}}{}{R}_{3}^{\kappa}(y,x)$ , $\prescript{\mathbf{z}}{}{R}_{4}^{\alpha}(x,y)=\prescript{\mathbf{z}}{}{R}_{3}^{\alpha}(x,y)\wedge\prescript{\mathbf{z}}{}{R}_{3}^{\alpha}(y,x)$ . Notice that $\prescript{\mathbf{z}}{}{D}_{4}$ is the set of all elements appearing in cycles of length $2^{k}$ in $\prescript{\mathbf{z}}{}{R}^{\kappa}$ , which is nonempty by our assumptions. Then it is straightforward to verify that $(R_{4},D_{4},\{\Lambda\})$ is a mighty tuple V’. Since the relation $\prescript{\mathbf{z}}{}{R}_{4}^{\alpha}$ is symmetric, we can apply Lemma 58 to derive a mighty tuple I. ∎

7.5 Classification for constraint languages with all constants

Lemma 25.

Suppose $\Gamma\supseteq\{x=a\mid a\in A\}$ is a set of relations on $A$ . Then the following conditions are equivalent:

1.

$\Gamma$ q-defines a mighty tuple I;
2.

$\Gamma$ q-defines a mighty tuple II;
3.

there exist an equivalence relation $\sigma$ on $D\subseteq A$ and $B,C\subsetneq A$ such that $B\cup C=A$ and $\Gamma$ q-defines the relations $(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in B))$ and $(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in C))$ .

Proof.

Let us prove that 1 implies 3. By Lemma 48 there is a mighty tuple I’ $(Q_{0},D_{0},B_{0},C_{0},\Delta)$ q-definable from $\Gamma$ . We derive the required relations in several steps.

Get rid of unnecessary parameters. Choose $\mathbf{z}_{0}=(b_{1},\dots,b_{|A|})$ satisfying condition 6 of a mighty tuple I. Then choose any $\delta_{0}\in\prescript{\mathbf{z}_{0}}{}{\Delta}$ . We get rid of parameters $\mathbf{z}$ and $\delta$ substituting the corresponding values: $Q_{1}=\prescript{\mathbf{z}_{0}}{\delta_{0}}{}Q_{0}$ , $D_{1}=\prescript{\mathbf{z}_{0}}{\delta_{0}}{}D_{0}$ .

Notice that we do not care about the sets $B_{0}$ and $C_{0}$ anymore because they are not necessary for the case with all constant relations. Later we only change $Q_{1}$ and $D_{1}$ .

Every equivalence relation must be trivial. Assume that for some $\alpha$ the binary relation $Q_{1}^{\alpha}$ is different from $D_{1}\times D_{1}$ and $Q_{1}^{\kappa}$ . Then choose $b\in D_{1}$ such that $\{b\}+Q_{1}^{\alpha}$ is not an equivalence class of $Q_{1}^{\kappa}$ and not $D_{1}$ . Put $D_{2}=\{b\}+Q_{1}^{\alpha}$ . Notice that $D_{2}$ is a unary relation q-definable from $\Gamma$ . Put $Q_{2}^{\mathbf{x}}(y_{1},y_{2})=Q_{1}^{\mathbf{x}}(y_{1},y_{2})\wedge(y_{1}\in D_{2})\wedge(y_{2}\in D_{2})$ to reduce the domain to $D_{1}$ . We can repeat this while some of the equivalence relation $Q_{2}^{\alpha}$ is not $D_{2}\times D_{2}$ and not $Q_{2}^{\kappa}$ . Thus, we assume that $Q_{2}^{\alpha}$ is either $D_{1}\times D_{1}$ or $Q_{2}^{\kappa}$ for any $\alpha$ .

Find appropriate $B$ and $C$ . Here we use the idea from the proof of Lemma 55 but for a much easier case. Let $\sigma_{1},\dots,\sigma_{N}$ be the set of all injective mappings from $\{1,2,\dots,|A|\}$ to $\{1,2,\dots,|A|^{2}\}$ . Let

U_{n}^{x_{1},\dots,x_{|A|^{2}}}=Q_{2}^{x_{\sigma_{1}(1)},\dots,x_{\sigma_{1}(|A|)}}+Q_{2}^{x_{\sigma_{2}(1)},\dots,x_{\sigma_{2}(|A|)}}+\dots+Q_{2}^{x_{\sigma_{n}(1)},\dots,x_{\sigma_{n}(|A|)}}

Since at least $|A|$ elements in the set $x_{1},\dots,x_{|A|^{2}}$ are equal, there exists $i\in\{1,2,\dots,N\}$ such that $x_{\sigma_{i}(1)}=x_{\sigma_{i}(2)}=\dots=x_{\sigma_{i}(|A|)}$ . Since $Q_{2}^{\forall}=D_{2}\times D_{2}$ the relation $U_{N}^{x_{1},\dots,x_{|A|^{2}}}$ is equal to $\prescript{\mathbf{z}}{\delta}{D}\times\prescript{\mathbf{z}}{\delta}{D}$ . Consider maximal $n$ such that $U_{n}^{\alpha}\neq D_{2}\times D_{2}$ for some $\alpha$ . Put $L=U_{n}$ and $R^{x_{1},\dots,x_{|A|^{2}}}=Q_{2}^{x_{\sigma_{n+1}(1)},\dots,x_{\sigma_{n+1}(|A|)}}$ . We know that $L^{\alpha}\neq D_{2}\times D_{2}$ for some $\alpha$ , $R^{\alpha}\neq D_{2}\times D_{2}$ for some $\alpha$ , but $L^{\alpha}+R^{\alpha}=D_{2}\times D_{2}$ for every $\alpha$ . Let $B_{0}\subseteq A^{|A|^{2}}$ be the set of all $\alpha$ such that $L^{\alpha}=D_{2}\times D_{2}$ , $C\subseteq A^{|A|^{2}}$ be the set of all $\alpha$ such that $R^{\alpha}=D_{2}\times D_{2}$ . Thus, we have $B,C\subsetneq A^{m}$ , where $m=|A|^{2}$ , such that $B\cup C=A^{m}$ , $L^{\mathbf{x}}(y_{1},y_{2})=(y_{1},y_{2}\in D_{2})\wedge(Q_{2}^{\kappa}(y_{1},y_{2})\vee(\mathbf{x}\in B))$ , and $R^{\mathbf{x}}(y_{1},y_{2})=(y_{1},y_{2}\in D_{2})\wedge(Q_{2}^{\kappa}(y_{1},y_{2})\vee(\mathbf{x}\in C))$ .

Reduce the arity of $B$ and $C$ . Let $B,C,L$ and $R$ be the relations of the minimal arity satisfying the above properties, that is, $B,C\subsetneq A^{m}$ for some $m$ , $B\cup C=A^{m}$ , $L^{\mathbf{x}}(y_{1},y_{2})=(y_{1},y_{2}\in D_{2})\wedge(Q_{2}^{\kappa}(y_{1},y_{2})\vee(\mathbf{x}\in B))$ , and $R^{\mathbf{x}}(y_{1},y_{2})=(y_{1},y_{2}\in D_{2})\wedge(Q_{2}^{\kappa}(y_{1},y_{2})\vee(\mathbf{x}\in C))$ . If $m=1$ then $L$ and $R$ are two relations we needed to define. Thus, we assume that $m>1$ . Put

	$\displaystyle L_{0}^{x_{1}}(y_{1},y_{2})$	$\displaystyle=\forall x_{2}\dots\forall x_{m}L^{x_{1},\dots,x_{m}}(y_{1},y_{2})$
	$\displaystyle R_{0}^{x_{1}}(y_{1},y_{2})$	$\displaystyle=\forall x_{2}\dots\forall x_{m}R^{x_{1},\dots,x_{m}}(y_{1},y_{2})$
	$\displaystyle B_{0}(x_{1})$	$\displaystyle=\forall x_{2}\dots\forall x_{m}B(x_{1},\dots,x_{m})$
	$\displaystyle C_{0}(x_{1})$	$\displaystyle=\forall x_{2}\dots\forall x_{m}C(x_{1},\dots,x_{m})$

Consider two cases:

Case 1. $B_{0}\cup C_{0}=A$ . Then $L_{0}$ and $R_{0}$ are ternary relations satisfying all the required properties, which contradicts our assumption about the minimality of $m$ .

Case 2. $B_{0}\cup C_{0}\neq A$ . Choose $a\in A\setminus(B_{0}\cup C_{0})$ . Put

	$\displaystyle L_{1}^{x_{2},\dots,x_{m}}(y_{1},y_{2})$	$\displaystyle=L^{a,x_{2}\dots,x_{m}}(y_{1},y_{2})$
	$\displaystyle R_{1}^{x_{2},\dots,x_{m}}(y_{1},y_{2})$	$\displaystyle=R^{a,x_{2}\dots,x_{m}}(y_{1},y_{2})$
	$\displaystyle B_{1}(x_{2},\dots,x_{m})$	$\displaystyle=B(a,x_{2},\dots,x_{m})$
	$\displaystyle C_{1}(x_{2},\dots,x_{m})$	$\displaystyle=C(a,x_{2},\dots,x_{m})$

Notice that $B_{1},C_{1}\subsetneq A^{m-1}$ , $B_{1}\cup C_{1}=A^{m-1}$ , and the relations $L_{1}$ and $R_{1}$ again satisfy all the required properties but have smaller arity, which contradicts our assumptions.

2 implies 1 by Lemma 16. It remains to prove that 3 implies 1. Suppose we have $B,C\subsetneq A$ , an equivalence relation $\sigma$ on $D$ and two q-definable relations $L(y_{1},y_{2},x)=(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in B))$ and $R(y_{1},y_{2},x)=(y_{1},y_{2}\in D)\wedge(\sigma(y_{1},y_{2})\vee(x\in C))$ . Let us define a mighty tuple II $(Q,D^{\prime},B^{\prime},C^{\prime})$ . Choose two elements $b$ and $c$ from $D$ that are not equivalent modulo $\sigma$ . Put $\prescript{\mathbf{z}}{\delta}{D^{\prime}}(y)=D(y)$ , and

	$\displaystyle\prescript{\mathbf{z}}{}{Q}^{x_{1},x_{2}}(y_{1},y_{2})$	$\displaystyle=\exists y\;L(y_{1},y,x_{1})\wedge R(y,y_{2},x_{2})$
	$\displaystyle\prescript{\mathbf{z}}{}{B^{\prime}}(y)$	$\displaystyle=\exists y^{\prime}\forall x\;(y^{\prime}=b)\wedge L(y,y^{\prime},x)$
	$\displaystyle\prescript{\mathbf{z}}{}{C^{\prime}}(y)$	$\displaystyle=\exists y^{\prime}\forall x\;(y^{\prime}=c)\wedge L(y,y^{\prime},x)$

Notice that the parameter $\mathbf{z}$ is fictitious. Since $B\cup C=A$ , we have $\prescript{\mathbf{z}}{}{Q}^{\forall}=D\times D$ . Since $B\neq A$ and $C\neq A$ , we have $\prescript{\mathbf{z}}{}{Q}^{\forall\forall}=\sigma$ . Thus, $(Q,D^{\prime},B^{\prime},C^{\prime})$ is a mighty tuple II, which completes the proof. ∎

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Π2P\Pi_{2}^{P} vs PSpace Dichotomy for the Quantified Constraint Satisfaction Problem

Abstract

1 Introduction

1.1 Complexity of the QCSP

Theorem 1 ([16]).

Theorem 2 ([22, 23]).

1.2 Reduction to CSP

Question 1.

Question 2.

Question 3.

2 Main Results

2.1 Π2P\Pi_{2}^{P} vs PSpace Dichotomy

Theorem 3.

Theorem 4.

2.2 What is inside Π2P\Pi_{2}^{P}?

Theorem 5.

Question 4.

Question 5.

Question 6.

Question 7.

2.3 PSpace-complete languages

Theorem 6.

Theorem 7.

2.4 Idea of the proof

2.5 Are there other complexity classes?

2.6 Structure of the paper

3 The most general PSpace-hard constraint language

Lemma 8.

Sketch of the proof:.

4 Π2P\Pi_{2}^{P}-complete constraint language

Lemma 9.

Proof.

Lemma 10.

Proof.

5 Proof of the main result

5.1 Necessary definitions and notations

5.2 Universal subset

Lemma 11 (proved in Section 6.2).

5.3 Induced CSP Instances

Lemma 12.

Proof.

Lemma 13.

Proof.

5.4 Mighty tuples

Theorem 14.

Lemma 15.

Lemma 16.

Lemma 17.

5.5 Reductions

5.6 Proof of Theorems 3, 6, and 7

Theorem 18.

Theorem 19.

Theorem 20.

Theorem 21.

Theorem 22.

Proof.

Theorem 23.

Proof.

Theorem 24.

Proof.

Lemma 25.

Conjecture 1.

6 Finding a solution of ℐR\mathcal{I}_{R}

6.1 Definitions

6.2 Auxiliary statements

Lemma 11.

Proof.

Lemma 26 ([21], Lemma 5.6).

Lemma 27.

Proof.

Lemma 28.

6.3 Finding a 1-consistent reduction

Lemma 29.

Lemma 30.

Proof.

Lemma 31.

Proof.

Lemma 32.

Proof.

Corollary 33.

$\Pi_{2}^{P}$ vs PSpace Dichotomy for the Quantified Constraint Satisfaction Problem

2.1 $\Pi_{2}^{P}$ vs PSpace Dichotomy

2.2 What is inside $\Pi_{2}^{P}$ ?

4 $\Pi_{2}^{P}$ -complete constraint language

6 Finding a solution of $\mathcal{I}_{R}$