Disjunctions of Two Dependence Atoms

Fröhlich, Nicolas; Kolaitis, Phokion G.; Meier, Arne

doi:10.4230/LIPIcs.CSL.2026.10

Disjunctions of Two Dependence Atoms

Nicolas Fröhlich

Leibniz Universität Hannover, Germany Phokion G. Kolaitis

University of California Santa Cruz, CA, USA
IBM Research, San Jose, CA, USA Arne Meier

Leibniz Universität Hannover, Germany

Abstract

Dependence logic is a formalism that augments the syntax of first-order logic with dependence atoms asserting that the value of a variable is determined by the values of some other variables, i.e., dependence atoms express functional dependencies in relational databases. On finite structures, dependence logic captures $\mathrm{NP}$ , hence there are sentences of dependence logic whose model-checking problem is $\mathrm{NP}$ -complete. In fact, it is known that there are disjunctions of three dependence atoms whose model-checking problem is $\mathrm{NP}$ -complete. Motivated from considerations in database theory, we study the model-checking problem for disjunctions of two unary dependence atoms and establish a trichotomy theorem, namely, for every such formula, one of the following is true for the model-checking problem: (i) it is $\mathrm{NL}$ -complete; (ii) it is $\mathrm{L}$ -complete; (iii) it is first-order definable (hence, in $\mathrm{AC^{0}}$ ). Furthermore, we classify the complexity of the model-checking problem for disjunctions of two arbitrary dependence atoms, and also characterize when such a disjunction is coherent, i.e., when it satisfies a certain small-model property. Along the way, we identify a new class of 2CNF-formulas whose satisfiability problem is $\mathrm{L}$ -complete.

Keywords and phrases:

Dependence logic, coherence, model-checking, complexity, functional dependencies

Funding:

Nicolas Fröhlich: The author appreciates funding by the German Research Agency (DFG) under the grant ME2479/3-1 and project id 511769688.

Arne Meier: The author appreciates funding by the German Research Agency (DFG) under the grant ME2479/3-1 and project id 511769688.

Copyright and License:

2012 ACM Subject Classification:

Theory of computation

\rightarrow

Problems, reductions and completeness ; Theory of computation

\rightarrow

Logic ; Theory of computation

\rightarrow

Logic and databases

Related Version:

Full Version: https://arxiv.org/abs/2508.16146 [6]

Acknowledgements:

We thank the anonymous reviewers for their valuable comments.

DOI:

10.4230/LIPIcs.CSL.2026.10

Event:

34th EACSL Annual Conference on Computer Science Logic (CSL 2026)

Editors:

Stefano Guerrini and Barbara König

Series and Publisher:

Leibniz International Proceedings in Informatics, Schloss Dagstuhl – Leibniz-Zentrum für Informatik

1 Introduction

Notions of dependence and independence are ubiquitous in several different areas of mathematics and computer science, including linear algebra, probability theory, and database theory. Dependence logic $\mathcal{D}$ , developed by Väänänen in [14], is a formalism for expressing and studying such notions. The basic building blocks of dependence logic are the dependence atoms, which, in effect, express functional dependencies on database relations. Recall that functional dependencies are integrity constraints on databases asserting that the values of some attribute of a database relation is determined by the values of some of the other attributes of that relation (see [1]). For example, if $R(\mathrm{course},\mathrm{dpt},\mathrm{inst})$ is a database relation with attributes for courses, departments, and instructors, then the functional dependency $(\mathrm{course}\rightarrow\mathrm{dpt})$ asserts that each course is offered by only one department, while the functional dependency $(\mathrm{inst}\rightarrow\mathrm{course})$ asserts that each instructor teaches only one course. In dependence logic $\mathcal{D}$ , these two functional dependencies are captured by the dependence atoms $\operatorname{dep}(\mathrm{course},\mathrm{inst})$ and $\operatorname{dep}(\mathrm{inst},\mathrm{course})$ , respectively. The formulas of $\mathcal{D}$ are obtained by combining dependence atoms, atomic formulas, and negated atomic formulas using conjunction $\wedge$ , disjunction $\lor$ , existential quantification $\exists$ , and universal quantification $\forall$ .

Even though the syntax of $\mathcal{D}$ resembles that of first-order logic $\mathsf{FO}$ , the semantics of $\mathcal{D}$ is second-order. Instead of single assignments of values to variables, the semantics of $\mathcal{D}$ uses sets of assignments, called teams, that represent database relations on which the dependence atoms are interpreted. The second-order character of the semantics of $\mathcal{D}$ is already manifested in the semantics of disjunction: a structure $\mathcal{M}$ and a team $T$ satisfy an $\mathcal{D}$ -formula of the form $\varphi_{1}\lor\varphi_{2}$ (in symbols, $(\mathcal{M},T)\models\varphi_{1}\lor\varphi_{2}$ ) if there are two teams $T_{1}$ and $T_{2}$ such that $T=T_{1}\cup T_{2}$ , $(\mathcal{M},T_{1})\models\varphi_{1}$ , and $(\mathcal{M},T_{2})\models\varphi_{2}$ . In terms of expressive power and as regards sentences, $\mathcal{D}$ is known to have the same expressive power as existential second-order logic $\mathsf{ESO}$ [10, 14]. From this result and Fagin’s Theorem [4], it follows that, on the class of all finite structures, the sentences of dependence logic can express precisely all decision problems in $\mathrm{NP}$ . In particular, there are $\mathcal{D}$ -sentences that can express $\mathrm{NP}$ -complete problems; in other words, the model-checking problem for such sentences is $\mathrm{NP}$ -complete.

In view of the preceding state of affairs, it is natural to ask: are there syntactic restrictions such that the model-checking problem for $\mathcal{D}$ -formulas obeying these restrictions is tractable? Rather surprisingly, Kontinen [8] showed that the model-checking problem can be $\mathrm{NP}$ -complete even for disjunctions of three dependence atoms. Concretely, consider the $\mathcal{D}$ -formula $\operatorname{dep}(x,y)\lor\operatorname{dep}(u,v)\lor\operatorname{dep}(u,v)$ . In [8], it is shown that the following problem is $\mathrm{NP}$ -complete: given a team $T$ of arity $4$ , does $T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(u,v)\lor\operatorname{% dep}(u,v)$ ? In other words, are there three teams $T_{1},T_{2},T_{3}$ such that $T=T_{1}\cup T_{2}\cup T_{3}$ , $T_{1}\models\operatorname{dep}(x,y)$ , $T_{2}\models\operatorname{dep}(u,v)$ , and $T_{3}\models\operatorname{dep}(u,v)$ ? On the tractability side, Kontinen [8] showed that the model-checking problem for disjunctions of two dependence atoms is always in $\mathrm{NL}$ , i.e., it is always solvable in non-deterministic logarithmic space. This was achieved via a logspace reduction of the model-checking problem for a disjunction of two dependence atoms to $\mathrm{2SAT}$ , the satisfiability problem for $2$ -CNF formulas, which is well known to be in $\mathrm{NL}$ .

Contributions.

In this paper, we carry out a systematic investigation of the disjunctions of two dependence atoms. The motivation for this investigation is twofold. First, while the model-checking for disjunctions of two dependence atoms is always in $\mathrm{NL}$ , there are such disjunctions for which the model-checking is $\mathrm{NL}$ -complete, while for others it is $\mathsf{FO}$ -definable (hence also in uniform $\mathrm{AC^{0}}$ ). So, one main aim is to pinpoint the exact complexity of the model-checking problem for a given disjunction of two dependence atoms. Second, disjunctions of dependence atoms capture natural database integrity constraints that have not been considered earlier. To make this point, let us consider the disjunction $\operatorname{dep}(\mathrm{course},\mathrm{dpt})\lor\operatorname{dep}(\mathrm% {inst},\mathrm{dpt})$ . The disjunct $\operatorname{dep}(\mathrm{course},\mathrm{dpt})$ expresses the functional dependency that each course is offered by only one department, while the disjunct $\operatorname{dep}(\mathrm{inst},\mathrm{dpt})$ expresses the functional dependence that each instructor is affiliated with only one department. It is unlikely that either of these two functional dependencies holds in any university, since it is typically the case that some courses are cross-listed by two departments and some instructors have joint appointments in two or more departments. Now, the disjunction $\operatorname{dep}(\mathrm{course},\mathrm{dpt})\lor\operatorname{dep}(\mathrm% {inst},\mathrm{dpt})$ expresses the constraint that the database relation $R(\mathrm{course},\mathrm{dpt},\mathrm{inst})$ can be split into two parts such that the functional dependency $\operatorname{dep}(\mathrm{course},\mathrm{dpt})$ holds in the first part and the functional dependency $\operatorname{dep}(\mathrm{inst},\mathrm{dpt})$ holds in the second part. This is a more relaxed integrity constraint, and it is quite plausible that it holds in many universities. Yet, this type of integrity constraint has not been investigated in database theory. The reason is that the largest collection of integrity constraints in databases studied in the past is the collection of embedded implicational dependencies [5], each of which is $\mathsf{FO}$ -definable, while, as our result will imply, the disjunction $\operatorname{dep}(\mathrm{course},\mathrm{dpt})\lor\operatorname{dep}(\mathrm% {inst},\mathrm{dpt})$ is not $\mathsf{FO}$ -definable. We believe that disjunctions of dependence atoms deserve to be studied as database integrity constraints in their own right; the work reported here makes a first step in this direction by focusing on the model-checking problem for such formulas.

We classify the complexity of the model-checking problem for disjunctions of two dependence atoms by establishing a trichotomy theorem: for every $\mathcal{D}$ -formula $\varphi$ that is the disjunction of two dependence atoms, one of the following three statements holds: (i) the model-checking problem for $\varphi$ is $\mathrm{NL}$ -complete; (ii) the model-checking problem for $\varphi$ is $\mathrm{L}$ -complete, where $\mathrm{L}$ is the class of all decision problems solvable in deterministic logarithmic space; (iii) the model-checking problem for $\varphi$ is $\mathsf{FO}$ -definable, hence in uniform $\mathrm{AC^{0}}$ , where $\mathrm{AC^{0}}$ is the class of all decision problems solvable by constant-depth, polynomial-size circuits.

As an illustration of this trichotomy theorem, consider again the database relation $R(\mathrm{course},\mathrm{dpt},\mathrm{inst})$ with data about courses, departments, and instructors. Consider also the dependence atoms $\operatorname{dep}(\mathrm{course},\mathrm{inst})$ , $\operatorname{dep}(\mathrm{course},\mathrm{dpt})$ , $\operatorname{dep}(\mathrm{inst},\mathrm{dpt})$ , $\operatorname{dep}(\mathrm{inst},\mathrm{course})$ that express, respectively, the functional dependencies $(\mathrm{course}\rightarrow\mathrm{inst})$ , $(\mathrm{course}\rightarrow\mathrm{dpt})$ , $(\mathrm{inst}\rightarrow\mathrm{dpt})$ , $(\mathrm{inst}\rightarrow\mathrm{course})$ . Our trichotomy theorem classifies the complexity of the model-checking for all possible disjunctions of these four dependence atoms. In particular, it implies that the following statements are true:

$\blacksquare$

The model-checking problem for $\operatorname{dep}(\mathrm{course},\mathrm{dpt})\lor\operatorname{dep}(\mathrm% {inst},\mathrm{dpt})$ is $\mathrm{NL}$ -complete.
$\blacksquare$

The model-checking problem for $\operatorname{dep}(\mathrm{inst},\mathrm{course})\lor\operatorname{dep}(% \mathrm{course},\mathrm{inst})$ is $\mathrm{L}$ -complete.
$\blacksquare$

The model-checking problem for $\operatorname{dep}(\mathrm{course},\mathrm{dpt})\lor\operatorname{dep}(\mathrm% {course},\mathrm{inst})$ is $\mathsf{FO}$ -definable, hence it is in uniform $\mathrm{AC^{0}}$ .

As a byproduct of the preceding complexity-theoretic classification, we also identify a collection of $2$ -CNF formulas for which the satisfiability problem is $\mathrm{L}$ -complete.

The preceding results are about unary dependence atoms $\operatorname{dep}(x,y)$ , where $x$ and $y$ are variables. The syntax of dependence logic allows also for higher-arity dependence atoms, i.e., for dependence atoms of the form $\operatorname{dep}(x_{1},\ldots,x_{n},y)$ with $n>1$ ; such atoms assert that the value of the variable $y$ is determined by the values of the variables $x_{1},\ldots,x_{n}$ . We leverage out results about the model-checking problem for disjunctions of two unary dependence atoms to establish a dichotomy theorem about the model-checking problem for disjunctions of two higher-arity dependence atoms: this problem is either $\mathrm{NL}$ -complete or $\mathsf{FO}$ -definable.

We complement these complexity-theoretic classifications of the model-checking problem with a structural classification of disjunctions of two dependence atoms. Specifically, we determine whether a given disjunction of two dependence atoms is a coherent or an incoherent $\mathcal{D}$ -formula. As defined in [8], a $\mathcal{D}$ -formula $\varphi$ is coherent if there is an integer $k$ such that a team $T$ satisfies $\varphi$ if and only if every sub-team $T^{\prime}$ of $T$ of size $k$ satisfies $\varphi$ . If no such $k$ exists, then $\varphi$ is incoherent. We show that a disjunction of two dependence atoms is coherent if and only if the model-checking for this formula is $\mathsf{FO}$ -definable. Furthermore, for each coherent disjunction of two dependence atoms, we determine its coherence level, i.e., the smallest $k$ that establishes the coherence of that disjunction. Our findings for disjunctions of two unary dependence atoms are summarized in Table 1.

The work reported here paves the way for further exploration of the model-checking problem for the quantifier-free fragment of dependence logic and, perhaps more importantly, it paves the way for further interaction between dependence logic and database theory.

Table 1: Complexity and coherence of disjunctions of two unary dependence atoms.

Formula	Coherence	Model-checking
$\operatorname{dep}(x,y)\lor\operatorname{dep}(z,u)$	incoherent (Prop. 8)	$\mathrm{NL}$ -complete (Prop. 10)
$\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$	incoherent (Thm. 25^⋆)	$\mathrm{NL}$ -complete (Thm. 13^⋆)
$\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z)$	incoherent (Thm. 25^⋆)	$\mathrm{NL}$ -complete (Thm. 13^⋆)
$\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$	incoherent (Thm. 25^⋆)	$\mathrm{L}$ -complete (Thm. 16)
$\operatorname{dep}(x,y)\lor\operatorname{dep}(x,y)$	coherent (level 3) (Prop. 8)	$\mathsf{FO}$ (Prop. 9)
$\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$	coherent (level 4) (Thm. 22)	$\mathsf{FO}$ (Prop. 9)

Due to space constraints, full proofs of results marked with a “ $\star$ ” in the body of the paper are given in the Appendix.

2 Preliminaries

This section contains some basic material about dependence logic. For additional material, we refer the reader to the monograph [14].

A vocabulary is a tuple $\tau=(R_{1},\ldots,R_{m})$ of relation symbols, each of which has a specified natural number $r_{i}$ as its arity. A $\tau$ -structure is a tuple $\mathcal{M}=(A,R_{1}^{\mathcal{M}},\ldots,R_{m}^{\mathcal{M}})$ , where $A$ is a set, called the domain of $\mathcal{M}$ , and each $R_{i}^{\mathcal{M}}$ is an $r_{i}$ -ary relation on $A$ .

Let $\operatorname{Var}$ be a countably infinite set of variables. The syntax of dependence logic $\mathcal{D}$ over a vocabulary $\tau$ extends the syntax of first-order logic $\mathsf{FO}$ in negation normal form with dependence atoms as additional atomic formulas. A dependence atom of arity $n$ is an expression of the form $\operatorname{dep}(y_{1},\ldots,y_{n},z)$ , where $y_{1},\ldots,y_{n},z$ are distinct variables in $\operatorname{Var}$ and $n\geq 0$ . If $n=1$ , we say that the dependence atom is unary. Intuitively, a dependence atom asserts that the value of the variable $z$ depends on the values of the variables $y_{1},\ldots,y_{n}$ . In particular, if $n=0$ , we have a dependence atom of the form $\operatorname{dep}(z)$ , which asserts that $z$ is a constant value, since it depends on no other variables; such atoms are called constancy atoms.

The formulas of dependence logic $\mathcal{D}$ are defined by the following Backus-Naur form:

\psi{\coloneqq}\,x_{k}{=}x_{\ell}\mid R_{i}(x_{1},\dots,x_{r_{i}})\mid\lnot R_% {i}(x_{1},\dots,x_{r_{i}})\mid\operatorname{dep}(y_{1},\ldots,y_{n},z)\mid\psi% {\land}\psi\mid\psi{\lor}\psi\mid\exists x\psi\mid\forall x\psi,

where $x_{k},x_{\ell},x_{1},\ldots,x_{r_{i}}$ are (not necessarily distinct) variables in $\operatorname{Var}$ , $R_{i}$ is a relation symbol in $\tau$ of arity $r_{i}\in\mathbb{N}$ , and $y_{1},\ldots,y_{n},z$ are distinct variables in $\operatorname{Var}$ .

Recall that the semantics of $\mathsf{FO}$ -formulas are given using a structure $\mathcal{M}$ and an assignment, i.e., a mapping from $\operatorname{Var}$ to the domain of $\mathcal{M}$ . Due to the presence of dependence atoms, however, the semantics of $\mathcal{D}$ -formulas are given using a structure $\mathcal{M}$ and a set of assignments.

Definition 1.

Let $X$ be a set of variables (i.e., $X\subseteq\operatorname{Var}$ ) and let $A$ be a set.

$\blacksquare$

An assignment with domain $X$ and range $A$ is a mapping $s\colon X\to A$ .
$\blacksquare$

A team with domain $X$ and range $A$ is a set $T$ of assignments with domain $X$ and range $A$ . We will write $\operatorname{Dom}(T)$ to denote the domain $X$ of the team $T$ . Similar $\operatorname{Rng}(T)$ denotes the range $A$ of the team $T$ .
$\blacksquare$

If $T$ is a team with domain $X$ and $Y\subseteq X$ , then the restriction of $T$ on $Y$ is the team $T\upharpoonright Y=\{\,s\upharpoonright Y\mid s\in T\,\}$ , where $s\upharpoonright Y$ is the restriction of the assignment $s$ to $Y$ .

Before giving the semantics of $\mathcal{D}$ -formulas, we need to introduce two operations on teams.

Definition 2.

Let $\operatorname{Var}$ be the set of all variables and let $A$ be a set.

$\blacksquare$

If $s\colon\operatorname{Var}\to A$ is an assignment with domain $\operatorname{Var}$ and range $A$ , $x$ is a variable, and $a$ is an element of $A$ , then $s^{x}_{a}\colon\operatorname{Var}\to A$ is the assignment such that $s^{x}_{a}(x)=a$ and $s^{x}_{a}(y)=s(y)$ , for every variable $y\not=x$ .
$\blacksquare$

If $T$ is a team with domain $\operatorname{Var}$ and range $A$ , $f\colon T\to A$ is a function, and $x$ is a variable, then the supplement team $T^{x}_{f}$ is the team $\{s^{x}_{f(s)}\mid s\in T\}.$
$\blacksquare$

If $T$ is a team with domain $\operatorname{Var}$ and range $A$ , and $x$ is a variable, then the duplicate team $T^{x}_{A}$ is the team $\{s^{x}_{a}\mid x\in\operatorname{Var}\nobreak\ \mbox{{\rm and}}\nobreak\ a\in A\}$ .

We are now ready to define the semantics of dependence logic $\mathcal{D}$ . In doing so, the supplement team will be used to define the semantics of existential quantification, while the duplicate team will be used to define the semantics of universal quantification.

Definition 3.

Let $\tau=(R_{1},\ldots,R_{m})$ be a vocabulary and let $\mathcal{M}=(A,R_{1}^{\mathcal{M}},\ldots,R_{m}^{\mathcal{M}})$ be a $\tau$ -structure. The satisfaction relation $(\mathcal{M},T)\models\varphi$ , where $T$ is a team with domain $\operatorname{Var}$ and range $A$ , and $\varphi$ is a formula of dependence logic $\mathcal{D}$ , is defined by induction on the construction of $\varphi$ and simultaneously for all teams with domain $\operatorname{Var}$ and range $A$ as follows:

$\blacksquare$

$(\mathcal{M},T)\models x_{k}=x_{l}$ if for all $s\in T$ , we have $s(x_{k})=s(x_{l})$ ;
$\blacksquare$

$(\mathcal{M},T)\models R_{i}(x_{1},\ldots,x_{r_{i}})$ if for all $s\in T$ , we have $(s(x_{1}),\ldots,s(x_{r_{i}}))\in R_{i}^{\mathcal{M}}$ ;
$\blacksquare$

$(\mathcal{M},T)\models\neg R_{i}(x_{1},\ldots,x_{r_{i}})$ if for all $s\in T$ , we have $(s(x_{1}),\ldots,s(x_{r_{i}}))\not\in R_{i}^{\mathcal{M}}$ ;
$\blacksquare$

$(\mathcal{M},T)\models\operatorname{dep}(y_{1},\ldots,y_{n},z)$ if for all $s_{1},s_{2}\in T$ with $s_{1}(y_{j})=s_{2}(y_{j})$ for $1\leq j\leq n$ , we have that $s_{1}(z)=s_{2}(z)$ ;
$\blacksquare$

$(\mathcal{M},T)\models\varphi_{1}\land\varphi_{2}$ if $(\mathcal{M},T)\models\varphi_{1}$ and $(\mathcal{M},T)\models\varphi_{2}$ ;
$\blacksquare$

$(\mathcal{M},T)\models\varphi_{1}\lor\varphi_{2}$ if there are teams $T_{1}$ , $T_{2}$ with $T=T_{1}\cup T_{2}$ and $(\mathcal{M},T_{i})\models\varphi_{i}$ , $i=1,2$ ;
$\blacksquare$

$(\mathcal{M},T)\models\exists x\varphi$ if there is a function $f\colon T\to A$ such that $(\mathcal{M},T^{x}_{f})\models\varphi$ , where $T^{x}_{f}$ is the supplement team associated with $T$ , $x$ , and $f$ ;
$\blacksquare$

$(\mathcal{M},T)\models\forall x\varphi$ if $(\mathcal{M},T^{x}_{A})\models\varphi$ , where $T^{x}_{A}$ is the duplicate team associated with $T$ and $x$ .

The semantics of disjunction in dependence logic are quite different from those in first-order logic. In particular, unlike first-order logic, a disjunction of the form $\varphi\lor\varphi$ need not be equivalent to $\varphi$ . The following examples illustrates this state of affairs.

Example 4.

Consider the $\mathcal{D}$ -formulas $\operatorname{dep}(x,y)$ and $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,y)$ . Let $\mathcal{M}=(\{0,1\})$ be a structure over the empty vocabulary and let $T=\{s,t\}$ be a team with two $s$ and $t$ assignments such that $s(x)=0,s(y)=0$ and $t(x)=0,t(y)=1$ . Clearly $(\mathcal{M},T)\not\models\operatorname{dep}(x,y)$ . Nonetheless, $(\mathcal{M},T)\models\operatorname{dep}(x,y)\lor\operatorname{dep}(x,y)$ , since $(\mathcal{M},\{s\})\models\operatorname{dep}(x,y)$ and $(\mathcal{M},\{t\})\models\operatorname{dep}(x,y)$ .

The next proposition, whose proof can be found in [14], states two important properties of dependence logic. In what follows, if $\varphi$ is a $\mathcal{D}$ -formula, then $\operatorname{free}(\varphi)$ is the (finite) set of all free variables of $\varphi$ .

Proposition 5 ([14, Lemma 3.27, Prop. 3.10]).

Let $\mathcal{M}=(A,R_{1}^{\mathcal{M}},\ldots,R_{m}^{\mathcal{M}})$ be a $\tau$ -structure and let $\varphi$ be a $\mathcal{D}$ -formula.

(Locality): Let $X$ be a set such that $\operatorname{free}(\varphi)\subseteq X\subseteq\operatorname{Var}$ . If $T_{1}$ , $T_{2}$ are teams with domain $\operatorname{Var}$ and range $A$ such that $T_{1}\upharpoonright X=T_{2}\upharpoonright X$ , then $(\mathcal{M},T_{1})\models\varphi$ if and only if $(\mathcal{M},T_{2})\models\varphi$ .
(Downward Closure): Let $P$ and $T$ be teams with domain $\operatorname{Var}$ and range $A$ such that $P\subseteq T$ . If $(\mathcal{M},T)\models\varphi$ , then also $(\mathcal{M},P)\models\varphi$ .

The Locality property tells that the satisfaction relation $(\mathcal{M},T)\models\varphi$ depends only on the restriction of the team $T$ on the set of the free variables of $\varphi$ . In particular, if $\psi$ is a $\mathcal{D}$ -sentence (i.e., a $\mathcal{D}$ -formula with no free variables), then $(\mathcal{M},T)\models\psi$ if and only $\mathcal{M},\{\emptyset\})\models\psi$ , where $\{\emptyset\}$ is the team consisting of the empty assignment.

Väänänen [14] showed that $\mathcal{D}$ -sentences have the same expressive power as the sentences of existential second-order logic $\mathsf{ESO}$ . Combined with Fagin’s theorem [4], this result implies that, on the class of all finite structures, $\mathcal{D}$ -sentences express precisely all $\mathrm{NP}$ problems. Kontinen and Väänänen [11] showed that $\mathcal{D}$ -formulas (with free variables) have the same expressive power as the downward closed formulas of existential second-order logic $\mathsf{ESO}$ (see [11] for the precise statement of this result).

In view of the Locality property, from now on we will only consider teams on finite domains, as long as, in each case, the domain of the team considered contains the free variables of the $\mathcal{D}$ -formula at hand.

Coherence.

In the following, we present the definition of a small-model property for dependence logic.

Definition 6 ( $k$ -coherence, [8, Def. 3.1]).

Let $\varphi(x_{1},\dots,x_{n})$ be a quantifier-free $\mathcal{D}$ -formula. We say that $\varphi$ is $k$ -coherent if for all structures $\mathcal{M}$ and teams $T$ of range $\mathcal{M}$ such that $\operatorname{free}(\varphi)\subseteq\operatorname{Dom}(T)$ , the following are equivalent:

(1.)

$(\mathcal{M},T)\models\varphi$ .
(2.)

For all $k$ -element sub-teams $S\subseteq T$ it holds that $(\mathcal{M},S)\models\varphi$ .

Notice that for the above equivalence, the direction “(1.) $\Rightarrow$ (2.)” is always true due to $\mathcal{D}$ being downwards closed (see Proposition 5).

Definition 7 (coherence-level, incoherence).

The coherence-level of $\varphi$ is the least natural number $k\in\mathbb{N}$ such that $\varphi$ is $k$ -coherent; if such a $k$ does not exist, we call $\varphi$ incoherent.

Clearly, first-order atomic formulas have a coherence-level of $1$ and dependence atoms a coherence-level of $2$ .

Proposition 8.

The following statements are true:

(1.)

The formula $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,y)$ has coherence-level $3$ ; [8, Prop. 3.8]
(2.)

The formula $\operatorname{dep}(x,y)\lor\operatorname{dep}(z,u)$ is incoherent. [8, Thm. 3.11]

We will use the shortcut $\operatorname{Rel}(T)\coloneqq\{(s(x_{1}),\dots,s(x_{n}))\mid s\in T\}$ , when the team $T$ is given as a relation. The following proposition shows that every quantifier-free $k$ -coherent formula of dependence logic is first-order definable (and, hence, is in uniform $\mathrm{AC^{0}}$ ).

Proposition 9 ([8, Thm. 4.9]).

Suppose $\varphi(x_{1},\dots,x_{n})\in\mathcal{D}$ is a quantifier-free $k$ -coherent formula over a vocabulary $\tau$ . Then there is a sentence $\varphi^{*}\in\mathsf{FO}(\tau\cup\{R\})$ , where $R$ is $n$ -ary, such that for all $\tau$ -structures $\mathcal{M}$ and for all teams $T$ of domain $\{x_{1},\dots,x_{n}\}$ , we have that $(\mathcal{M},T)\models\varphi(x_{1},\dots,x_{n})$ if and only if $\operatorname{Rel}(T)\models\varphi^{*}(R)$ .

3 Model-checking

The model-checking problem is a fundamental decision problem arising in every logic. Informally, this problem asks if a finite structure satisfies a formula of the logic at hand. Actually, according to the taxonomy introduced by Vardi [15], there are two versions of the model-checking problem, the combined complexity version and the data complexity version. In the first version, the input consists of a formula and a finite structure, while in the second version the formula is fixed and the input consists of just a finite structure.

Here, we focus on the data complexity version of the model-checking problem for dependence logic $\mathcal{D}$ . Specifically, every $\mathcal{D}$ -formula $\varphi$ gives rise to the following decision problem.

Problem:	$\mathrm{MC}(\varphi)$ – the model-checking problem for $\varphi\in\mathcal{D}$
Input:	A finite $\tau$ -structure $\mathcal{M}=(A,R_{1}^{\mathcal{M}},\ldots,R_{m}^{\mathcal{M}})$ and a team $T$ with domain a finite set $X$ and range $A$ such that $\operatorname{free}(\varphi)\subseteq X\subseteq\operatorname{Var}$ .
Question:	Does $(\mathcal{M},T)\models\varphi$ ?

As mentioned in the introduction, the results in [9, 14] imply that $\mathrm{MC}(\varphi)$ is in $\mathrm{NP}$ for every $\mathcal{D}$ -formula $\varphi$ , and that there are $\mathcal{D}$ -formulas $\psi$ for which $\mathrm{MC}(\psi)$ is an $\mathrm{NP}$ -complete problem. Furthermore, the intractability of the model-checking problem is even true for some disjunctions of three dependence atoms; for example, $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(u,v)\lor% \operatorname{dep}(u,v))$ is $\mathrm{NP}$ -complete, as shown by Kontinen [8].

Note that if $\varphi$ is a $\mathcal{D}$ -formula in which no relation symbols from the vocabulary $\tau$ occur, then the satisfaction relation $(\mathcal{M},T)\models\varphi$ depends only on the team $T$ and not on the $\tau$ -structure $\mathcal{M}$ . From this point on, we will focus on disjunctions of two dependence atoms, hence no relation symbols from the vocabulary $\tau$ occur in such dependence logic formulas. For this reason, we will drop $\mathcal{M}$ from the satisfaction relation and, instead, write $T\models\varphi$ , where $\varphi$ contains no relation symbols from $\tau$ . Thus, if $\varphi$ is such a $\mathcal{D}$ -formula, then the model-checking problem $\mathrm{MC}(\varphi)$ asks: given a team $T$ with domain a finite set $X$ such that $\operatorname{free}(\varphi)\subseteq X\subseteq\operatorname{Var}$ , does $T\models\varphi$ ? The following result yields a sufficient condition for the model-checking problem to be in $\mathrm{NL}$ , the class of all decision problems solvable in non-deterministic logarithmic space.

Proposition 10 ([8, Thm. 4.12]).

Suppose $\varphi$ and $\psi$ are $2$ -coherent quantifier-free $\mathcal{D}$ -formulas with no relation symbols. Then there is a logarithmic-space reduction from the model-checking problem $\mathrm{MC}(\varphi\lor\psi)$ to $\mathrm{2SAT}$ . Consequently, $\mathrm{MC}(\varphi\lor\psi)$ is in $\mathrm{NL}$ .

We now describe the reduction of $\mathrm{MC}(\varphi\lor\psi)$ to $\mathrm{2SAT}$ in some detail, since it will be relevant later on (Section 3.2) in the present paper.

Given a team $T=\{s_{1},\dots,s_{k}\}$ , go through all two-element subsets $\{s_{i},s_{j}\}$ of $T$ , and construct a set $C$ of two-variable clauses, as follows:

$\blacksquare$

If $\{s_{i},s_{j}\}\not\models\varphi$ , then $(x_{i}\lor x_{j})\in C$ .
$\blacksquare$

If $\{s_{i},s_{j}\}\not\models\psi$ , then $(\lnot x_{i}\lor\lnot x_{j})\in C$ .

Inuitively, these two clauses encode the property that the team $\{s,t\}$ satisfies neither $\varphi$ nor $\psi$ . Then let $\Theta_{T}$ be the $2$ CNF-formula $\bigwedge_{\theta\in C}\theta$ . It can be shown that $T\models\varphi\lor\psi$ if and only if $\Theta_{T}$ is satisfiable (see [8] for a detailed proof).

Since every dependence atom is a $2$ -coherent formula, we obtain the following result.

Corollary 11.

If $\varphi$ and $\psi$ are dependence atoms, then $\mathrm{MC}(\varphi\lor\psi)$ is in $\mathrm{NL}$ .

3.1 NL-Completeness

This section will cover the cases in which the model-checking problem for disjunctions of two unary dependence atoms is complete for nondeterministic logarithmic space.

Proposition 12 ([8, Thm. 4.14]).

The problem $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(z,w))$ is $\mathrm{NL}$ -complete.

We will outline the reduction, constructed in the proof of this result, because the next result (Theorem 13) is based on this construction. The full proof of Proposition 12 can be found in [8]. The intuition of the construction is that the team encodes the clauses of the formula. The split ensures a consistent assignment on the left side, while the right side allows to “buffer” one unsatisfied literal per clause. Consider a $\mathrm{2SAT}$ instance $\theta(p_{1},\dots,p_{m})=\bigwedge_{i\in I}E_{i}$ , with $E_{i}=(A_{i_{1}}\lor A_{i_{2}}),i\in I$ , where $A_{i_{1}},A_{i_{2}}$ are literals. For each conjunct $E_{i}$ , create a team $T_{E_{i}}=\{s_{i_{1}},s_{i_{2}}\}$ with domain $\{x,y,z,w\}$ , where variable $x$ ranges over $\{p_{1},\dots,p_{m}\}$ ; variable $y$ ranges over the truth values $\{0,1\}$ ; variable $z$ ranges over the indices $i\in I$ , thus $z$ denotes the clause $E_{i}$ ; and variable $w$ ranges over the values $\{0,1\}$ . Variables $z$ and $w$ ensure that at least one of the assignments from each $T_{E_{i}}$ go into the subset of $T$ that eventually encodes an assignment satisfying $\theta$ . For example, the team $T_{E_{i}}$ for a clause $(p_{k}\lor\lnot p_{j})$ is the one on the left in Figure 1. Finally $T=\bigcup_{i\in I}T_{E_{i}}$ .

$\begin{array}[]{ccccc}\hline\cr\hline\cr T_{E_{i}}&x&y&z&w\\ \hline\cr s_{i_{1}}&p_{k}&1&i&0\\ s_{i_{2}}&p_{j}&0&i&1\\ \hline\cr\hline\cr\end{array}\qquad\quad\begin{array}[]{cccc}\hline\cr\hline% \cr T_{E_{i}}&x&y&z\\ \hline\cr s_{i_{1}}&p_{k}&i&1_{k}\\ s_{i_{2}}&p_{j}&i&0_{j}\\ \hline\cr\hline\cr\end{array}\qquad\begin{array}[]{cccc}\hline\cr\hline\cr T_{% E_{i}}&x&y&z\\ \hline\cr s_{i_{1}}&i&p_{k}&1\\ s_{i_{2}}&i&p_{j}&0\\ \hline\cr\hline\cr\end{array}$

Figure 1: Teams for

(p_{k}\lor\lnot p_{j})

in Proposition 12 (left) and Theorem 13 (middle/right).

Theorem 13 ( $\star$ ).

The following statements are true:

(1.)

The problem $\mathrm{MC}(\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z))$ is $\mathrm{NL}$ -complete.
(2.)

The problem $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z))$ is $\mathrm{NL}$ -complete.

Proof sketch.

Membership follows from Proposition 10. For hardness, we proceed as in the proof of Proposition 12, that is, we transform a 2CNF formula into a team, additionally, we require that the clauses do not contain the same variable twice. For the first case, $\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ , we use $T_{E_{i}}$ shown in the middle of Figure 1. For the second case, $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z)$ , we use $T_{E_{i}}$ depicted in the right of Figure 1. Furthermore, let again $I$ be the set of clause indices and $m$ be the number of propositions in the 2CNF formula $\theta$ .

(1.) Variable $x$ ranges over $\{p_{1},\dots,p_{m}\}$ , variable $y$ ranges over $I$ , and variable $z$ ranges over $\{\,0_{k},1_{k}\mid 1\leq k\leq m\,\}$ , that is, each proposition has its own truth values. The dependence atom $\operatorname{dep}(x,z)$ ensures the existence of a consistent assignment that satisfies $\theta$ , while $\operatorname{dep}(y,z)$ allows only one of $s_{i_{1}}$ and $s_{i_{2}}$ to be on its “side of the split.”

(2.) Variable $x$ ranges over $I$ , variables $y$ ranges over $\{p_{1},\dots,p_{m}\}$ , and variable $z$ ranges over $\{0,1\}$ . Here, the atom $\operatorname{dep}(y,z)$ ensures the existence of a consistent assignment that satisfies $\theta$ , while $\operatorname{dep}(x,y)$ ensures only one of $s_{i_{1}}$ and $s_{i_{2}}$ is on its “side of the split.” $\hfill\blacktriangleleft$

3.2 L-Completeness

This section will cover the cases in which the model-checking problem for disjunctions of two unary dependence atoms is complete for logarithmic space.

If $T$ is a team with domain $X$ and $x$ is a variable in $X$ , then we will use the notation $\operatorname{Rng}_{x}(T)\coloneqq\operatorname{Rng}(T\upharpoonright\{x\})$ to denote the set of values in the range of $T$ that $x$ takes.

Definition 14.

Let $T$ be a team with domain $\{x,y\}$ such that $\operatorname{Rng}_{x}(T)\cap\operatorname{Rng}_{y}(T)=\emptyset$ . The undirected graph of $T$ is $G_{T}\coloneqq(V_{T},E_{T})$ , where

\displaystyle V_{T}

\displaystyle=\operatorname{Rng}(T),\qquad E_{T}=\bigl\{\,\{s(x),s(y)\}\,\big|% \,s\in T\,\bigr\}.

Notice that, by definition, $G_{T}$ is a bipartite graph. The next lemma provides a graph-based interpretation of the condition under which a team $T$ satisfies the formula $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .

Lemma 15 ( $\star$ ).

Let $T$ be a team with domain $\{x,y\}$ such that $\operatorname{Rng}_{x}(T)\cap\operatorname{Rng}_{y}(T)=\emptyset$ . The following statements are equivalent:

(1.)

$T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .
(2.)
$G_{T}$ can be transformed into a directed graph $G^{\prime}_{T}=(V_{T},E^{\prime}_{T})$ such that
1. (a)
  
  each node has at most one out-going edge, and
2. (b)
  
  for each $\{u,v\}\in E_{T}$ either $(u,v)\in E^{\prime}_{T}$ or $(v,u)\in E^{\prime}_{T}$ .
(3.)

For each connected component $C=(V_{C},E_{C})$ of $G_{T}$ , we have that $|E_{C}|\leq|V_{C}|$ .

A visual example for $G_{T}$ and the equivalence of (1.) and (2.) are depicted in Figure 2.

$\begin{array}[]{ccc}\hline\cr\hline\cr T&x&y\\ \hline\cr s_{1}&0&\hat{0}\\ s_{2}&0&\hat{1}\\ s_{3}&1&\hat{0}\\ s_{4}&1&\hat{1}\\ \hline\cr\hline\cr\end{array}$ $\leftrightsquigarrow$ $\begin{matrix}\includegraphics{dagpub-standalone-combined-2026-01-06-10-41-31-% page-1.pdf2svg.svg}\end{matrix}$ $\leftrightsquigarrow$ $\begin{matrix}\includegraphics{dagpub-standalone-combined-2026-01-06-10-41-31-% page-2.pdf2svg.svg}\end{matrix}$ $\leftrightsquigarrow$ $\begin{array}[]{ccc}\hline\cr\hline\cr T&x&y\\ \hline\cr s_{1}&0&\hat{0}\\ s_{4}&1&\hat{1}\\ \hline\cr s_{2}&0&\hat{1}\\ s_{3}&1&\hat{0}\\ \hline\cr\hline\cr\end{array}$

Figure 2: Example depicting the construction of

G_{T}

from team

T

on the left. Followed by the transformation into directed graph

G^{\prime}_{T}

, where each node has at most one out-going edge. This induces the split

\{s_{1},s_{4}\}\models\operatorname{dep}(x,y)

and

\{s_{2},s_{3}\}\models\operatorname{dep}(y,x)

of

T

shown on the right.

Theorem 16.

The problem $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x))$ is in $\mathrm{L}$ .

Proof.

Algorithm 1 decides $T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ requiring only logarithmic space.

We assume the input to the algorithm to be the graph $G_{T}$ , which is possible since its edge set is, depending on the used encoding, the same as the team $T$ . The correctness follows immediately from Lemma 15, i.e., Algorithm 1 returns true if and only if $|E_{C}|\leq|V_{C}|$ for all connected components $C$ of $G_{T}$ . The algorithm iterates over all nodes $v$ in $G_{T}$ and checks that the size of the corresponding connected component (so its number of nodes) is not smaller than its number of edges. The size count, $c$ , is achieved by counting the number of reachable nodes. In parallel, the sizes of their respective neighborhoods are added up in a different counter, $d$ . Since this counts each edge twice, halving $d$ and comparing to $c$ suffices.

The amount of space required by the algorithm is as follows. Computing the number of nodes in a connected component can be done in logarithmic space in $|G_{T}|$ (which is in $O(|T|)$ ) using undirected reachability (which is well-known to be in $\mathrm{L}$ by Reingold’s theorem [13]). The algorithm stores an index of the current node $v$ and an index of an auxiliary node $u$ , hence both are, by binary encoding, in $O(\log|T|)$ . Furthermore, two counters are used: one to count the size of the connected component, $c$ , and another to count the number of edges in the connected component, $d$ . Both counters are, again, in $O(\log|T|)$ . Storing these values together requires only logspace in the input size. The claimed result applies, as $\mathrm{L}^{\mathrm{L}}=\mathrm{L}$ (that is to say, $\mathrm{L}$ is low for itself). $\hfill\blacktriangleleft$

Algorithm 1 Model-checking

\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)

.

Lemma 17 ( $\star$ ).

Let $T$ be a team with domain $\{x,y\}$ . If the size $|T|>|\operatorname{Rng}_{x}(T)|+|\operatorname{Rng}_{y}(T)|$ , then $T\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .

In the following, we will utilize the definition of first-order reductions as defined by Immerman [7] and Dahlhaus [3].

Theorem 18.

The problem $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x))$ is $\mathrm{L}$ -hard under first-order reductions.

Proof.

We reduce from the complement of $\mathrm{UFA}$ (which is known to be $\mathrm{L}$ -complete [2]):

Problem:	$\mathrm{UFA}$ – Undirected Forest Accessibility
Input:	acyclic undirected graph $G=(V,E)$ , nodes $u$ and $v$ .
Question:	is there is a path between $u$ and $v$ ?

$\begin{array}[]{ccc}\hline\cr\hline\cr T_{\{a,b\}}&x&y\\ \hline\cr s_{1}&a&b\\ s_{2}&b&a\\ \hline\cr\hline\cr\end{array}\qquad\begin{array}[]{ccc}\hline\cr\hline\cr T_{% \{\top,u\}}&x&y\\ \hline\cr s_{1}&\top&u\\ s_{2}&u&\top\\ \hline\cr\hline\cr\end{array}\qquad\begin{array}[]{ccc}\hline\cr\hline\cr T_{% \{\top,v\}}&x&y\\ \hline\cr s_{1}&\top&v\\ s_{2}&v&\top\\ \hline\cr\hline\cr\end{array}\qquad\begin{array}[]{ccc}\hline\cr\hline\cr T_{% \{\bot,u\}}&x&y\\ \hline\cr s_{1}&\bot&u\\ s_{2}&u&\bot\\ \hline\cr\hline\cr\end{array}\qquad\begin{array}[]{ccc}\hline\cr\hline\cr T_{% \{\bot,v\}}&x&y\\ \hline\cr s_{1}&\bot&v\\ s_{2}&v&\bot\\ \hline\cr\hline\cr\end{array}$

Figure 3: Team

T_{\{a,b\}}

for edge

\{a,b\}\in E

and teams

T_{\{\top,u\}},T_{\{\top,v\}},T_{\{\bot,u\}},T_{\{\bot,v\}}

.

Suppose $\langle G=(V,E),u,v\rangle$ is an instance of $\mathrm{UFA}$ . We will construct a team $T$ , such that $T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ if and only if there is no path between $u$ and $v$ in $G$ .

For each undirected edge $\{a,b\}\in E$ , create a team $T_{\{a,b\}}$ with two assignments $s$ and $t$ , such that $s(x)=t(y)=a$ and $s(y)=t(x)=b$ . Next, let $\top$ and $\bot$ be two elements such that $\{\top,\bot\}\cap V=\emptyset$ . Create four more teams $T_{\{\top,u\}},T_{\{\top,v\}},T_{\{\bot,u\}}$ , and $T_{\{\bot,v\}}$ as depicted in Figure 3. Now, $T$ is the union $\bigcup_{\{a,b\}\in E}T_{\{a,b\}}\cup\{T_{\{\top,u\}},T_{\{\top,v\}},T_{\{\bot% ,u\}},T_{\{\bot,v\}}\}$ . The team $T$ can be constructed using an $\mathsf{FO}$ query $I\colon\mathrm{STRUC}[\tau_{g}]\to\mathrm{STRUC}[\tau_{2t}]$ , where $\tau_{g}=(E^{2})$ is the vocabulary of graphs with edge relation $E$ and $\tau_{2t}=(T^{2})$ is the vocabulary of teams with domain of size two, encoded via the binary relation $\operatorname{Rel}(T)$ (see, Proposition 9).

Now, we will show that $\langle G=(V,E),u,v\rangle\in\overline{\mathrm{UFA}}$ if and only if $T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .

“ $\Rightarrow$ ”: Assume that there is no path between $u$ and $v$ in $G$ . Then take $u$ and $v$ as roots for trees corresponding to their acyclic connected components. For all other connected components choose any node as root. We partition $T$ into two teams as follows:

	$\displaystyle T_{1}$	$\displaystyle=\{s\in T\mid s(y)\text{ is the parent node of }s(x)\}$
		$\displaystyle\cup\{s\in T\mid(s(x),s(y))=(u,\bot)\}\cup\{s\in T\mid(s(x),s(y))% =(v,\top)\}$
		$\displaystyle\cup\{s\in T\mid(s(x),s(y))=(\bot,v)\}\cup\{s\in T\mid(s(x),s(y))% =(\top,u)\}$
	$\displaystyle T_{2}$	$\displaystyle=T\setminus T_{1}=\{s\in T\mid(s(x),s(y))=(t(y),t(x)),t\in T_{1}\}$

Observe that $T_{2}$ contains the “reverse directions” of $T_{1}$ (see Figure 4).

Figure 4: (Left): Example

\mathrm{UFA}

instance. (Right): Visualization of partition into

T_{1}

(solid lines) and

T_{2}

(dashed lines).

Now, $s(x)\neq t(x)$ for all $s,t\in T_{1}$ , because each $s(x)\in V$ only has one parent $s(y)\in V$ in their corresponding trees (the cases containing $\top$ or $\bot$ satisfy this condition by construction of $T_{1}$ , as they appear only once for $x$ ; and $u, v$ are roots, so do not have parents themselves). Thus, $T_{1}$ trivially satisfies $\operatorname{dep}(x,y)$ . Since the construction and definition of $T_{2}$ are symmetric, $s(x)\neq t(x)$ for all $s,t\in T_{2}$ . This yields $T_{2}\models\operatorname{dep}(y,x)$ .

“ $\Leftarrow$ ”: For the other direction, assume the contraposition, i.e., that there is a path between $u$ and $v$ in $G$ . Let $C=(V_{C},E_{C})$ be the connected component with $\{u,v\}\subseteq V_{C}$ . We will focus now on the sub-team for $C$ together with the special assignments (shown in Figure 3) and argue why this sub-team cannot be split to satisfy $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ (this would imply that the full team cannot be split as well). For that purpose, let $T_{C}$ be the union $\bigcup_{\{a,b\}\in C}T_{\{a,b\}}$ of the teams created by the edges in $C$ . Observe that

|T_{C}|=2|E_{C}|=2(|V_{C}|-1)\text{ and }|\operatorname{Rng}_{x}(T)|=|% \operatorname{Rng}_{y}(T)|=|V_{C}|,

i.e., the size of the team is twice the number of edges in $C$ , because each edge adds two assignments by definition. The number of edges in $C$ is then one less than the number of vertices in $C$ . The sizes of the domains of $x$ and $y$ are the same as the number of nodes in $C$ .

Next, let $T_{C}^{\prime}=T_{C}\cup\{T_{\{\top,u\}},T_{\{\top,v\}},T_{\{\bot,u\}},T_{\{% \bot,v\}}\}$ , then

|T_{C}^{\prime}|=2|E_{C}|+8=2(|V_{C}|-1)+8\text{ and }|\operatorname{Rng}_{x}(% T)|=|\operatorname{Rng}_{y}(T)|=|V_{C}|+2,

because eight new assignments are added to the team and two new elements to the domains. Thus, we get

|T_{C}^{\prime}|=2(|V_{C}|-1)+8=2|V_{C}|+6>2|V_{C}|+4=|\operatorname{Rng}_{x}(% T)|+|\operatorname{Rng}_{y}(T)|

and, by Lemma 17, we have that $T_{C}^{\prime}\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ . Therefore $T\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ is true, as $T_{C}^{\prime}\subseteq T$ and due to downwards closure. $\hfill\blacktriangleleft$

A restricted case of Krom-satisfiability.

In the sequel, we will show how our results also establish the $\mathrm{L}$ -completeness of the satisfiability problem restricted to a particular type of 2CNF formulas, namely, those that are monotone, transitive, and dual-free.

Definition 19.

Let $\phi$ be a 2CNF formula with variables $\operatorname{Var}(\phi)=\{x_{1},\dots,x_{n}\}$ , such that:

(1.)

Every clause is monotone, i.e., it is of the form $(x_{i}\lor x_{j})$ or of the form $(\bar{x}_{i}\lor\bar{x}_{j})$ . The first type is called positive, the second type negative.
(2.)

No two “dual” clauses $(x_{i}\lor y_{j})$ , $(\bar{x}_{i}\lor\bar{x}_{j})$ appear in $\phi$ .
(3.)

Transitivity holds: if $(x_{i}\lor x_{j})$ and $(x_{j}\lor x_{k})$ are clauses of $\phi$ , then so is $(x_{i}\lor x_{k})\in\phi$ ; furthermore, if $(\bar{x}_{i}\lor\bar{x}_{j})$ and $(\bar{x}_{j}\lor\bar{x}_{k})$ are clauses of $\phi$ , then so is $(\bar{x}_{i}\lor\bar{x}_{k})\in\phi$ .

We call such 2CNF formulas monotone transitive dual-free and abbreviate the corresponding satisfiability problem by $\mathrm{mtdf\text{-}2SAT}$ .

Lemma 20 ( $\star$ ).

The problems $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x))$ and $\mathrm{mtdf\text{-}2SAT}$ are equivalent via first-order reductions.

Proof sketch.

Notice how the 2CNF formulas in the reduction of Proposition 10 are monotone, transitive and dual-free, if we have that $\varphi=\operatorname{dep}(x,y)$ and $\psi=\operatorname{dep}(y,x)$ . This gives a one-to-one correspondence, thus the other directions is simply the inverse. $\hfill\blacktriangleleft$

Lemma 20 together with Theorem 16 and Theorem 18 yield the following result.

Theorem 21.

Satisfiability of monotone transitive dual-free 2CNF formula is $\mathrm{L}$ -complete under first-order reductions.

4 Coherence

This section is devoted to identifying the coherent disjunctions of two unary dependence atoms and determining their coherence-level.

$\begin{array}[]{c}\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)\\ \begin{array}[]{ccccc}\hline\cr\hline\cr T&x&y&z\\ \hline\cr s_{1}&1&1&1\\ s_{2}&1&1&2\\ s_{3}&1&2&1\\ s_{4}&1&2&2\\ \hline\cr\hline\cr\lx@intercol\hfil(a)\hfil\lx@intercol\end{array}\end{array}% \quad\begin{array}[]{c}\operatorname{dep}(x)\lor\operatorname{dep}(y,z)\\ \begin{array}[]{cccc}\hline\cr\hline\cr T&x&y&z\\ \hline\cr s_{1}&1&1&1\\ s_{2}&1&1&2\\ s_{3}&2&2&1\\ s_{4}&2&2&2\\ \hline\cr\hline\cr\lx@intercol\hfil(b)\hfil\lx@intercol\end{array}\end{array}% \quad\begin{array}[]{c}\operatorname{dep}(x)\lor\operatorname{dep}(x,y)\\ \begin{array}[]{cccc}\hline\cr\hline\cr T&x&y\\ \hline\cr s_{1}&1&1\\ s_{2}&1&2\\ s_{3}&2&1\\ s_{4}&2&2\\ \hline\cr\hline\cr\lx@intercol\hfil(c)\hfil\lx@intercol\end{array}\end{array}% \quad\begin{array}[]{c}\operatorname{dep}(y)\lor\operatorname{dep}(x,y)\\ \begin{array}[]{cccc}\hline\cr\hline\cr T&x&y\\ \hline\cr s_{1}&1&1\\ s_{2}&2&1\\ s_{3}&3&2\\ s_{4}&4&2\\ \hline\cr\hline\cr\lx@intercol\hfil(d)\hfil\lx@intercol\end{array}\end{array}$

Figure 5: Counterexamples of 3-coherence for formulas used in the proofs of Thms. 22 and 24.

Theorem 22.

The formula $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ has coherence-level $4$ .

Proof.

We first show that $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ is not 3-coherent. Consider the team $T$ in Figure 5 ( $a$ ). This team does not satisfy $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ . To see this, consider the two maximal subsets, $T_{1}=\{s_{1},s_{2}\}$ and $T_{2}=\{s_{3},s_{4}\}$ , that satisfy $\operatorname{dep}(x,y)$ . We cannot split $T=T_{1}\cup T_{2}$ , because $T_{1}\not\models\operatorname{dep}(x,z)$ and $T_{2}\not\models\operatorname{dep}(x,z)$ .

Every 3-element subset of $T$ satisfies $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ , because either $T_{1}$ or $T_{2}$ is now a singleton (overlapping splits not need to be considered), thus trivially satisfying the dependence atom $\operatorname{dep}(x,z)$ . Therefore $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ is not 3-coherent.

Next, we show that if $T$ does not satisfies $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ , then there is a sub-team $T^{\prime}\subseteq T$ of size $|T^{\prime}|\leq 4$ that does not satisfies $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ either. Let $T$ be a team with $T\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ . If $|T|\leq 4$ , then $T^{\prime}=T$ . Otherwise assume without loss of generality that all assignments in $T$ agree on $x$ . Two assignments with different values for $x$ trivially satisfy both atoms. Thus, we can make this assumption by splitting $T$ into sub-teams that agree on $x$ , and then consider each sub-team separately.

Now, let $T^{\prime}=\{s,u,v,t\}\subseteq T$ , such that the following constraints hold: $s(y)\neq t(y)$ and $s(z)\neq t(z)$ ; $u(y)\neq s(y)$ and $u(z)\neq t(z)$ ; $v(z)\neq s(z)$ and $v(y)\neq t(y)$ . We first show that $T^{\prime}\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ is true and then that such a $T^{\prime}\subseteq T$ always exists (see Figure 6). By the constraints on $y$ , we have that $\{s,v\}$ and $\{u,t\}$ are the two maximal sub-teams that satisfy $\operatorname{dep}(x,y)$ . Therefore at least one of the two sub-teams must also satisfy $\operatorname{dep}(x,z)$ for the disjunction to be true. But $\{s,v\}\not\models\operatorname{dep}(x,z)$ , because $s(x)=v(x)$ and $s(z)\neq v(z)$ ; analogously $\{u,t\}\not\models\operatorname{dep}(x,z)$ . Thus $T^{\prime}\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ is true.

Let $s\in T$ and assume no $t\in T$ meets the constraints above. Then $T$ can be split via $T_{1}=\{w\in T\mid w(y)=s(y)\}\models\operatorname{dep}(x,y)$ and $T_{2}=\{w\in T\mid w(z)=s(z)\}\models\operatorname{dep}(x,z)$ . Next, assume there are $s, t$ and $u$ , but no $v$ . Then we split $T$ via $T_{1}=\{w\in T\mid w(y)=t(y)\}\models\operatorname{dep}(x,y)$ and $T_{2}=\{w\in T\mid w(z)=s(z)\}\models\operatorname{dep}(x,z)$ . The last case is analogous. This is a contradiction to $T\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ , hence such a sub-team must always exists.

$\begin{array}[]{cccc}\hline\cr\hline\cr T&x&y&z\\ \hline\cr s_{1}&0&0&0\\ s_{2}&0&0&1\\ s_{3}&0&1&2\\ s_{4}&0&2&0\\ s_{5}&0&2&1\\ \hline\cr\hline\cr\end{array}$ $\leadsto$ $\begin{matrix}\includegraphics{dagpub-standalone-combined-2026-01-06-10-41-31-% page-5.pdf2svg.svg}\end{matrix}$ $\begin{array}[]{cccc}\hline\cr\hline\cr T&x&y&z\\ \hline\cr s_{1}&0&0&0\\ s_{2}&0&0&2\\ s_{3}&0&1&1\\ s_{4}&0&2&1\\ s_{5}&1&2&0\\ \hline\cr\hline\cr\end{array}$ $\leadsto$ $\begin{matrix}\includegraphics{dagpub-standalone-combined-2026-01-06-10-41-31-% page-6.pdf2svg.svg}\end{matrix}$

Figure 6: Two examples for the construction in Theorem 22. Edge label denote the variable values where the assignments from the vertices differ, e.g., edge

\{s_{1},s_{3}\}

with label

y

means that

s_{1}(y)\neq s_{3}(y)

. On the left we have that

s_{1}

corresponds to

s

,

s_{3}

to

u

and

v

, and

s_{5}

to

t

from the construction.

$\hfill\blacktriangleleft$ As a special case of the previous lemma (ignore the $x$ column), we deduce the following corollary.

Corollary 23.

The formula $\operatorname{dep}(y)\lor\operatorname{dep}(z)$ has coherence-level $4$ .

Theorem 24 ( $\star$ ).

Each of the following formulas has coherence-level $4$ :
(1.) $\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ ; (2.) $\operatorname{dep}(x)\lor\operatorname{dep}(x,y)$ ; (3.) $\operatorname{dep}(y)\lor\operatorname{dep}(x,y)$ .

Proof sketch.

For the lower bound consider the teams (b)–(d) in Figure 5. The upper bound works with a similar construction as in the proof of Theorem 22. $\hfill\blacktriangleleft$

Finally, we turn towards the three incoherent cases. For these formulas, there exists no $k\in\mathbb{N}$ , such that the respective formula is $k$ -coherent. Notice that the incoherence result can here be deduced from the fact that $\mathsf{FO}\subsetneq\mathrm{L}$ . Nevertheless, we provide independent proofs for these results in the appendix.

Theorem 25 ( $\star$ ).

The following three formulas are incoherent:
(1.) $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ ; (2.) $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z)$ ; (3.) $\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ .

By combining the results from Section 3 and 4, we obtain the following complete classification for the complexity of the model-checking of disjunctions of two dependence or constancy atoms.

Theorem 26.

The following statements are true for the model-checking problem of disjunctions of two unary dependence or constancy atoms:

(1.)

If $\varphi$ is one of the formulas $\operatorname{dep}(x,y)\lor\operatorname{dep}(z,u)$ , $\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ , $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z)$ , then the problem $\mathrm{MC}(\varphi)$ is $\mathrm{NL}$ -complete.
(2.)

The problem $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x))$ is $\mathrm{L}$ -complete.
(3.)

In all other cases, the problem $\mathrm{MC}(\varphi)$ is in $\mathsf{FO}$ , hence also in uniform $\mathrm{AC^{0}}$ .

5 Disjunctions of higher-arity dependence atoms

In this section, we consider disjunctions of two dependence atoms whose arity may be higher than one. We leverage the preceding results about disjunctions of two unary dependence atoms to classify the complexity of disjunctions of two dependence atoms of higher arity.

Before stating and proving the main result of this section, we present two lemmas. The idea in the next lemma is to encode a subset of the domain by a tuple and also change the range appropriately.

Lemma 27.

Consider a $\mathcal{D}$ -formula of the form $\operatorname{dep}(x_{1},\dots,x_{n},y)\lor\operatorname{dep}(u_{1},\dots,u_{m% },v)$ . For every team $T=\{s_{1},\dots,s_{k}\}$ with domain $\{x_{1},\dots,x_{n},y,u_{1},\dots,u_{m},v\}$ and range $A$ , there is a team $T^{\prime}$ with domain $\{x^{\prime},y,u_{1},\dots,u_{m},v\}$ and range $A^{n}\cup A$ , such that

T\models\operatorname{dep}(x_{1},\dots,x_{n},y)\lor\operatorname{dep}(u_{1},% \dots,u_{m},v)\quad\mbox{iff}\quad T^{\prime}\models\operatorname{dep}(x^{% \prime},y)\lor\operatorname{dep}(u_{1},\dots,u_{m},v).

Proof.

Consider the assignments $s^{\prime}_{i}$ , $1\leq i\leq k$ , with domain $\{x^{\prime},y,u_{1},\ldots,u_{m},y\}$ and range $A^{n}\cup A$ defined as follows: $s^{\prime}_{i}(x^{\prime})=(s(x_{1}),\ldots,s(x_{n}))$ and $s^{\prime}_{i}(w)=s_{i}(w)$ , if $w$ is one of the variables $y,u_{1},\ldots,u_{m},v$ . Furthermore, consider the team $T^{\prime}=\{s^{\prime}_{1},\dots,s^{\prime}_{k}\}$ be the team, which has domain $\{x^{\prime},y,u_{1},\dots,u_{m},v\}$ and range $A^{n}\cup A$ .

Assume first that $T\models\operatorname{dep}(x_{1},\dots,x_{n},y)\lor\operatorname{dep}(u_{1},% \dots,u_{m},v)$ via $T=T_{1}\cup T_{2}$ , that is, $T_{1}\models\operatorname{dep}(x_{1},\dots,x_{n},y)$ and $T_{2}\models\operatorname{dep}(u_{1},\dots,u_{m},v)$ . Let $T_{1}^{\prime}=\{s_{i}^{\prime}\in T^{\prime}\mid s_{i}\in T_{1}\}$ and $T_{2}^{\prime}=\{s_{i}^{\prime}\in T^{\prime}\mid s_{i}\in T_{2}\}$ . Then the following hold:
$\displaystyle T_{1}\models\operatorname{dep}(x_{1},\dots,x_{n},y)$ $\displaystyle\nobreak\ \mbox{iff}\nobreak\ \forall s,t\in T_{1}\colon s(x_{1})% =t(x_{1}),\dots,s(x_{n})=t(x_{n})\Rightarrow s(y)=t(y)$ $\displaystyle\nobreak\ \mbox{iff}\nobreak\ \forall s,t\in T_{1}\colon(s(x_{1})% ,\dots,s(x_{n}))=(t(x_{1}),\dots,t(x_{n}))\Rightarrow s(y){=}t(y)$ $\displaystyle\nobreak\ \mbox{iff}\nobreak\ \forall s^{\prime},t^{\prime}\in T_% {1}^{\prime}\colon s^{\prime}(x^{\prime})=t^{\prime}(x^{\prime})\Rightarrow s^% {\prime}(y)=t^{\prime}(y)\nobreak\ \mbox{iff}\nobreak\ T_{1}^{\prime}\models% \operatorname{dep}(x^{\prime},y).$

Furthermore, $T_{2}^{\prime}\models\operatorname{dep}(u_{1},\ldots,u_{m},v)$ , since $T_{2}\upharpoonright\{u_{1},\dots,u_{m},v\}=T^{\prime}_{2}\upharpoonright\{u_{% 1},\dots,u_{m},v\}$ . Thus, if $T\models\operatorname{dep}(x_{1},\dots,x_{n},y)\lor\operatorname{dep}(u_{1},% \dots,u_{m},v)$ , then $T^{\prime}\models\operatorname{dep}(x^{\prime},y)\lor\operatorname{dep}(u_{1},% \dots,u_{m},v)$ . The converse is established using a similar argument. $\hfill\blacktriangleleft$

By applying Lemma 27 twice, we obtain the following result.

Corollary 28.

Consider a $\mathcal{D}$ -formula of the form $\operatorname{dep}(x_{1},\dots,x_{n},y)\lor\operatorname{dep}(u_{1},\dots,u_{m% },v)$ . For every team $T=\{s_{1},\dots,s_{k}\}$ with domain $\{x_{1},\dots,x_{n},y,u_{1},\dots,u_{m},v\}$ and range $A$ , there is a team $T^{\prime}$ with domain $\{x^{\prime},y,u^{\prime},v\}$ and range $A^{n}\cup A^{m}\cup A$ , such that

T\models\operatorname{dep}(x_{1},\dots,x_{n},y)\lor\operatorname{dep}(u_{1},% \dots,u_{m},v)\quad\mbox{iff}\quad T^{\prime}\models\operatorname{dep}(x^{% \prime},y)\lor\operatorname{dep}(u^{\prime},v).

The next lemma establishes the coherence of one of the simplest higher-arity disjunctions.

Lemma 29.

The formula $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ has coherence-level 4.

Proof.

To show that $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ is not 3-coherent, notice that this formula is equivalent to $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,u)$ on every team $T$ that is constant on $y$ , i.e., $s(y)=t(y)$ holds, for all $s,t\in T$ . Thus, the supplement team $T_{f}^{y}$ given by $T$ in Figure 5 (a) (and with suitable variable renamings) and constant function $f(s)=a$ is a counterexample to the 3-coherence of $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ .

To show that $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ is 4-coherent, consider an arbitrary team $T$ with domain $\{x,z,y,u\}$ . Without loss of generality, we may assume that $T$ is constant on $x$ . This is so because $T$ can be split into sub-teams $T_{a_{i}}=\{s\in T\mid s(x)=a_{i}\in A\}$ that are constant on $x$ . If $T$ falsifies $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ , then at least one of these sub-teams also falsifies $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ . Indeed, if every sub-team $T_{a_{i}}$ satisfied $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ , then their union $T$ would also satisfy $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ , since every team $T^{\prime}$ consisting of two assignments $s\in T_{a_{j}}$ and $t\in T_{a_{k}}$ with $a_{j}\not=a_{k}$ satisfies both dependence atoms (because $s(x)=a_{j}\neq a_{k}=t(x)$ ).

Now, since $T$ is constant on $x$ , whether or not $T\models\operatorname{dep}(x,z)$ depends only on the values of $z$ and so $\operatorname{dep}(x,z)$ acts like the constancy atom $\operatorname{dep}(z)$ . Similarly, whether or not $T\models\operatorname{dep}(x,y,u)$ , depends only on the values of $y$ and $u$ , and so $\operatorname{dep}(x,y,u)$ acts like the unary dependence atom into $\operatorname{dep}(y,u)$ . Therefore, we have that $T\models\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ if and only if $T\models\operatorname{dep}(z)\lor\operatorname{dep}(y,u)$ . As the latter disjunction was shown to be 4-coherent in Theorem 24, we have that $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ is 4-coherent as well. $\hfill\blacktriangleleft$

We are now ready to state and prove a classification theorem for the complexity of the model-checking problem for disjunctions of dependence atoms of higher arities. Note that, when a dependence atom $\operatorname{dep}(y_{1},\ldots,y_{n},u)$ of higher arity is considered, then, by the semantics of dependence logic $\mathcal{D}$ , the order $y_{1},\ldots,y_{n}$ of the variables is immaterial.

Theorem 30.

Let $\operatorname{dep}(x_{1},\ldots,x_{m},z)$ and $\operatorname{dep}(y_{1},\ldots,y_{n},u)$ be two dependence atoms such that $n\geq m$ and $n>1$ (i.e., at least one of them is non-unary), and some of the variables of the first atom may coincide with some of the variables of the second. Then the following hold:

(1.)

If neither the set $\{x_{1},\ldots,x_{m}\}$ is contained in the set $\{y_{1},\ldots,y_{n}\}$ nor the set $\{y_{1},\ldots,y_{n}\}$ is contained in the set $\{x_{1},\ldots,x_{m}\}$ , then the problem

$\mathrm{MC}(\operatorname{dep}(x_{1},\ldots,x_{m},z)\lor\operatorname{dep}(y_{% 1},\ldots,y_{n},u))$

is $\mathrm{NL}$ -complete.
(2.)

Otherwise (i.e., if $\{x_{1},\ldots,x_{m}\}\subseteq\{y_{1},\ldots,y_{n}\}$ ), the problem

$\mathrm{MC}(\operatorname{dep}(x_{1},\ldots,x_{m},z)\lor\operatorname{dep}(y_{% 1},\ldots,y_{n},u))$

is in $\mathsf{FO}$ , hence also in uniform $\mathrm{AC^{0}}$ .

Proof.

For the first part, membership in $\mathrm{NL}$ follows from Proposition 10. We show $\mathrm{NL}$ -hardness for the special case $z=u$ , by exhibiting a logarithmic-space reduction from the problem $\mathrm{MC}(\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z))$ (see Theorem 13). Let $\hat{x}$ be a variable in $\{x_{1},\ldots,x_{m}\}$ but not in $\{y_{1},\ldots,y_{n}\}$ ; similarly, let $\hat{y}$ be a variable in $\{y_{1},\ldots,y_{n}\}$ but not in $\{x_{1},\ldots,x_{m}\}$ . Given a team $T$ , we construct in logarithmic space a team $T^{\prime}$ as follows: for every $s\in T$ , we put in $T^{\prime}$ an assignment $s^{\prime}$ , such that $s^{\prime}(z)=s(z)$ , $s^{\prime}(x_{i})=s(x_{i})$ if $x_{i}=\hat{x}$ and $a$ otherwise, likewise $s^{\prime}(y_{i})=s(y_{i})$ if $y_{i}=\hat{y}$ and $a$ otherwise, where $a$ is some fixed element in the range of $T$ . Then the following equivalences hold:

	$\displaystyle T\models\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$
	$\displaystyle\mbox{iff}\quad\exists T_{1},T_{2}\colon T=T_{1}\cup T_{2},T_{1}% \models\operatorname{dep}(x,z)\text{ and }T_{2}\models\operatorname{dep}(y,z)$
	$\displaystyle\mbox{iff}\quad\exists T_{1},T_{2}\colon T=T_{1}\cup T_{2},% \forall s_{1},t_{1}\in T_{1}\colon s_{1}(x)=t_{1}(x)\Rightarrow s_{1}(z)=t_{1}% (z)\text{ and }$
	$\displaystyle\quad\forall s_{2},t_{2}\in T_{2}\colon s_{2}(y)=t_{2}(y)% \Rightarrow s_{2}(z)=t_{2}(z)$
	$\displaystyle\mbox{iff}\quad\exists T^{\prime}_{1},T^{\prime}_{2}\colon T^{% \prime}=T^{\prime}_{1}\cup T^{\prime}_{2},\forall s^{\prime}_{1},t^{\prime}_{1% }\in T^{\prime}_{1}\colon s^{\prime}_{1}(\hat{x})=t^{\prime}_{1}(\hat{x})% \Rightarrow s^{\prime}_{1}(z)=t^{\prime}_{1}(z)\text{ and }$
	$\displaystyle\quad\forall s^{\prime}_{2},t^{\prime}_{2}\in T^{\prime}_{2}% \colon s^{\prime}_{2}(\hat{y})=t^{\prime}_{2}(\hat{y})\Rightarrow s^{\prime}_{% 2}(z)=t^{\prime}_{2}(z)$
	$\displaystyle\mbox{iff}\quad\exists T^{\prime}_{1},T^{\prime}_{2}\colon T^{% \prime}=T^{\prime}_{1}\cup T^{\prime}_{2},T^{\prime}_{1}\models\operatorname{% dep}(x_{1},\dots,x_{m},z)\text{ and }T^{\prime}_{2}\models\operatorname{dep}(y% _{1},\dots,y_{n},z)$		( $\star$ )
	$\displaystyle\mbox{iff}\quad T^{\prime}\models\operatorname{dep}(x_{1},\dots,x% _{m},z)\lor\operatorname{dep}(y_{1},\dots,y_{n},z).$

Equivalence $(\star)$ holds, because if $x_{i}\neq\hat{x}$ , then $s^{\prime}(x_{i})=t^{\prime}(x_{i})=a$ (and similarly for the $y_{i}$ ’s).

For the second part, we distinguish two cases, namely whether $m=n$ or $m<n$ .

Case “ $m=n$ ”:: The formula $\operatorname{dep}(x_{1},\dots,x_{m},z)\lor\operatorname{dep}(x_{1},\dots,x_{m% },u)$ has coherence-level 4, because, via Corollary 28, the model-checking problem of $\operatorname{dep}(x_{1},\dots,x_{m},z)\lor\operatorname{dep}(x_{1},\dots,x_{m% },u)$ is equivalent to that of $\operatorname{dep}(x,y)\lor\operatorname{dep}(x,z)$ . Thus, by Proposition 9 and Theorem 22, we have that $\mathrm{MC}(\operatorname{dep}(x_{1},\dots,x_{m},z)\lor\operatorname{dep}(y_{1% },\dots,y_{n},u))$ is in $\mathsf{FO}$ .
Case “ $m<n$ ”:: Assume $y_{m+1},\dots,y_{n}$ are the variables that do not occur in the first dependence atom $\operatorname{dep}(x_{1},\dots,x_{m},z)$ . Let $x=(x_{1},\dots,x_{m})$ and $y=(y_{m+1},\dots,y_{n})$ . Then, by Corollary 28, the disjunction $\operatorname{dep}(x_{1},\ldots,x_{m},z)\lor\operatorname{dep}(x_{1},\ldots,x_% {m},y_{m+1},\dots,y_{n},u)$ is equivalent to the disjunction $\operatorname{dep}(x,z)\lor\operatorname{dep}(x,y,u)$ , which is 4-coherent by Lemma 29. Thus by Proposition 9, we have that $\mathrm{MC}(\operatorname{dep}(x_{1},\dots,x_{m},z)\lor\operatorname{dep}(y_{1% },\dots,y_{n},u))$ is in $\mathsf{FO}$ .

$\hfill\blacktriangleleft$

Observe that, whenever $\mathrm{MC}(\varphi)$ was shown to be in $\mathsf{FO}$ , the proof of Theorem 30 actually showed that $\varphi$ is coherent. Combined with the earlier results about unary dependence atoms, we obtain the following characterization of coherence of disjunctions of dependence atoms.

Corollary 31.

Let $\varphi$ be a disjunction of two dependence atoms of arbitrary arities. Then $\varphi$ is coherent if and only if $\mathrm{MC}(\varphi)$ is first-order definable.

6 Outlook

In this paper, we carried out a systematic investigation of the model-checking problem for disjunctions of two dependence atoms. The work reported here suggests several different directions for future research, including the following:

1.

Is it possible to classify the computational complexity of the model-checking problem for all quantifier-free formulas of dependence logic $\mathcal{D}$ ? In particular, is there a dichotomy theorem between $\mathrm{NP}$ -completeness and polynomial-time solvability for this problem? Note that it is conceivable that every problem in $\mathrm{NP}$ is polynomial-time equivalent to the model-checking problem of a quantifier-free $\mathcal{D}$ -formula. In that case, such a dichotomy theorem would be impossible (unless $\mathrm{P}=\mathrm{NP}$ ), since by Ladner’s theorem [12], if $\mathrm{P}\not=\mathrm{NP}$ , then there are problems in $\mathrm{NP}$ that are neither $\mathrm{NP}$ -complete nor are in $\mathrm{P}$ .
2.

By Proposition 9, if $\varphi$ is a coherent, quantifier-free $\mathcal{D}$ -formula, then $\varphi$ is equivalent to an $\mathsf{FO}$ -formula $\varphi^{*}(T)$ that includes a relation symbol $T$ for the team. We conjecture that the converse is true, which would imply that coherence coincides with first-order definability for quantifier-free $\mathcal{D}$ -formulas. Our results confirm this conjecture for the case in which $\varphi$ is a disjunction of two dependence atoms.
3.

Study the implication problem for disjunctions of two dependence atoms, i.e., given a finite set $\Sigma$ of disjunctions of two dependence atoms and a disjunction $\psi$ of two dependence atoms, does $\Sigma$ logically imply $\psi$ ? For functional dependencies (i.e., single dependence atoms), the implication problem is solvable in polynomial time; moreover, there is a set of simple axioms, known as Armstrong’s axioms [1, p. 186], for reasoning about functional dependencies. As argued in the introduction, disjunctions of two dependence atoms form a natural class of database dependencies that have not been studied in their own right thus far. Investigating the implication problem for disjunctions of two dependence atoms will enhance the interaction between dependence logic and database theory.

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Appendix A Appendix

Theorem 13 ( $\star$ ). [Restated, see original statement.]

The following statements are true:

(1.)

The problem $\mathrm{MC}(\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z))$ is $\mathrm{NL}$ -complete.
(2.)

The problem $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z))$ is $\mathrm{NL}$ -complete.

Proof.

Membership follows directly from Proposition 10. We now give the proof of $\mathrm{NL}$ -hardness for (1.). The proof works analogous for (2.).

We show $\mathrm{2SAT}\leq^{\mathrm{log}}_{m}\mathrm{MC}(\operatorname{dep}(x,z)\lor% \operatorname{dep}(y,z))$ . Suppose $\theta(p_{1},\dots,p_{m})$ is an instance of $\mathrm{2SAT}$ of the form $\bigwedge_{i\in I}E_{i}$ , with conjuncts $E_{i}=(\ell_{i_{1}}\lor\ell_{i_{2}})$ , where $\ell_{i_{1}},\ell_{i_{2}}$ are different literals, i. e., $\ell_{i_{1}}\neq\ell_{i_{2}}$ . This can easily be archived by replacing conjuncts of the form $(\ell\lor\ell)$ with $(\ell\lor q)$ and $(\ell\lor\lnot q)$ , where $q\not\in\{p_{1},\dots,p_{m}\}$ is a new proposition.

We will construct a team $T$ , such that the following are equivalent:

$\blacksquare$

$T\models\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ .
$\blacksquare$

$\theta(p_{1},\dots,p_{m})$ is satisfiable.

For each conjunct $E_{i}$ , we create a team $T_{E_{i}}$ with two assignments $s_{i_{1}}$ and $s_{i_{2}}$ of domain $\{x,y,z\}$ , that encode the literals of $E_{i}$ . To this end, variable $x$ ranges over the propositions $\{p_{1},\dots,p_{m}\}$ , variable $y$ ranges over clause (index) $I$ , and variable $z$ ranges over $\{\,0_{k},1_{k}\mid 1\leq k\leq m\,\}$ , such that each proposition has its own two truth values. This is necessary to ensure that $z$ has two different values in each $T_{E_{i}}$ . Thus, the dependence atom $\operatorname{dep}(y,z)$ makes sure we have to choose at least one of the assignments from each $T_{E_{i}}$ into the subset of $T$ , that will satisfy $\operatorname{dep}(x,z)$ , thereby encoding the assignment that satisfies $\theta$ .

Each literal $\ell_{i_{j}}$ in $E_{i}$ , gives rise to one assignment. For example, the team $T_{E_{i}}$ for a clause $(p_{k}\lor p_{j})$ is the one in the middle in Figure 1.

Now $T$ is the union $\bigcup_{i\in I}T_{E_{i}}$ . Suppose $\theta(p_{1},\dots,p_{m})$ is satisfiable. Then there exists an assignment $F\colon\{p_{1},\dots,p_{m}\}\to\{0,1\}$ , that satisfies $\theta(p_{1},\dots,p_{m})$ . We write $F(p_{k})_{k}$ to denote the truth values $\{0_{k},1_{k}\}$ used by variable $z$ . For example, if $F(p_{k})=0$ , then $F(p_{k})_{k}=0_{k}$ . Define the partition of the team $T$ into two sets in the following way:

\displaystyle T_{1}

\displaystyle=\{s\in T\mid s(x)=p_{k}\text{ and }s(z)=F(p_{k})_{k}\},\qquad T_% {2}=T\setminus T_{1}

The assignments in $T$ that agree with the assignment $F$ are chosen to $T_{1}$ . Since $F$ evaluates $\bigwedge_{i\in I}E_{i}$ true, it satisfies every conjunct $E_{i}$ . Thus $T_{1}$ contains at least one assignment from each $T_{E_{i}}$ . Thus there will be at most one tuple from each $T_{E_{i}}$ left to $T_{2}$ . This $T_{2}$ trivially satisfies $\operatorname{dep}(y,z)$ since all tuples in $T_{2}$ disagree on $y$ , i. e., for all $s,t\in T_{2}$ it is true that $s(y)\neq t(y)$ , if $s\neq t$ .

Next we will show that $T_{1}$ satisfies $\operatorname{dep}(x,z)$ . Let $s,t\in T_{1}$ , such that $s(x)=t(x)=p_{k}$ . Then by the definition of $T_{1}$ it holds that $s(y)=F(p_{k})_{k}=t(y)$ holds. Thus $T_{1}\models\operatorname{dep}(x,z)$ .

The other direction: Suppose $T\models\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ . Then there is a partition of $T$ into $T_{1}$ and $T_{2}$ , such that $T_{1}\models\operatorname{dep}(x,z)$ and $T_{2}\models\operatorname{dep}(y,z)$ . We will define the assignment $F\colon\{p_{1},\dots,p_{m}\}\to\{0,1\}$ in the following way:

$\blacksquare$

If there exists $s\in T_{1}$ , such that $s(x)=p_{k}$ , then $F(p_{k})_{k}=s(z)$ .
$\blacksquare$

Otherwise, if for all $s\in T_{1}$ it holds $s(x)\neq p_{k}$ , then $F(p_{k})=1$ .

Let us show that $F$ is a function, which satisfies $\theta(p_{1},\dots,p_{m})$ :

1.

Clearly, $\operatorname{Dom}(F)=\{p_{1},\dots,p_{m}\}$ and $\operatorname{Rng}(F)=\{0,1\}$ .
2.

$F$ is a function: Let $p_{k}\in\{p_{1},\dots,p_{m}\}$ . Suppose there exists $s,t\in T_{1}$ , such that $s(x)=t(x)=p_{k}$ holds. Since $T_{1}\models\operatorname{dep}(x,z)$ holds, it follows that $s(z)=t(z)$ holds. Suppose there are no $s\in T_{1}$ , such that $s(x)=p_{k}$ . Then by definition of $F$ it holds that $F(p_{k})=1$ .
3.

$F$ satisfies $\theta(p_{1},\dots,p_{m})$ : Note that $y$ is constant and $z$ is assigned different values by each tuple in each $T_{E_{i}}$ . Thus $T_{1}$ contains at least one of the tuples from each $T_{E_{i}}$ . Let $s\in T_{E_{i}}$ , such that $s\in T_{1}$ . Recall, that in every assignment the values $s(z)$ encodes the truth value of $s(x)$ such that $E_{i}$ is satisfied. Since $s$ agrees with $F$ , it holds that $F(\ell_{i_{j}})_{i_{j}}=s(z)$ , which implies that $F(E_{i})=1$ .

Each conjunct $E_{i}$ , of $\theta$ gives rise to a constant size team of two assignments with domain $\{x,y,z\}$ . Thus the team $T$ can be constructed in $\mathrm{L}$ for each $\theta$ . $\hfill\blacktriangleleft$

Lemma 15 ( $\star$ ). [Restated, see original statement.]

Let $T$ be a team with domain $\{x,y\}$ such that $\operatorname{Rng}_{x}(T)\cap\operatorname{Rng}_{y}(T)=\emptyset$ . The following statements are equivalent:

(1.)

$T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .
(2.)
$G_{T}$ can be transformed into a directed graph $G^{\prime}_{T}=(V_{T},E^{\prime}_{T})$ such that
1. (a)
  
  each node has at most one out-going edge, and
2. (b)
  
  for each $\{u,v\}\in E_{T}$ either $(u,v)\in E^{\prime}_{T}$ or $(v,u)\in E^{\prime}_{T}$ .
(3.)

For each connected component $C=(V_{C},E_{C})$ of $G_{T}$ , we have that $|E_{C}|\leq|V_{C}|$ .

Proof.

We will prove the equivalences as follows.

(1.) $\Rightarrow$ (2.)

Assume, w.l.o.g., $T=T_{1}\uplus T_{2}$ , such that $T_{1}\models\operatorname{dep}(x,y)$ and $T_{2}\models\operatorname{dep}(y,x)$ . Then the directed graph $G^{\prime}_{T}=(V_{T},E^{\prime}_{T})$ with $E^{\prime}_{T}=\{\,(s(x),s(y))\mid s\in T_{1}\,\}\cup\{\,(s(y),s(x))\mid s\in T% _{2}\,\}$ has properties (a) and (b).

For (a), wrongly assume, w.l.o.g., that there are two edges $e_{1}=(a,b)$ and $e_{2}=(a,b^{\prime})$ with $b\neq b^{\prime}$ and $a,b,b^{\prime}$ values in $\operatorname{Rng}(T)$ (the same argument could be made for the second component being $a$ and the first components being different). Now, these edges correspond to assignments $s(x)=a$ , $s(y)=b$ , and $s^{\prime}(x)=a$ , $s^{\prime}(x)=b^{\prime}$ . Both $s,s^{\prime}\in T_{1}$ and would violate $\operatorname{dep}(x,y)$ which is a contradiction.

For (b), notice that each $\{u,v\}$ corresponds to a single assignment $s\in T$ . By assumption, $s$ is either in $T_{1}$ or $T_{2}$ , whence it has to appear in $E^{\prime}_{T}$ in either direction.

(2.) $\Rightarrow$ (1.)

Let $G^{\prime}_{T}=(V_{T},E^{\prime}_{T})$ be a directed graph, where each node has at most one out-going edge. Define $T_{1}=\{s\in T\mid(s(x),s(y))\in E^{\prime}_{T}\}$ and $T_{2}=\{s\in T\mid(s(y),s(x))\in E^{\prime}_{T}\}$ . Then, $T_{1}\models\operatorname{dep}(x,y)$ , because each $s(x)$ appears only once by assumption, so the dependency is not violated. By analogous arguments we have that $T_{2}\models\operatorname{dep}(y,x)$ .

(2.) $\Rightarrow$ (3.)

We make a case distinction. If each node in $G^{\prime}_{T}$ has exactly one out-going edge, then $|V_{T}|=|E^{\prime}_{T}|\overset{\text{\tiny(b)}}{=}|E_{T}|$ and likewise $|E_{C}|=|V_{C}|$ for each connected component $C$ of $G_{T}$ . Otherwise, if some nodes have no out-going edge, then $|E_{C}|<|V_{C}|$ .

(3.) $\Rightarrow$ (2.)

A connected component in $G_{T}$ has at least $|V_{C}|-1$ edges (otherwise it would not be connected). So there are only two cases:

(i)

If $|E_{C}|=|V_{C}|-1$ , then $C$ is a (spanning-)tree. Define $G^{\prime}_{T}$ by orienting the edges from the leaves to the root, i.e., bottom-up. Since each node has one parent, each node has one outgoing edge (except for the root which has no outgoing edge).
(ii)

If $|E_{C}|=|V_{C}|$ , then $C$ contains exactly one cycle $\mathcal{C}$ , because $C$ is connected and has one more edge than its spanning-tree. Choose any orientation of $\mathcal{C}$ (clockwise or counter-clockwise). Afterwards, all nodes in $C\setminus\mathcal{C}$ (hence outside of the cycle $\mathcal{C}$ ) form tree-like connected components and we can proceed as in (i): There is one node directly connecting to $\mathcal{C}$ which will become the root, and its outgoing edge connects to a node in the cycle $\mathcal{C}$ .

$\hfill\blacktriangleleft$ See 17

Proof.

Assume $T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ and $|T|>|\operatorname{Rng}_{x}(T)|+|\operatorname{Rng}_{y}(T)|$ , then there exists a split $T=T_{1}\cup T_{2}$ such that $T_{1}\models\operatorname{dep}(x,y)$ and $T_{2}\models\operatorname{dep}(y,x)$ . Clearly, $|T_{1}|\leq|\operatorname{Rng}_{y}(T)|$ and $|T_{2}|\leq|\operatorname{Rng}_{y}(T)|$ , because otherwise each of $T_{1}$ and $T_{2}$ would not satisfy the respective dependence atom. That is true, because then there would be a pair $s,t\in T_{1}$ with $s(x)=t(x)$ and $s(y)\neq t(y)$ , respectively $s,t\in T_{2}$ with $s(y)=t(y)$ and $s(x)\neq t(x)$ . Furthermore, we have that $|T|\leq|T_{1}|+|T_{2}|$ . As a result, we get $|T|\leq|T_{1}|+|T_{2}|\leq|\operatorname{Rng}_{x}(T)|+|\operatorname{Rng}_{y}(% T)|$ , which is a contradiction to $|T|>|\operatorname{Rng}_{x}(T)|+|\operatorname{Rng}_{y}(T)|$ . $\hfill\blacktriangleleft$

Definition 19. [Restated, see original statement.]

Let $\phi$ be a 2CNF formula with variables $\operatorname{Var}(\phi)=\{x_{1},\dots,x_{n}\}$ , such that:

(1.)

Every clause is monotone, i.e., it is of the form $(x_{i}\lor x_{j})$ or of the form $(\bar{x}_{i}\lor\bar{x}_{j})$ . The first type is called positive, the second type negative.
(2.)

No two “dual” clauses $(x_{i}\lor y_{j})$ , $(\bar{x}_{i}\lor\bar{x}_{j})$ appear in $\phi$ .
(3.)

Transitivity holds: if $(x_{i}\lor x_{j})$ and $(x_{j}\lor x_{k})$ are clauses of $\phi$ , then so is $(x_{i}\lor x_{k})\in\phi$ ; furthermore, if $(\bar{x}_{i}\lor\bar{x}_{j})$ and $(\bar{x}_{j}\lor\bar{x}_{k})$ are clauses of $\phi$ , then so is $(\bar{x}_{i}\lor\bar{x}_{k})\in\phi$ .

We call such 2CNF formulas monotone transitive dual-free and abbreviate the corresponding satisfiability problem by $\mathrm{mtdf\text{-}2SAT}$ .

We call a set $\{v_{1},\dots,v_{k}\}$ of variables in $\phi$ a (variable-)clique, if for each pair $v_{i},v_{j},i\neq j$ it is true that either $(v_{i}\lor v_{j})$ or $(\lnot v_{i}\lor\lnot v_{j})$ is a clause in $\phi$ . Note that, by the transitivity (3.), the variables of a $\mathrm{mtdf\text{-}2SAT}$ instance appear in at most two cliques, once positive and once negative. Thus we can represent $\phi$ , by listing its positive and negative cliques. See 20

Proof.

We present reductions between $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x))$ and $\mathrm{mtdf\text{-}2SAT}$ (for illustrations, see Figure 7). Let $\tau_{t}=(T^{3})$ be the vocabulary of structures where $T$ encodes a team such that $T(i,j,k)$ is true if and only if $s_{i}(x_{j})=a_{k}$ , where $i, j, k$ are natural numbers. Note that this differs from the encoding $\operatorname{Rel}(T)$ , used in the proof of Theorem 18, in that it explicitly contains the index of assignments. This additional information is necessary for this reduction. Then $\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x))\colon\mathrm{% STRUC}[\tau_{t}]\to\{0,1\}$ is the Boolean query that is true if and only if $T$ is a valid team and $T\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .

Further let $\tau_{pn}=(P^{2},N^{2})$ be the vocabulary of structures for $\mathrm{mtdf\text{-}2SAT}$ , where $P(v,c)$ encodes positive variables that occur in the same clique $c$ and $N(v,c)$ encodes negative variables. Then $\mathrm{mtdf\text{-}2SAT}\colon\mathrm{STRUC}[\tau_{pn}]\to\{0,1\}$ is the Boolean query that is true if and only if $P$ and $N$ encode a mtdf-2CNF formula $\phi$ and there is some assignment $\mathfrak{I}$ such that $\mathfrak{I}\models\phi$ .

The first-order reduction $I_{\textrm{tpn}}:\mathrm{STRUC}[\tau_{t}]\to\mathrm{STRUC}[\tau_{pn}]$ is as follows:

I_{\textrm{tpn}}\coloneqq\lambda_{vc}\langle\texttt{true},\psi_{P},\psi_{N}% \rangle,\qquad\psi_{P}(v,c)\coloneqq T(v,0,c),\qquad\psi_{N}(v,c)\coloneqq T(v% ,1,c),

where true means the universe is the same. We show that $A\in\mathrm{MC}(\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x))% \Leftrightarrow I_{\textrm{tpn}}(A)\in\mathrm{mtdf\text{-}2SAT}$ . Assume $A$ encodes a team $T$ such that $T=T_{1}\cup T_{2}$ , where $T_{1}\models\operatorname{dep}(x,y)$ and $T_{1}\models\operatorname{dep}(y,x)$ . Then the assignment $\mathfrak{I}$ with $\mathfrak{I}(x_{i})=1$ , if $s_{i}\in T_{1}$ and $\mathfrak{I}(x_{i})=0$ , if $s_{i}\in T_{2}$ satisfies the formula encodes by $I_{\textrm{tpn}}(A)$ . This is because, in a positive clique, at most one variable is false. Thus, in all clauses, at least one variable is true. The opposite holds for negative cliques.

Now for the reduction in the other direction. Consider the following first-order reduction $I_{\textrm{pnt}}\colon\mathrm{STRUC}[\tau_{pn}]\to\mathrm{STRUC}[\tau_{t}]$ given by:

I_{\textrm{pnt}}\coloneqq\ \lambda_{sxa}\langle\texttt{true},\psi_{T}\rangle,% \qquad\psi_{T}(s,x,a)\coloneqq\ (x=0\to N(s,a))\land(x=1\to P(s,a)).

Notice that $I_{\textrm{pnt}}$ is simply the inverse of $I_{\textrm{tpn}}$ , i. e., $I_{\textrm{pnt}}=I^{-1}_{\textrm{tpn}}$ . $\hfill\blacktriangleleft$

\begin{array}[]{ccc}\hline\cr\hline\cr T&x&y\\ \hline\cr s_{0}&0&0\\ s_{1}&0&1\\ s_{2}&0&2\\ s_{3}&1&1\\ s_{4}&2&1\\ s_{5}&2&2\\ \hline\cr\hline\cr\end{array}

$\leftrightsquigarrow$ $T(0,0,0)$ , $T(0,1,0)$ , $T(1,0,0)$ , $T(1,1,1)$ , $T(2,0,0)$ , $T(2,1,2)$ , $T(3,0,1)$ , $T(3,1,1)$ , $T(4,0,2)$ , $T(4,1,1)$ , $T(5,0,2)$ , $T(5,1,2)$ , $\leftrightsquigarrow$ $P(0,0)$ , $N(0,0)$ , $P(1,0)$ , $N(1,1)$ , $P(2,0)$ , $N(2,2)$ , $P(3,1)$ , $N(3,1)$ , $P(4,2)$ , $N(4,1)$ , $P(5,2)$ , $N(5,2)$ , $\leftrightsquigarrow$ $(x_{0}\lor x_{1})$ , $(x_{1}\lor x_{0})$ , $(x_{1}\lor x_{2})$ , $(x_{2}\lor x_{1})$ , $(x_{0}\lor x_{2})$ , $(x_{2}\lor x_{0})$ , $(x_{4}\lor x_{5})$ , $(x_{5}\lor x_{4})$ , $(\lnot x_{1}\lor\lnot x_{3})$ , $(\lnot x_{3}\lor\lnot x_{1})$ , $(\lnot x_{3}\lor\lnot x_{4})$ , $(\lnot x_{4}\lor\lnot x_{3})$ , $(\lnot x_{1}\lor\lnot x_{4})$ , $(\lnot x_{4}\lor\lnot x_{1})$ , $(\lnot x_{2}\lor\lnot x_{5})$ , $(\lnot x_{5}\lor\lnot x_{2})$ ,

Figure 7: Example reduction for Lemma 20. Team

T

gets encodes via structure

\langle\{0,1,2,3,4,5\},T^{3}\rangle

that gets mapped to the structure

\langle\{0,1,2,3,4,5\},P^{2},N^{2}\rangle

which encodes a mtdf-2CNF formula

\phi

; and vice versa. Notice how the split

\{s_{1},s_{3},s_{5}\}\models\operatorname{dep}(x,y)

and

\{s_{0},s_{2},s_{4}\}\models\operatorname{dep}(y,x)

corresponds to the assignment

\mathfrak{I}(x_{1})=\mathfrak{I}(x_{3})=\mathfrak{I}(x_{5})=1

and

\mathfrak{I}(x_{0})=\mathfrak{I}(x_{2})=\mathfrak{I}(x_{4})=0

.

Theorem 24 ( $\star$ ). [Restated, see original statement.]

Each of the following formulas has coherence-level $4$ :
(1.) $\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ ; (2.) $\operatorname{dep}(x)\lor\operatorname{dep}(x,y)$ ; (3.) $\operatorname{dep}(y)\lor\operatorname{dep}(x,y)$ .

Proof.

We first demonstrate that $\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ , $\operatorname{dep}(x)\lor\operatorname{dep}(x,y)$ and $\operatorname{dep}(y)\lor\operatorname{dep}(x,y)$ are not 3-coherent.

(1.) Consider the team $T$ in Figure 5 ( $b$ ). We have that $T\not\models\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ which is straightforward to see; consider for example the splits $\{s_{1},s_{2}\}\models\operatorname{dep}(x)$ or $\{s_{3},s_{4}\}\models\operatorname{dep}(x)$ and notice that in both cases $\operatorname{dep}(y,z)$ is falsified by $\{s_{3},s_{4}$ } and $\{s_{1},s_{2}\}$ respectively.

Now, all 3-element sub-teams have a split $T^{\prime}=T_{1}\cup T_{2}$ , where $T_{1}$ is of size two and $T_{1}\models\operatorname{dep}(x)$ and $T_{2}$ is of size one and therefore satisfies any dependence atom.

(2.) Not $3$ -coherent, because of the team in Figure 5 ( $c$ ).

(3.) Not $3$ -coherent, because of the team in Figure 5 ( $d$ ).

For coherence it suffices to only consider $\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ . The other two cases directly follow from the first.

We show that if $T\not\models\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ , then there must be a sub-team $T^{\prime}\subseteq T$ with $|T^{\prime}|\leq 4$ and $T^{\prime}\not\models\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ . Start with a pair of assignments $s_{1},s_{2}\in T$ for which $\{s_{1},s_{2}\}\not\models\operatorname{dep}(y,z)$ holds, i. e., $s_{1}(y)=s_{2}(y)$ and $s_{2}(z)\neq s_{2}(z)$ . Such a pair has to exist in $T$ to falsify the disjunction. Now, there are two cases for the assignment of $x$ .

Case 1:: The assignments $s_{1}$ and $s_{2}$ have the same value for $x$ , i. e., $s_{1}(x)=s_{2}(x)$ . Then for $T$ to falsify the disjunction, there must be a second pair $\{s_{3},s_{4}\}\not\models\operatorname{dep}(y,z)$ with $s_{3}(x)\neq s_{1}(x)$ and $s_{4}(x)\neq s_{1}(x)$ . but then $T^{\prime}=\{s_{1},s_{2},s_{3},s_{4}\}$ and $T^{\prime}\not\models\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ . To see this, assume there is a split $T_{1}\cup T_{2}=\{s_{1},s_{2},s_{3},s_{4}\}$ with $T_{1}\models\operatorname{dep}(x)$ and $T_{2}\models\operatorname{dep}(y,z)$ . If $s_{1}\in T_{1}$ , then $s_{3}\in T_{2}$ , but $s_{4}$ cannot be in $T_{1}$ or $T_{2}$ . If $s_{1}\in T_{2}$ , then $s_{2}\in T_{1}$ , therefore $s_{3}$ and $s_{4}$ must be in $T_{2}$ which is not possible. Therefore no split exists.
Case 2:: The assignments $s_{1}$ and $s_{2}$ have different values for $x$ , i. .e, $s_{1}(x)\neq s_{2}(x)$ . Then the pair satisfies neither $\operatorname{dep}(x)$ nor $\operatorname{dep}(y,z)$ , so in any potential split $s_{1}\in T_{1}$ implies $s_{2}\in T_{2}$ and vice versa. Since the original team falsified the disjunction, there must be an assignment $s_{3}\in T$ that can neither be in $T_{1}$ nor $T_{2}$ if $s_{1}\in T_{1}$ and $s_{2}\in T_{2}$ . In the same way there must be an assignment $s_{4}\in T$ that is not in $T_{1}$ or $T_{2}$ for $s_{1}\in T_{2}$ and $s_{2}\in T_{1}$ . Therefore $T^{\prime}=\{s_{1},s_{2},s_{3},s_{4}\}$ also has no split satisfying $\operatorname{dep}(x)\lor\operatorname{dep}(y,z)$ .

$\hfill\blacktriangleleft$

Theorem 25 ( $\star$ ). [Restated, see original statement.]

The following three formulas are incoherent:
(1.) $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ ; (2.) $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z)$ ; (3.) $\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ .

Proof.

(1.) We show that for all even $n\in\mathbb{N}$ , there is a team $T$ of size $|T|=n+1$ with $T\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ , but for all $n$ -element sub-teams $T^{\prime}\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ holds.

$\begin{array}[]{ccc}\hline\cr\hline\cr T_{n}&x&y\\ \hline\cr s_{0}&1&\frac{n}{2}\\ s_{1}&1&1\\ s_{2}&1&2\\ s_{3}&2&2\\ s_{4}&2&3\\ \vdots&\vdots&\vdots\\ s_{n-1}&\frac{n}{2}&\frac{n}{2}\\ s_{n\phantom{+0}}&\frac{n}{2}&1\\ \hline\cr\hline\cr\end{array}\qquad\begin{array}[]{ccc}\hline\cr\hline\cr T^{% \prime}&x&y\\ \hline\cr s_{0}&1&\frac{n}{2}\\ s_{3}&2&2\\ \vdots&\vdots&\vdots\\ s_{n\phantom{+0}}&\frac{n}{2}&1\\ \hline\cr s_{1}&1&1\\ s_{2}&1&2\\ \vdots&\vdots&\vdots\\ s_{n-1}&\frac{n}{2}&\frac{n}{2}\\ \hline\cr\hline\cr\end{array}\qquad\qquad\begin{array}[]{cccc}\hline\cr\hline% \cr T_{n}&x&y&z\\ \hline\cr s_{0}&1&\frac{n}{2}&3\\ s_{1}&1&1&1\\ s_{2}&1&2&2\\ s_{3}&2&2&1\\ s_{4}&2&3&2\\ \vdots&\vdots&\vdots&\vdots\\ s_{n-1}&\frac{n}{2}&\frac{n}{2}&1\\ s_{n\phantom{+0}}&\frac{n}{2}&1&2\\ \hline\cr\hline\cr\end{array}\qquad\begin{array}[]{cccc}\hline\cr\hline\cr T^{% \prime}&x&y&z\\ \hline\cr s_{0}&1&\frac{n}{2}&3\\ s_{3}&2&2&1\\ \vdots&\vdots&\vdots&\vdots\\ s_{n\phantom{+0}}&\frac{n}{2}&1&2\\ \hline\cr s_{1}&1&1&1\\ s_{2}&1&2&2\\ \vdots&\vdots&\vdots&\vdots\\ s_{n-1}&\frac{n}{2}&\frac{n}{2}&1\\ \hline\cr\hline\cr\end{array}$

Figure 8: Team

T_{n}

in the proof of Theorem 25, and constructed split of an example sub-team

T^{\prime}=T_{n}\setminus\{s_{4}\}

. On the left for

\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)

and on the right for

\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)

.

Let the team $T_{n}$ be as depicted in Figure 8. We show that it is impossible to split $T_{n}=T_{1}\cup T_{2}$ such that $T_{1}\models\operatorname{dep}(x,y)$ and $T_{2}\models\operatorname{dep}(y,x)$ . Start with assignment $s_{1}$ and choose $s_{1}\in T_{1}$ . Next, $s_{2}\in T_{2}$ is the only choice, because $T_{1}=\{s_{1},s_{2}\}\not\models\operatorname{dep}(x,y)$ . Then $s_{3}\in T_{1}$ , because $T_{2}=\{s_{2},s_{3}\}\not\models\operatorname{dep}(y,x)$ , but $T_{1}=\{s_{1},s_{3}\}\models\operatorname{dep}(x,y)$ . Continue this procedure until all assignments but $s_{0}$ are either in $T_{1}$ or $T_{2}$ . Now, $s_{0}$ cannot be in $T_{1}$ , because $\{s_{0},s_{1}\}\not\models\operatorname{dep}(x,y)$ and it cannot be in $T_{2}$ , because $\{s_{0},s_{n-2}\}\not\models\operatorname{dep}(y,x)$ . If instead we chose $s_{1}\in T_{2}$ and repeated the process backwards, that is continued with $s_{n}$ , then $\{s_{0},s_{2}\}\not\models\operatorname{dep}(x,y)$ and $\{s_{0},s_{n-1}\}\not\models\operatorname{dep}(y,x)$ . Since $s_{1}$ has to be in either $T_{1}$ or $T_{2}$ , we can conclude that no split exists and $T_{n}\not\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .

Next, we show that all sub-teams of size $n$ satisfy the disjunction. The sub-team $T_{n}\setminus\{s_{0}\}\models\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ given the split described above. Otherwise, let $T^{\prime}=T_{n}\setminus\{s_{i}\}$ and consider the split above where $s_{1}\in T_{1}$ . We define a new split for $T^{\prime}$ as follows

	$\displaystyle T^{\prime}_{1}$	$\displaystyle=\{s_{j}\in T_{1}\mid 1<j<i\}\cup\{s_{j}\in T_{2}\mid j>i\}\cup\{% s_{0}\}$
	$\displaystyle T^{\prime}_{2}$	$\displaystyle=\{s_{j}\in T_{2}\mid 1<j<i\}\cup\{s_{j}\in T_{1}\mid j>i\}\cup\{% s_{1}\}$

Now, $T^{\prime}_{1}\models\operatorname{dep}(x,y)$ and $T^{\prime}_{2}\models\operatorname{dep}(y,x)$ .

We will present the case that $s_{i}\in T_{1}$ ; the case $s_{i}\in T_{2}$ follows analogously. If $s_{i}\in T_{1}$ , then $s_{i+1}$ has to be in $T_{2}$ as shown above. Now, since $s_{i}\not\in T^{\prime}$ , it is possible that $s_{i+1}\in T^{\prime}_{1}$ . From this $s_{i+2}\in T^{\prime}_{2},s_{i+3}\in T^{\prime}_{1},\dots$ follows immediately. This chain leads to $s_{n}\in T^{\prime}_{1}$ , which allows $s_{1}\in T^{\prime}_{2}$ , which in turn makes $s_{0}\in T^{\prime}_{1}$ possible. Therefore we have that $T^{\prime}_{1}\models\operatorname{dep}(x,y)$ and $T^{\prime}_{2}\models\operatorname{dep}(y,x)$ .

We have shown that for every even $n\in\mathbb{N}$ we can construct a team such that $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ is not $n$ -coherent, that is $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ is incoherent.

(2.) Extend the team $T_{n}$ in (1.) with variable $z$ , such that $s(z)=s(x)$ for all $s\in T_{n}$ . Then incoherence of $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,z)$ follows immediately from the incoherence of $\operatorname{dep}(x,y)\lor\operatorname{dep}(y,x)$ .

(3.) For $\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ extend $T_{n}$ with variable $z$ such that if $s(x)=s^{\prime}(x)$ or $s(y)=s^{\prime}(y)$ , then $s(z)\neq s^{\prime}(z)$ for all $s,s^{\prime}\in T$ . For example see team $T_{n}$ in Figure 8. It is easy to check, that the arguments of (1.) also hold true in such a team for $\operatorname{dep}(x,z)\lor\operatorname{dep}(y,z)$ . $\hfill\blacktriangleleft$

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Disjunctions of Two Dependence Atoms

Abstract

Keywords and phrases:

Funding:

Copyright and License:

2012 ACM Subject Classification:

Related Version:

Acknowledgements:

DOI:

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Editors:

Series and Publisher:

1 Introduction

Contributions.

2 Preliminaries

Definition 1.

Definition 2.

Definition 3.

Example 4.

Proposition 5 ([14, Lemma 3.27, Prop. 3.10]).

Coherence.

Definition 6 (k-coherence, [8, Def. 3.1]).

Definition 7 (coherence-level, incoherence).

Proposition 8.

Proposition 9 ([8, Thm. 4.9]).

3 Model-checking

Proposition 10 ([8, Thm. 4.12]).

Corollary 11.

3.1 NL-Completeness

Proposition 12 ([8, Thm. 4.14]).

Theorem 13 (⋆).

Proof sketch.

3.2 L-Completeness

Definition 14.

Lemma 15 (⋆).

Theorem 16.

Proof.

Lemma 17 (⋆).

Theorem 18.

Proof.

A restricted case of Krom-satisfiability.

Definition 19.

Lemma 20 (⋆).

Proof sketch.

Theorem 21.

4 Coherence

Theorem 22.

Proof.

Corollary 23.

Theorem 24 (⋆).

Proof sketch.

Theorem 25 (⋆).

Theorem 26.

5 Disjunctions of higher-arity dependence atoms

Lemma 27.

Proof.

Corollary 28.

Lemma 29.

Proof.

Theorem 30.

Proof.

Corollary 31.

6 Outlook

References

Appendix A Appendix

Theorem 13 (⋆). [Restated, see original statement.]

Proof.

Lemma 15 (⋆). [Restated, see original statement.]

Proof.

Proof.

Definition 19. [Restated, see original statement.]

Proof.

Theorem 24 (⋆). [Restated, see original statement.]

Proof.

Theorem 25 (⋆). [Restated, see original statement.]

Proof.

Definition 6 ( $k$ -coherence, [8, Def. 3.1]).

Theorem 13 ( $\star$ ).

Lemma 15 ( $\star$ ).

Lemma 17 ( $\star$ ).

Lemma 20 ( $\star$ ).

Theorem 24 ( $\star$ ).

Theorem 25 ( $\star$ ).

Theorem 13 ( $\star$ ). [Restated, see original statement.]

Lemma 15 ( $\star$ ). [Restated, see original statement.]

Theorem 24 ( $\star$ ). [Restated, see original statement.]

Theorem 25 ( $\star$ ). [Restated, see original statement.]