On Homogeneous Models of Fluted Languages
Abstract
We study the fluted fragment of first-order logic which is often viewed as a multi-variable non-guarded extension to various systems of description logics lacking role-inverses. In this paper we show that satisfiable fluted sentences (even under reasonable extensions) admit special kinds of “nice” models which we call globally/locally homogeneous. Homogeneous models allow us to simplify methods for analysing fluted logics with counting quantifiers and establish a novel result for the decidability of the (finite) satisfiability problem for the fluted fragment with periodic counting. More specifically, we will show that the (finite) satisfiability problem for the language is Tower-complete. If only two variable are used, computational complexity drops to NExpTime-completeness. We supplement our findings by showing that generalisations of fluted logics, such as the adjacent fragment, have finite and general satisfiability problems which are, respectively, - and -complete. Additionally, satisfiability becomes -complete if periodic counting quantifiers are permitted.
Keywords and phrases:
Fluted Fragment, Fluted Logic, Fluted Fragment with Periodic Counting, Adjacent Fragment, Adjacent Fragment with Counting, Adjacent Fragment with Periodic Counting, Counting Quantifiers, Periodic Counting Quantifiers, Decidable Fragments of First-Order Logic2012 ACM Subject Classification:
Theory of computation Complexity theory and logicAcknowledgements:
The author would like to thank Dr. Ian Pratt-Hartmann and Dr. Bartosz Jan Bednarczyk for their valuable feedback.Funding:
This work is supported by the NCN grant 2018/31/B/ST6/03662.Editors:
Jörg Endrullis and Sylvain SchmitzSeries and Publisher:

1 Introduction
The fluted fragment (denoted ) is a fragment of first-order logic in which, roughly put, variables appear in predicates following the order in which they were quantified. For illustrative purposes, we translate the sentence “Every conductor nominates their favorite soloist to play at every concert” into this language as follows:
(1) |
As a non-example, the sentences axiomatising transitivity, symmetry and reflexivity of a relation are not in the fluted fragment.
The fluted fragment is a member of argument-sequence logics – a family of decidable (in terms of satisfiability) fragments of first-order logic which also includes the ordered [11, 13], forward [2] and adjacent [4] fragments. The fluted fragment in particular is decidable in terms of satisfiability even in the presence of counting quantifiers [19] or a distinguished transitive relation [22]. Surprisingly, the satisfiability problem for under a combination of the two not only retains decidability but also has the finite model property [24]. We refer the reader to [23] for a survey.
In this paper we will mostly be concerned with what we call the fluted fragment with periodic counting (denoted ). We remark that periodic counting quantifiers generalise standard (threshold) counting quantifiers which have been an object of intensive study as an extension for the fluted fragment in the past few years [19, 24]. Under this new formalism, we are allowed to write formulas requesting an even number of existential witnesses. As an example, we can express sentences as “Every orchestra hires an even number of people to play first violin” in our language (see (2)).
The origins of flutedness trace back to the works of W. V. Quine [26]. It is, however, the definition given by W. C. Purdy (in [25]) that has become widespread and will be the one we use. The popularity of Purdy’s idea of flutedness is not without cause, at least when keeping the field of description logics in mind. Indeed, after a routine translation, formulas of the description logic are contained in the two-variable sub-fragment of . This is even the case when is augmented with role hierarchies, nominals and/or cardinality restrictions (possibly with modulo operations). We refer the reader to [12] for more details. In terms of expressive power, closely parallels – a new formalism with counting constrains expressible in quantifier-free Boolean algebra with Presburger arithmetic (see [16, 1]). Thus, noting that the guarded fragment with at least three variables becomes undecidable under counting extensions [8], and that the guarded fluted fragment has “nice” model theoretic properties such as Craig interpolation [3], fluted languages emerge as perfect candidates for generalising description logics in a multi-variable context.
In Sections 3 and 4 we establish that classes of models of satisfiable -sentences always contain a “nice” structure in which elements behave (in a sense that we will make clear) homogeneously. Utilising this behaviour we will show that the fluted fragment extended with periodic counting quantifiers has a decidable satisfiability problem. Intriguingly, even though periodic counting quantifiers generalise standard counting quantifiers, our methodology allows us to avoid Presburger quantification, which was required to establish decidability of satisfiability for with standard counting [19].
In section 5 we show that the satisfiability problems for the fluted fragment with counting extensions become undecidable when minimal syntactic relaxations are allowed. More precisely, the section will culminate with a result showing that the finite satisfiability problem for the 3-variable adjacent fragment with counting is -complete. Additionally, the general satisfiability problem will be shown to be -complete when 4 variables are used, and -complete if periodic counting is allowed. Denoting the adjacent fragment as , we provide a brief survey of complexity and undecidability standings in Table 1.
The work in this paper is closely related to [6] in which decidability of satisfiability is established for the two-variable fragment with periodic counting (denoted ) but without a sharp complexity-theoretic bound. Our homogeneity conditions, which stem from lack of inverse relations in fluted logics, allow us to establish NExpTime-completeness for both the finite and general satisfiability problems of .
2 Preliminaries
We use to denote the set of integers , and to denote along with the first infinite cardinal; i.e. . By picking some we write for the linear set . In the extended integers , the cardinal is the maximum element under the canonical ordering “” and
-
;
-
for all ; and
-
for all .
A linear Diophantine inequation is an expression of the form where are constant values taken form , is a vector of variables, and “” is any of the relations “” (each interpreted as one would assume). It is known that when the cardinal is disallowed, a solution for a set of such inequations may be found in NPTime [17]. The picture does not change when is permitted as a solution and/or constant. Indeed, we may reduce the problem of finding a solution over to that of finding it over as follows. First guess which variables should be mapped to and which should have a finite value. Then, check that each inequation featuring a variable assigned holds and discard them. What will be left is a system of inequations with constants in and in variables assumed to be finite. See [20, Ch 7.4] for greater detail. We will allow systems of inequations to contain disjunctions.
By a word we mean a tuple , where for each , . In case , we often write instead of . By we mean the reversal of ; i.e. . If and are words we write for the concatenation of the two. We use the terms “tuple” and “word” interchangeably.
Now, take some structure and an -tuple of elements from . Suppose for some first-order formula . Fixing we extend the syntax of first-order logic with (threshold) counting quantifiers and periodic counting quantifier . Semantically, if and only if . Similarly, if and only if . We refrain from further generalisation to ultimately periodic counting quantifier (as in [6]) as they can be expressed as a disjunction of formulas using periodic counting quantifiers . Thus, a sentence such as “Every orchestra hires an even number of people to play first violin” may be written in a language with periodic counting:
(2) |
Formally, the fluted fragment with periodic counting is the union of sets of formulas defined by simultaneous induction as follows:
-
(i)
any atom , where is a contiguous subsequence of and is a predicate of arity , is in ;
-
(ii)
is closed under Boolean combinations;
-
(iii)
if , then is in for every .
We write for the set of all fluted formulas with periodic counting and define the -variable fluted fragment with periodic counting to be the set . (Here is the set of first-order formulas that do not use more than variables). We will implicitly restrict attention to signatures which feature no function and/or constant symbols, and where “” is always interpreted as the canonical equality relation. Lastly, we use interchangeably with whenever convenient.
Variables in fluted logics convey no meaningful information. Indeed, since the arity of every predicate in is fixed and each atom features a suffix of the variable quantification order, we may employ what we call variable-free notation. As an example, formulas (1) and (2) may be (respectively) written as
(3) | |||
(4) |
without ambiguity (up to a shift of variable indices). Writing for , we will employ variable-free notation extensively throughout subsequent sections.
Fix a first-order formula with periodic counting quantifiers. We assume that numeric values are encoded in binary and write for the number of symbols used in . We point out that the signature of (which we write as for brevity) is no larger than .
Now, let be a formula in . We say that is in normal-form if it takes the following shape
(5) |
where are sets of indices, each is a quantifier-free formula in variables, each is a quantifier-free formula in variables, and each is a linear set. Using standard rewriting techniques we prove the following in Appendix A.
Lemma 1.
Suppose is an -sentence. Then, we may compute, in polynomial time, an equisatisfiable normal-form -sentence .
Notice that the negation before the periodic counting quantifier in the second conjunct of (5) is not moved-inwards. This is done deliberately so as to avoid computing complements of linear sets, which may be of exponential size as a function of .
Now, let be a structure over a finite signature . We say that an atom is -fluted if it features a suffix of as arguments. The fluted atomic -type of (denoted ) is then the set of -fluted (possibly negated) atoms over with arity no greater than that are satisfied by in . We invite the reader to view the tuple as emitting , and as absorbing . Write for the fluted -type obtained by deleting entries in of arity greater than and decrementing variable indices by 1. We say that a fluted -type is an endpoint of a fluted -type if . We define to be the set of all fluted -types over .
The fluted -profile of (denoted ) is a function mapping to the number of times emits . Formally, If is a quantifier-free -formula, we write just in case . Clearly, if and only if .
In the sequel we will assume that all structures are countable. This comes with no loss of generality as is subsumed by the countable fragment of infinitary logic ; a language for which the downward Löwenheim-Skolem Theorem holds (folklore, see [15, p. 69]). Note that, in , the finite model property fails as is an axiom of infinity.
3 The Two-Variable Sub-fragment
In this section we restrict attention to the two-variable fluted fragment with periodic counting. To achieve decidability of (finite) satisfiability we first specify what kind of “nice” models we will be looking for. Take any -structure and . For convenience, we write for the set of all elements with . We say that is globally homogeneous in if for each . That is to say, is globally homogeneous in if, for each , the number of fluted 2-types emitted is the same for each element in . The structure is globally homogeneous if each is globally homogeneous in .
For the remainder of the section fix some normal-form -sentence . We claim that if is a satisfiable, then it is satisfiable in a globally homogeneous model. To see this, take some structure and a pair of elements that realised the fluted 2-type in . Since consists of fluted formulas, it does not feature atoms of the form and . Referencing only, we lack information to deduce what formulas are satisfied by the pair in . On the other hand, if and only if for each quantifier-free -formula . Thus, if we were to in any way alter the fluted 2-type of in , the set of quantifier-free -formulas satisfied by in would not change.
Taking a step back, take any and recall that is the set of all elements realising the 1-type in . We redefine 2-types emitted by elements of in in such a way that makes globally homogeneous in the rewiring of . Let us fix any and write . The element and profile picked will serve as an “example” of how the rest of should form fluted 2-types with other elements of the model. Taking any and we see that it is impossible to prohibit from emitting the fluted 2-type to via any normal-form -formula. We thus allow to impersonate in by rewiring the fluted 2-type to be for each and, additionally, by resetting to be the 2-type and to be . Clearly, only fluted 2-types emitted by were reconsidered in this procedure, thus pairs in satisfy the same quantifier-free -formulas as before. To see that still models we need only show that does not violate and for each and . By our rewiring procedure, we have that . Thus,
Having already argued that and for each and we therefore conclude that .
Since the only 2-types redefined are those emitted by , we can run this construction in parallel for each element in . Clearly, this renders globally homogeneous in the rewired model. Since elements in are left untouched by our rewiring, we can repeat this procedure for each thus proving the following:
Lemma 2.
Suppose is a satisfiable normal-form -sentence. Then, is satisfiable in a globally homogeneous model.
In globally homogeneous structures elements realising the same fluted 1-type are, in a sense, stripped away of their individuality as they all realise the same fluted 1-profile. When the globally homogeneous structure is clear from context, we can unambiguously write for the fluted 1-profile realised by each element of (for ).
When considering the (finite) satisfiability problem for normal-form -sentences such as , we will confine ourselves to the search of globally homogeneous models. More precisely, we will produce a system of linear Diophantine inequations that has a solution over if and only if is satisfiable in a globally homogeneous model. For this purpose, let , , , and be sequences of variables. Intuitively, the value assigned to will represent the number of elements realising the fluted 1-type ; – the number times the 2-type is emitted by an element realising ; and with and acting as periodic counters for elements realising when considering linear sets and . To be more precise, when given a satisfying assignment for , we will build a globally homogeneous with
(6) |
On the other hand, we will have that any globally homogeneous model gives rise to a solution of with the following assignments for all , , and :111 In case (resp. ) is , we allow (resp. ) to take any integer value.
(7) |
We will proceed first by showing that the latter assignment satisfies our (yet to be defined) system of inequations .
When considering any model we have that the domain is non-empty. The following is thus trivially satisfied by the assignment :
() |
Additionally, picking any element (for any ) and picking any , we have that the number of fluted 2-types emitted by to must be . Assuming that is globally homogeneous, we may fixate on the fact that the shared profile of has exactly witnesses for fluted 2-types with the endpoint , or, more formally, . Thus, the assignment satisfies the following set of inequations:
() |
Of course, under the supposition that for some and such that , we have that . Clearly, must be a member of the linear set . Thus, there is a period counter such that . Turning the equation around we get that . Recalling that for all , we have that the following is satisfied by our assignments:
() |
On the other hand, supposing for some and with such that , we have that , thus leaving outside of the linear set . Notice that this happens when the following conditions are met:
-
1.
; or
-
2.
and for each (which only happens when ); or
-
3.
and ; or
-
4.
for some we have .
Note that the listed conditions are exhaustive. We translate the requirements 1–4 into four functions which map fluted 1-types paired with indices in to linear equations. The functions are then defined as follows:
-
,
-
,
-
,
-
.
Since adheres to at least one of the four conditions, we write the following clauses for eligible fluted -types:
() |
To verify that is indeed a satisfying assignment for , we need only consider case 4. For this assume that for some . We can thus write . Since , we conclude that thus satisfying .
Reflecting on the semantics of the equality predicate, we see that for any given in any globally homogeneous model there is exactly one such that features the non-negated equality predicate and . More precisely, and the endpoint of is . We have that our assignment respects this condition and thus satisfies the following inequations:
() |
Finally, we forbid the periodic counters and from taking the value :
() |
Putting everything together, we have engineered a system of equations that is satisfied by extracting relevant cardinalities (see (7)) from homogeneous models of .
Lemma 3.
Suppose is a globally homogeneous model. Then, is satisfiable.
We now move on to the converse direction:
Lemma 4.
Suppose is satisfiable. Then, there is a globally homogeneous model .
Proof.
Suppose that has a satisfying assignment. To avoid notational clutter, we identify the solution vectors of variables in as themselves. We will build a globally homogeneous model over the domain Intuitively, we wish that elements of realise the fluted 1-type . We thus assign for all . Recalling that is the set of all elements that realise the 1-type in , we have that is of cardinality as required by (6). By , the domain is non-empty.
Picking any we now move on to the assignment of fluted 2-types. Take any and let , i.e. is the set of all fluted -types containing the fluted 1-type as an endpoint. By , we have that , thus . Now, pick some element . In case , equation prohibits fluted 2-types that feature the (non-negated) equality literal. We set fluted 2-types between and elements of in any way that results in for each (n.b. the exact configuration of fluted 2-types between and elements of is irrelevant as the fluted 2-type of for any is not set in this process). In case notice that by there is exactly one such that (i) , (ii) , and with (iii) . By again, we have that . We therefore set the fluted 2-types between and for each as in the case before and, additionally, specify that .
By repeating the fluted 2-type assignment for each element and fluted 1-type we are guaranteed that elements in (where ) realise the fluted 1-profile as required by (6). The resulting structure is clearly globally homogeneous.
We now claim that the resulting structure is a model of . Indeed, take any with and suppose for some . Let . By equations and , the sum is a member of the linear set . Since the element is of fluted 1-type , we have that it realises the profile in . By our construction, for each . Thus, which is equivalent to as required.
On the other hand, suppose for some . We claim that . To see this, let be the set . Writing we take note of equations and , and conclude that one of the following conditions must be true:
-
1.
is smaller than the minimal element of ; or
-
2.
and ; or
-
3.
and ; or
-
4.
is in between two consecutive elements of .
Whichever case it may be, we have that . Again, recalling that for each , we conclude that which is equivalent to saying .
Given an -sentence we present a decision procedure for the (finite) satisfiability problem. Compute a normal-form formula from and write the linear Diophantine equations . Now, guess a solution vector which can be done in non-deterministic polynomial time as a function of . If is indeed a solution for , accept, otherwise, reject. In the case of the finite satisfiability problem, prohibit from being a solution in . Correctness of the procedure follows from the fact that, by Lemma 2, if satisfiable then it is satisfiable in a globally homogeneous model. Thus, by Lemma 3, if is satisfiable, then has a solution. On the other hand, by Lemma 4, if has a solution, then is satisfiable.
Noting that the satisfiability problem for is NExpTime-hard [21], and that is bounded by a polynomial function on the number of different fluted 1- and 2-types (of which there are many), we conclude the following:
Theorem 5.
The (finite) satisfiability problem of is NExpTime-complete.
4 More Than Two Variables
We now generalise our results on homogeneity and decidability of satisfiability for higher-arity formulas of . Thus, throughout this section, we will be working in the -variable sub-fragment of , where is fixed.
Firstly, we lift our homogeneity conditions to the multivariable setting. Suppose is a -structure and take any -tuple from and . Let be the set . We say that is -homogeneous in if for each and all we have that . That is to say, absorbs the same fluted -type from each -tuple that realises the fluted -type . If each is -homogeneous in , then we say that the -tuple is homogeneous in . Finally, if each -tuple is homogeneous in , then we say that is locally -homogeneous.
When considering satisfiable normal-form -sentences we can, without loss of generality, confine ourselves to locally -homogeneous structures. To see this fix some normal-form -sentence and take and . Proceeding similarly as before Lemma 2 we see that the fluted -type of does not impact the satisfaction of quantifier-free -formulas by . Thus, redefining the fluted -types emitted by will not alter the satisfaction of quantifier-free -formulas by other tuples. Notice again that it is impossible to prohibit by a normal-form formula. Thus, by setting for each and repeating the procedure for each , we will have that is -homogeneous in . Clearly as the tuples rewired by this procedure (i.e. with ) now emit to a given witness if and only if does. The rewired tuples thus satisfy the same exact -formulas with at most 1 quantifier as does. Repeating the procedure for all and we will have the following:
Lemma 6.
Suppose is a satisfiable normal-form -sentence. Then, is satisfiable in a locally -homogeneous model.
Using local -homogeneity coupled with variable reduction techniques prevalent in studies of fluted logics (see [21]), we will establish a decidability result for the (finite) satisfiability problem of . More specifically, fixing a normal-form -sentence we will compute a normal-form -sentence that is satisfiable in structures holding just enough information to build locally -homogeneous models for . To aid motivation, we fix to be any locally -homogeneous model of . We shall construct whilst also expanding into . Note that the construction depends exclusively on the syntactic properties of .
First, set . Take to be a sequence of fresh -ary predicate symbols. In we decorate -tuples over with just in case we have that . That is to say, remembers which -tuples can be extended (by appending an element to the left) to realise the fluted -type . It is clear that models the following:
() |
Proceeding similarly, let be a sequence of new -ary predicates. Intuitively, we will have if in it is the case that absorbs the fluted -type emitted from for some . Notice that, by local -homogeneity, if absorbs from some with , then it absorbs from for all . Thus, by our construction, is the unique predicate amongst satisfied by in . Clearly, models:
() |
Again taking and any suppose for some . In case is non-empty (thus guaranteeing ), we pick any and write . Since is -homogeneous in , the exact element in we pick has no effect on . By our construction, is then exactly the set of element such that . Since it is then immediate that models the following:
() |
Similar observations follow whenever for some . This time, however, the cardinality of must be outside the set . Clearly, models:
() |
Writing we have shown the following:
Lemma 7.
Suppose is a locally -homogeneous model. Then, can be extended to a model of .
Lemma 8.
Suppose . Then, we can construct a locally -homogeneous model of over the same domain.
Proof.
Supposing is satisfiable we take any model . Now, let be the model but with the predicates in and removed from the signature. We proceed by expanding into a locally -homogeneous model of the original sentence .
Fix and take some . Supposing that , by we have that . Taking any we observe that the conjuncts of enforce the following:
-
if for some , then can absorb the fluted -type ,
-
satisfies at least one of the predicates , and
-
satisfies at most one of the predicates .
We can then safely set for each , where is taken from the subscript of the unique that satisfies in . By repeating the above procedure for all and tuples we will obtain the desired structure .
To verify that is a model of we first claim that For this purpose, fix some and , and suppose , where . Recall that by . Then, gives us . Taking any we have, by our construction, that if and only if . Thus, , which is equivalent to saying . Repeating the argument for each and will yield the required result.
Let us take stock of the previous three lemmas. Take to be an -sentence. Without loss of generality, assume that it is in normal-form (Lemma 1). By Lemma 6, is satisfiable if it is satisfiable in a locally -homogeneous model. By computing the formula we have, by Lemmas 7 and 8, that is (finitely) satisfiable if and only if is. Noting that the (finite) satisfiability problem for is in NExpTime (Theorem 5) and that can be constructed from in polynomial time in regards to the number of different fluted - and -types (of which there are ), we conclude the following:
Theorem 9.
The (finite) satisfiability problem for is in -NExpTime.
Noting that the (finite) satisfiability problem for is -NExpTime-hard [21], we see that no elementary function can encapsulate the complexity of (finite) satisfiability for . We thus conclude our section having reached our initial goal:
Theorem 10.
The (finite) satisfiability problem for is Tower-complete.
5 Counting With Reversed Relations
In this section we will show that relaxing the syntactic restrictions of the fluted fragment with counting yields undecidability of satisfiability. We define the language to be but with the addition of atoms with reversed variable sequences. More formally, if is an -atom, then and are -atoms. The language () is then the obvious extension of with (periodic) counting quantifiers. Clearly, the languages , , and are subfragments of the adjacent fragment with the appropriate counting extensions. We will use counting quantifiers and with the meanings “there is exactly one element s.t. …” and “there is at most one element s.t. …” along side periodic counting qunatifiers. For simplicity, we do away with variable-free notation and use variable sequences of and in place of .
We proceed by reducing Hilbert’s 10th problem to the finite satisfiability problem of . Let be a system of Diophantine equations. We assume that each equation is of one of the following (simple) forms: (i) , (ii) , or (iii) , where are mutually disjoint variables. Clearly, no loss of generality occurs as any (non-simple) Diophantine equation can be rewritten into the simpler form by introducing new variables. For each we will define a formula depending on the form that takes. Then, will be the advertised formula that is finitely satisfiable if and only if has a solution over . We specify that the signature of includes (1) unary predicates for each variable of in , (2) binary predicates for each of the form (ii), and (3) ternary predicates for each of the form (iii). We will not assume that the equality predicate is available. In the sequel we will argue that if , then has a solution with for each variable . (And, of course, the converse as well). For technical reasons we wish for the sets and with to be disjoint. Denoting for the set of variables in , we first define which clearly has the required effect. We proceed by taking in turn.
Suppose first that is of the form (i) . We ensure that every model of will have by defining to be
Now, supposing that is of the form (ii) , we define with the intent that models of will have being a bijection between and (i.e. ):
Lastly, if is of the form (iii) , then (defined just below) will guarantee that forces to be a bijection between and (i.e. ):
It is then straightforward to show (see Appendix B) that is finitely satisfiable iff has a solution over . Noting that the problem of finding solutions to Diophantine equations over is -complete, and that is an -sentence that is satisfiable if and only if is finitely satisfiable, we conclude the following:
Theorem 11.
The finite satisfiability problem for is -hard. If periodic counting is permitted, then so is the general satisfiability problem.
We note that one can use the same type of argument as above when reducing from the problem of solving Diophantine equations over to the general satisfiability problem of . Such an approach, however, is not fruitful for determining undecidability as the problem of finding solutions to over is in NPTime [14]. Thus, to show undecidability of general satisfiability, we resort to the tiling problem. Take with . We will produce and -sentence that is satisfiable if and only if tiles the infinite plane in accordance to the horizontal () and vertical () constraints. Additionally, we will argue that one can append additional conjuncts to and thus obtain a reduction from the tiling problem with a designated tile recurring infinitely often on the first column (see [10] for details about the problem). All-in-all, such reductions will guarantee that is -hard, whilst making hard for .
Take to be unary and , to be binary predicate symbols. We define the canonical -grid to be a -structure over the domain with the following extensions:
-
,
-
, and
-
.
We say that a structure is a -grid if restricted to elements and the signature is isomorphic to the canonical -grid. More leniently, is grid-like if it contains a homomorphic embedding of . It is well known that the satisfiability problem posed over subclasses of grid-like structures is undecidable for even inexpressive logics such as . The following is almost immediate:
Lemma 12.
The satisfiability problem for posed over subclasses of grid-like structures is -hard. The satisfiability problem for posed over subclasses of -grids is -hard.
The lemma above is, of course, the “easy” part of a much larger reduction. The axiomatisation of grid-like structures and -grids is where the expressive power of and is needed. Before writing the advertised formulas, we build the motivating structure we will be looking for in three steps. Suppose is the canonical -grid. Letting , be binary and be unary predicate symbols we define the graphed expansion of to be the structure over the domain with the following extensions:
-
,
-
for each ,
-
if and only if ,
-
, and
-
.
Intuitively, , when restricted to , is the canonical -grid. Notice that each has elements in that are -successors and elements that are -successors. In other words, the coordinates of are explicitly encoded in as the out-degrees of and respectively. (In the future, we will simply speak of - and -degree with “out” being left implicit). We invite the reader to regard as the set of extra elements which help encode positions of grid elements. Notice that the singleton is not featured in any binary relations (most notably, and ). This is deliberate as it will act as a spare part in the constructions to come.
We now define the mapped expansion of , where itself is the graphed expansion of . For this, we introduce quaternary , , , and ternary predicates , , whilst setting the following extensions:
-
-
-
-
-
and
-
.
Recall that each sends -edges to each . Fixing some with , we have that injectively maps -edges originating from to -edges of . On the other hand, is a bijection between the -edges of with the spare part and -edges of . The relation simply remembers which -edges of are mapped to -edges of via . Relations , and act similarly.
Lastly, We say that is the ordered expansion of , where itself is the graphed and mapped expansion of , if the signature contains two additional relations and which we will define to be total orders over . For motivational purposes, we will forget that grid elements and are pairs of natural numbers and instead focus on the - and -degrees of the elements. In we will have that
-
if and only if the -degree of is no more than that of , and
-
if and only if the -degree of is no more than that of .
(Again, the - and -degrees encode the horizontal and vertical positions of the element).
We will now write the sentence one conjunct at a time. At a high level, the conjuncts simply state facts about the graphed mapped and ordered expansion of . In the sequel we will argue that the satisfaction of by is sufficient to deduce that the structure in question is grid-like. For readability, we will be using variable sequences and instead of . Strictly speaking, the formulas to be defined are not (reverse) fluted, but can be made such by moving quantifiers inwards.
Fix to be as described above. We first capture some graphed properties. Recall that in there is a single spare part element . Noting that this element is not part of the grid () we have that models:
() |
Additionally, recall that the spare part has no incoming edges - or -edges in . Thus, also models:
() |
Moving to grid elements, we see that there is a single element in with no - or -degree. Thus, is a model of:
() |
Additionally, each element in has a single - and -successor. Thus, models:
() |
For the next two conjuncts fix . Notice that the -successor has an -degree that is larger by 1 when compared to its predecessor . We can thus map the -edges from to the set of -edges from taken together with the spare part element bijectively. This is exactly how the extension to in is set up. Noting that -successors have analogous properties we conclude that models the following sentences:
() | |||
() |
Recall that for grid elements we have if and only if the -degree of is no more than that of . Fixing and its -successor we see that . That is, and its -successor have the same -degrees. Thus, models:
() |
Now, recall that the ordering () is total on . Thus, models:
() |
Taking with recall that the elements have -edges mapped injectively by . Thus, if and only if is an injection between the -edges of and that of . We capture the “only-if” direction of the dependency with the sentences and , whilst the “if” direction is handled by and .
Still holding the supposition that , we write a sentence ensuring that each -edge from is mapped to some single -edge from :
() |
With the next sentence we require that each -edge from is a witness (in regard to ) to at most a single -edge from :
() |
It is easy to verify that this is indeed how is set up. Noting that and behave symmetrically we conclude .
For the converse direction of the implication take any and recall that if and only if there is some for which . In other words, remembers which -edges from are featured in a mapping (by ) with -edges from . We axiomatise this relationship as follows:
() |
Utilising we can then test if each -edge of is mapped to some -edge of and, if that is indeed the case, require that the grid elements be related via accordingly. We do just that with :
() |
Noting that , and behave similarly, we have that .
Lastly, notice that there are no two grid elements that have the same - and -degrees. Thus, models the uniqueness requirement as given by :
() |
Notice that by stepping inside the realm of periodic counting, we may capture the fact that in there are no transfinite positions by defining the sentence limiting the - and -degrees of elements to finite values:
() |
Recalling that we have showed the following:
Lemma 13.
The graphed, mapped and ordered expansion of the canonical -grid is a model of .
We proceed with the other direction as follows:
Lemma 14.
Suppose . Then is a grid-like structure. In addition, if , then is an -grid.
Proof.
Suppose first that . Notice that by there is exactly one element that satisfies in and, by , has no incoming - and -edges. This will be our spare part element in the argument to come. Now, take any element such that (i.e. is a grid element) with finite - and -degree. Such an element is guaranteed to exist by . Then, gives us that has an -successor and a -successor . Notice that, by each -edge originating from along with the spare part is paired with exactly one edge -edge from in . That is to say, writing and , we have that for each there is exactly one such that . The reverse is established by . Clearly, there is a bijection between and thus making the -degree of one greater than that of . By we have that and . We first fixate on the fact that . Writing and we have, by , that for each there is exactly one such that . By , for each there is at most a single such that . We may thus regard as being an injection between -edges of and that of . Then, again by , and the fact that , we have that is an injection between the -edges from and that of . By the Cantor-Schröder-Bernstein Theorem, their -degrees are thus equal. A symmetric argument holds for the - and -degree of .
Now, let and be, respectively, the -sucessor of and the -sucessor of promised by . Using the same arguments as in the paragraph above, it is easy to see that the -degrees of and coincide; and so do the -degrees. We claim that . By we need only show that is equal to with respect to the orderings and as we already have that and by . Fixating on -edges first, we have, by , that and are comparable by in some way. Suppose, without loss of generality, that . Writing and we have, by , that for each there is exactly one such that , and, by , for each there is at most one such that the same holds. That is to say, is an injection between -edges originating from and -edges from . Notice that since and both have an equal and finite -degree, we can conclude that is a bijection between the edges. Using this, we have that for each by . Clearly, the antecedents of are met and thus as required. Repeating the argument for we indeed have (by ) that thus closing the grid.
By repeating the argument above on element in with finite - and -degree we conclude that contains a homomorphic embedding of the canonical -grid thus making it grid-like.
Supposing, in addition, that we have that each element in has a finite - and -degree. We may thus unambiguously identify these elements as the pair of their -degree and -degree . Hence, the structure restricted to elements in and signature is isomorphic to the canonical -grid thus making an -grid as required.
Combining Lemmas 13 and 14 we have that is a satisfiable -sentence modeled exclusively by (some non-empty subclass of) grid-like structures, whilst is a satisfiable -sentence that is modeled only by (some non-empty subclass of) -grids. Combining the observation above with Lemma 12 we have the following:
Theorem 15.
The satisfiability problem for is -hard. The same problem for is -hard.
Note that the adjacent fragment with periodic counting is a fragment of the constructive fragment of , which has a -complete satisfiability problem [10, 15]. We conclude the section by reformulating results of Theorems 11 and 15 in terms of the adjacent fragment:
Corollary 16.
The finite and general satisfiability problems for the adjacent fragment with counting are, respectively, - and -complete. If periodic counting is permitted, then the general satisfiability problem turns to be -complete.
6 Discussion
In this paper we utilised the homogeneity property of satisfiable -sentences to establish a decision procedure for the (finite) satisfiability problem of the new language. With this methodology we not only gained a better understanding of models of fluted formulas, but also managed to establish decidability of (finite) satisfiability using simpler methods when compared to Presburger quantifiers discussed in previous literature [19, 24].
Reflecting on global homogeneity we see that, as opposed to local homogeneity, the rewiring in Lemma 2 is impacted by the presence of the (in)equality atom. More precisely, the semantics of predicates of arity at most do not interfere in the rewiring for local -homogeneity. Because of this we may establish a more general result for local homogeneity of fluted sentences with semantic extensions. Consider a signature that is split into symbols with a fixed interpretation (e.g. transitive relation, reversed relation, etc.) and standard predicate symbols with no fixed meaning. Furthermore, suppose that the maximum arity of any symbol in is at most . Then, Lemma 6 implies the following:
Corollary 17.
Suppose is a fluted, normal-form, -variable sentence (possibly with periodic counting) over the signature , and where . Then if is satisfiable it is satisfiable in a locally -homogeneous model.
The same cannot be said about -variable sentences when , and thus establishing an analogue to global homogeneity (as was done for with equality in Lemma 2) requires case-by-case consideration. Nonetheless, we believe our approach could not only be used to simplify existing decidability procedures for satisfiability (e.g. for with a transitive relation and counting [24]) but to also expand on expressiveness of fluted languages.
We make use of Corollary 17 when (briefly) analysing the language formed by combining and . That is to say, is with -atoms and allowed. The following is almost immediate. (See Appendix C for the proof).
Theorem 18.
The (finite) satisfiability problem for is decidable.
Combining the result above with Corollary 16, and noting that more expressive counting quantifiers render the two-variable fragment undecidable [7, 5], one can argue that the language is on the edge of decidability (for satisfiability). There are, however, other maximal fragments with counting that have a decidable satisfiability problem. The reader might have noticed that, in Section 5, we did not consider the general satisfiability problem for the 3-variable adjacent fragment with counting (whilst noting that the finite variant is undecidable in Theorem 11). Surprisingly, the problem was recently shown to be in by the current author. The details, however, are beyond the scope of this article. We conclude the paper by outlining the following problems which, to the best of our knowledge, are open.
-
1.
What is the complexity of satisfiability for the 3-variable adjacent fragment with counting?
-
2.
Is the satisfiability problem for the adjacent fragment with periodic counting -hard?
-
3.
Is the satisfiability problem for the guarded adjacent fragment with counting decidable?
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Appendix A Preliminaries
Lemma 1.
Suppose is an -sentence. Then, we may compute, in polynomial time, an equisatisfiable normal-form -sentence .
Proof.
We start by assuming that contains no universal quantifiers. No loss of generality follows this supposition as every formula of the form is equivalent to . Writing , take any subformula of , where is quantifier-free. Supposing there are free variables in , let be a fresh predicate of arity . We write as and define to be but with replaced by . Clearly, . Conversely, if , we may expand to by setting if for each . Then, as required. Processing and subsequent sentences in the same way, we are left with a sentence composed solely of proposition letters and sentences . The conjunction of the aforementioned sentences is then (after rearrangement) of the required form.
Appendix B Counting With Reversed Relations
Claim.
Suppose is an instance of Hilbert’s 10th problem and is computed as described as above Theorem 11. Then, is finitely satisfiable if and only if has a solution over .
Proof.
Suppose . We claim that is a satisfying assignment for . Thus, again taking in turn, we have that if is of the form (i) , then . If takes the form (ii) , we then claim that is a bijection between and having cardinality . Indeed, by the first conjunct of we have that each element in is paired with a single element in ; the converse is establishes by the second conjunct. Thus, clearly, . But since we have that . This coupled with our initial assumption that gives us as required. Lastly, suppose takes the form (iii) . By the first conjunct of for each there is a single such that . Hence, gives rise to a function Thus, writing in place of , we claim that is a bijection between and and has cardinality . It is easily seen that is surjective from either the second or third conjunct of . To establish injectivity suppose . But, by the second conjunct of , we have that thus . Notice that this establishes that is the only element in for which, under the assignment , , Combining this fact with the third conjunct of we have that, again under the assignment , , thus making . We finish our argument by noting that as required.
Conversely, suppose has a solution. We define to be the satisfying assignment for and construct a model of as follows. For each variable set be a set of distinct elements and set the domain of to be where each for . Clearly, . If was constructed from of the form (i) , then as required by . On the other hand, if is of the form (ii) , we have that by setting to be a bijection between and (this can be done as and, by initial assumption, ). Lastly, if is (iii) , then . Thus, index elements of as , elements of as and elements of as . Clearly, by setting we have that thus concluding the proof.
Appendix C Discussion
Theorem 18.
The (finite) satisfiability problem for (i.e. the combination of and ) is decidable.
Proof.
Let be the -variable sub-fragment of . Taking some sentence we proceed by induction on the number of variables. If , then is a sentence in the two-variable fragment with periodic counting which is known to have a decidable (finite) satisfiability problem [6]. Now, set and suppose that the (finite) satisfiability problem for is decidable. We may assume (by allowing to contain binary predicates of the form and ) that is in normal-form (5). Defining to be the set of predicates of arity no more than 2, we have that if is satisfiable then, by Corollary 17, it is satisfiable in an -homogeneous model. Applying the variable reduction outlined in Lemmas 7 and 8 we will then have the required result.