New Techniques for Universality in Unambiguous Register Automata
NNew Techniques for Universality in UnambiguousRegister Automata
Wojciech Czerwiński ! University of Warsaw
Antoine Mottet ! ˇ Department of Algebra, Faculty of Mathematics and Physics, Charles University in Prague
Karin Quaas ! University of Leipzig
Abstract
Register automata are finite automata equipped with a finite set of registers ranging over thedomain of some relational structure like ( N ; =) or ( Q ; < ). Register automata process words overthe domain, and along a run of the automaton, the registers can store data from the input wordfor later comparisons. It is long known that the universality problem, i.e., the problem to decidewhether a given register automaton accepts all words over the domain, is undecidable. Recently, weproved the problem to be decidable in 2-ExpSpace if the register automaton under study is over( N ; =) and unambiguous, i.e., every input word has at most one accepting run; this result was shortlyafter improved to 2-ExpTime by Barloy and Clemente. In this paper, we go one step further andprove that the problem is in ExpSpace, and in PSpace if the number of registers is fixed. Our proofis based on new techniques that additionally allow us to show that the problem is in PSpace forsingle-register automata over ( Q ; < ). As a third technical contribution we prove that the problem isdecidable (in ExpSpace) for a more expressive model of unambiguous register automata, where theregisters can take values nondeterministically, if defined over ( N ; =) and only one register is used. Theory of computation → Automata over infinite objects
Keywords and phrases
Register Automata, Data Languages, Unambiguity, Unambiguous, Univer-sality, Containment, Language Inclusion, Equivalence
Funding
Wojciech Czerwiński : Supported by the European Research Council (ERC) grant LIPA,grant agreement No 683080.
Antoine Mottet : This author has received funding from the ERC under the European Union’sHorizon 2020 research and innovation programme (grant agreement No 771005).
Karin Quaas : Funded by the Deutsche Forschungsgemeinschaft (DFG), project 406907430.
Acknowledgements
We thank Lorenzo Clemente, Sławomir Lasota, and Radosław Piórkowski forinspiring discussions on URA.
Certainly, determinism plays a central role in the research about computation models. Re-cently, a lot of active research work [1, 5, 2, 15, 13] is devoted to its weaker form: unambiguity .A system is unambiguous if for every input word there is at most one accepting run. Un-ambiguous systems exhibit elegant properties; in particular many natural computationalproblems turn out to be easier compared to the general case. A prominent example isthe universality problem for finite automata, i.e. , the problem of deciding whether a givenautomaton accepts every input word. It is in PTime [17] and even in NC [18] in theunambiguous case, as opposed to PSpace-completeness in the general case.In his seminal overview article about unambiguity, Colcombet [4] states some very naturalconjectures about unambiguous systems that are so fundamental that one can be surprisedthat they are still open. An example conjecture, motivated by the fact that the universality a r X i v : . [ c s . F L ] F e b New Techniques for Universality in Unambiguous Register Automata problem for unambiguous finite automata is in PTime, was that for every unambiguous finiteautomaton the complement of its language can be accepted by another unambiguous finiteautomaton with at most polynomial size with respect to the size of the original automaton.This conjecture was surprisingly resolved negatively by Raskin [16], who provided a familyof automata where a blowup Θ( n log log log( n ))) ) is unavoidable. Still, a lot of other naturalquestions remain unresolved. Some of them are not algorithmic (as the above one), whileothers ask for the existence of faster algorithms in the unambiguous case.Usually one cannot hope for designing more efficient algorithms for the emptiness problem,as it is often easy to transform a nondeterministic system to a deterministic (and thusunambiguous) system which has empty language if and only if the accepted language of theoriginal system is empty. Indeed, it is often sufficient to change the labelling of every transitionof the system to its unique transition name. This transformation preserves the emptinessproperty, but not much more. Therefore there is a hope that the unambiguity assumption mayresult in faster solving of problems like universality, equivalence and language containment.Recently there was a substantial amount of research in this area [10, 3, 6, 13, 5, 1]. Theconsidered problem is often the universality problem. Indeed, the universality problem isprobably the easiest nontrivial problem for which there is a hope to obtain an improvementin the unambiguous case. Equivalence and containment are often not much harder, eventhough sometimes a bit more involved techniques are needed.For register automata , this line of research was started in [13]. Register automata (RA,for short) extend finite automata with a finite set of registers that take values from an infinitedata domain for later comparisons. More detailed, RA are defined over a relational structure,like ( N ; =) or ( Q ; <, =); they process finite words over the domain of the relational structure,and the registers can store values from the input word for comparing them using the relationsprovided by the relational structure. In the more expressive model of register automata with guessing (GRA) the registers can even take arbitrary values. In [13] it is shown thatfor unambiguous RA (URA) over ( N ; =) the containment problem is in 2-ExpSpace and inExpSpace for a fixed number of registers. Without the unambiguity assumption, this problemis known to be much harder. Concretely, the universality problem is undecidable as soon asthe automaton uses two registers [11, 14, 8], and Ackermann-complete in the one-registercase [9]. In the case of GRA even the one-register case is undecidable. The result for URAin [13] was improved by Barloy and Clemente [1] who have shown that the problem is in2-ExpTime and in ExpTime for a fixed number of registers, using very different tools such aslinear recursive sequences in two dimensions.
Our contribution
Our result improves statements of Barloy and Clemente [1] even further. We provide threeresults shown by two different techniques. Our first technique is to show that in some casesone can assume that only a linear or exponential number of different configurations canbe reached via an input word. This claim immediately provides an improved upper boundcompared to [1]. ▶ Theorem 1.
The containment problem L ( A ) ⊆ L ( B ) is in ExpSpace , if A is an RA and A proof for undecidability can be done using a reduction from the undecidable reachability problem forMinsky machines, following the lines of the proof of Theorem 5.2 in [7]. The nondeterministic guessingcan be used to express that there exists some decrement for which there is no matching precedingincrement. . Czerwiński, A. Mottet, K. Quaas 3 B is a URA over ( N ; =) . The containment problem is in PSpace on inputs A , B both havinga bounded number of registers. This approach can also be applied to unambiguous one-register automata over ( Q ; <, =). ▶ Theorem 2.
The universality problem for one-register
URA over ( Q ; <, =) is in PSpace . However, we will see that the techniques for URA do not work for unambiguous GRA(GURA), not even in the one-register case. In that case we solve the universality problem,and even the containment problem, with the use of more sophisticated analysis. In short, weshow that we can modify the set of reachable configurations such that it becomes small andequivalent in some sense, which also allows us to obtain a more efficient algorithm. ▶ Theorem 3.
The containment problem L ( A ) ⊆ L ( B ) is in ExpSpace , if A is a GRA over ( N ; =) and B is a one-register GURA over ( N ; =) . We recently learned that, independently from our work, Bojańczyk, Klin and Moermanclaim a yet unpublished result about orbit-finite vector spaces, which implies an ExpTimealgorithm for GURA and PSpace complexity if the number of registers is fixed. However, webelieve that our contribution does not only provide an improved complexity of the consideredproblem, but also techniques that can be useful in future research on unambiguous systems.
In this section, we define register automata , introduced by Kaminski et al [11, 12]. We startwith some basic notions used throughout the paper. We use Σ to denote a finite alphabet,and N and Q denote the set of non-negative integers and rational numbers, respectively.Given a, b ∈ N with a ≤ b , we write [ a, b ] to denote the set { a, a + 1 , . . . , b } .A relational structure is a tuple D = ( D ; R , . . . , R k ), where D is an infinite domain, and R , . . . , R k are binary relations over D , and we assume that R k is the equality relation. Inthis paper, we are mainly interested in the relational structures ( N ; =) of the non-negativeintegers with equality, and ( Q ; <, =) of the rationals with the usual order and equalityrelations.A data word is a finite sequence ( σ , d ) . . . ( σ k , d k ) ∈ (Σ × D ) ∗ . If Σ = { σ } is a singletonset, we may write d · d · . . . · d k shortly for ( σ, d )( σ, d ) . . . ( σ, d k ). We use ε to denote theempty data word. A data language is a set of data words. We use data( w ) to denote the set { d , . . . , d k } of all data occurring in w .Let D ⊥ denote the set D ∪ {⊥} , where ⊥ ̸∈ D . We let ⊥ ̸ = d for all d ∈ D , and ⊥ is incomparable with respect to ≤ to all d ∈ D . We use boldface lower-case letters like a , b , . . . , u . . . to denote tuples in D n ⊥ , where n ∈ N . Given a tuple a ∈ D n ⊥ , we write a i forits i -th component, and data( a ) denotes the set { a , . . . , a n } ⊆ D ⊥ of all data occurring in a . Let R = { r , . . . , r n } be a finite set of registers . A register valuation is a mapping u : R → D ⊥ ; we may write u i as shorthand for u ( r i ). Let D R⊥ denote the set of all registervaluations. A register constraint over D and R is defined by the grammar ϕ ::= true | R ( t , t ) | ¬ ϕ | ϕ ∧ ϕ where R is a binary relation symbol from the relational structure D , and t i ∈ { } ∪ { r, ˙ r | r ∈ R} . Here r refers to the currentvalue of the register r , and ˙ r refers to the future value of the register r . We use Φ( D , R ) New Techniques for Universality in Unambiguous Register Automata to denote the set of all register constraints over D and R . The satisfaction relation | = on D R⊥ × D × D R⊥ is defined by structural induction as follows. We only give some atomic cases;the other cases can be derived easily. We have ( u , d, v ) | = ϕ if ϕ is of the form true , ϕ is of the form R ( r i , D | = R ( u i , d ), ϕ is of the form R ( ˙ r i , r i ) and D | = R ( v i , u i ), ϕ is of the form R ( ˙ r i , D | = R ( v i , d ).For example, ϕ := ¬ ( r = ∧ ( ˙ r = r ) is a register constraint over ( N ; =) and R = { r } , andwe have (1 , , | = ϕ , whereas (1 , , ̸| = ϕ .It is important to note that only register constraints of the form ˙ r = r and ˙ r = r . In absence of such a register constraint, the register r cannondeterministically take any of infinitely many data values from D , with the followingrestrictions: the register constraint ¬ ( ˙ r = r is differentfrom the current input datum, so that r may take any datum in D except for the inputdatum. Likewise, the register constraint ¬ ( ˙ r = r ) requires that r takes any datum in D except for the current value of r . Register automata that allow for such nondeterministic guessing of future register values are also called register automata with guessing . Formally, aregister automaton with guessing (GRA) over D and Σ is a tuple A = ( R , L , ℓ init , L acc , E ),where R is a finite set of registers, L is a finite set of locations, ℓ init ∈ L is the initial location, L acc ⊆ L is the set of accepting locations, E ⊆ L × Σ × Φ( D , R ) × L is a finite set of edges.If every edge of A contains some constraint of the form ˙ r = r or ˙ r = r ∈ R ,so that the future value of every register is uniquely determined, then we simply speak of register automata (RA, for short), i.e. , register automata without guessing. If the numberof registers of a GRA (RA, respectively) is fixed to k ∈ N , then we speak of k -GRA ( k -RA,respectively).A state of A is a pair ( ℓ, u ) ∈ L × D R⊥ , where ℓ is the current location and u is the currentregister valuation. Abusing notation a bit, we usually write ℓ ( u ) instead of ( ℓ, u ). The state ℓ init ( u init ), where u init maps every register r ∈ R to ⊥ , is called the initial state , and astate ℓ ( u ) is called accepting if ℓ ∈ L acc . Given two states ℓ ( u ) and ℓ ′ ( u ′ ) and some inputletter ( σ, d ) ∈ Σ × D , we postulate a transition ℓ ( u ) σ,d −−→ A ℓ ′ ( u ′ ) if there exists some edge( ℓ, σ, ϕ, ℓ ′ ) ∈ E such that ( u , d, u ′ ) | = ϕ . A run of A on the data word ( σ , d ) . . . ( σ k , d k ) is asequence ℓ ( u ) σ ,d −−−→ A ℓ ( u ) σ ,d −−−→ A . . . σ k ,d k −−−→ A ℓ k ( u k ) of such transitions. We say thata run as above starts in ℓ ( u ); similarly, the run ends in ℓ k ( u k ). A state ℓ ( u ) is reachable in A if there exists a run that ends in ℓ ( u ). A run is initialized if it starts in the initial state,and a run is accepting if it ends in some accepting state. A data word w is accepted from ℓ ( u ) if there exists an accepting run on w that starts in ℓ ( u ). The data language accepted by A , denoted by L ( A ), is the set of data words that are accepted from the initial state.A GRA is unambiguous if for every input data word w there is at most one initializedaccepting run. Note that unambiguity is a semantic condition; it can be checked in polynomialtime [4]. We write GURA and URA to denote unambiguous GRA and RA, respectively. ▶ Example 4.
Let us study the behaviour of the 1-GRA depicted in Figure 1. The GRA isover ( N ; =) and the singleton alphabet Σ = { σ } (we omit the letter σ from all transitionsin the figure). Suppose the first input letter is d . In order to satisfy the constraint of . Czerwiński, A. Mottet, K. Quaas 5 ℓ ℓ ℓ ¬ ( ˙ r = r = rr = r = r ¬ ( r = Figure 1
A 1-GURA ∗ undecidable [7] in ExpSpace (Th. 7) in PSpace (Th. 7) ∗ undecidable [12]1 undecidable in ExpSpace (Th. 3) Table 1
Universality over ( N ; =) the transition from ℓ to ℓ , the automaton has to nondeterministically guess some datum d ′ ̸ = d and store it into its register r . Being in the state ℓ ( d ′ ), the automaton can onlymove to the accepting location ℓ if the next input datum is equal to d ′ (indicated by theconstraint r = d ); for every other input letter, the automaton satisfies the constraint ¬ ( r = d )and stays in ℓ , and it keeps the register value to satisfy the constraint ˙ r = r . In this way,the automaton accepts the language { d · . . . · d k | ∀ k ≥ ∀ ≤ i < k. d i ̸ = d k } . Notethat the automaton is unambiguous: for every input data word there is only one acceptingrun. We remark that the accepted data language cannot be accepted by any RA (withoutguessing) [12]. Hence, GRA are more expressive than RA.In this paper, we study the universality problem : given a GRA A , is A universal, i.e. ,does L ( A ) = (Σ × D ) ∗ hold? In Table 1, we give an overview of the decidability status forregister automata over ( N ; =), in bold the new results for unambiguous register automatathat we present in this paper. For many computational models, a standard approach for solving the universality problem isto explore the (potentially infinite) state space of the automaton under study. Starting fromthe initial state, the basic idea is to input one letter after the other, and keep track of the sets of states that are reached, building a reachability graph whose nodes are the reachedsets of states (per input letter). The key property of this state space is that it containssufficient information to decide whether the automaton under study is universal: this is thecase if, and only if, every node of the graph contains an accepting state. Let us formalizethis intuition for register automata.Fix a k -GRA A = ( R , L , ℓ init , L acc , E ) over D and Σ, for some k ∈ N . A configuration of A is a subset of L × D k ⊥ . The set C init , denoting the singleton set containing the initial stateof A , is a configuration, henceforth called the initial configuration . Let C be a configuration,and let ( σ, d ) ∈ (Σ × D ). We use Succ A ( C, ( σ, d )) to denote the successor of C on the input ( σ, d ), formally defined bySucc A ( C, ( σ, d )) := { ℓ ( u ) | ∃ ℓ ′ ( u ′ ) ∈ C ℓ ′ ( u ′ ) σ,d −−→ A ℓ ( u ) } . In order to extend this definition to data words, we define inductively Succ A ( C, ε ) := C and Succ A ( C, w · ( σ, d )) := Succ A (Succ A ( C, w ) , ( σ, d )). We say that a configuration C is reachable in A by the data word w if C = Succ A ( C init , w ); we say that C is reachable in A if there exists some data word w such that C is reachable in A by w . We say that C is coverable if there exists some C ′ ⊇ C such that C ′ is reachable in A . Given a configuration C , we use data( C ) to denote the set { d i ∈ D k ⊥ | ∃ ℓ ∈ L , ≤ i ≤ k ℓ ( d , . . . , d k ) ∈ C } of dataoccurring in C . Notice that every configuration reachable in an RA (without guessing) isnecessarily finite. In contrast, the configuration { ℓ ( d ′ ) | d ′ ∈ N , d ′ ̸ = d } is reachable in the New Techniques for Universality in Unambiguous Register Automata
GRA in Figure 1 by the single-letter data word ( σ, d ). If C = { ℓ ( u ) } is a singleton set, thenwe may, in slight abuse of notation, omit the curly brackets and write ℓ ( u ).We say that a configuration C is accepting if there exists ℓ ( u ) ∈ C such that ℓ ∈ L acc ;otherwise we say that C is non-accepting . Clearly, A is universal if, and only if, everyconfiguration reachable in A is accepting. This suggests to reduce the universality problem toa reachability problem for the state space corresponding to the given input GRA. However,the state space of a GRA is infinite, in two different aspects.First of all, the state space is infinitely branching , as each of the infinite data in D may giverise to a unique successor configuration. The standard approach for solving this complicationis to abstract from concrete data, using the simple observation that, e.g. , the data word 3 · ℓ (4) if, and only if, 5 · ℓ (2). This isformalized in the following.A partial isomorphism of D ⊥ is an injective mapping π : D → D ⊥ with domain dom( π ) := D ⊆ D such that if ⊥ ∈ D then π ( ⊥ ) = ⊥ . Let π be a partial isomorphism of D ⊥ and let C be a configuration such that data( C ) ⊆ dom( π ). We define the configuration π ( C ) := { ℓ ( π ( d ) , . . . , π ( d k ))) | ℓ ( d , . . . , d k ) ∈ C } ; likewise, if { d , . . . , d k } ⊆ dom( π ), we define thedata word π ( w ) = ( σ , π ( d )) . . . ( σ k , π ( d k )). We say that two pairs ⟨ C, w ⟩ and ⟨ C ′ , w ′ ⟩ areequivalent with respect to π , written ⟨ C, w ⟩ ∼ π ⟨ C ′ , w ′ ⟩ , if π ( C ) = C ′ and π ( w ) = w ′ . If w = w ′ = ε , we may write C ∼ π C ′ . We write ⟨ C, w ⟩ ∼ ⟨ C ′ , w ′ ⟩ if ⟨ C, w ⟩ ∼ π ⟨ C ′ , w ′ ⟩ forsome partial isomorphism π of D ⊥ . ▶ Proposition 5.
Let A be a GRA . If ⟨ C, w ⟩ ∼ ⟨ C ′ , w ′ ⟩ , then Succ A ( C, w ) ∼ Succ A ( C ′ , w ′ ) . Secondly, there can be infinitely many reachable configurations even up to the equivalencerelation ∼ . As an example, consider the GURA in Figure 1. For every n ≥
1, the configuration C n := { ℓ ( d ′ ) | d ′ ∈ N \{ d , . . . , d n }} ∪ { ℓ ( d n ) } with pairwise distinct data values d , . . . , d n is reachable by the data word d · d · . . . · d n , and C n ̸∼ C n ′ for n ̸ = n ′ . There are similarexamples also for URA, cf. [13].In order to obtain our results, we will prove that one can solve the reachability problemfor the state space of A by focussing on a subset of configurations reachable in the automatonunder study. The concrete methods are different for URA and GURA, however, for bothmodels we will take advantage of Proposition 5 and its simple consequence (cf. [13]). ▶ Corollary 6.
Let A be a GRA . If ⟨ C, w ⟩ ∼ ⟨ C ′ , w ′ ⟩ and Succ A ( C, w ) is non-accepting(accepting, respectively), then Succ A ( C ′ , w ′ ) is non-accepting (accepting, respectively). ( N ; =) In this section, we study the complexity of the universality problem for URA over therelational structure ( N ; =). We prove the following theorem. ▶ Theorem 7.
The universality problem isin
PSpace for k - URA for any fixed k ∈ N ,in ExpSpace for
URA . We start by showing that we can assume URA to have a specific form that simplifies thecoming proofs. Given some k -URA A , we say that A is pruned if for every state ℓ ( u ) thatis reachable in A there exists a data word w that is accepted from ℓ ( u ), and u i ̸ = u j forall 1 ≤ i < j ≤ k , i.e. , no datum appears more than once in u . The proof of the followingproposition is simple and omitted. . Czerwiński, A. Mottet, K. Quaas 7 ▶ Proposition 8.
For every k - URA one can compute in polynomial time an equivalent pruned k - URA . In the following we always assume that a k -URA is pruned, even if we do not explicitlymention it. For simplicity, we also assume that the alphabet of the URA we consider aresingletons. The techniques we develop can be easily lifted to the more general case where Σis not a singleton.We introduce some constants that bound from above the number of states with the samelocation occurring in a configuration reachable in a universal URA. Let A be a k -URA. Fora configuration C of A , define M C ∈ N to be the maximal number M such that in C thereare M different states with the same location. Define M A ∈ N ∪ {∞} to be the supremum of M C , for C ranging over all the configurations C reachable in A , if A is a universal k -URA, i.e. , L ( A ) = D ∗ . In the sequel, we show that M A < ∞ . In order to do so, for k ∈ N , define M k ∈ N ∪ {∞} to be the supremum of all the M A , for A ranging over pruned and universal k -URA. The main technical result of this section is showing that M k is finite and moreoverupper-bounded by an exponential function of k .Let n be the number of locations of A . First observe that showing M k ∈ N easily impliesthe existence of a NPSpace algorithm deciding whether A is universal. Indeed, if M k < ∞ ,then every configuration C reachable in A has size at most n · M k , as otherwise C containsmore than M k states with the same location. Thus, in order to decide whether A is notuniversal, we can apply the following algorithm:By Corollary 6, A is not universal iff A does not accept some data word ( σ , d )( σ , d ) . . . ,where d i ∈ { , . . . , i } for all i .Guess, letter by letter, an input data word ( σ , d )( σ , d ) . . . , where d i ∈ { , . . . , i } .For each i ≥
1, define C i := Succ A ( C i − , ( σ i , d i )), where C = C init .If for some i ≥
1, the configuration C i is not accepting or its size exceeds n · M k , weknow that A is not universal.Otherwise we keep the configuration in the space linear with respect to n and count thelength of the word. If the length exceeds the number of possible configurations, then thisrun is not accepting. The length counter can be also kept in linear space.The above is hence a PSpace-algorithm for deciding non-universality for k -URA. By Savitch’stheorem, there also exists one for deciding universality for k -URA. Moreover, if we show that M k is exponential in k , then the above algorithm works in space exponential with respect to k , so is in ExpSpace even without fixing the number of registers k . Therefore, in order toshow Theorem 7, it is enough to prove that M k is bounded by some exponential function of k . The rest of this Section is devoted mainly to showing the following lemma. ▶ Lemma 9. M k ≤ ( k · k · k !) k . A short Ramsey argument given below shows that M k is finite for all k , however onlygiving a doubly-exponential bound. Before starting the proof, we remark that these techniquesalone cannot be used to lower the complexity of the universality problem for k -URA or forURA even more. This is because M k ≥ k !, which is the subject of the following lemma. ▶ Lemma 10. M k ≥ k ! . Proof.
We define a family of pruned universal k -URA ( A k ) k ≥ over Σ = { σ } such that M A k ≥ k !. Consider the following part of a pruned universal k -URA A k (shown for the case k = 3): New Techniques for Universality in Unambiguous Register Automata . . . ℓ r = r = r = . . . ℓ ′ ∈{ r ,r ,r } ∈{ r ,r ,r } / ∈ { r , r , r } / ∈{ r ,r ,r } / ∈ { r , r , r } The rest of the automaton makes sure that the configuration { ℓ ( u ) | u ∈ { , . . . , k } k is a permutation } ∪ { ℓ ′ (1 , . . . , k ) } is reachable in A k by the k -letter data word 1 · · . . . · k ( e.g. , each ℓ ( u ) is reached by apath storing the input data in a different order). By taking the (disjoint) union with anunambiguous automaton accepting every data word of length < k and every k letter wordthat has a repeated data value, we obtain a universal automaton. ◀ Our main tool to prove Lemma 9 is a structural observation, which delivers an understandingof how reachable configurations in universal k -URA can look like. Before diving into it wepresent an intuition by the following example. ▶ Example 11.
Let C be a configuration reachable in some universal 2-URA A over Σ = { σ } by some data word w , and assume that C contains three states ℓ (1 , ℓ (3 ,
4) and ℓ (5 , ℓ . We will argue that this is impossible. Assume that from ℓ (1 , · · · · ℓ (3 , · · ℓ (5 , · ·
7, where 8 and9 are fresh data values, that is, they do not occur in w . Since A is universal, the dataword 8 · · ℓ ′ , d , d ) in C . The set { d , d } has onlytwo elements, and so the intersection with at least one of the sets { , } , { , } and { , } must be empty. For instance, assume that { d , d } ∩ { , } = ∅ and ( d , d ) = (3 , ⟨ ℓ ′ (3 , , · · ⟩ ∼ ⟨ ℓ ′ (3 , , · · ⟩ , so that by Corollary 6, the data word 1 · · ℓ ′ (3 , w · · · A . Below we generalise this reasoning, in particular tothe case where some registers in the reached states keep the same value ( i.e. , not all aredifferent, as 1, 2, 3, 4, 5 and 6 in the above example). However the intuition stays the same.We say that a set of tuples T ⊆ D m ⊥ is m-full (or simply full if m is clear from the context)if there exists a set of indices I ⊆ [1 , m ] such that:all the tuples in T are identical in indices from I , namely for all i ∈ I and all t , t ′ ∈ T wehave t i = t ′ i ;all the data values occurring in tuples in T on indices outside I are different, namelyfor all i ̸∈ I , all j ∈ { , . . . , m } , and all t , t ′ ∈ T we have t i ̸ = t ′ j unless both t = t ′ and i = j . Note that in particular, this condition applies to the case t ′ = t , and thus t i ̸ = t j whenever i ̸∈ I and j ∈ { , . . . , m } are different.The following are examples of full sets:a 4-full set is the set of 4-tuples (1 , , , , (1 , , , , (1 , , , I = { , } ;a 5-full set containing one tuple (3 , , , , I ⊆ [1 ,
5] works here;a 2-full set containing tuples (2 , , (3 , , (4 , , (5 , I = { } .For a location ℓ and set of tuples T ⊆ D k ⊥ we write ℓ ( T ) = { ℓ ( t ) | t ∈ T } . The followinglemma delivers the key observation, which uses the notion of k -full sets. . Czerwiński, A. Mottet, K. Quaas 9 ▶ Lemma 12. If A is a pruned universal k - URA , then there exists no configuration C reachable in A such that ℓ ( T ) ⊆ C for some location ℓ ∈ L and some k-full set of tuples T ⊆ D k ⊥ of size more than k . Proof.
Let A be a pruned universal k -URA, and suppose towards contradiction that thereexists a configuration C reachable in A such that ℓ ( T ) ⊆ C for some location ℓ and some k -full set T ⊆ D k ⊥ of size more than k . Let w be the data word such that C = Succ A ( C init , w ).Assume without loss of generality that the indices on which tuples from T are identical are I = { , . . . , n } for some n ≤ k . Let us choose some k + 1 tuples from T , let the i -th of itbe of the form t i = ( c , . . . , c n , o i , . . . , o im ), where n + m = k . We call the c j the common data values and the o ij the own data values of t i . A is pruned and ℓ ( t ) is reachable in A , sothere must exist a data word w ∈ (Σ × D ) ∗ that is accepted from ℓ ( t ). Without loss ofgenerality we can assume that w does not contain the own data values of any of the othertuples t , . . . , t k +1 . Indeed, if this is the case, we can replace synchronously all occurrencesof such a data value by a fresh data value not occurring in data( w ); the resulting data wordis still accepted from ℓ ( t ). For every i ∈ [2 , k + 1], let w i be the word w in which foreach j ∈ [1 , m ] the own data value o j is replaced by the data value o ij . Clearly, for every i ∈ [2 , k + 1] ⟨ ℓ ( t ) , w ⟩ ∼ ⟨ ℓ ( t i ) , w i ⟩ , so that by Corollary 6 the data word w i is acceptedfrom ℓ ( t i ).Let us now consider the data word w fresh that is obtained from w by replacing synchron-ously every occurrence of every o j is by some fresh data value each. As A is universal, alsothe data word w fresh needs to be accepted from some state in C . Let q fresh = ℓ ′ ( e , . . . , e k )be the state in C from which w fresh is accepted. Notice that we do not enforce ℓ ̸ = ℓ ′ ,similarly e i may be equal to some of the c i or o ji , but this does not have an effect on ourreasoning. For each tuple t i = ( c , . . . , c n , o i , . . . , o im ), let the set of its own data values be O i = { o i , . . . , o im } . By assumption all the sets O , . . . , O k +1 are pairwise disjoint. As thereare k + 1 of them, we know that at least one of them is disjoint from the set of data valuesin the state q fresh , namely with E = { e , . . . , e k } . Assume without loss of generality that O ∩ E = ∅ . This however means that ⟨ q fresh , w ⟩ ∼ ⟨ q fresh , w fresh ⟩ , so that by Corollary 6 w is also accepted from q fresh . In consequence, there are at least two accepting runs over w from configuration C , one from ℓ ( t ) and one from q fresh . Hence there are at least twoinitialized accepting runs over w · w . This is a contradiction to the unambiguity of A . ◀ We give here the argument showing that M k is bounded by some doubly-exponentialfunction in k . We show this argument in order to illustrate the techniques, which neededto be refined in our proof of Lemma 9. First recall that the Ramsey number R m ( n ) is thesmallest number of vertices k of the graphs such that any clique of k vertices with its edgescoloured on m different colours contain a monochromatic subgraph G of n vertices, namelysuch that all the edges in G are of the same colour. It can be shown by induction that R m ( n )is finite, and indeed its growth is bounded by 2 n O ( m ) . For the definition of k -full set considerpage 8. ▶ Proposition 13.
Every set T ⊆ D k ⊥ of size at least R k +1 (4 k ( k + 1)! + 1) contains a k-fullsubset of size at least k + 1 . Proof.
Construct a graph with vertices being tuples from T and edge between t and t ′ becoloured by the number of data values that t and t ′ have in common. Clearly the colourbelongs to the set { , . . . , k } , so there are k + 1 colours. Because | S | ≥ R k +1 (4 k ( k + 1)! + 1)we know from Ramsey’s theorem that there are at least 4 k ( k + 1)! + 1 tuples such that everyintersection is of the same size - assume this size to be m . Let S be a set of 4 k ( k + 1)! + 1 such tuples and let s ∈ S be one of them. Let s = ( d , . . . , d k ). Divide all the other tuples s ′ into (cid:0) km (cid:1) · m ! sets depending on which m data values from { d , . . . , d k } belong to s ′ (thereare (cid:0) km (cid:1) options), on which positions they are located in s ′ (also (cid:0) km (cid:1) options) and in whichorder ( m ! options). It is easy to see that (cid:0) km (cid:1) · m ! ≤ k k !, as (cid:0) km (cid:1) ≤ k and m ! ≤ k !. Wedivide 4 k ( k + 1)! tuples (we omit s ) into at most 4 k k ! sets, so by the pigeonhole principleat least one of them contains at least k + 1 elements: let these elements be s , . . . , s k +1 .Notice now that the tuples s , . . . , s k +1 form a k -full set: indeed on positions on which theyhave the m shared data they are identical and on the other positions all the data values aretotally different. Thus T contains a k -full set of size k + 1, which finishes the proof. ◀ By refining the reasoning, we obtain the following result that directly implies Lemma 9,when setting B = n = k . ▶ Lemma 14.
Every set T ⊆ D n ⊥ of size at least ( B · n · n !) n + 1 contains an n -full subsetof size bigger than B . Proof.
Let us denote by D B,n the maximal size of the set of n -tuples such that any n -fullsubset has size at most B ; in other words, D B,n is the least integer such that if X is a set of n -tuples of size D B,n + 1, then X contains an n -full subset of size B + 1. Our aim is to showthat D B,n ≤ ( B · n · n !) n . We show it by induction on n .For the induction base assume n = 1. Then any set of data values is a full set, so clearly D B, ≤ B ≤ B · · D B,m ≤ ( B · m · m !) m for all m < n and consider some set T ⊆ D n ⊥ of n -tuples. Assume that T contains no n -full subset of size bigger than B . Pick some tuple t = ( d , . . . , d n ) ∈ T . We first show that there can be at most 4 n · n ! · D B,n − tuples in T whose data intersect data( t ). Let us denote N = 4 n · n ! · D B,n − . Let S be the set of thosetuples, assume towards contradiction that the size of S exceeds the bound N . For each tuple s ∈ S there are at most 2 n − s ) ∩ data( t ), so by the pigeonhole principlethere are more than 2 n · n ! · D B,n − tuples which have the same set data( s ) ∩ data( t ). Thosedata values can occur in tuples from S on at most 2 n different sets of indices, and in at most n ! different orders, so by the pigeonhole principle more than D B,n − tuples from S havethe same data values shared with t on the same indices. After ignoring the indices sharedwith t at most n − B among these tuples, which leads to the contradiction withassumption that for more than N tuples from T their data intersects data( t ).Therefore we know that all the tuples but the mentioned N ones have data disjointwith data( t ). Let use denote t = t and T to be the set of tuples with data disjoint fromdata( t ). Let t ∈ T . We now repeat the argument for t similarly as for t and get thatthere are at most N tuples with data intersecting data( t ). Repeating this argument we geta sequence of tuples t , t , . . . , t m such that for each i ̸ = j we have data( t i ) ∩ data( t j ) = ∅ .After adding each tuple t j to the sequence we define the set T j +1 of elements, which havedisjoint data with all the tuples t , . . . , t j . As long as T j +1 is nonempty we can continue theprocess. It is easy to see that | T j +1 | ≥ | T j | − N . Assume now towards contradiction that D B,n > ( B · n · n !) n , which implies that D B,n > ( B · n · n !) · D B,n − = B · N . We can seenow that | T B | >
0, which means that we can construct tuples t , t , . . . , t B , t B +1 such thatfor each i ̸ = j we have data( t i ) ∩ data( t j ) = ∅ . This however means that { t , . . . , t B +1 } is afull set of size B + 1, which is more than B . This contradicts the assumption, which showsthat D B,n ≤ ( B · n · n !) n and finishes the proof. ◀ . Czerwiński, A. Mottet, K. Quaas 11 We can now apply a reduction from containment to universality provided by Barloy andClemente (Lemma 8 in [1]) to obtain Theorem 1 from the introduction. ( Q ; <, =) In this section, we prove Theorem 2 by using the techniques developed in the precedingsection. Let us define constants M O k for k -URA with order similarly as M k for k -URA. Themain technical lemma is the following; Theorem 2 follows. ▶ Lemma 15. M O = 1 . Proof.
Towards contradiction suppose that for some pruned universal 1-URA A with orderthere is a configuration C reachable in A by a data word w pref ∈ (Σ × Q ) ∗ , such that ℓ ( d ) , ℓ ( d ) ∈ C for some location ℓ and data values d < d . Because A is pruned, thereexists a data word w ∈ (Σ × Q ) ∗ that is accepted from ℓ ( d ). Without loss of generalitywe can assume that w does not contain any data in ( d , d ]. Indeed, if w contains somedatum in ( d , d ], then we can replace it synchronously by some datum greater than d ,while taking care that that the relative order of all data in w is preserved, so that, for theresulting data word w , we have ⟨ ℓ ( d ) , w ⟩ ∼ ⟨ ℓ ( d ) , w ⟩ . By Corollary 6, the resulting dataword w is also accepted from ℓ ( d ). Notice that for similar reasons, also the data word w obtained from w by replacing every occurrence of d by d is accepted from ℓ ( d ). Now,if w does not contain d , then w = w . Hence w is accepted from both ℓ ( d ) and ℓ ( d ),contradiction to unambiguity of A . So let us assume w contains d . Pick some data value d fresh that is fresh, i.e. , it does not occur in w pref , and additionally d < d fresh < d . Weclearly can choose such a fresh data value, as there are infinitely many rational numbersbetween d and d and only finitely many of them occur in w pref . Let w fresh be the wordobtained from w by synchronously replacing every occurrence of d by d fresh . The word w fresh is accepted from some configuration in C , let it be ℓ ′ ( d ′ ). Notice now that if d ′ < d fresh ,then ⟨ ℓ ′ ( d ′ ) , w fresh ⟩ ∼ ⟨ ℓ ′ ( d ′ ) , w ⟩ , so that ℓ ′ ( d ′ ) accepts also w by Corollary 6; in the othercase, i.e. , if d ′ > d fresh , then we have ⟨ ℓ ′ ( d ′ ) , w fresh ⟩ ∼ ⟨ ℓ ′ ( d ′ ) , w ⟩ , so that ℓ ′ ( d ′ ) also accepts w . Therefore in the first case automaton A has two accepting runs over w pref · w and inthe second case over w pref · w . This is a contradiction to the unambiguity of A . ◀ The following lemma shows that our techniques by itself are not sufficient to solve thecase of 2-URA with order. ▶ Lemma 16. M O = ∞ . Proof.
For all n ≥
1, consider the configuration C n := { ℓ ′ (1 , n ) , ℓ (1 , , . . . , ℓ ( n − , n ) } ,which is for all n ≥ ℓℓ ′ ˙ r = r = r = r < , ˙ r = r > , ˙ r = r < r = r > r = r > ∨ r ≤ r ≤ < r r ≥ ∨ r ≤ r < < r ˙ r = r < < r ˙ r = ¬ ( r = ) The state ℓ ′ ( d , d ) keeps track of the first two distinct data read, with d < d . It isresponsible for accepting any datum d outside the interval [ d , d ). The state ℓ ( x, y ) issuch that d ≤ x < y ≤ d and it is responsible for accepting every datum d ′ in theinterval [ x, y ). Moreover, if d ′ ∈ ( x, y ), then ℓ ( x, y ) splits into ℓ ( x, d ′ ) and ℓ ( d ′ , y ). Theautomaton ensures that no two intervals [ x, y ) , [ x ′ , y ′ ) overlap, and that all the intervals [ x, y )present in a configuration cover the interval [ d , d ), thus the automaton is unambiguousand universal. ◀ ( N ; =) In this section, we aim to prove the decidability of the universality problem for the moreexpressive model of GURA. Let us first argue that the techniques developed in Section 4 donot work for GURA. ▶ Example 17.
One can easily construct a universal 1-GURA with reachable configuration C containing ℓ (0) , ℓ (1), and ℓ ′ × { n ∈ N | n ̸ = 0 , } . If from both ℓ and ℓ ′ there are outgoingedges with constraint r = n is acceptedfrom C . In particular, the word 0 is accepted from ℓ (0), but we cannot replace 0 by some fresh datum to obtain a contradiction as in Example 11.The example shows that we need more sophisticated methods to solve the universalityproblem. Moreover, and in contrast to the result for RA, we cannot rely on the reductionfrom containment to universality by Barloy and Clemente [1], as it holds for RA withoutguessing only. We hence present a direct proof for containment as stated in Theorem 3. Theidea is based on exploring a sufficiently big part of the infinite synchronized state space ofboth automata A and B , following the approach in [13]. The main difference with [13] lies inthe complications that arise due to the fact that a configuration of a GURA may be infinite . For the rest of this section, let A = ( R A , L A , ℓ A init , L A acc , E A ) be a GRA with R A = { r , . . . , r m } , and let B = ( R B , L B , ℓ B init , L B acc , E B ) be a GURA with a single register r .We aim to reduce the containment problem L ( A ) ⊆ L ( B ) to a reachability problem in( S , ⇒ ) where: S is the set of synchronized configurations ( ℓ ( d ) , C ), where ℓ ( d ) ∈ ( L A × N R A ⊥ ) is a singlestate of A , and C is a configuration of B ,( ℓ ( d ) , C ) ⇒ ( ℓ ′ ( d ′ ) , C ′ ) if there exists a letter ( σ, d ) ∈ (Σ × N ) such that ℓ ( d ) σ,d −−→ A ℓ ′ ( d ′ ),and Succ B ( C, ( σ, d )) = C ′ .We define S init := ( ℓ A init ( v init ) , C init ) to be the initial synchronized configuration of A and B . We say that a synchronized configuration S ′ is reachable from S if there is a ⇒ -pathfrom S to S ′ . S is reachable if it is reachable from S init . Call a synchronized configuration( ℓ ( d ) , C ) bad if ℓ ∈ L A acc is an accepting location and C is non-accepting, i.e. , ℓ ′ ̸∈ L B acc forall ( ℓ ′ , u ) ∈ C . Thus, a bad synchronized configuration is reachable iff L ( A ) ̸⊆ L ( B ).We extend the equivalence relation ∼ defined in Section 3 to synchronized configurationsin a natural manner, i.e. , given a partial isomorphism π of N ⊥ such that data( d ) ∪ data( C ) ⊆ dom( π ), we define ( ℓ ( d ) , C ) ∼ π ( ℓ ( d ′ ) , C ′ ) if π ( C ) = C ′ and π ( d ) = d ′ . We shortly write S ∼ S ′ if there exists a partial isomorphism π of N ⊥ such that S ∼ π S ′ . Clearly, an analogonof Corollary 6 holds for this extended relation. In particular, we have the following: ▶ Proposition 18.
Let
S, S ′ be two synchronized configurations of ( S , ⇒ ) such that S ∼ S ′ .If S reaches a bad synchronized configuration, so does S ′ . . Czerwiński, A. Mottet, K. Quaas 13 The support of a configuration C of B is the set supp( C ) of data d ′ such that at least oneof the following two conditions holds: ℓ ( d ′ ) ∈ C for some ℓ ∈ L such that ( { ℓ } × D ) ∩ C is finite, ℓ ( d ′ ) ̸∈ C for some ℓ ∈ L such that ( { ℓ } × D ) ∩ C is cofinite.Note that supp( C ) ⊆ data( w ) whenever C = Succ A ( C init , w ).Let S = ( ℓ ( d ) , C ) be a synchronized configuration, and let a, b ∈ supp( C ) be two datavalues in the support of C . We say that a and b are indistinguishable in S , written a ≡ S b ,if a, b ̸∈ data( d ) and { ℓ ∈ L | ℓ ( a ) ∈ C } = { ℓ ∈ L | ℓ ( b ) ∈ C } .Given a configuration C of B , we define for every datum d ∈ N the sets C + d := { ℓ ( d ) ∈ L × { d } | ℓ ( d ) ∈ C and data( C ∩ ( { ℓ } × N )) is finite } C − d := { ℓ ( d ) ∈ L × { d } | ℓ ( d ) ̸∈ C and data( C ∩ ( { ℓ } × N )) is infinite } . We give here an example for the definition of C + d and C − d . ▶ Example 19.
Let C = { ℓ (0) , ℓ (1) } ∪ { ℓ ( d ) | d ∈ N \{ , }} ∪ { ℓ ( d ) | d ∈ N \{ , }} .Then C +0 = { ℓ (0) } C +1 = { ℓ (1) } C +2 = ∅ C − = { ℓ (0) } C − = { ℓ (1) , ℓ (1) } C − = { ℓ (2) } We say that a configuration C is essentially coverable if for every two ℓ ( u ) , ℓ ′ ( u ′ ) ∈ C , theset { ℓ ( u ) , ℓ ′ ( u ′ ) } is coverable. ▶ Proposition 20.
Let C be an essentially coverable configuration, and let b ∈ supp( C ) .Then (( C \ C + b ) ∪ C − b ) is essentially coverable, too. Proof.
Let ℓ ( c ) , ℓ ′ ( c ′ ) ∈ (( C \ C + b ) ∪ C − b ). If ℓ ( c ) , ℓ ′ ( c ′ ) ∈ C \ C + b , then { ℓ ( c ) , ℓ ′ ( c ′ ) } is coverableby essential coverability of C . Suppose ℓ ( c ) , ℓ ′ ( c ′ ) ∈ C − b . By definition of C − b , c = c ′ = b .Pick some value e ∈ N \{ b } such that ℓ ( e ) , ℓ ′ ( e ) ∈ C . Note that such a value e must exist, asby definition of C − b , the sets data(( { ℓ } × N ) ∩ C ) and data(( { ℓ ′ } × N ) ∩ C ) are cofinite, andhence their intersection is non-empty. By essential coverability of C , { ℓ ( e ) , ℓ ′ ( e ) } is coverable.There must thus exist some data word w such that { ℓ ( e ) , ℓ ′ ( e ) } ⊆ Succ( ℓ init ( ⊥ ) , w ). Let π be any partial isomorphism satisfying π ( e ) = b and whose domain contains data( w ). Clearly, { ℓ ( b ) , ℓ ′ ( b ) } ⊆ Succ( ℓ init ( ⊥ ) , π ( w )), and hence { ℓ ( b ) , ℓ ′ ( b ) } is coverable. Finally, suppose ℓ ( c ) ∈ C \ C + b and ℓ ′ ( c ′ ) ∈ C − b . By definition, we have c ̸ = b = c ′ . Since data(( ℓ ′ × N ) ∩ C )is cofinite, there is d ̸ = c such that ℓ ′ ( d ) ∈ C . By essential coverability of C , there exists adata word w such that { ℓ ( c ) , ℓ ′ ( d ) } ⊆ Succ( ℓ init ( ⊥ ) , w ). By picking a partial isomorphism π such that π ( d ) = c ′ and π ( c ) = c , we obtain that { ℓ ( c ) , ℓ ′ ( c ′ ) } ⊆ Succ( ℓ init ( ⊥ ) , π ( w )), whichconcludes the proof. ◀ The following is the main technical result of this section. ▶ Proposition 21.
Let S = ( ℓ A ( d ) , C ) be a synchronized configuration of A and B such that C is essentially coverable, and let a ̸ = b be such that a, b ∈ supp( C ) and a ≡ S b . Then S reaches a bad configuration in ( S , ⇒ ) if, and only if, S ′ := ( ℓ A ( d ) , ( C \ C + b ) ∪ C − b ) reaches abad configuration in ( S , ⇒ ) . Proof. ( ⇐ ) Suppose there exists some data word w such that there exists an accepting runof A on w that starts in ℓ A ( d ), and Succ B ( C \ C + b ∪ C − b , w ) is non-accepting. We assume inthe following that Succ B ( C + b , w ) is accepting; otherwise we are done. Let ℓ + ( b ) ∈ C + b bethe unique state such that Succ B ( ℓ + ( b ) , w ) is accepting. In the following, we prove that we can without loss of generality assume that w does not contain any a ’s. Pick some a ′ ∈ N such that a ′ ̸∈ data( w ) ∪ supp( C ) ∪ data( d ). Let π be the isomorphism defined by π ( a ) = a ′ , π ( a ′ ) = a , and π ( d ) = d for all d ∈ N ⊥ \{ a, a ′ } . Then ⟨ ℓ A ( d ) , w ⟩ ∼ π ⟨ ℓ A ( d ) , π ( w ) ⟩ (as a ̸∈ data( d ) by a ≡ S b ), and ⟨ ℓ + ( b ) , w ⟩ ∼ π ⟨ ℓ + ( b ) , π ( w ) ⟩ . By Corollary 6, there exists anaccepting run of A on π ( w ) that starts in ℓ A ( d ), and Succ B ( ℓ + ( b ) , π ( w )) is accepting. Weprove that Succ B ( ℓ ( c ) , π ( w )) is non-accepting, for every ℓ ( c ) ∈ C \ { ℓ + ( b ) } ∪ C − b : first, let ℓ ( c ) ∈ C \ { ℓ + ( b ) } . By essential coverability of C , { ℓ + ( b ) , ℓ ( c ) } is coverable. By unambiguityof B , Succ B ( ℓ ( c ) , π ( w )) must be non-accepting. Second, let ℓ ( c ) ∈ C − b . But then c = b , andhence ⟨ ℓ ( c ) , w ⟩ ∼ π ⟨ ℓ ( c ) , π ( w ) ⟩ . By assumption, Succ B ( ℓ ( c ) , w ) is non-accepting, so that byCorollary 6, Succ B ( ℓ ( c ) , π ( w )) is non-accepting, too. Note that π ( w ) indeed does not containany a ’s. We can hence continue the proof assuming that w does not contain any a ’s.Next, we prove that if we replace all b ’s occurring in w by some fresh datum not occurringin supp( C ) ∪ data( w ) ∪ data( d ), we obtain a data word that guides S to a bad synchronizedconfiguration. Formally, pick some datum b ′ ̸∈ data( w ) ∪ supp( C ) ∪ data( d ), and let π be theisomorphism defined by π ( b ) = b ′ , π ( b ′ ) = b , and π ( d ) = d for all d ∈ N ⊥ \{ b, b ′ } . Note that π ( w ) does not contain any a ’s or b ’s. Clearly, ⟨ ℓ A ( d ) , w ⟩ ∼ π ⟨ ℓ A ( d ) , π ( w ) ⟩ . By Corollary6, there still exists an accepting run of A on π ( w ) that starts in ℓ A ( d ). We prove thatSucc B ( C, π ( w )) is non-accepting. Let ℓ ( c ) ∈ C . We distinguish three cases. Let c ̸∈ { b, b ′ } . Then ⟨ ℓ ( c ) , w ⟩ ∼ π ⟨ ℓ ( c ) , π ( w ) ⟩ . Since Succ B ( ℓ ( c ) , w ) is non-accepting byassumption, so that by Corollary 6 also Succ B ( ℓ ( c ) , π ( w )) is non-accepting. Let c = b . By a ≡ C b , the state ℓ ( a ) is in C and ℓ ( a ) , π ( w ) ∼ ℓ ( c ) , π ( w ) since a and c do not appear in w . By essential coverability of C , { ℓ ( a ) , ℓ ( c ) } ⊆ C is coverable. Byunambiguity of B , we obtain that Succ B ( ℓ ( c ) , π ( w )) is non-accepting. Let c = b ′ . Note that ⟨ ℓ ( b ) , w ⟩ ∼ π ⟨ ℓ ( b ′ ) , π ( w ) ⟩ . Recall that b ′ ̸∈ supp( C ). This impliesthat data( C ∩ ( { ℓ } × N ⊥ )) is cofinite. We distinguish two cases. b ∈ data( C ∩ ( { ℓ } × N ⊥ )), i.e., ℓ ( b ) ∈ C . But note that ℓ ( b ) ̸∈ C + b by cofiniteness ofdata( C ∩ ( { ℓ } × N ⊥ )). Hence ℓ ( b ) ∈ C \{ ℓ + ( b ) } . b ̸∈ data( C ∩ ( { ℓ } × N ⊥ )), i.e., ℓ ( b ) ∈ C − b .In both cases, we have proved above that Succ( ℓ ( b ) , w ) is non-accepting. By ⟨ ℓ ( b ) , w ⟩ ∼ π ⟨ ℓ ( b ′ ) , π ( w ) ⟩ and Corollary 6, Succ B ( ℓ ( b ′ ) , π ( w )) is non-accepting, too.Altogether we have proved that Succ B ( C, π ( w )) is non-accepting, while there exists someaccepting run of A on π ( w ) starting in ℓ A ( d ). This concludes the proof for the ( ⇐ )-direction.( ⇒ ) Suppose there exists some data word w such that there exists some acceptingrun of A on w starting in ℓ A ( d ), and Succ B ( C, w ) is non-accepting. We assume in thefollowing that Succ B ( C \ C + b ∪ C − b , w ) is accepting; otherwise we are done. Let ℓ − ( b ) be astate in C − b such that Succ B ( ℓ − ( b ) , w ) is accepting. Pick some datum a ′ ∈ N ⊥ such that a ′ ̸∈ data( w ) ∪ supp( C ) ∪ data( d ). Let π be the isomorphism defined by π ( b ) = a , π ( a ) = a ′ , π ( a ′ ) = b , and π ( d ) = d for all d ∈ N \{ a, b, a ′ } . Clearly, ⟨ ℓ A ( d ) , w ⟩ ∼ π ⟨ ℓ A ( d ) , π ( w ) ⟩ , so thatby Corollary 6, there exists some accepting run of A on π ( w ) starting in ℓ A ( d ). We provethat Succ B ( C \ C + b ∪ C − b , π ( w )) is non-accepting. Let ℓ ( c ) ∈ C \ C + b ∪ C − b . We distinguish thefollowing cases: Let c = a , i.e., ℓ ( a ) ∈ C . By a ≡ S b , we also have ℓ ( b ) ∈ C . Note that ⟨ ℓ ( b ) , w ⟩ ∼ π ⟨ ℓ ( a ) , π ( w ) ⟩ . Note that ℓ ( b ) ̸ = ℓ − ( b ). By assumption, Succ B ( ℓ ( b ) , w ) is non-accepting. ByCorollary 6, Succ B ( ℓ ( a ) , π ( w )) is non-accepting, too. Let c ̸ = a . Note that also ⟨ ℓ − ( b ) , w ⟩ ∼ π ⟨ ℓ − ( a ) , π ( w ) ⟩ . Recall that Succ B ( ℓ − ( b ) , w )is accepting. By Corollary 6, Succ B ( ℓ − ( a ) , π ( w )) is accepting. We prove below that { ℓ − ( a ) , ℓ ( c ) } is coverable. By unambiguity of B , this directly implies that Succ B ( ℓ ( c ) , π ( w ))is non-accepting. . Czerwiński, A. Mottet, K. Quaas 15 Recall that { d ∈ N | ℓ − ( d ) ∈ C } is cofinite. Pick some datum d ∈ N \{ c } such that ℓ − ( d ) ∈ C . We distinguish two cases.Assume ℓ ( c ) ∈ C \ C + b . Since C is essentially coverable, the set { ℓ − ( d ) , ℓ ( c ) } is coverable.Hence there must exist some data word u such that { ℓ − ( d ) , ℓ ( c ) } ⊆ Succ B ( ℓ init ( ⊥ ) , u ).Let π ′ be a partial isomorphism satisfying π ′ ( d ) = a , π ′ ( a ) = d , and π ′ ( e ) = e for all e ∈ data( u ) ∪ { c } . Then { ℓ − ( a ) , ℓ ( c ) } ⊆ Succ B ( ℓ init ( ⊥ ) , π ′ ( u )), hence { ℓ − ( a ) , ℓ ( c ) } iscoverable.Second suppose ℓ ( c ) ∈ C − b , i.e., c = b . This implies that { e ∈ N | ℓ ( e ) ∈ C } is cofinite.Pick some datum e ∈ N \{ d } such that ℓ ( e ) ∈ C . Since C is essentially coverable,the set { ℓ − ( d ) , ℓ ( e ) } is coverable. Hence there must exist some data word u suchthat { ℓ − ( d ) , ℓ ( e ) } ⊆ Succ B ( ℓ init ( ⊥ ) , u ). Let π ′ be a partial isomorphism satisfying π ′ ( d ) = a , π ′ ( a ) = d , π ′ ( b ) = e , π ′ ( e ) = b , and π ′ ( f ) = f for all f ∈ data( u ). Then { ℓ ( b ) , ℓ − ( a ) } ⊆ Succ B ( ℓ init ( ⊥ ) , π ′ ( u )), hence { ℓ ( c ) , ℓ − ( a ) } is coverable.Altogether we have proved that Succ B (( C \ C + b ) ∪ C − b , π ( w )) is non-accepting, while there isan accepting run of A on π ( w ) starting in ℓ A ( d ). This finishes the proof for the ( ⇒ )-direction,and thus the proof of the Proposition. ◀ As in [13], Proposition 21 is enough to obtain an ExpSpace algorithm deciding containment,proving Theorem 3.
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