136 and I. Walukiewicz Given two types and we denote by the delayed type which assigns to a letter the type A type is reachable from a type denoted if holds for some context This relation is a quasiorder and we use for the accompanying equivalence relation. The following simple lemma is given without a proof: Lemma 2. If is a subtree of then If then The following lemma shows that for TL(EF)-definable languages, the relation is a congruence with respect to the function Lemma 3. If and then Proof. Since a TL(EF)-definable language satisfies it is sufficient to prove the case where Let be a context such that and let be a context such that All these contexts exist by assumption. Let be a tree of type and let be a tree of type Consider the two sequences of trees and defined by induction as follows: By a simple induction one can prove that for all By Lemma 2, for all Since there are only finitely many signatures, there must be some such that Consequently, by Lemma 1, the delayed types and are equal. We are now ready to show that the language L is typeset dependent. Let and be two trees with the same typeset. If this typeset is empty, then both trees have one node and, consequently, the same delayed type. Otherwise one can consider the following four types, which describe the sons of and We need to prove that By assumption that the typesets of and are equal, both and occur in nonroot nodes of and both and occur in nonroot nodes of Thus holds for some and similarly for and The result follows from the following case analysis: TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. Characterizing EF and EX Tree Logics 137 for some By assumption we must have for some Hence By Lemma 3 we get As from Lemma 2 we obtain Similarly one proves the equality for some As in the case above. A short analysis reveals that if neither of the above holds then and for some Therefore and and an application of Lemma 3 yields the desired result. 4.2 A EF-Admissible Language Is TL(EF)-Definable We now proceed to the most difficult part of the proof, where a defining TL(EF) formula is found based only on the assumption that the properties P1 to P4 are satisfied. We start by stating a key property of EF-admissible languages which shows the importance of neutral letters. Lemma 4. If the delayed type of a tree is then its every proper subtree with delayed type has the root label in Proof. Consider some proper subtree of delayed type and its root label Let be the brother of the node and let be its delayed type and label, respectively. Obviously By property P3 we get and consequently As is a partial order by P1 and since holds by definition, we get Hence belongs to Note that if the trees and have delayed type then so does the tree for any because is a partial order. In particular, the above lemma says that nodes with delayed type form cones whose non-root elements have labels in Formulas Defining Delayed Types. A delayed type is definable if there is some TL(EF) formula true in exactly the trees of delayed type The construction of the formulas will proceed by induction on the order. The first step is the following lemma: Lemma 5. Let be a delayed type such that all types are definable. For every delayed type there is a TL(EF) formula such that: The proof of this lemma is omitted here. We would only like to point out that some effort is required, since the formula is not allowed to use the EX operator. We will use this lemma to construct a formula defining For the rest of Section 4.2 we fix the delayed type and assume that every delayed type is definable by a formula TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. 138 and I. Walukiewicz The first case is when has no neutral letters. By Lemma 4, in a tree of delayed type both sons have delayed types smaller than since there are no neutral letters for In this case we can set The correctness of this definition follows immediately from Lemma 5. The definition of is more involved when the set of neutral letters for is not empty. The rest of Section 4.2 is devoted to this case. Consider first the following formula: The intention of this formula is to spell out evident cases when the delayed type of a node cannot be The first disjunct says that there is a descendant with a delayed type and a label that prohibit its ancestors to have type The second disjunct says that the type of the node is not but the types of all descendants are This formula works correctly, however, only when some assumptions about the tree are made. These assumptions use the following definition: a tree satisfies the property if Lemma 6. Let be a tree where holds for all This tree satisfies if and only if Proof. The left to right implication was already discussed and follows from the assumptions on the formulas used in and from Lemma 5. For the right to left implication, let with describing delayed types and labels of the nodes 0 and 1 which correspond to the left and right sons of the root. We consider three cases: This is impossible because and hold, so the labels must belong to and thus and Since holds, the label belongs to If the inequality were true (which is not necessarily implied by our as- sumption that then by property P3 we would have a contradiction with Therefore we have and hence the first disjunct of holds. The case where and is symmetric. In this case the second disjunct in the definition of must hold by Lemma 5. Let stand for and consider the formula This formula will be used to express the property. We use as the non-strict version of AG, i.e. is an abbreviation for the formula TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. Characterizing EF and EX Tree Logics 139 Lemma 7. A tree satisfies iff holds for all Proof. By induction on the depth of the tree If satisfies because it satisfies then obviously holds for all Otherwise we have By induction assumption, holds for all But then, by Lemma 6, This, together with gives and hence Let be such that holds for all By induction assumption, we have We need to prove that satisfies If holds, then satisfies and we are done. Otherwise, as holds, and Hence, by Lemma 6, satisfies the second disjunct in Since the type of a tree can be computed from its delayed type and root label, the following lemma ends the proof that every EF-admissible language is TL(EF)definable: Lemma 8. Every delayed type is definable. Proof. By induction on the depth of a delayed type in the order If has no neutral letters then the defining formula is as in (1). Otherwise, we set the defining formula to be Let us show why has the required properties. By Lemma 7, If then we get using Lemma 6 and (2). For the other direction, if then clearly holds in By Lemma 4, holds for all therefore satisfies by (2), and then the formula holds by Lemma 6. 5 TL(EX, EF) The last logic we consider in this paper is TL(EX, EF). As in the previous sections, we will present a characterization of TL(EX, EF)-definable languages. For the rest of the section we fix an alphabet along with a L and will henceforth omit the L qualifier from notation. Recall the type reachability quasiorder along with its accompanying equiv- alence relation which were defined on p. 136. The class of a type is called here its strongly connected component and is denoted We extend the relation to SCCs by setting: TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. 140 and I. Walukiewicz We use the standard notational shortcuts, writing when but not similarly for Let be some SCC and let The of a tree is the tree whose domain is the set of nodes in at depth at most and where a node is labeled by: if is at depth smaller than if is at depth and ? otherwise. Let denote the set of possible The intuition behind the of is that it gives exact information about the tree for types which are smaller than while for other types it just says “I don’t know”. The following definition describes languages where this information is sufficient to pinpoint the type within the strongly connected component Definition 6. Let The language L is if every two trees and with types in and the same view have the same type. The language is if it is for every SCC and it is SCC-solvable if it is for some It turns out that SCC-solvability is exactly the property which characterizes the TL(EX, EF)-definable languages: Theorem 3. A regular language is TL(EX, EF)-definable if and only if it is SCC-solvable. The proof of this theorem will be presented in the two subsections that follow. 5.1 An SCC-Solvable Language Is TL(EX, EF)-Definable In this section we show that one can write TL(EX, EF) formulas which compute views. Then, using these formulas and the assumption that L is SCC-solvable, the type of a tree can be found. Fix some such that L is Let be the set of possible that can be assumed in a tree of type By assumption on L being we have: Lemma 9. Let be a tree such that The type of is if and only if its belongs to the set The following lemma states that views can be computed using TL(EX, EF). We omit the simple proof by induction. Lemma 10. Suppose that for every type there is a TL(EX, EF) formula defining it. Then for every and every there is a formula satisfied in exactly the trees whose is We define below a set of views which certainly cannot appear in a tree with a type in a strongly connected component TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. Characterizing EF and EX Tree Logics 141 Observe that is a set of The following lemma shows that the above cases are essentially the only ones. Lemma 11. For a tree and an SCC the following equivalence holds: Proof. Both implications follow easily from Fact 9 if one considers the maximal possible node satisfying the right hand side. The following lemma completes the proof that L is TL(EX, EF)-definable. Lemma 12. Every type of L is TL(EX, EF)-definable. Proof. The proof is by induction on depth of the type in the quasiorder Consider a type and its SCC By induction assumption, for all types there is a formula which is satisfied in exactly the trees of type Using the formulas and Lemma 10 we construct the following TL(EX, EF) formula (recall that is the non-strict version of AG defined on page 138): By Lemma 11, a tree satisfies if and only if Finally, the formula is defined: The correctness of this construction follows from Fact 9. 5.2 A TL(EX, EF)-Definable Language Is SCC-Solvable In this section, we are going to show that a language which is not SCC-solvable is not TL(EX, EF)-definable. For this, we introduce an appropriate Ehrenfeucht- Fraïsé game, called the EX+EF game, which characterizes trees indistinguishable by TL(EX, EF)-formulas. The game is played over two trees and by two players, Spoiler and Duplicator. The intuition is that in the EX+EF game, the player Spoiler tries to differentiate the two trees using moves. The precise definition is as follows. At the beginning of the game, with the players are faced with two trees and If these have different root labels, Spoiler wins. If they have the same root labels and Duplicator wins; otherwise the game continues. Spoiler first picks one of the trees with Then he chooses whether to make an EF or EX move. If he chooses to make EF move, he needs to choose some non-root node and Duplicator must respond with a non-root node of the other tree. If Spoiler chooses to make an EX move, he picks a son of the root in TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. 142 and I. Walukiewicz and Duplicator needs to pick the same son in the other tree. If a player cannot find an appropriate node in the relevant tree, this player immediately looses. Otherwise the trees and become the new position and the game is played. The following lemma is proved using a standard induction: Lemma 13. Duplicator wins the EX+EF game over and iff and satisfy the same EX+EF formulas of modality nesting depth For two types we define an to be a multicontext C such that there are two valuations of its holes giving the types and The hole depth of a multicontext C is the minimal depth of a hole in C. A multicontext C is for an SCC if it has hole depth at least and is an for two different types Lemma 14. L is not SCC-solvable if and only if for some SCC and every it contains multicontexts which are for Proof. A context exists for if and only if L is not The following lemma concludes the proof that no TL(EX, EF) formula can recognize a language which is not SCC-solvable: Lemma 15. If L is not SCC-solvable then for every there are trees and such that Duplicator wins the EX+EF game over and Proof. Take some If L is not SCC-solvable then, by Lemma 14, there is a multicontext C which is for some SCC Let be the holes of C, let be the appropriate valuations and the resulting types. We will use this multicontext to find trees and such that Duplicator wins the EX+EF game over and Since all the types used in the valuations and come from same SCC, there are contexts and such that This means there are two contexts and with holes each, such that: 1) and agree over nodes of depth less than when all holes of are plugged with we get the type and 3) when all holes of are plugged with we get the type These are obtained by plugging the appropriate “translators” and into the holes of the multicontext C. Let be some tree of type The trees for are defined by induction as follows: By an obvious induction, all the trees have type and all the trees have type As there exists a context D[] such that and (or the other way round). TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. Characterizing EF and EX Tree Logics 143 To finish the proof of the lemma, we will show that Duplicator wins the EX+EF game over the trees The winning strategy for Duplicator is obtained by following an invariant. This invariant is a disjunction of three properties, one of which always holds when the game is about to be played: 1. 2. 3. The two trees are identical; The two trees are and for some The two trees are and for The invariant holds at the beginning of the first round, due to 2, and one can verify that Duplicator can play in such a way that it is satisfied in all rounds. Item 2 of the invariant will be preserved in the initial fragment of the game when only EX moves are made, then item 3 will hold until either the game ends or item 1 begins to hold. 6 Decidability In this section we round up the results by showing that our characterizations are decidable. Theorem 4. It is decidable in time polynomial in the number of types if a lan- guage is: TL(EX)-definable; TL(EF)-definable; TL(EX,EF)-definable. Proof. Using a simple dynamic algorithm, one can compute in polynomial time all tuples such that for some context C[], and Using this, we can find in polynomial time: Whether L contains an The and relations on types. Since the delayed type of a tree depends only on the types of its immediate subtrees, the number of delayed types is polynomial in the number of types. The relation on delayed types can then be computed in polynomial time from the relation Having the relations and one can check in polynomial time if L is EF-admissible. TEAM LinG Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. 144 and I. Walukiewicz This, along with the characterizations from Theorems 1 and 2, proves decid- ability for TL(EX) and TL(EF). The remaining logic is TL(EX, EF). By Theorem 3, it is enough to show that SCC-solvability is decidable. In order to do this, we give an algorithm that detects if a given SCC admits bad multicontexts of arbitrary size, cf. Lemma 14. Fix an SCC We define by induction a sequence of subsets of if and either there is a pair a type and a letter such that and or there are pairs and a letter such that and The sequence is decreasing so it reaches a fix-point in no more than steps. The following lemma yields the algorithm for TL(EX, EF) and con- cludes the proof of Theorem 4: Lemma 16. admits bad multicontexts of arbitrary size iff Corollary 1. If the input is a CTL formula or a nondeterministic tree automa- ton, all of the problems in Theorem 4 are E XPTIME - complete. Proof. Since, in both cases, the types can be computed in time at most expo- nential in the input size, the E XPTIME membership follows immediately from Theorem 4. For the lower bound, one can use an argument analogous to the one in [17] and reduce the E XPTIME -hard universality problems for both CTL [3] and nondeterministic automata [13] to any of these problems. 7 Open Problems The question of definability for the logics TL(EX), TL(EF) and TL(EX, EF) has been pretty much closed in this paper. One possible continuation are logics where instead of EF, the non-strict modality is used. The resulting logics are weaker than their strict counterparts (for instance the language is not definable in and therefore decidability of the their definability problems can be investigated. Another question is what happens if we enrich these logics with past quantification (there exists a point in the past)? This question is particularly relevant in the case of TL(EX, EF), since the resulting logic coincides with first- order logic with two variables (where the signature contains and two binary successor relations). Finally, there is the question for CTL. 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Kingdom, and supported by the German Academic Exchange Service (DAAD) Supported by the European Research Training Network “Games” P Gardner and N Yoshida (Eds.): CONCUR 2004, LNCS 3170, pp 14 6–1 60, 2004 © Springer-Verlag Berlin Heidelberg 2004 Please purchase PDF Split-Merge on www.verypdf.com to remove TEAM watermark this LinG Message-Passing Automata Are Expressively Equivalent to EMSO Logic 147... reception bound to can be decrypted (as expressed by the guard * Supported by the Swiss National Science Foundation, grant No 21-65180.1 P Gardner and N Yoshida (Eds.): CONCUR 2004, LNCS 3170, pp 16 1–1 76, 2004 © Springer-Verlag Berlin Heidelberg 2004 Please purchase PDF Split-Merge on www.verypdf.com to remove TEAM watermark this LinG 162 J Borgström et al Fig 1 Equivalences If the equation holds, then no... without autoconcurrency It would also be interesting to have logics that capture formalisms such as locally- and globally-synchronized HMSCs and related automata models [5] References 1 R Alur, K Etessami, and M Yannakakis Inference of message sequence charts In 22nd International Conference on Software Engineering ACM, 2000 2 R Alur and M Yannakakis Model checking of message sequence charts In CONCUR 1999,... 1999 (MSC99) Technical report, ITU-TS, Geneva, 1999 8 D Kuske Asynchronous cellular automata and asynchronous automata for pomsets In CONCUR 1998, volume 1466 of LNCS, 1998 9 D Kuske Regular sets of infinite message sequence charts Information and Computation, 187:8 0–1 09, 2003 10 P Madhusudan Reasoning about sequential and branching behaviours of message sequence graphs In ICALP 2000, volume 2076 of... Morin Recognizable sets of message sequence charts In STACS 2002, volume 2285 of LNCS Springer, 2002 13 M Mukund, K Narayan Kumar, and M Sohoni Synthesizing distributed finitestate systems from MSCs In CONCUR 2000, volume 1877 of LNCS Springer, 2000 14 A Muscholl and D Peled Message sequence graphs and decision problems on Mazurkiewicz traces In MFCS 1999, volume 1672 of LNCS Springer, 1999 15 W Thomas... and P Zafiropulo On communicating finite-state machines Journal of the ACM, 30(2), 1983 4 M Droste, P Gastin, and D Kuske Asynchronous cellular automata for pomsets Theoretical Computer Science, 247( 1–2 ), 2000 5 B Genest, A Muscholl, H Seidl, and M Zeitoun Infinite-state high-level MSCs: Model-checking and realizability In ICALP 2002, volume 2380 of LNCS Springer, 2002 6 J G Henriksen, M Mukund, K... straightforward way, obeying the usual assumption that capture of bound names is avoided by implicit for example, replaces all free occurrences of in P by F, renaming bound names in P where needed 3 Semantics – Concrete and Symbolic Concrete Semantics The concrete semantics we use is similar to the one used in [BDP02], apart from that we do not have a let-construct in the language Because of this, input and . “Games”. P. Gardner and N. Yoshida (Eds.): CONCUR 2004, LNCS 3170, pp. 14 6–1 60, 2004. © Springer-Verlag Berlin Heidelberg 2004 TEAM LinG Please purchase PDF Split-Merge. Notes in Computer Science, pages 2–2 6, 2002. Fixed points vs. infinte generation. In Logic in Computer Science, pages 40 2–4 09, 1988. M. Otto. Eliminating