 So I'm going to present a joint work with Markus Lorei from the University of Leipzig and My talk is going to be about the model checking problem. So in general the model checking problem can be stated as follows so if you have a class of structure C and a logic L Then you're given a structure S in this class and a formula phi in the logic and you ask whether The given structure satisfies the formula This is basically The combined complexity of the model checking problem One can also ask the question whether For a fixed system So if you in advance fix a system and you're only given a formula then you would like to answer the model checking problem This is the so-called expression complexity and then you can do the same Vice versa where you fix the formula and you have the structure as input then you Have the data complexity of the model checking problem So in this talk, I'm going to consider so the the class of structures that we're going to consider are in general infinite and All well all the the logics that this talk is going to be about. I mean the slides are going to be about our formula fragments of monadic second-order logic and Yes, so the L theory So if you have a logic L the L theory of a structure is basically the set of all formulas that satisfy The given structure so you fix the structure and the theory of the structure is the set of formulas that satisfy it okay, so first of all I would like to give some examples of infinite state systems and Come to grant rewrite systems. These are the classes that we have studied So first of all I would like to recall pushdown systems So and there are different definitions of pushdown systems This is the one that is perhaps the most suitable for for the setting So a pushdown system is a finite object. It is give basically a finite set of rewrite rules So you have finite words I denoted them by you and V and you have a finite set of rewrite rules the left-hand side and the right-hand side are both finite words and Yeah, and what you do with them so They induce an infinite transition system and the set of states of this infinite transition system are basically the set of all words and So the set of all finite words and when can when can you go from one word to another word? Well, it's basically prefix rewriting. Okay, so if you have a rule saying You goes to V by the label L then you have an edge from UW to VW for all suffixes W Okay, so this is prefix rewriting and pushdown systems are very established class of infinite structures and so I guess Most of you know this class and I would like to now define the ground tree rat systems They can be defined as a little extension as a as the same kind of extension as pushdown systems But you just replace a finite words by finite trees Finite ranked trees so what is a ground tree rewrite system instead of having finite words you have a finite set of Rules where the left-hand side and the right-hand side are both finite ranked trees such as the following ones for example So I assume you have a ranked alphabet saying that a is a binary as a unary symbol B is a binary symbol and the symbol C is a Nullary symbol then the left-hand side is a tree over this alphabet and the right And the the tree on the right-hand side is also this alphabet and again What are what what is the infinite graph that is described by such a finite for formalism? Well again this the set of vertices of a set of all finite trees and when can you go from one tree to another well you just You you you take a huge tree and you find a subtree that that matches the left-hand side of some rule You apply this rule and you replace this subtree By the right-hand side of the rule so you do this exactly once okay, not not for all occurrences of it of the subtree the number of leaves is the right hand Yes the bottom yet, it's not it's so basically as for words it's the prefix basically starts at the bottom Yeah, so it's like bottom bottom rewriting Okay, let me give an example of a of a ground tree rewrite system. So the system is the set of finite rule the finite set of rules assume we have Nullary symbol X and we have a unary symbol X prime and assume the following rule that is basically leaf rewriting So this is this rule rewrites a leaf X to a tree of size of size one of size two X prime goes to X and the same you do with Y Okay, so another question is what? What is the infinite transition system that is induced by this finite set? Assume we have an additional you binary symbol, okay, let's call the binary symbol Z So let's look at this tree here. So this is the initial state. Let's go. Let's assume. This is the initial state So Z XY so we have a root where Z is we have a we have two children X and Y and Now the question is from this initial state What are what is the the infinite graph induced by those two rules? Well, first of all you can apply the first rule. So you go to some new state. So The new state would be So you start with Z X and Y and then you go with with X you go to Z X prime X and the right-hand side is left alone and this Is untouched and basically this this This note here basically corresponds to to this word for to this tree and if you apply the other rule you get to go to another to the Same tree, but where the Y is expanded and if you apply this rule on and on you basically get the infinite tree Ah, sorry the the infinite grid, sorry and it is well known that the infinite Grid has an undecidable MSO theory. So we know that there is already a fixed ground tree, right? Graph whose MSO theory is undecidable Okay So let me recall what is known about ground tree at systems So first of all as we just saw They do not have a decidable MSO theory in general and this lower bound basically already holds for a CTL So basically, what is the intuition if you have a if you have a Already with a CTL formula you can express basically you can express The run of a of a two-counter machine basically you have two two branches like these two and with a CTL formula You can express the that those two branches communicate Okay, so but what is known on the on the positive side? So if they if you're given a regular set of Trees and you would like to know what are the set of trees that are reachable from the set then this set is again effectively regular and moreover The set of so you given again if you're given a set of regular trees and you would like to know what is the set of recurrently reachable states trees then this is also effectively regular from by looting and well Then there is a fragment of CTL which is decidable so if you restrict your until formulas to be let's say trivial on the left-hand side and you get these EF and EG formulas and Just from the latter two results. You can basically design while at the logic a logic EF. So Exists finally and exist globally such that this logic is decidable has a decidable model checking problem Okay Then it is well known that the first-order theory of Grantry rewrite graphs is decidable and this was shown by a douche and he saw by agree by a tree tree automata techniques, but these up the upper bound is Well decidable by a tower of exponentials Okay, on the other hand one can see decidability of ground tree of the first-order theory of ground tree rewrite graphs by the fact that they are tree automatic so Since they are tree automatic it follows from result but looms are that their first-order theory is decidable more over their First-order theory plus reachability is decidable So if you if you looked at ground tree write graphs and haven't have a reachability relation as a plug in predicate then The first-order theory is also decidable Okay, but on the other hand it is easy to it's clear that they have unbounded degree. So the degree Of these graphs is finite, but unbounded So the reason why I'm mentioning this is basically because there is a result that says okay if the if the if you have a tree automatic graph and Which is additionally has bound a degree then the first-order theory is elementarily decidable But in our setting basically these graphs do not have the property to be bounded Therefore one needs a different approach to decide the first-order theory in elementary time Okay so to so our main result is basically that the first so the First-order theory of grantry write graphs is complete for this complexity class a time so Two to the polly end and so it's basically the set of all problems solvable by an alternating Turing machine running in this time double exponential time, but Whose a number of alternations is Polynomally bounded so such a Turing machine if I allow this Turing machine to run in two full exponential time Then the number of alternations could be two full exponential, but if we restrict this to polynomially many alternations, then you get this complexity class and this is Natural complexity class it has complete problems So on the one hand we show that if both are of both the system and the formula as part of the input Then the model checking problem can be solved in this complex isn't this complexity class so and the the concrete polynomial is well is the linear number of alternations that is not that surprising since the number of alternations is basically the depends on the formula only and But the the running time is two-fold. Yeah, two to the two to the n where n is the input size So the upper bound is basically for the combined complexity and the lower bound we show for a fixed grant rewrite system So there is already a fixed grant rewrite system whose first order theory is hard for this complexity class under lock space reductions Okay, so we're in this talk. I'm mainly going to focus on the upper bound Perhaps it is worth mentioning for this complexity class that Pressburg arithmetic, so the first order theory of natural numbers with addition is complete for this class It's known to be complete for this class. So the model checking problem for granted systems With respect to first order sentences is has the same complexity Okay, so let me Mainly talk about the upper bound so So let us assume we have a grant rewrite system given and the formula phi. So we call this Grantry units system. This is the finite object But let's call G the induced graph and let's assume that phi is the first order sentence Then we're going to solve them model checking problem in two steps So basically in the first step we compute a letter rewriting system This is a notion that is well, I just call it this way for this talk. Yeah, just do I hope it's it is the right choice and So I we define a letter rewriting system so we go away from the tree read systems you go to a letter rewriters and I'll explain this in a minute and But the the size of the resulting letter rewrite system is doubly exponential in the original input size and we Modify the the input first order formula By to a formula phi prime such that the following holds well our initial problem our initial graph Satisfies the formula if and only if the Computed letter rewrite system satisfies the constructed formula by prime and in a second step we solve the model checking problem for this For this resulting letter read system by using standard techniques Okay, I'm mainly I'm going to talk about the the first step in the this translation Okay so What is first of all? What is a letter rewrite system? It's very simple object. You have a finite alphabet. You have an binary relation on this finite alphabet and Now this again induces an infinite transition system. What what is the infinite transition system it Consists the vertices consists of all words over this finite alphabet And when can you go from one word to another word? Well, you just look at the word you a V Okay, you have a you have a prefix and a suffix and then you just replace this you apply one rule to one letter in the word Okay, so you you replace this a So from the word you a V you replace the a by a B Provided well the rule exists. Okay, very simple we might exist and I denotes the resulting infinite transition system by well, I Should call this s prime. Sorry. This is the graph s prime well consists of all words and the given letter rewriting So one thing that our translation does a thing that our translation will do We're gonna compute a concrete a letter rewrite system and what is the set of all letters in This computed letter rewrites system while it consists of the set of ranked trees of We have a grant you a system so we can define the set of all ranked trees of exponential size, okay? That's all you should remember So in but concretely, it's it's the size of the grant rewrites system times four to the quantifier rank of the formula But just think of we compute a finite alphabet consisting of all small trees small is exponential Okay Then now so this set is basically doubly exponential. They are doubly exponential many trees of exponential size Okay, so we have a doubly exponential big alphabet and with this and this is the alphabet that we're going to consider This so and on these trees basically so basically on these letters You can you can define the letter rewrite systems the complete letter rewrite system Okay good, so so let R be the quantifier rank of the formula and The intuition is basically that if you have a formula that has quantifier rank I So it's a smaller quantifier rank then such a formula then can cannot touch any tree of size The size of the ground to that system times four to the I So what I mean if you have a if you have a huge enough tree Then a formula cannot touch the tree and cannot touch the know the leaf of a tree Just because the tree is too big. That's as simple as that Okay, so again, let me recall that this alphabet that we're going to talk about is gamma and It consists of the rank trees of exponential size and on this Finite set we're going to define two subsets so we have Exponential big trees we're going to define small trees and medium trees Okay So for each quantifier. Well, you can choose for each quantifier rank. You did we did basically define the set of all small trees the small trees are of size at Most size of the ground to that system times four to the I okay. These are via small trees And what are medium trees medium trees are trees that are not small but Look as follows. Well, you have a you have a Root and below you all you only have small trees But the resulting tree is not small okay, you can have at most to give exactly one symbol here and The whole tree is not small. So it's a bit bit bigger than small. Okay Okay, so we have The exponential trees and we have two subsets the small trees and the medium trees So let's assume our given Formula phi is as follows with in pre-nix normal form Okay, and we're going to construct. I promise to give a new formula phi prime And now the question is how does this formula phi prime look? well, it's Phi prime is just the homomorphic translation of phi, but the quantifiers are relativized So instead of talking about the original formula talks about the existence of trees Yes, there exists a tree such that something holds the resulting formula phi prime talks about over words, right talks about Properties over the words in the word in the letter rewrite system But I Additionally restrict those quantifications to only quantify. So if I have a quantifier qi X I then I restrict this quantifier to only range over the small trees so the one letter words it's a lot one letter word or Some word Where exactly one letter is a medium tree and the rest of the the prefix and the suffix are basically small trees okay, and Additionally, I can talk about the neighborhood if I already have quite if I already have quantified some some elements Then I can talk about the neighborhood of these elements Yes, so the lemma so I'm not going to talk about the proof the whole proof But I just want to give you a rough intuition so we the original quantria to quantry read system satisfies the formula even only if the letter read systems satisfies the modified formula so the proof sketches basically by a n-phrase arguments so assume you have you have Already selected in the in the n-phrase game. You have already selected some words in the letter rewrite systems but those words we require to have That that they satisfy this local Neighborhood property that just talked about yes so either small trees or these small trees medium small trees or in the neighborhood and Assume that we have so a partial isomorphism from the words that were selected in the game and to some trees that are selected in the infinite system then and This neighborhood talks about well about a neighborhood of a certain radius namely four to the I plus one and the claim is that this Well, this election of nodes can be extended in the next step So can be extended to a further partial isomorphism for any move of spoiler But whether neighborhood is well, it's become smaller Okay, you divide the neighborhood before you had a neighborhood of four to the I plus one and now you want to have a smaller neighborhood of Four to the I So how does the game work? So what how can we what is the the intuition behind the reduction? So assume spoiler chooses so I'm only going to talk about the trend the the step from the ground the under the assumption that Spoiler makes a move in the ground tree right system and now the question is how does duplicator? Respond in the letter rewrite system. So assume that spoiler chooses some tree in the I round in the ground tree right system and The first assumption is assumed that The chosen tree is already close to a tree that has already been selected before close in a certain in the sense that it isn't in this radius Three times four to the I so just assume is the total the chosen tree by by spoilers close to the already chosen trees Then one can show since one this extended isomorphism the partial isomorphism only has to talk about the the neighborhood of size four to the I that the partial isomorphism can be extended in such a way that you just apply it the the same choice as Since that since the tree is not far away one can choose the The same strategy as as given by the induction hypothesis so now the question is what what happens if Spoiler chooses a tree that is far away from all previously chosen trees So assume that the tree is not in it four times three times four to the I neighborhood Well, then we make a case distinction. Is this tree small or is the tree big in case that the chosen tree is small? well, then well, then it is small then we just Then duplicate a response by a small tree by one letter. Yeah, I promise that it has to be a letter or so the response has to be either a small tree or and small a sequence of small trees and one medium tree and Then a sequence of small trees again in this case when Spoiler chooses a small tree already then I can answer with a small tree, but what happens if spoiler answers with well starts the game with With a tree of bigger size. Well, then if you if you have a tree of that is bigger than S times four to the I Then you know That there's a lead at least one place in this in this huge tree That is not small Okay, and this occurrence this exists this occurrence we call this occurrence of a subtree that is not too well That is that violates being small. We call this subtree. It's it's rooted somewhere at age age is some symbol and We have the subtree that is not small, but it's medium and Then we additionally select on all we do the following go We select maximally big sub trees that are small Okay, and now how does the the spoiler? Respond And the duplicate how does it how does he respond? Well, you just collect all those small sub trees you put here the medium tree and All the remaining small trees, so he he responds by a sequence of Trees where all of them are small except for exactly one so in this way we showed that you can reduce the Model checking problem for ground tree that systems to an two-fold exponentially bigger Laterary right system Let me come to the lower bound So the low bound is that there is already a fixed ground tree that system or ground tree at graph whose first order theory is hard for The complex the complexity class to to the poly n with polynomial number of alternations Well instead of proving hardness for this complexity class We just let me just sketch the proof for the two next time lower bound So for the two next time lower bound there Assume some tiling problem. Okay, assume some fixed tiling problem such that's deciding the following question You're given an initial coloring of the tiling So you've given an initial coloring of the first the lower parts of the of the tile And the question is whether you can expand this this given coloring to a 2 to the n times 2 to the n tiling I Want to yes, then The point is that there is already a fixed tiling system for which this is too next-time hard and the fixed ground tree At graph will depend on this fixed tiling system So what is the intuition? Well This is the grid The 2 to the n times 2 to the n grid and then there's one point ij and we're going to represent one point in this two-fold exponential big grid by such a tree such a tree is Exponentially big and at the leaves you have a zero zero zero zero zero ones and this tree basically represents a by well a Binary number and this binary number Should encode this i and j so one such a tile tree corresponds to one position in to one point in this grid and now What what do we now want to define what we want to define grid trees what are grid trees? They are just huge trees such that each small subtree is a tile tree Okay, and Additional what the local the local properties hold that are given by the tiling system So a grid tree is a small Well, it is a huge tree such that all small subtree are tile trees Each grid element appears exactly once you don't we don't want to talk about more than one Good element exactly one and it satisfies the vertical and horizontal matching relations Okay, and then Alema shows that you can define over a fixed grant read system that basically modifies modifies little sub trees of such a tile tree So one can come up with a with a first for first order formula. That is not too big big that can talk about the existence of a grid tree that satisfies the initial coloring okay, and Asking whether the initial coloring can be extended to a bigger coloring just boils down to asking for the existence of A grid tree that satisfies the initial coloring Okay, so I would like to conclude with Well, one further lower bound. It's basically so if you do the following little extension if you and This I claim that there's already a fixed grant rewrite graph But with a unary predicate that is given by a regular tree language And then the first order theory is non elementary. So there's already a fixed grant read system With a fixed regular language such that the first order theory is non elementary You can think of it the intuition is basically that this fixed Regular tree language can simulate basically the computation of a of a bottom-up tree automaton and therefore you can easily Reduce from the satisfiability problem Over finite words for first order logic as An interesting and perhaps a bit ambitious open problem, but it's the following So it's the by simulation It's the following problem decidable given two grant rewrite graphs Are they by similar? Yeah, so there is currently not known to be decidable the best known lower bound is extra time and Which also corresponds to the same lower bound for push-down systems, but there's nothing known Okay, that's all thanks for your attention