 Okay, so let me first recall some of the things we saw yesterday. So the first result I presented yesterday was that the small model property in for for two variable fragment of first of the logic. In fact, in our construction, we started from an arbitrary model of a formula phi and constructed a new model, whose domain was a subset of the domain of the original model. The one types of elements were retained from the original model and the only only some two types have been exchanged. It's modified. So in fact, this construction guarantees also some some additional properties that the set of one types realized in both structures unidentical. All the types which are realized in the original structure are also realized in this new small model and each non-royal type recall that the type is royal if it is realized in a structure exactly once. So each non-royal type from from the original structure is realized in this new structure at least twice. So this is this was the first the first property we obtained yesterday. Another simple result about two variable logic, which will be useful today. It's a very simple thing that if we have an arbitrary model of a formula phi, then we and and we take an arbitrary non-royal one type realized in this model, then we can add additional copy of this of this one type to the model. So this is the second result from yesterday and result. Very simple observation in fact. And then we considered two variable guarded fragment with equivalence relations in guards. Let me recall the normal form of the formulas which we are interested in. So we may say that there exists an element satisfying something. We may impose restrictions on connections between pairs of connected elements. So we may say that for all x, y if they are connected either by an equivalence relation which is which equivalence relation relations are denoted by E i or by a relation which is not required to be equivalence. Such relations are denoted by R i or similarly. So for connected elements we may impose some additional restrictions on connections between them. And there are formulas which would say that some elements satisfying some atomic properties require the so-called witnesses. Yes, for all x, if x satisfies some atomic formula, then there exists y connected to this x either by E i or equivalence relation or non-equivalence relation and this connection should satisfy some properties. In my today's talk, yesterday we considered the general satisfiability problem for two variable guarded fragment with equivalence guards. Today we will consider the finite satisfiability problem and to simplify the presentation I will not consider the these two types of conjuncts. It has those contracts which may require witnesses connected not necessarily equivalently as to an element. So in fact what we can say, we can only say about require witnesses inside equivalence classes. This simplification is not crucial. It is very easy to add these two types of conjuncts. It will be just left as an exercise. One another thing is notation I sometimes use by phi phi with superscript ek. I denote the fragment of this formula which is related to which imposes something in equivalence classes of the relation ek. So it contains all the formulas which have ek as guards plus those universal formulas guarded by non-equivalence relations because they also may restrict they also may have influence on the shape of E 1 class here because in this E i class or E k class because in this E k class of course we may have another binary connections. One important restriction on our language is that we allow these equivalence relations only as guards. So this is why we call this variant with equivalence guards. So E i relations cannot appear in these formulas psi. But in the formulas psi we may have additional realizations of additional we may use those non-equivalence symbols. Another thing I we observed yesterday is that in fact there is a close relation between models for formulas for two variable logic and equivalence classes in models for formulas of two variable guarded fragment of equivalence guards. In a single equivalence class in a model of GF 2 plus E g formula we may we may do everything we do in the first order logic yes for example if we have in first order with two variables if we have a two variable formula not necessarily guarded then and if we if we translate it to the guarded fragment by using a dummy equivalence guard say E 1. E 1 is a symbol which which is not a member of the signature of this formula. And simply we add this we add guards to this quantifiers here then we obtain a formula whose equivalence classes have to be models of this formula. So this there is a close relationship between these two things and what we do yesterday for general satisfiability we started from an arbitrary model of GF 2 plus E g formula and unraveled it into a trilyke way providing starting from from some element and then providing classes for this element in such a way that any pair of elements is connected in this trilyke model at most by one equivalence relation. This is the place where we use this restriction on on the usage of equivalence symbols yes they can be they can appear only in gas so we can do such things we can unravel the unravel the model then we can we can make the classes exponentially bounded and we can make this model regular. So and this this implies that there is a simple a simple decision procedure for for this problem working in nondeterministic exponential time. Okay, so today I will talk about the finite satisfiability problem as you see starting from an arbitrary model even finite we usually will usually end in in in an infinite model. So such approach is not very promising here and instead of three unravelings we will describe the properties of models by using systems of equivalence systems of linear equations. Let let me first show you that there are some differences in in this satisfiability and finite satisfiability problems. For example the case of satisfiability, general satisfiability, we obtained a property that every satisfiable formula has a model with exponentially bounded classes yes. So so this property cannot be obtained in the case of finite models. A very simple example here will construct a formula delta n of land polynomial in n such that it's every model will will contain a full binary tree of depth to the n and then we say we will say that all leaves of the of this tree are members of the same class. The number of leaves will be double exponential so so it will imply that there exists a double exponential class. Okay, so so the models will look like this. We will have an element we say that there exists an element satisfying unary predict unary relation root and for each and and and we would think that each of the elements encodes one binary number from from the set zero up to two to the n minus one. Such and such such numbers can be encoded and coded in the usual way using unary predicates. And I explained yesterday that we can count up to two to the n. We can say that a pair of elements encodes numbers which differ by one. So we may say that for every element from an even level of the tree there are two elements on the next level connected to this element by ret relation, E1 relation. One of them is in L, one in not in L to have to ensure that there are really two such elements. And from odd levels we require two elements connected by blue relation E2. And we want to enforce that models really are really complete binary trees. Yes, so we have to ensure that there is no such situation that two elements from the same level say this gray element and this gray element use the same element on the next level as as as the witnesses for example this one. And to do this we say that elements from say an even level like even level this is zero one two so for example from the second level that there are not two elements on this level which are connected by ret relation. Yes, such reusing of a witness would cause that because of transitivity of the ret relation that the two elements the gray elements would become related by this ret relation. So we simply say that if there are two elements related by ret relation E1 and they belong to an even level to check if level is even or what we just look at least significant bit then they are really the same element yes they cannot be distinct. So this ensures that we have really a complete binary tree and then we have to say that all leaves of this tree are related by this blue relation here we cannot say directly yes for all elements if x y if x is a leaf and y is a leaf then connect them by blue relation because this formula would not be guarded yes but we can for example say that each leaf is connected by a blue relation to a root. We have two to the two to the two to the n leaves unfortunately we cannot enforce in the guarded fragment there is exactly one root because it is impossible we always take always in the guarded fragment can take another copy of the same model and this is this is a model so the situation in models maybe like this that we have several roots each of them with a complete binary tree and some of the leaves connect to the first root some of the leaves connect to the second root but of course we always have a ratio one root per two to the two to the n leaves so at least one of the classes must be of size at least that the exponential of course if we if we allow for infinite models then we can and then we can always construct classes exponentially but since we simply instead of taking this root we can construct a new one and then go on but in finite models we can enforce doubly exponentially bounded classes so what in the in the remaining part of the talk I will use a notion of accounting type and the accounting type the purpose of counting types is to count the number of realizations of one types in equivalence classes so formally accounting type is a function from the set of one types over the signature of a formula phi of the formula far we consider and this function returns natural numbers and we usually we we think in we think about this in this way that if we have a class c and say e i class c equivalence class c in in a model a then its counting time simply says for for a given one type discounting its counting type says how many times this type is realized in this class how many elements of this one type we have in this class we will say that accounting type is admissible for e i classes with respect to to this this formula phi if there exists an e i class c of realizing exactly this counting type which satisfies this part of the formula related related to e i yes so in other words accounting type is admissible for for e i classes if we can build e i classes which from local point of view are allowed to appear in models of of phi and admissibility of accounting type is a simple property we can simply for for a given counting type we simply guess the structure and perform model checking so this is an it is not not hard to check if if accounting type is admissible so to explain my approach i will first show you a very simple very simple idea so in a simplified case so the simplified case is that we assume that we have a bounded number of admissible counting types this may not be true in general is because in these counting types we do not know a bound on on on the values which are values of this function of the function yes so potentially we may have infinitely many admissible counting types but let us assume that there is only some bounded number of them for example we may assume that we are interested in satisfiability in models with exponentially bounded classes only yes then then the number of counting types will be bounded doubly exponentially yes if we have exponentially many possible exponentially many elements then the values of this function are all values of the functions which are counting types are also exponentially bounded so that there might there may be at most doubly exponentially many such classes and so and we construct we we construct a system of equations for each counting type which is admissible for e i class we introduce a variable an unknown yes so we have unknowns for e i e one classes unknowns for e two classes for types if you want classes for types of e two classes and so on and what we say by our equations we only say an obvious thing which which has to be satisfied in every model that each one for each one type t and each equivalence relation e i e j there is this type is realized exactly the same number of times in e one classes and in e two classes this is an obvious thing because every element is a member of an e one class is a member of e two class yes e i e j in general so this is the obvious obvious thing which should be satisfied in the model and we claim that this is this condition is that this is if and only if so if we we claim that phi has a model if and only if the constructed system of equations has a non-negative integer solution okay let me show you an example assume that we have one possible we have two two equivalence relations e one red and e two blue and there are three possible admissible counting types for e one classes one with yellow and and one yellow and one green one type blue and red one just a singleton class with green relation and admissible and there is only once admissible type for blue classes with red blue two yellow and two green elements so what our system of equation says it simply says that from the point of view of red one types the number of realizations of x in the model should be equal to the number of relations of b because these are the only only types which in which red element appear similar equation is obtained from the if we write an equation from the point of view of blue one type from the point of view of yellow one type we see that every realization of x requires two realizations of a yes because we have two two yellow types in in this x type of class and from the point of view of green elements of green elements every realization of x requires two classes with green elements so it may be a or c so this is why we say two x is equal to a a plus c and obviously if there is a model using this admissible counting types then this model has to satisfy then then this system of equation has a solution in the opposite direction consider an arbitrary solution for example we may say we may take the solution with one realization of x two realizations of a is equal to two b is equal to one and c is equal to zero this is a solution to the system and now let us see that we can construct a model with exactly this with exactly such types of classes but not with exactly this number of realizations of these classes because sometimes for example this solution suggests exactly one realization of this this blue class blue type of this type of blue class however there are some red classes with two elements in in in our language we may enforce that we may have formulas which enforce that every pair of classes blue class and the red class have at most one element in common in fact we obtained this property in our construction yesterday but this property can be can be enforced for example we may say that every pair of elements which is connected by blue relation is also connected by an auxiliary symbol b which is not required to be equivalent and similarly every pair of elements which is connected by red relation satisfies is not connected by b yes then we cannot intersect two classes in more than one element yes but then because because then this pair of elements should be connected by b and not connected by b simultaneous so in this case it may happen that we we won't be able to construct a model with just one equivalence class of this type yes because this class requires two may require two two different classes of type x but what can we do what what we are sure after when we have the solution of this system we may let us define the notion of the base set for this solution the base set of elements the base set of elements contains as many elements of particular one types as is suggested by the solution so the solution suggests that there will be one class of type x so in the base set and no no other blue classes yes so in the suggested set of one types is one red one blue two yellow and two green yes similarly we may we may do it from the point of view of red classes the suggested solution is a is equal to two b is equal to one c is equal to zero so we have two green elements because a is equal to two yellow elements four no sorry no two green two yellow one red one one blue yes because b is equal to one of course the base sets computed using solutions for blue classes and red classes have to be equal because we constructed this is our system in such a way and what we know about this base set our the solution of the system guarantees that this base set can be divided into proper into admissible e1 classes and into admissible e2 classes independently yes so what we do we simply take some number of copies it's six copies in this case we have we have six elements in this set so we take six copies and we arrange them into in a grid in this case in such a way that every column of this grid and every uh every horizontal line in this grid every row in this grid is a copy of this base set such an arrangement can be easily obtained just by starting from an from from the call the first column and putting this base set in an arbitrary order and then just shifting the picture one row down yes and and now we can divide columns into blue classes and uh rows into one classes exactly as suggested by this solution yes so after this construction every element has two admissible classes constructed yes e blue and and red so everything is okay remember that that that for simplicity i removed those formulas which require non non equivalent witnesses this approach of course can be can be easily generalized if we have more than two equivalence relations then we simply construct multi-dimensional grid grids in this way okay so this is this is the the idea and the problem is that we do not know a bound on the number of counting types this may be infinite so in particular we would we would have infinitely many variables yes so it would be a problem instead of this we will use some approximations of this of these counting types and before I define this approximation let me let me first just just formalize one one lemma which we will use so uh we say that we say that a counting type safely extends another counting type uh if it preserves the number of uh one types which are realized zero on one types in this real estate which are suggested by this type yes so if if theta of t is of t is equal zero or one then theta prime is also equal zero of or one if theta of t is greater than one then theta prime of t is simply greater or equal to theta of course this definition has a close connection to the notion of this royal types royal types in models of photo formulas yes simply uh that if we have a class of type theta and then then we can extend it to a safely extend it to a class of type theta prime which still from local point of view is a proper class for for for the given formula and this is this recall lemma lemma 10 is this lemma about adding adding single realizations of non-royal types to fo2 classes and this is just a simple application of this lemma yes if we have uh if we have a class c is just a model which is an equivalence class of ei this it satisfies this fragment of the formula related to ei then if we take a counting type theta prime which extends the counting type of of c and then we can construct a model of this formula for ei which has exactly time by adding the required elements as many times as required so this is one thing and another thing is the notion of m counting type so instead of using exact counting type we'll use some approximations we'll count up to some m m counting type m is a parameter is a function of of the type take from from the set of one types to to the set zero one up to m and the interpretation is that theta if theta of t t is equal to m then it means then then in the class there is at least m elements of the of type t m minus 1 m minus 2 1 and 0 means exactly m means at least now lemma 4 was is this this small model construction for fo2 formula and using this this construction we may prove that there exists m exponentially bounded in this in the length of the formula such that if we have a class admissible for for ei classes a class sorry a class satisfying phi of phi with superscript ei and its m counting type is theta then there exists a class c prime which has exact time equal to this theta yes so if there is m realizations at least m realizations then in this new structure there will be exactly m realizations and this is the same construction as in on this lemma for okay so these are two two auxiliary lemmas and now what how we construct the system of this time the system of inequalities so we introduce variables this time for m counting types which are admissible for ei so again we check if such a if there is a class with exact counting type we've given exact counting type like like this with at most m realizations yes of every one type what is the number of of these m counting types it is doubly exponential in the size of the formula phi as we observed later because we consider the similar situation what will be the our procedure we will guess the set the set of one types which will be realized in the model of of the given formula phi and write such some some inequalities there will be two kinds of inequalities the first kind of inequalities are just inequalities which ensure that elements of these quest one types will appear in the model and the second the second one will the second kind of inequalities will say something about royal types so now let us let us redefine this no notion royal we say that a one type t is royal for ei classes with respect to the formula phi of course if every counting type admissible for ei has at most one realization of this type yes so such a royal type a type is royal for e1 classes which if a type is royal for e1 classes say then we probably won't be able to add additional realizations of this type to the given class given a visible class and what kind of equations we write if t is royal for ei classes then for every other equivalence relation we write inequality which say that t is realized in is suggested by by m counting types for ei classes at most as many times as for ej classes okay this may be unclear I suppose so let us observe let us let us look at an example assume again assume that we have two admissible m counting types so this means that in in our model we allow for such classes and these classes may maybe further extended by this class may be extended by yellow elements this may be extended by green elements this one in fact also can be extended by yellow elements but green elements in yellow elements because we know a pattern which allows to connect to two yellow element inside a class and we have three admissible m counting types for red classes and what what we say in our system of equations so a red from the point of view of red element red element is royal for both red classes and blue blue classes and red classes in each class it appears at least at most once so what we say we want to guarantee that the number of realizations of oh the definition the definition from the previous slice implies that we should construct two inequalities which will produce such a such an equality yes if we realize such a type then we have to realize type type a so x x plus y should be equal to a yes from the point of view of blue elements the blue yellow elements of course yellow elements are royal only for e2 classes oh sorry e2 for for red classes it should be one for red classes because in each each such red class there is at most one realization of yellow element for blue classes yellow is not royal yes because we have a class admissible class with admissible type of class with two yellow elements so this is this is the second kind of an equal the first kind of inequalities are those inequalities which say that our uh three three one types assume that we guess that there will be these three one types in the model and we ensure that they will be in a model so we say that x plus y is greater or equal to one x plus y is greater or equal to one so so and x plus y y is greater or equal to one this this ensures that in blue classes we will have red yellow and green and the same we do for red classes okay an example solution is one x is equal to one y is equal to one a is equal to two b is four and c is zero now how to construct a model again we will define the base set for for this solution but this time the base set will be defined in a slightly more complicated way namely how many elements of red type will have in the base set we look how many types times these elements these red elements are suggested in by our solution in blue classes so twice yes and how many in red classes no also twice in this case so to the base set we take maximum maximum from these two numbers in this case two yes but for yellow yellow classes we have in blue classes yellow elements are suggested three times and in red classes they are suggested one two and and four six times so to the base set we take six yellow elements so in our case the base set is like this this is a single line is just a base set yes so a base set contains two red elements six yellow elements and five green elements again five five is the number suggested by blue classes here and and red classes suggest four four green elements so we take five and what we know what we know what is guaranteed by this solution the solution guarantees the system was created in such a way that the solution that the base set may be divided not divided in the base set we may distinguish one copy of class of type x and one copy of type y we have enough elements of appropriate colors for these here is this e1 class and here is sorry this blue class and here is this another blue class a blue class of type x class of type y and similarly for e2 classes we have enough elements to to have two classes of type a two a and four classes of type b they are they are they are presented here yes this is because in our base set we also we always took this this greater value of elements suggest from those suggested by e1 classes or e2 classes of course we may have some elements left yes in this we have three elements if we divide class divide the base set into e1 classes we have three yellow elements left if we divide into e2 classes we have one green elements left left it means without its own class but using our lemma we can simply add these elements to a class which contains which contains yellow elements because we are sure that those remaining elements are not have not have types which are not royal because for royal types types we always have a quality yes and all of them will be taken to or royal elements are taken to these classes suggested by the solution the elements which which remain must have non-royal types and if they have no royal type then they can be adjoined to classes containing realizations of these types using our our lemma and we ensure that there will be such classes by by this inequalities this kind of inequalities similarly the green element this green element may be adjoined for example to this class and to connect these two to green elements we will use this pattern from admissible from admissible class of type c which is not realized in our model at all but we have a pattern of connections of connection okay so so this is one implication yes if we have if our system of equations has a solution then we can reconstruct the model in the opposite direction we use in the opposite direction we use or maybe I will show you the slides in the opposite direction you use this lemma which says that for every class can be approximated by a class realizing exactly its M counting type so if we have a formula and its model then we take a solution to the system of equations the number of real is if x is a type is a M counting type of a class yes so we check how many times this counting type is realized in the model this M counting type is realized in the model how many how many times how many classes have exactly this number of realizations of one types counted up to M yes so for example a class with one red two yellow three green and for example four green elements is counted as a class of M counting type x yes and maybe I will skip the skip skip the detail argumentation but but this theorem this this lemma about approximations allows us to prove that the system of equation is okay so let us now think about bounds of on the size of models we can obtain this this way and the computational complexity of the problem so this lemma is taken from the paper from the paper by calvaneza but in fact it is built on the classical results on system of equations by papadimitrio so let us assume that we have a system of linear inequalities or equations with n unknowns m inequalities or qualities and n unknowns and coefficients are absolute value of a coefficient is bounded by a then this system has a non-negative integer solution of course we are interested in non-negative integer solutions yes because these are numbers of realizations of classes so it has to be natural numbers so the system has such a solution then it also has a solution whose in which values are bounded by such such such a number it is not hard to see that in our case this this number is doubly exponential because m the number of of inequalities the number of inequalities depends on number of one types we construct two inequalities for each one type we have n we have doubly exponential number of unknowns because unknowns are m types of classes m m counting type of types of classes and coefficients are exponential because in each such m counting type we have at most exponentially many realizations of each each atomic type yes so so in fact um we have doubly exponentially bounded solution so doubly exponentially bounded number of types of classes but in each in each in each class we will we will to each class we initially put put only uh exponential number of elements so so the whole number the total number of elements will be doubly exponential and this this is this theorem so so uh our construction from the beginning of this talk when we constructed a model of double exponential size was essentially optimal every finite satisfiable formula we have a model bounded doubly exponential which follows from from this construction okay but we want to obtain a non-deterministic exponential bound on the side on the complexity of the satisfiability problem and this theorem suggests a simple procedure just guess a model double exponential size and check that this it is a model yes but this procedure is not in doubly exponential non-deterministic time well how to obtain uh seeing the exponential time uh observe that our equations are of of very simple form they are just c one x one plus c two x two plus plus c three x three and so on greater or equal to to a number b which in our case is either zero or one all of our inequalities can be rewritten to such such a form yes so it is not hard to see that instead of solving such a system of equations over integers we can solve it over over real numbers yes at this moment we are not interested in constructing the structure for for formula phi but only checking if this formula if this if this if this model exists yes so we only want to solve the system of equations we guess the set of one types which will be realized and this is exponentially of exponentially bounded information and then construct a system of equations and try to try to solve it yes but instead of solving it over integers let's solve it over real numbers of course if there is a solution over real numbers then we can multiply this solution by the product of denominators in all for all in the solution for all variables and we obtain integer numbers then and those numbers obviously will satisfy these inequalities because they are very simple for greater equal or something yes if if those real now if we multiply this side by by some number then of course by by some positive number of course then the relation will be still retained okay so we may solve the system over over real numbers instead of integer numbers and we can use a simple fact from linear algebra because what is the shape of our our system of equations we have exponentially many equality inequalities and doubly exponentially many variables instead of of course of course such a system if such a system of inequalities has a solution then we also can construct a system of of equalities which we have a solution so let us think about equalities and in linear algebra we know that from linear algebra we know that the rank of these metrics is at most 2 to the n yes so we can we can move all the variable variables we can find a non-zero maximal non-zero minor we can move all the in all the variables not belonging to its minor to the right side of the system and substitute them for zero yes and we will have we will still have a solution in which only only those elements only those variables will be non-zero yes so the theorem is the theorem the observation is that if we have such a system of equation over real numbers then we can always find which has a solution then we can always find a solution in which only exponentially many unknowns has non-zero variables yes and all the others are zeros this can be done for real numbers not for integers yes because we require that this this system is over to do to do this thing we want this system to be over a field no not just over the rig so what is our algorithm now we guess we guess which types will which which variables will be non-zero in our solution so we guess the set of one types and we guess the set of those m counting types which will correspond to non-zero variables and now we have only and we construct the system but using only those variables yes so we have a system with 2 to the n equations and 2 to the n variables we solve it over real so we can do it in deterministic time though deterministic polynomial time so we finally get it is not important that this is deterministic because we have a non-nondeterministic algorithm because we guess something yes so fine we guess exponential number of information construct a system of equation and solve it all of this in exponential time okay okay so I have a comment here that this technique of describing models by system of equations or inequalities is very often used has been used in many many many examples I just mentioned two things which are related to our to our talk because they consider two variable logics in fact yes the first the first two papers are about description logics with number constraints yes description logic with number constraints the second one just two variable logic with counting quantifiers or one one variable logic with counting quantifiers so for example one variable logic with counting quantifiers it it has quantifiers of the form for all x there exists at least m m x such that or there exists at most m x some such that yes so in this case this this systems of equalities are even more more natural I think so and it was used by Ian this approach was used by Ian Prathartman to produce an easy an easy proof of of a simple proof of the two variable of decidability of two variable god of two variable fragment with counting not not of decidability but of next time upper complexity bounds the previous the previous proof also by Ian Prathartman was much more much more complicated and the previous results implied only decidability without without precise complexity bounds okay in this description logics description logics are also based on two variable logics and and these numerical restrictions correspond to counting quantifiers okay so maybe I have two more two more things to show maybe I will start from the third one the second I would just state the second one yes so the the statement of this next theorem is that the satisfiability problem for two variable logic or even to the guarded fragment with equivalence relations which can be used outside guards in our in our construction we we use the the fact that equivalence relations are only in guards so if we allow them outside guards then we can show undecidability using three equivalence relations so this this is one thing and to show that we are very very close to the border between decidable and unsidable we can also show that two variable logic with full two variable logic with two equivalence relations is decidable yes so so three equivalence relations are undecidable and two are decidable and let's I am not going to to to show you details of the theorem but I would like to show you some crucial differences in this situation and in in our situation two variable guarded fragment with equivalence guards so what are what are the differences that this problem is much harder yes what are the differences the first thing is that we of course we may enforce non-empty intersections of e1 and e2 classes because we may simply say that for example there exists an element for every x satisfying p there exists an element y connected to to this x by e1 and connected by e2 yes there are no there are no restrictions on the pattern of there are no guards in fact here necessary so this is the first thing another thing we may say that some types of classes regardless of this notion of type but some types of classes are realized exactly once in our model for example we may say that every pair of every element satisfying r are contained in one e1 class yes and we cannot cannot when we build our models we cannot take a copy of these classes because this is forbidden we could also always do it in in the guarded fragment here we won't be able to do it of course a simpler way of way of doing is just saying that there is exactly one element satisfying p in the whole model yes which which is which can be done without equivalence relations then then both its classes will be unique in some in some sense it appears that we may even if we consider general satisfiability problem we may even even enforce infinite classes this time remember that in the case of two variable guarded fragment with equivalence guards we had we could always build models with exponentially bounded classes in the in the general it is also much harder to deal with with free witnesses with those witnesses which which lie not not in the equivalence class of a given element so there are some some some complications in fact we can enforce um triply exponentially models of three triple exponential size this time using this using this this this fact that we may have some non-trivial intersection of classes this is the picture the same picture we saw previously this time we consider those bullets to be non not just individual elements but intersections containing two to the n elements each and having such an intersection we may encode in such an intersection a number from the set two to zero up to two to the two to the n because we have simply we have two to the n elements and we may use an additional binary unary symbol which works as zero one yes and we have a binary representation of a number from from the set zero up to two to the two to the n and having in f of two we can we can say that two intersections connected by say blue relation encode values which differ by one this is not obvious but but not not not difficult just just a complicated pattern of quantifiers some some are using a variables and it can be done so so we may say we may enforce we may enforce full binary trees of doubly exponential depths and then we may again say as similar as previously we may say that all leaves are in one blue class for example this time we do not even need to to use this route we may simply say that if element is a is a leaf we can be checked somehow then if a pair of elements in both elements from from this from this pair are leaves then they are connected by blue relation so we may have models of triply exponential size and classes of triply exponential size what is the idea of of the proof so so the first thing is natural we do not want to deal with these intersections so instead we will treat them as individual elements but to do this we have to prove that every model every formula which is satisfiable is satisfiable in a model with small intersections and it can be done so so so now such small intersections we have we have so we have exponential bounded intersections so there is two to the two to the doubly exponential number of isomorphism types of possible possible intersections yes and we will treat them as elements but we have but now we have two to the two to the two to the n possible types of elements and we proceed similarly as in the case of two parable guarded fragments with equivalence guards of course there is a lot of a lot of details some of them are quite interesting some of them are quite annoying but but finally the procedure is to quest some information of doubly exponential size and construct a system of linear inequalities but this time the system is larger it has this is one thing that is larger of course it has two to the two to the n variables and two to the two to the two to the n sorry and doubly exponentially many very equations or inequalities and triply exponentially many variables so the situation is similar as previously as previously but but the problem is the problem is that we may hey that we do not have this nice form of inequalities so of course of course we may solve we may solve this system in triply exponential type triply exponential non deterministic time obtaining three next time upper bound on the on the complexity but we want to obtain two next time one exponential lower and the difference in comparison with this guarded fragment is that we will require equations of the form x plus y plus z is not greater than than one these are these that they are because we may say that they appear because we may say that some types of classes appear exactly once in the model so we simply say types of types of classes of our elements so we may have something like this so maybe sometimes we have that x is equal exactly once maybe sometimes we'll have the x plus y plus z is not greater than one and this trick we've taken real solution and multiplying it by denominators will not work because it will not preserve not necessarily preserve this kind of inequalities but it appears so we have to we have to solve the system of equation over integers this time and it appears that there is a similar theorem not not not that simple like like this one but eisenbrant and spawning could show that if such a system of equation has such a system of equation with not too many equations but many variables has a has a solution that it has a solution in which only polynomially many with respect to the number of equations polynomially many variables is non zero so exactly what what we what we want yes but this is slightly this is not this is also not not very difficult theorem but not that obvious as in the case of real numbers so we proceed analogously yes we we guess which variables would be non zero construct the system of of equations ignoring those unnecessary variables so we will have doubly exponentially many variables and we have to solve this system over integers but this the problem of solving system over over integers is in mp so so everything is okay we finally finish in doubly exponential non deterministic time a lower bound bound can be also can be also shown uh it was shown by ian pat hartman using using a similar trick to to the one uh this this one we can't this trick we can't get to construct construct a grid of 2 to the n of doubly exponential size so the problem is too next time complete okay so i have some i have some more time so i will go back to to this undecidability uh theorem okay so now i will try to explain why two variable guarded fragment becomes undecidable if we allow equivalence relations outside guards so i will use a reduction from from domino systems a domino system is just a tuple consisting of set of domino tiles and some restrictions on horizontal and vertical connection and the problem is to tile a grid of grid n times n preserving those those constraints yes so so we simply say these constraints say which pairs of domino types may may be horizontal neighbors which types of domino types may may be vertical