 Zelo, najbolj je, da je vse formula, ki je skupnila grids. Grids in tudi tudi. Danes, da se povrste 2 varjezje, je zelo povrste, ki je skupnila, tez konstrajenje. Proma je, da se skupnila grids, In ovo je tukaj. Všeč nekaj smo boželi? Vojte 1 formula. A formula eta. Takim, da, Kaj bi prejnil najbolj bolj na večen izgledov. Začunila. Prvana je 2 predikatje. H is horizontal relation for horizontal neighborhood and V for vertical. So we expand this language by some unary predicates, unary predicates and three equivalence relations. And we want this standard grid, we have to be able to expand this standard grid to a model of alpha. This is one property, which we want. And another property is that each model should satisfy the property that for every x there exists horizontal, vertical, vertical, etc. And the third thing is that it should satisfy such... That the relation H should be... In every model, relation H should be complete over V. So if we have such a pattern in a model, then here we need H connections. If we have these two properties, then from every model we will be able to extract a grid. And it is enough for our purposes. Ok, so having such a formula, there would be no other problems. So the problem now is just to construct this formula. And here is the expansion of our grid. So how we expand the grid? We... You probably can't see those numbers here, but this H is not the grid relation. The grid relation is not presented on this picture. And here we have unary predicates H0, H1, H2, H3. And again... I don't remember, no. Something is wrong here or something is invisible here. I think there are three types. Zero, one, two... Three types, sorry. Zero, one, two, zero, one, two, zero, one, two. On both edges, yes? And we divide... We cover this n times n grid with tiles which represent equivalence classes. So these black tiles represent e2 classes, these gray represent e1 classes, and white represent e3 classes. The desired division into these classes should have such a property that it is regular. And, of course, two tiles of the same color are not objective. In fact, we treat this pair of squares as a single tile. Having... There are simple patterns than this one, but I have this one on the picture. Observe that such properties cannot be obtained with just two equivalence relations, yes? And, of course, these tiles have to be of bounded size. In our case, each tile covers at most nine elements. So they have to be of some constant size, bounded by a constant. And having three relations we can obtain. Having two, it is impossible. Of course, it is much, much easier with four, let's say this is one color, second, third, fourth, and then we repeat the pattern. But with three it is also obvious how to construct such an expansion. And now what will be said by our formula? So our formula captures some properties of this expansion of the standard grid. So it says that there exists the element of coordinate 00, and it explicitly says that every element is a vertical and a horizontal successor. So this is this initial fragment. We have a formula which aximatizes grid relation h, which has the following shape. It simply says that if a pair of elements is connected by h, then this element satisfies some formula, which here we have an example. This formula says what are allowed one type, coordinates of x and y and connection by equivalence relations. For example, if we have nine such formulas, each for the possible coordinate here, possible combination of coordinates here. For example, phi h 00 tells us about 00 is here. This is an element of type 00. So it says that if there is h connection, which starts at this point, then the second point should have coordinates 10, and the connection between them should be only e1, equivalence connection. We have similar formulas for all possible combinations of coordinates. Analogus formula, of course we may have more than one equivalence relation here. Sometimes we will have two, as in this case, for example, this is 01, then the next element is connected by e1 and e2 to this excel. We write a similar formula for vertical connections, and then we write a group of formulas which will ensure this property, this completeness of h over v. As we will say that if we have some elements that are connected by one of the equivalence relations and have appropriate types, then they are connected by h. Let me show you how it works. In our model, we know that there is elements of coordinate 00 and that it has vertical and horizontal successor. This is stated explicitly in our initial formulas. Probably this picture is, again, not visible. So maybe I will write it on the blackboard. I will draw something on the blackboard. So we know that there exists an element of coordinate 00. It has horizontal successor and vertical successor. Horizontal successor, oh, and maybe I will just show you formulas. This formula enforces coordinates 01 on this element. It says that if a pair of elements is connected by h, then an appropriate combination of coordinates should appear. If we have here 00, then there is only one possible formula, one, zero, for this. We know also that these two elements are connected by e1. Similarly, this element will have coordinates 01 and will be connected, probably there is no example for this, and this will be connected by also only e1. Then, again, this element should have successor, horizontal successor, and it will have coordinates 02, vertical successor, which will have coordinates 01. The connections will be enforced, I will follow this picture. Here we will have e2 and e3, and here we will have only e1. And now what happens? These three elements belong to the same e1 class. So in fact there is e1 connection here. And using one of those formulas, again it is not presented, the appropriate formula is not presented on this slide, but we have a formula, which says that if there is a pair of elements connected by e1, one of them coordinates 01, one of them coordinates 01, then they are also connected by picture, or by black relation. Sorry, by h. They are connected by h, because this is written in this formula. But in the next step, we know that every pair of elements connected by h, having appropriate coordinates, is also connected by e2. So we will have this e2 connection here. And we will continue this procedure. We will build a new vertical horizontal successor, and we will always be able to conclude that those elements, those relation h is complete over. So this is just a sketch of the proof, of course, but this is the main idea. In this argument we use only transitivity of this relation ei. OK, so now I would like to go to conclusions. These were all the results I wanted to present. Some in detail, some in some sketches. Let me just summarize some results and tell you about some related results. So regarding two variable logics with equivalence relations, we have the following interesting small hierarchy, let us say. The satisfiability problem for two variable guarded fragment of equivalence guards is complete for non-deterministic exponential time. I presented this proof in details of finite and general case. I also sketched the ideas for the proof that fo2 with two equivalence relations is to next time complete. There is one intermediate case, two variable guarded fragment with two equivalence relations, but with these equivalences allowed outside guards. It appears that it is in 2x-type. The reason for this is that in this language we cannot say that a class is realized, a type of a class is realized exactly once, for example. This simplifies things. It simplifies things and allows to build three-like models in fact, three-like unravelings. And for example, we can use alternating Turing machines to check existence of such models, working in exponential space. What else? If we consider fo2, two variable logic with some specific number of equivalence relations, then in case of one equivalence relation the satisfiability problem is next time complete and we can show finite model property, exponential model property. So, simply the proof is just take proof, the first step is to show that every satisfiable formula has a model with small classes and then choose only some number of these classes to construct a model. This is quite simple. Two relations are to next time complete as I tried to explain you in some sketches. And three are undecidable, which I also want sketched to you. Another related result, instead of equivalence relations we may consider transitive relations which are more interesting and more general in fact, because having a transitive relation in two variable logic we can say that it's an equivalence. In the guarded fragment with transitive guards it is not that obvious because you cannot say for all x, y if x, y, then T, y, x because then you use this transitive relation outside the guard, but it is also possible to enforce transitive relation is an equivalence relation. So what are the results? Guarded fragment with transitive guards is decidable and to x time complete so it is harder than equivalence guards. F02 with two transitive relations is undecidable and the proof also works for guarded fragment if we allow those transitive symbols outside guards. And it is interesting that the case of F02 with one transitive relation is up to my knowledge open. I tried to solve this at least two or three times and it is quite difficult. The combinatorial nature of the models is quite complicated. So up to my knowledge this is open. One transitive relation in F02 a simple question we do not know even if it is decidable. Another possible kind of relations which may be worth investigating are the linear orders. They are also natural of course. And F02 with one linear order is decidable. It was showed by Martin Otto sometime ten years ago and it is quite easy construction. If we have two linear orders and the situation is much more complicated there are results by there is a result by Spentik and Zelme but this is for the restricted case in which we have only two binary predicates and those linear orders and no other binary predicates and their argument works only for finite visibility. Up to my knowledge the general cases is also open here. There are also some other related results to this case when we allow for some successor relations also some combinations they are also by Thomas Selme by Amal Def Manuel there are several papers and the linear order leads to decidability. Maybe one one more thing which is worth mentioning here is work by by the group Mikolaj Bojanček, Ankem Ušel, Thomas Spentik, Kuksegu Femme, Klerdevi they they considered the so called data words and also data trees which are not on this slide data words are words over a finite alphabet whose positions are naturally ordered by a linear order so we have a linear order and every position carries so called data value and these data values are represented by equivalence relations so two nodes are equivalent if they carry the same data value. This is motivated by XML and it appears that this is decidable if we allow for one linear order this natural linear order induce successor relation and equivalence relation so this is a combination of linear equivalences but this is only non-elementary upper bound is known here and it will be very difficult to improve it it becomes simpler if we do not allow this successor relation I don't remember the complexity but it is much, much simpler but again in this setting there is no other binary predicates that is just three binary predicates mentioned here and only unary predicates in fact these are just words ok, I think of course there are much more results but I concentrated on those related to satisfiability in restricted classes of models restricted by some requirements of equivalence relations for equivalence relations, transitive relations or linear orders ok, thank you very much