 Pomeni se je, da vse vse poločje, ki je ukljeno grez. Grez in tudi v zelo, pa posledajte obroženje tega zovana. Pomeni je, da vse doprestrati grez in ko dopljemo to. Kako so namnešli? Mnošli formula eta, nekaj nekaj standard je, da so ga n-etak več, m PMMzv tudi model eta. Zrbeče, ker ta standard bili 2 predikat. H, v svetu horizontal, svoj bolj, za hrojzontarje vso knightsi i vrti. Blah smo in začeli pri warno, traje pri kliklje, in v tih equivalenja. Danes smo iz selectu. Počeli smo podgebiti starf.bez momentov, kako smo počeli. Enše nekaj popričjave je, da vse modeli zbih vzestavuješ še samopratilj, da pri vse mači postah pošliHex z hodnji z težkou, še vse vse vse vse krati, in vso pošluši, da sem vse, ki se zapotržaš vse in še in vse auto je vse vse importovan na vstah, kako bi se venomje pošli spunjati. To, da sili sem vse konnečno, to vse bo več vse, ki je zašel vse krati na vstah, potem se potrebno izdajš, da sva vse postrže vse pristo, z vsem modelom, da smo vse možemo pošličiti vzelo. To je nekaj pošličit. Tako, tako, nekaj ne bo nekaj pošličit, zato je to pošličit, da je to pošličit. In taj je vzelo, da je to vzelo. Zato, če je, da smo vzelo? Bratube, prezent. Nisem tudi počke ligi, kaj si nekaj pogledaj, ali tudi h je ne. Adverbo. Vzelo je res page, pa ima h štev izvennjevega h0, h1, h2, h3. Zobu je, da ne bo. Nebo, bolj. Mi se da se ti tudi trebili in učinili. Tudi trebili. Tudi trebili. Tudi trebili. Tudi trebili. V nekaj objev. Alež... ...vež... ...zapravljamo... ...zapravljamo to n. Vsega. Tajle to odličaj. Vsega. Vsega. S tem način včeliči taj nekaj držav, taj grej, taj nekaj e1 včeliči in taj nekaj je e3 včeliči. The desired division into this classes should have such a property that it is regular. And, of course, two tiles of the same color are not object. In fact, we treated these pairs of squares as a single tile. tako počke. Očori, vse če glori nekaj nezaj. Tudi imaš tudi izmaj tudi in življ so počke. Prejtvoj, če neke mali proprenje z dve obsačke. Mte tudi tudi 3... In zdaj tudi počki še nesel počke pošnev. Čedno tudi počke zdi nove, nekaj ne znane. ... ... ... ... ... ... ... včešlji se dogron, z kaj uvijelj na zašličenje vsega. A različno je zašličen početanje tudi pri zašličenju. Zato je tukaj, da tukaj se načinje zašličenje zašličenje za 0.0 in izgleda, da početanje zašličenje zašličenje zašličenje zašličenje za šloj. Tukaj je zašličenje zašličenje. Serim više, ki je o stemitrstvo Voskrtima hr familymenje, kako jaz šešal. Seltj je, kako je oeleta v vseh hr. Vseh Satisva vseh vseh vseh vseh vseh vseh vseh vseh kako je, tja je tje, ko se izderjajo s monetima, ko je ovarjski drčas nebo zrčo in okljavanje vseh vseh vseh. If we have nine such formulas for possible coordination. For example, hOO tells us about zero zero is here. here is another element of type 00. So it says, that if there is h connection that it starts at this point, then the second point should have coordinates one, zero, and the connection between them should be only one equivalence connection. And we have similar formulas for all possible combinations of coordinates. An analogous 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 0, 1, then the next element is connected by e1 and e2 to this x element. We write a similar formula for vertical connections, and then we write a group for 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. And let me show you how it works. So in our model we know that there is elements of coordinate 0, 0, and that it has vertical and horizontal success. This is stated explicitly in our initial formulas. Ok, 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 0, 0. It has horizontal successor and vertical successor. Horizontal successor, oh, and maybe I will just show you formulas. This formula enforces coordinates 1, 0 on this element. It says that if a pair of elements is connected by h, then an appropriate combination of coordinates should appear. So if we have here 0, 0, then there is only one possible formula, 1, 0 for this. And we know also that these two elements are connected by e1. Similarly, this element will have coordinates 0, 1, and will be connected, oh, probably there is no example for this, but this will be connected by also only e1. Then, again, this element should have successor, horizontal successor, and it will be in, it will have coordinates 2, 0, vertical successor, which will have coordinates 1, 1. 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 of those kind, 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 of coordinates 0, 1, one of them of coordinates 1, 1, 1, and they are also connected by picture, oh, by black relation, e to s e, 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 have this e2 connection here. And we will continue this procedure. Yes, we will build a new vertical horizontal successors, 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 of 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 nondeterminist exponential time. I presented this proof in detail, in both 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 time. 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 the 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. Simply the proof is just 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. Two relations are two next time complete as I tried to explain you in some sketches. And three are undecidable, which I also want to sketch 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. With transitive guards it is not that obvious, because you cannot say for all x, y, if tx, y, then tyx, because then you use this transitive relation outside the guard, but it is also possible to enforce that transitive relation is an equivalence relation. So what are the results? Guard with transitive guards is decidable and two 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 these two or three times and it is quite difficult. The combinatorial nature of the models is quite complicated. 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 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, those linear orders and no other binary predicates and their argument works only for finite satisfiability. Up to my knowledge the journal cases is also open here. There are also some other related results and to this case when we allow for some successor relations also some combinations they are also by Thomas Zelme by Amal Def Manuel, there is several papers about this. There certainly are other leads to decidability. Maybe one more thing which is worth mentioning here is work by the group Thomas Schventz, 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 a so-called data value and these data values are represented by equivalence relations. 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 is not a linear order induced successor relation and equivalence relation. 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 and 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 but it is similar to unary. Of course there is much more results of variable logic 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. Thank you very much. That's all.