Speaker is Richard Statman, who in his short abstract promised us a somewhat eccentric survey to on algebraic and combinatorial structures intrinsic to functional programming. So please start. Yeah, thank you. So as a result of some of the really useful remarks of the referees, I'm going to concentrate on just one aspect on the groups. And there's a long list of references in the extended abstract. So I won't present references in this talk. The picture here, of course, Bosch's vagabond son. But I want to talk about among other things, the son of the vagabond, group. So this is the usual presentation of lambda calculus, which we all assume numbers of times already in this conference. The main thing I want to emphasize here is that data will be used throughout. It's rather important to talk about even my heads. So we will have data almost all of the time. And of course, here we have combinators. And the particular selection of combinators I've used are actually churches, BCKIW. And of course, for equivalents, we need strong reduction and the combinatory axioms that would be handled. Now, way back when, it's got to be in the 1930s, church observed that lambda calculus forms a monoid with a product equal B, the combinator, and the identity, the usual identity. And if we're interested only in the combinatory version, we have to get enough axioms for for a conversion, really, to make a monoid plus. We have the following axioms for B. Written this way, they're sort of incomprehensible. But in this way, where B is thought of as a product in the monoid, so B distributes over product. And we have this interesting axiom about how B commits with B itself. Now, another way of looking at this last axiom is remember that B can be distributed in the other direction. So one way of thinking of this is if I have a stack of B's preceded by another stack of B's which is shorter, then I can commute them this way by increasing the second stack by one. It turns out that these axioms are precisely the axioms you need in order to generate a normal form, which shows that if we look at the monoid generated by B and I alone, just a couple of combinations of B and I, using those axioms, we can put things in a normal form and we can prove that that monoid itself, I'll go one step in advance. You can prove, and I'll say why in a minute, that it's a cancellation monoid. And it generates then a group. And that group turns out to be exactly Richard Thompson's group F. One of the so-called three vagabond groups, they were discovered by Thompson in 1962 in connection with trying to prove some theorems about groups and the decidability of the war problem groups. It was really challenging in that respect to prove the undecidability of the war problem groups and has no apparent connection with land accountants at all. Now, what is Thompson's group? Well, I'll give a nice description of the group later. Let me remark that the reference to Cannon Floyd, I don't know the third author. Cannon Floyd is a very, very good survey article on Thompson's groups. And even there, they go to some lengths to present group F in a very complicated manner. We'll see this in much simpler representation later. At any rate, the easiest way to understand it is simply to give it as an infinitely generated group that satisfies this axon or this collection of axons, which say that you get the n plus first generator by conjugating the nth generator by any of the generators k less than n. And this holds for all k less than n, and that's where the axon's coming. It's not simply a definition of it n plus first generator. It says all of these are equal. That's Thompson's group. So that's the group one. It comes up in a very funny way. And the bi monoid is very small fragment of land accountants. Now, here we talk, the bi monoid generates F, so a little description of how that works. So these are subscripts. Super scripts, subscripts. Yes, I apologize from the way I write slides. I'm sort of a dinosaur with respect to creating slides. And these are the kind of slides you would have created 30 years ago. I would have created 30 years ago. I'm not very good at detecting. I didn't want to try that. So here's v0 and vn plus 1 is simply applying v. And the subscripts are simply composing v with itself nine times. And it's very easy to write out what the normal forms you get are. Because after all, it's a well-known fact that if you take a particular combination of vn i, they're all normalizable. They all have simple types. They all give you proper combinators. That is all the numbers that you're up front. And what you get is every tree with the variables that they leave send exactly the right order. That's exactly what you get. And so it's very easy to see what these would do. Now, our axioms before then show us people put these in the following form. That is a stack like this and n1 and 2, so on and nk, where the superscripts here are non-increasing order. There are many combinatorial ways of describing what's going on here, but just write it that way. And indeed, it follows almost immediately that you get the homomorphism from f into the vi monon. Now, it was well known that f is simple, so we're really done. But we don't need that, since it's a cancellation monon, so it generates the group. In fact, that would be a proof that it's a cancellation monon. But at the time, I did this 40 years ago. I didn't know about f, so I had to go out and prove. Okay. Sorry, could you just go back one slide? Just back to the previous slide. And just to check, I mean, here the star is the star, is what you defined before? Yes, the star of b. And I write it this way because b is associative, which you have on a player. So that's, yeah, that's what the, remember that was that off-handed mark at the beginning of seven slides ago was, use enough aided to make sure you have a monoid. And so b, when used this way, is in fact associative. So I wrote this way from the monoid. And you said, you said that this builds a homomorphism from what to what? Yeah. This establishes the fact that you can take f, the group f in the generators, be careful. Okay. Move the inverse from one side to the other. So it's a monoid. And then we just showed that those axioms can be realized here. And that establishes only a homomorphism from f onto the b on monoid. What is the, what do you map the generator g of n? What, how do you interpret the generator? Yeah. Um, yeah. Oh, yes, I'm sorry. That's the next slide. Yeah, g of n is b to n. Okay. And that's the infinite, infinite, f can be generated because, finally, generated, generated by two generators. That's not true. But I use the infinite number of generators because it's just easier to grasp the axioms, because the axioms are pretty strong. And you can see, if we go back to these axioms, if you put that over here, okay, and make it a monoid equation, that's what we have here essentially. If you could stick, you can put a stack of these on top of both, stack of these here and stack of these there by this equation. So that's essentially it. Okay. Okay. Thank you. Ah, okay. So that's group one. That's, um, so the group f is generated by the non-monoid. It is not the group of the monoid, the group of the monoid itself. Now, yeah, I'm going to use that terminology again and again. If you have a monoid, then there is a set of convertible elements. That is elements that have right and left inverses. And of course, if they have right and left inverses, they must be the same. And that's called the group of the monoid. And that group, for a lot of monos, for most monos, I guess, that group is trivial. It's the identity. But in cases that we're going to be talking about, it's much better. So we'll be talking about the group of the monoid. Now, let me talk about one of my favorite subjects, surjective parent. Okay. So church started with something with delta function, but it only was for, yeah, about a little history of the surjective parent that you'd go with. It was only for beta normal ones. And none of them clocked through that surjective parent. Yeah, here it is. Pairing, surjective. It's not church loss. We'll do real show that it's consistent. If you're interested in this extension of the land accountants, very, I mean, it's the paradigm extension with a non-lift linear unit. Let's put it that way. Very nice, although much later than most, a very nice survey paper was stolen in 2006. I really like that. So I'm excited. What does that have to do with what we're talking about? Well, surjective pairing is part of this extension of the land accountants. Now, I would let me introduce the notion of cartesian monides. There are more monides that correspond to surjective pairing. So there, what we're going to do is we're just going to take a surjective pairing, actually, usual axons, think of the first order axons, think of a model of these first order axons, and now think of lifting surjective pairing with functions to use white-point-wise. So, pair of two functions is the point-wise, and they satisfy these axons. These are not surprising. If you get one extra thing, that is, that composition distributes over here. And now, if we put this in the land accountants using A-day, we get the fact that it's an alternate beat, except here's B, and we're using A-day all the time. Interestingly enough, this is a fragment of Bacchus Fp. So, we have another connection. Now, Bacchus Fp does not contain land accountants, but it contains a lot of axons. In fact, it contains at least all of these. So, here we have another connection to functional programming, which could not go through other factors. So, I first started talking about Cartesian monologues in 1996. It's not my idea of Cartesian monologues. Our land accountants, terminology, but their long list of references that we extended after. Now then, let us... So, here you go. Here's a first order theory. You've got that for a minute. Here are the axons, over enough axons for a monologue. I is an identity monologue associated with a binary operation, multiplication, and L-O-R constants pairing up a binary operation. It's in the universal theory. There's a pre-monologue, pre-objectives, pre-Cartesian monologue. So, where am I going to go with this? I'm going to show the view that this monologue, of course, it has a group. And what is the group of this monologue? It turns out that the group of this monologue is another one of Thomson's groups, the biggest of Thomson's groups, the Dagobon group itself, B. Kind of interesting. That means that B is part of Lambda Capitalism. It's rejected. So, let me talk a little bit about that. So, we have the pre-Cartesian monologue. Talk a little bit about how things go. And the reason why I want to do this is because I'm going to get a very nice compact description of the elements in the group B. And you'll want to compare that to the Cannon-Floyd representation that group there is using. You'll see at least the monologue, this is much more prestigious representation. So, all right, so we have a term. W is well written. It's a member of the sub-loan on our agenda. That is a spring of hours and hours, multiplied together by its associate of source, like this one. Or it's a pair of six columns. So, that's what I mean by well written. So, yeah, let me, I'm going to turn on the lights for a minute and I'll throw a picture on my back. So, what is a well-written monologue? Well, where's the spring? What is it? It's the spring. I'm going to trade, I'm going to use this tree, our spring is not so much. Now, how do you multiply two of these together by the distributor board? You take this and you put it up here. And this spring projects to the sub-tree that it wants to protect. And you get something on the sub-tree. I think that's almost good. The total is good. I have the total, so let's give it a go. Okay, that's good. So, that's what these trees look like. Now, the one thing I should say about well-written expressions is, we've only used these axons here. Oh, sorry, I forget three. We used one, two, and four. What about three? Now, three comes in at the end. Well-written expressions can be extracted, can be contracted and expanded by using these contractions. And that is sort of how these are used. We've got a left over here and a right over here. And they're followed by exactly the same and that's absolutely right. Now, we can contract that with it. Of course, I can do both of them, but what is this, why does this have to do with all of them? Well, if you do a little bit of classical rewriting, you can see that two expressions in the theory of Cartesian monologues, two terms, two elements in the theory of Cartesian monologues are equal. If not only if they have a common, they can reduce the common on the form. That common on the form looks like many as well-written things. And the equivalence relation on well-written expressions is simply this. These are beta reductions and this is a data equivalence of beta normal terms. Really, that's what's going on. So we have well-written expressions. A well-written expression can be what I was about. Just a well-written expression. Moreover, W is unique modular expressions and contractions. That's what I'm just saying. As I say, this is an elementary exercise in rewriting using those axons. Ah, okay. Now, second fact, it's easy to prove that a well-written expression, W, is a member of the group of the Cartesian monologues. Every line's got a group. If and only if W can be expanded and contracted or contracted to a different well-written expression, remember they're all equivalent, such that tree, the new one, W, has exactly two leading in leaves. It doesn't matter what the shape of the tree is. Only that the number of leaves is two leading more. And every string of L's and R's of that length N occurs. So what you have is you have a permutation of these strings stuck at the leaves. You can read it from left to right. Interesting thing is it has nothing to do with the shape of the tree, the number of leaves, and every string should occur. Now, we could think of the normal sequence of strings. You could view them as lexicraphic. So what you have is some permutation of these strings and length N from the lexicraphic one to some of them. Ah, yeah, okay. So, we'll turn that around. Here it is. Again, I say there's no restriction on the shape of the trees because F is somehow being hidden in the condition. The group F is hanging on. You can change any one tree to any other tree. Another simple consequence of this is that the group F's elements are going to be in order. That's kind of easy to construct. You just look at the shapes of these things. Vagabond Group is the most intuitively defined as a subgroup of homilimorphism that you can express. The system of those elements of or those homilimorphism that are step functions. Okay, so what we have is I'll see what I mean. I have to put a step function in quotation because they're not exactly step functions. So, canter space, there's a set of equal binary sequences. Product apology. It's a canter space apology. It's a very dispute kind of apology. Not exactly. It's a totally disconnected house. And the homilimorphism with the canter space, and I'll be more specific about what these things are. When I say step functions, I don't literally mean that over intervals you get a constant, but I mean that over intervals they are one-on-one and onto a specific other interval. That's pretty weird. Let me be more specific. I'm not that the canter space, the canter space is a specific piece by step function. That's what I mean by these step functions. If. Okay, so capital F here is a piece by step operator. If. Give it an infinite sequence. F looks at that infinite sequence one bit at a time. Then I find besides it, you're seeing a few bits. Okay, then it takes those new bits, cuts them off, and sticks on another sequence called V. So it's a continuous operation on canter space. Once it's got enough information, okay, when it maps, the way it maps is it shifts the shift. It cuts this off, and it shifts G, which is the end of this whole sequence by V. Moreover, for any other H, it does the same thing. It sees U, it shifts by U. So piecewise shift operators. This condition here is why I call them step functions, or, of course, they're hungry enough for some reason. Okay, they're continuous. So these are all the piecewise shift operators. The piecewise shift operators that are bijectual, they are, of course, one precisely the group, the vagabond group group. So here, F, G, and L are infinite binary sequences. Now, if we set the identity, we set left is shift by zero, right shift by a one. And now the pairing operation simply says, if I see a zero giving the first coordinate on a phi F, if I see a one giving the second coordinate on G, composition, ah, composition, well then, is universal. Now, under this definition, let's have a piecewise shift operators for the Cartesian model, right? And the group of this Cartesian model is B. I guess that works kind of fast, but next group, all right, Mr. Marie Angelo, Angelo, the Sonny's group, and intuitive description. Start with the group G, consider sequences of G, H, E, G, and the finite support, the permutations PQ, the natural numbers, the finite support, the finite product, this way. You all know about the hereditary permutations. So this is the abstract group that would go with the hereditary permutations. Yes, and this is all described in the diagram. There's a simpler, much simpler representation on trees, but I prefer today to stay with the hereditary permutations. The random terms, the two-sided inversions are about the hereditary permutations. That is Mr. Gonzalez's theorem. Group of the lambda calculus model, the entire model is originally described by Church, under beta-A conversion. It is the cosine hereditary permutations. Okay. Now, so we have two different groups. We've got D and it's subgroups. Oh, yes. I should mention before I go on, there is one other group. Oh, I think I'm talking about it. Yes, the other group, other part of the diagram is T, the grouping between F and P. And it would be interesting to find exactly any, if there's an interesting fragment of the lambda count that corresponds to T. Already, if you look at the hereditary permutations, they're all linear. Okay. So already in the linear lambda captains, you have the focus on D, hereditary permutations, so they're already, so it's going to have to be less than linear. And I don't know exactly where to concept. Now, we have a different... Can I just ask a question? Yes. Can I just ask a question? So, I mean, so I mean, just to recap, so the group F is generated from the BI monoid. Yes. And the group V is generated by, is it the extension of the BI monoid with surjective pairing? No. It's not just generated. It is. Once we go up to surjective pairing, okay, the group D is the group of that monoid. In the case of F, F is not the group of that monoid. The group of that monoid is pretty much the identity. You have to add the inverse. Yes. But once you have V, of course, you have group F as a subgroup. Yeah, okay. That's sort of a subtle difference. Yeah. Yeah, there are things in the BI monoid, which is not, you could add them. It doesn't have a group. In fact, I believe coronal bone group with something like that is on stage. I would have to go back a little bit, some of correspondence, but yeah, I think he actually thought about it. It certainly is not a classical monoid. Okay, so this group is familiar to hereditary primary patients. The group from the surjective pairing is not so clear. What's the connection between the two? Well, let me cite an observation of young girl on a crop, which is quite easy to prove. It's around 19 maybe. I don't know exactly when. Finally, it generates subgroups, and it's an honest group of clients. There are no elements of infinite order. It's very easy to put in fact, like, you know, these hereditary primary patients don't have types. It's an easy answer. Now, of course, as I pointed out before, there are elements of infinite order in the vagabond group. Therefore, the vagabond group cannot be isomorphically indebted onto the dishonest group. Now, both of these groups are subgroups of the group of the monoid of theta eta plus surjective pairing. That is lambda count, this is surjective pairing. So that group is really much bigger. Well, when you say, wait a minute, maybe, maybe the dishonest group is isomorphically indebted into the V. Interesting question. What is this group? V cannot be indebted onto the dishonest group, open problem. What is the group of lambda, theta eta plus surjective pairing? Let me add, I can't prove that you can't embed the dishonest group into V. I conjecture you can't. There is something of a parody consideration just I can describe. I can't get, I can't embed it because of the way that direct limit goes. Intuitively, it's something like this. When you do composition of hereditary primary patients, you do an alternation of right and left actions, group actions. And you are always flipping sides in order to see the composition work well. So there's a flipping of sides which can't be done in the other part. That's probably, that's probably a comprehensive question. Prove, prove my conjecture that dishonest group, group of primary patients cannot be isomorphically indebted into V. And open problem. What is the characterization of the group of lambda, theta eta plus surjective? I think that probably spoke through the question. Any questions? Yeah. So I wanted to ask whether there are any questions or comments and queers exactly on time if you missed the exact time. And thanks for giving another open problem. I have a few questions. One is about another open problem. Sorry. Which you mentioned in the references that you gave in your abstract for an old technical report of yours, combinators and the theory of partitions. Yes. So you looked at some problems associated to the BI monoid. And at the end of the paper you, so at the end of the, so you prove the theorem that unification is undecidable. Yes. But you said that you leave the decision problem for monoid equations open. And what did you mean by that? And is it still an open problem? If you, this was from 1988, so maybe it's a bit unfair to ask this question, but do. I don't have to look back. This is probably, I didn't know what the collection is. Notice that I haven't, I have not mentioned, oh, ah, yes. There was another slide. I'll get to that in a second. Yeah. I have students who have ordered. Okay. And it's going to be talking about the complexity of all these decision problems. I know some. Yes. Yes. Okay. That's what I thought. I think I don't know the answer to that. Maybe there will be. But what is the complexity of that, if I get into these things? Yeah, I thought, actually, I thought that was. I think that may have been settled. Yeah. I think I know the answer, actually. I think that was settled by Berger. By whom? By R. I think I'm pronouncing it correctly. Berger. By R. And I think the reference should be, reference should be in the standard abstract one, but it's not. It's an oversight. I think it's, yeah, I think it's calling on the time. I think what it has to do is change the representation. But yeah, let me add, yeah, there was a, an afterthought. There is a sort of application to group theory, which I thought was kind of interesting. Recently, you were, well, I would call it an application. I don't have group rules. Let me do what you can. Sort of. Recently, we were able to use the Cartesian model representation to show that the following problem is decidable. You give me a finally generated subgroup. I want a membership problem to finally generate. So you give me a finite number of elements of read and another element is this other element in the subgroup generated on the first set. And the Cartesian model representation would be chosen. In fact, the decidable problem, the problems of this sort of need could not. And I don't think that follows easily from opening up to knowing about you. So I hope that's a good question. Okay. Okay. Other, no, we won't, you want to ask some more questions? I have some more questions, but I mean, I mean, I'm interested in all of these groups. I mean, even so the first, the BI monoid generating the group. So B, these, the combinators B and I are related to planar lambda calculus that Alex talked about yesterday. Also Kasia talked about, but one, so, I mean, whereas the B, C and I, that combinator basis generates all of the inner terms, B and I do not generate all of the planar terms. And some monoid, some larger monoid that somehow contains all of the planar terms were by planar. I mean, terms where the variables are used in order. Good question. Good question. It's a very vague, vague question. Okay. So it's kind of another open problem or okay. Thanks. Are there any other comments or questions? We still have like two minutes before switching to another talk. Yes. This morning, we were presented another problem on what was called the whole property for B. Ask whether this property has to do with another, you know, could be presented another way. And it seems to me that it is somewhat connected with the problem of those groups generated by B. So it's not clear, but it's some question for the audience, not for the speaker, but to look at the connection between the open problem presented this morning and the talk of that morning, that's what you're saying that you may have the idea. No, to follow up, I just, I posted a link to Knacknoz slides from the open problem session. Yeah. And I mean, he has this decomposition and he was using this decomposition. So I mean, it is related, I think, his question to these kinds of questions. Okay, thank you. Okay. Thanks for this comment. Yeah, I could, you know, the combinatorial, combinatorial medium of this representation, something like this. He builds up a, he builds up a binary tree by starting from a carrot. That's what happens like, come on, just a branch. These are binary trees. There's no line in the binary. So, and then to get the next tree, you add a carrot song. And that's basically what these opportunities are doing. They're generating, so these are all proper combinatorials we have where the variables just are listed in, you know, in American order from that to right across the top of the tree. And we're just sticking out a carrot. Yeah, that's all this thing does. This is delayed, delayed composition, which is a simple stick in a carrot. So that's the combinatorial medium of this normal form, which is, all it says is that you can build up every binary tree. That's all it said. Algebraically, it corresponds to 10. See what I mean? That's the last one, but of course you're going to do this. That's the combinatorial medium. Okay, thank you. I think we have to stop at this point. So let us all thank the speaker. And of course, we can discuss it in Discord. And now we turn to the last slide in this session and


Thank you, sir. Good afternoon, sir. Good afternoon, gentlemen. Good afternoon. Congratulations. Thank you. President, how are you doing? Good. How are you? How are you? How are you? Well, I know we have a lot of time here. We have a team of the other states that have heard and carried the information and been convinced that I need to hear from you about the results and the acre that you would like to clean the dramatic. The first time I am telling you this, but today is the 20th of May. It's been a long, really long, and I should say first I believe that Tom Coling The leadership has been fully supported in Congress. I think we're on the brink of a good agreement. It is not the last word, as you have already said, that we will ever speak, no matter if that's a control, but I think it works well. I think Bob Bell and others will have other suggestions in the future on the upper seat. But this is, I think, a good sound undertaking. It's worthy of bipartisan support. I hope you feel so. And that we can move forward with this here. Jim? I agree. They were working hard back then. Thank you, sir. It's exactly that way. Bob? I want to comment. I'm fully and all the others who participated, they were considerably hard and long. They were very archivist and short-term, but they stayed with them by a door. Tom Foy doesn't serve her job always. This is an example of that. I think we've got a lot of work. Just to mention, with your leadership, we've been fully supported in the final results and we're in a long yet productive negotiation. One hour, Jim and Jim, we're going to have a new tradition, which I can support. I'll be pleased to work closely with all of you to gain the support of the Congress. The outset of your negotiations, I said I would prefer to have a bipartisan agreement with Congress on this budget, not really proceed with a sequester. And that remains my view, and even though all of us here around the table would likely have a somewhat different agreement than the one before us, this is a compromise that I think represents a good, safe effort. All right, Bob. Jim Wright, let me start with you. President, I agree with you. It's a bipartisan agreement. No one has got everything that we wanted, but I think that very fact demonstrates its value. And you can truly represent it as a bipartisan agreement. It's going to take the help of all of us, and I think that's it. We're going to need enthusiastic support. It's infinitely better. And I think that for both sides, and for all of us here to represent and what was happening otherwise, we're going to see the situation. A great point here is after act college, we'll all be able to get a majority in the Democrats. President, I think we all have to be very upbeat about this package. It's a result of a lot of hard work on the part of a great number of senators and House members on both sides, and here to represent this office. There's already some bad mouthing going around. It's not what everybody in the anyone in here would like to wipe for himself, I'm sure, but it's the legislative process at work.