2019-04-22 – Between Midrash and Logic – Lesson 12

Okay, let’s begin. So we’ll finish today around 2:20, something like that, because of Nitzotzot. I’ll do about half an hour of lecture. We’re also finishing the semester, so maybe this will be a good point to stop at: I’ll present the players, we’ll put the players on the board, and then next semester we can start working with them a bit. Okay, I’ll summarize a little of what I did last time, and then I’ll go over it again because it needs a bit of updating. We started with… I’ll start here because I’ll probably need the other boards too. We started with a table like this. We marked A and B, A and B, zero, one, one, and a question mark. That’s kal va-chomer. Let’s write that over here on the side, this is kal va-chomer. With fill zero, if I put a zero here—maybe let’s do it. With fill zero, then basically we have here zero, one, and one, zero, right? Meaning these are B and A, with no connection between them, right? Here it’s zero, one; here it’s one, zero. Neither column is larger than the other column in all the rows. There’s no simple relation between the two columns. Therefore there are no arrows between them. They’re independent, right? So basically this is alpha and this is beta. That’s the microscopic parameter. What happens with fill one? With fill one, we have… we put a one here. In that case, B is of course stronger than A in all the parameters, right? One and one—so one is stronger than zero, and one is like one—meaning it’s either greater or equal; B is greater than or equal to A in all the rows. There is an order relation between them, so the diagram is this. Here, when before it was zero, then there was no simple order relation between the columns. What happens in such a case? If this is… maybe I’ll mark it the other way. Wait… this is beta. Yes, fine, okay. So if this is alpha, this is two alpha, right? We said we take the simplest possible option. You could also have put here alpha and beta, right? Meaning that too would have been possible; alpha and beta is also stronger than alpha, but of course it’s simpler to put two alpha—we prefer one parameter. So that’s the simplest fill, therefore it’s clear that fill one is preferable, right? Therefore one is greater than zero. All right? Preferable to zero. Therefore kal va-chomer works—the correct fill is one. Okay? Now I move to the next case. The next case we saw was a refutation. Okay, don’t despair. A, B, C. A, B, refutation column—it’s the same thing, it doesn’t matter. Here we have zero, one, one, question mark, and zero, one. Right? Those are the data of the refutation. Once… so let’s write here: refutation of kal va-chomer. All right? That’s this table. Refutation of kal va-chomer, where we have fill one—what happens when we have fill one? As if one is written here. Once one is written here, then this column is the strongest column, right? It’s one-one. So therefore basically we have here B, and A and C both go into B—they are weaker than B. Between themselves there is no connection, right? Therefore the table is this, right? The diagram is this. Okay, that’s the form. Now if we want to fill it, then this is alpha, this is two alpha, and this is alpha-beta. Here there’s an example where we have to introduce another parameter, because if I were to put here, say, three alpha or four alpha, then there would be a relation between these two, but the datum is that there is no relation between these two. So we have to introduce another parameter here. Beta alone—why can’t it be here? Why does it have to be alpha and beta? Because then there’s no common denominator. So it has to be stronger than this, and therefore this is the natural solution. What happens in the situation of fill zero? In the situation of fill zero, then B and C collapse together, right? B and C look exactly the same, one and zero, right? And they are completely independent of A; there’s no arrow between them, right? So if we now want to make fill one, this is alpha and beta. Which is preferable? Alpha. No one is preferable—that’s the given fact, this is a refutation. Now here there was room for a bit of hesitation, because after all this model actually looks worse, right? It’s worse—why is it worse? Because alpha has to increase in valency, from alpha to two alpha. Here both alpha and beta are zero or one, okay? Whereas here beta is zero or one and alpha is zero, one, or two. So it’s more complicated. But we said that since we know that a refutation is something where the two fills must be equivalent, it’s therefore clear that valency apparently does not play a role. Okay, so I’m already writing a conclusion here: for this thing to be a refutation, valency does not play a role. Meaning if something has an advantage in valency, that changes nothing. Okay? Here by the way you also see something similar; here you see something a bit weaker, right? Here we see that if this thing has an advantage in terms of number of parameters—what in a minute I’ll call dimension, yes? Here this is dimension one, only alpha; here it’s both alpha and beta. The fact that here alpha gets three values—zero, one, or two—doesn’t bother me. So here the claim is that valency plays a less important role than the number of parameters. Here the claim is stronger: valency doesn’t play any role at all. Not only is it weaker—from here you see that it plays no role at all. Why doesn’t it play a role? The fact that this is more complicated—does that mean it can’t happen? I didn’t understand. We don’t know whether it’s one or zero, so why did we have the assumption, the initial thought, that we’d use the easier thing, the simpler thing? We always choose the simpler explanation. What do you mean? What is a presumption? We talked about this. What is a presumption? After three times that the ox gores, you assume it has a goring nature. Why? Maybe those three times were accidental? The assumption is that the simpler explanation is that the three times stem from a goring nature, and I choose the simpler explanation. Okay? Rabbi Chaim at the beginning of Chagigah, what I mentioned one of the previous times. Fine, so that’s regarding… notice that so far we’ve reached the conclusion that valency cannot play any role, right? There’s no choice—if it has any weight, we’re stuck with the refutation. If it had any weight, that would mean that one is actually less good than zero. Why does that bother me? A refutation—it’s fine if it’s zero, no? If zero is the better fill, why does that bother me? Exactly, because then it’s not a refutation; it’s a counterproof. I talked about that in the previous lecture, right? A refutation leaves both fills open, the question remains open. A counterproof proves that the correct fill is zero; that’s not a refutation. A refutation only says: you don’t have a proof that the fill is one. Therefore, perforce, valency plays no role at all. Now look what happens next. The next table—we haven’t done it yet, but it doesn’t matter, the idea is the same. Binyan av. What does the table of binyan av look like? Okay. If this is a frog and it’s also green, this is also a frog—the question is, what about it, is it also green or not? Right? That’s basically an analogy. Okay. This has mass and falls to the earth, so that one also has mass—does it also fall to the earth? Fine, that’s basically an analogy. What happens here? So let’s try to check it with the same technique. Fill zero—what does fill zero do here? A table like this, right? A is the stronger one, so if A is alpha, then B is two alpha. Again I choose the simplest solution, and not alpha and beta—there’s no point in adding a parameter. Why is B two alpha? Binyan av could be all kinds of things; we don’t know what the power relation is. Binyan av is always equal. No, it’s more than alpha—it has to be more than alpha, more. Why does it have to be more? It could also be alpha. If it were alpha, then why would a difference arise here? Can’t binyan av reveal that it’s alpha? No, because if it were alpha then no difference could arise here—what do you mean? I’m talking now about fill zero. We’re talking about fill zero. Ah, on the assumption that… the understanding doesn’t work, the price is that it really has to be alpha. Exactly what will now happen with fill one—that’s what you’re suggesting. That’s exactly the point. What happens with fill one? They collapse together, right? We have A and B collapsing together, and that’s exactly alpha, exactly what you created, that both of them are alpha, right? Meaning the price we pay for fill zero—why is it not the correct fill? Why is fill one the correct fill? Because with fill zero a more complicated model is obtained. Okay. And now we’re stuck. Because earlier we proved that valency plays no role, and now suddenly we see that this is worse than that—why? Only because of valency, because in terms of parameters the number is the same; in both there is one parameter. So if—just a second—if here we discover that valency plays no role, then binyan av is open, it’s a refutation: you can’t know whether the fill here is one or zero, so that doesn’t explain binyan av. And the other way around, if I decide that valency does play a role and therefore valency determines that one is the preferable fill, here I can still manage, because I can say that valency plays a role but it’s less significant than the number of parameters; and here there are two parameters, while here there is only one with higher valency, so this is still preferable to that. But here I’m stuck, because here in both fills there are two parameters and the entire difference is only valency, so if valency plays a role, then this should have been a counterproof and not a refutation. So something here is stuck. Yes. But how do I know whether now I play with this as two alpha or as A-B? Because I take the simplest thing—the minimum possible number of parameters, that’s the simplest model. That’s how we always do it. So here we’re stuck. Valency necessarily does play a role, and we have some sort of contradiction here. I’m just taking you through the exact path we followed in getting to the correct model. What that basically means is that we have no way to describe this whole range of inferences, of these types, even at the most basic level—we haven’t yet gone beyond two-by-three, a two-by-three table—and it already turns out there is no way to describe it only through valency and the number of parameters. We need more variables here, or more criteria of priority that will determine different priorities. I’ll just do a refutation of binyan av and then come back to say what the different priorities are. What is a refutation of binyan av? So we have a table like this: A, B, C—this, we said, is binyan av, and this is a refutation of binyan av. So we have here one one one question mark, and here we have zero and one. Right? That is basically a refutation of binyan av. Okay. Now let’s do the analysis of that. So under fill zero, what happens? B and C collapse together, right? And A is stronger than them. So this is the table, right? And under fill one, A and B collapse together, and they are stronger than C. That’s the table. What comes out here even without doing anything? Clearly this is an excellent refutation, right? It’s exactly the same model. There’s no difference between them; it’s a classic refutation. So if I say this is alpha and two alpha, then this too will be alpha and two alpha. All right? So basically everything works out; it’s only binyan av where we get stuck against the refutation of kal va-chomer. Binyan av versus refutation of kal va-chomer puts us into a problem—the model has to expand. There has to be another criterion of priority that takes into account things beyond valency or the number of parameters. It has to be—that’s a proof. Okay? What could it be? We started thinking what it actually could be here. So we’re basically trying to look—we looked at these two. We see here several… I want to get to the criteria, I’m going backward: I want to get to criteria such that in binyan av, fill one will be preferable, and in refutation of kal va-chomer the two fills will be equivalent. That’s basically my goal. Okay? So we need to focus on these two as we look for how to refine or improve our model. Now the natural criteria in this situation are topological criteria. By topological I mean that what’s described here is a graph, in mathematical language. And a graph has topology. In a minute I’ll come down a bit through the properties. A graph has topology, and it’s pretty clear that the topology of the graph also reflects, in some sense, simplicity versus complexity of the solution. Meaning: does this represent a table, right—this that fill is one, this that fill is zero? I’m asking which is the simpler table. The simpler table is the one whose representing graph is simpler. Now it’s very plausible that here this graph is simpler, right? This is the simplest graph possible: it’s a graph with one point. No arrows, no more points, nothing. As simple as can be—the most minimal graph there is. It will be preferable to anything. So it’s very plausible that the number of vertices in the graph—these are vertices—here there are three vertices, here there are two—the number of vertices in the graph will be one of the criteria. Right? Let’s see whether that doesn’t ruin other things for us. So now I’m starting to go through all of these and see what happens when I take into account also the number of parameters, also valency, and also the number of vertices. So look: the number of vertices here and here is equal, so nothing changes. The result here doesn’t change. The result here does change, right? This is simpler, right? It’s simpler, and valency might now offset it—say that it does play a role. Right? Valency could offset it against the number of parameters, and therefore this is a refutation, because there’s an advantage to this fill and an advantage to that fill. Why does valency play here? I’m just proposing suggestions. I’m saying, after all, this one has higher valency. Right? What? No, that’s not good. Why? It’s smaller… ah, sorry, right, you’re correct. So valency here won’t help at all—you saved me having to rule it out later. Fine. So valency doesn’t help us here at all. Basically what happens here with the refutation of kal va-chomer? Basically fill zero becomes preferable if we look at the number of parameters. But as we said, a refutation of kal va-chomer—a refutation is not a counterproof. It cannot be that fill zero is preferable. What happens here? Here one becomes better, that’s clear—we already saw that. What happens here? Nothing changes. So we’re stuck here, right? We’re basically stuck here. We already understood that the number of vertices in the graph is a criterion, but there have to be additional criteria. Now what criteria could come in here? Actually one is enough for me to fix the issue, I think. Let’s see. If I now talk about the connectivity of the graph. Connectivity means—you see here the graph is made up of two parts with no connection between them. They don’t talk to each other. Right? Why does that affect the graph—which is more connected? Because once there’s a connection between the parameters, there’s more logic to making an inference between them. Not necessarily; if there’s no connection between parameters, then by your theory the graph is simpler. No—you want to infer from known facts to an unknown fact. If the known facts are not connected to the known fact—sorry, to the sought fact—then it’s less plausible to infer. Like we talked about, if you remember, with love of jazz and love of literature. Someone who doesn’t like jazz likes literature; someone who likes jazz will surely like literature. What do you mean? Love of jazz and love of literature—this is alpha and this is beta—what’s the connection? All right? Meaning, the more connectivity there is in the graph, the better the graph is. Notice: does this ruin anything for us? Here it only improves our situation. Right? Zero becomes even worse. Here zero becomes worse, but on the other hand in terms of number of points it’s better. But that’s perfectly fine. It’s a refutation. What does a refutation mean? That there’s an advantage to one, there’s an advantage to zero, and therefore I can’t decide between them. So that’s fine, right? What happens here? Here the number of points prefers this; in terms of connectivity both are connected, so nothing changes, and likewise here too. So apparently this fixes everything, right? But for considerations that will become clear later, I’ll just present the full picture already now: we need one more criterion. One more topological criterion. We have two topological criteria—the number of vertices, the connectivity of the graph—and the third criterion is direction changes. Direction changes… all these, by the way, are known indices in graph topology. Meaning, we didn’t even start from this analysis. We started from graph theory, and only afterward came back here. In graph theory these are the most natural criteria for determining the simplicity of a graph. The simplicity of a graph is the number of vertices, the connectivity of the graph, and the direction changes. What do direction changes mean? Look: when I go along a path like this, along a path, and try to traverse the whole graph—here I have to reverse direction. Meaning here I go from here to here, and this movement is against the arrow. Here it’s with the arrow, here it’s against the arrow, and likewise here. So there’s a direction change here. What does that direction change indicate? That’s already what’s called the theory of directed graphs, understanding it from ordinary graphs. So in a directed graph, direction changes are a determining parameter—they determine the character or status of a graph. And a direction change basically means that there is no simple relation between the parameters. Meaning, if they were all arranged on one chain—C goes to B and B goes to A—there would be no direction change at all. Right? The hierarchy between them would be much simpler, and it would be more plausible to infer conclusions when the hierarchy in the system is very simple. When the hierarchies in the system are complicated, it’s less plausible to infer conclusions. So the directionality of graphs in our context is expressed as a non-simple hierarchy among the variables or among the results—these A, B, and C, yes, the results. There is a non-simple hierarchy among them. When there is a non-simple hierarchy among them, of course it’s less plausible to make an inference. So if so, now we basically have three topological criteria, and I define the general priority criterion. The general priority criterion says this: dimension—that is, how many parameters. Valency—we’ll soon see; for the moment it doesn’t play a role. And number of points. Four, connectivity. And five, direction changes. All these—or maybe let’s call them indices rather than parameters, because parameters are our alpha and beta. Topological indices, coming from graph topology—those three. The first two are not connected to graph topology. Now, once we’ve already seen—basically, I went through these basic inferences and saw that what was still missing for us was direction change. Will direction changes ruin things? Up to now, without direction changes, we saw that everything works out. Right? With direction changes, the only place they enter is here, right? But that’s not a problem, because here there’s an advantage in terms of number of points, and here there’s an advantage in terms of connectivity, so now it’s no longer decided. And here there’s an advantage in terms of direction changes. Here—sorry—this has the advantage in terms of direction changes. Right? But what does that mean for us? What can we learn from that? That when there are two advantages against one, that doesn’t matter—it’s still a refutation. And that’s very logical. Wait, is lack of connection preferable to a direction change? No, no—on the contrary, neither is more important than the other. On the contrary, you have no way to compare the indices. Once there is a preference by any of the five—leave aside number two, valency—one of the other four criteria gives preference to fill zero, and another one, or two, or three others give preference to fill one, then it’s a refutation. I’m just asking about direction change: if I have a direction change as against no connection, like in the refutation of kal va-chomer. Okay, so that’s equivalent. That’s equivalent. A direction change is a bad feature, because here you have to make a direction change; that goes against fill one. And no connection is even worse than a direction change, as it were? Why is it even worse? How did you decide that? I’m asking. This is one defect and that is one defect. You have a defect in this and a defect in that. Now the logic of these inferences also works like this. Understand: the logic of these inferences basically says—after all, in order to refute a kal va-chomer, it may be that in ten features the kal va-chomer holds. If one refutation is found, the kal va-chomer falls. Why? Because maybe that very feature is the one that determines the relevant law. Therefore the logic definitely also points to the result we got, namely: if we have preferences for both sides, it doesn’t matter how many preferences there are on each side—that is a refutation. Even if one side is preferable—here is where it usually shows up—say this thing is preferable in terms of number of points and also in terms of direction changes, because here there is a direction change and here there isn’t. So it has two advantages. Right? But this one has the advantage of—wait—number of points, direction changes, and in terms of connectivity. And in connectivity this is preferable to that. So this has two advantages and this has one advantage. Right? And alpha and beta—the dimension is of course the same in both, both are alpha and beta. So what does that mean? It’s a refutation. Okay? And I think that fits very well with the way these inferences operate. Meaning, it doesn’t matter how many advantages one side has; if there is one respect in which the other side is preferable, you can no longer prove anything, because maybe that’s what determines the law. That is exactly the logic of a refutation. Okay? So that is basically the general theory, and now the definition—two more points. The basic definition of how I apply this criterion—maybe I’ll erase valency because for now we don’t need it at all, it only confuses things. The general criterion is basically this: I fill in—I take a table, I want to decide what to do with it. I set fill zero and fill one. For each fill I draw the diagram and fill it, meaning I find the parameters, the model. Okay. And now I measure the two fills, after I have made the diagram and found the model, according to four parameters: dimension, number of points, connectivity, and direction changes. If one of the fills is preferable—there is preference only for one of them, no matter preference in what; there could be a preference in dimension, a preference in number of points, and in everything else they are equal—but there is no preference for the other side, there is only preference in one direction, then that fill is the correct fill. If there are preferences in both directions, no matter how many preferences there are, that is a refutation. Okay, that is basically the conclusion. One additional point that I still need to add in order to complete the picture is that we require—when we get to more complicated graphs we’ll see why this matters—we require that the increase in the number of parameters—this is the only place where valency enters. In our model, at least for now, valency enters only in this requirement: that when we increase the valency of one of the indices, we are allowed to do it only for one index. You don’t make a model where there is alpha, two alpha, three alpha, and beta, and two beta. If we need to go up to two beta, then add a parameter. Meaning, you don’t raise valency in two different parameters. Now we arrived at this ad hoc, meaning these criteria came out directly for us from graph theory. This criterion basically helped us fix one graph that didn’t work out—it was basically a hole. But it seems to me that it has a logic. And the logic basically says this: when you describe some graph—it can be a terrible graph, a terribly complicated table of ten by five by six by seven, doesn’t matter—think of a very, very large table, and there’s one empty cell, and you want to know what to write there, zero or one, all right? So you basically want to infer from the whole table to that specific cell. For that, there has to be some basic hierarchy shared by the whole table. Because otherwise, if it’s a collection of unrelated hierarchies, it’s like connectivity. If it’s a collection of unrelated hierarchies, then it’s not correct to take the whole table as a basis and infer from it the missing law. Now how do I represent that requirement, that there be one basic hierarchy for the whole table? So I say: let’s determine that only one parameter can rise to a higher valency—three, four, five—and then what will happen? That parameter will basically serve as the strength parameter. Meaning, if A, B—and let’s say that our solution for A is alpha, and our solution for B is beta and alpha, and our solution for B is beta and alpha, and for C it’s beta and two alpha, and for D it’s beta and three alpha, and so on. What does that basically mean? It basically means—or maybe, you know what, maybe we’ll even make this gamma if you want there to be another parameter. What does that basically mean? It basically means that there is some hierarchy shared by all the results. In terms of alpha, one stands above the other—meaning they have some common dimension, and the difference is only a difference of strength. Besides that, besides that, they have certain properties that are unique properties. There is a unique property in B and C that does not exist in D and does not exist in A; let’s call it beta. There is a property in D that does not exist in the others; let’s call it gamma. But there is also something shared by all of them, where the difference is only quantitative. Because if that were not so, then you would basically be talking about subgraphs that are unrelated. You would basically be talking about different hierarchies, and you would need to look only at part of the table, not the whole thing. If you want the whole table to be relevant for inferring the missing Jewish law, then you need to assume that there is something shared by all the variables in the table, that there is some simple relation between them. And that, I think, is the meaning of the requirement that valency rise only in one parameter. Okay, but again I’m saying: this of course came out ad hoc, meaning simply because this model explained everything except one table. In that table we tried to find a way out, and the way out that came out was this. Afterward we thought about why—what the meaning is of taking valency to be defined on one parameter—and it seems to me that this is one possible meaning of the matter. I already said last time that we are running here between the empirical and the a priori, meaning we look at tables and try to understand what could bring us to the correct results on the one hand, and on the other hand we need to look at whether there is logic in the criterion we arrived at. That is the a priori perspective. One needs to compare what happens in practice with what it makes sense should happen, and in that way build the model. Okay, I’ll stop here. What I want to do next semester is that we’ll start now with the topic of kiddushin, the topic of chuppah and kiddushin. This is the most complicated topic, I think, that I know in terms of these forms of inference. We’ll see—we’ll get to a table of I don’t remember how much anymore, seven by five, or eight by five, or nine—I already don’t remember how many. Step by step we’ll see how the topic develops within the framework of the model. And then after that I’ll come back to discuss a bit the meaning of the model itself, and then we’ll move on to other hermeneutic principles. Okay, so that’s for next semester.