2019-04-22 – Between Midrash and Logic – Lesson 14

Table of Contents

• The course of the passage up to the challenge to the small common denominator
• “A document proves it” and the expansion of the common denominator
• Building graphs, dimension, and changes of direction as the deciding factor
• Interpreting the parameters and identifying beta with benefit
• A challenge to the large common denominator: they exist even against one’s will
• Rav Huna and correcting a datum in the table as a “rescue”
• Returning to the model: equivalence, a fourth dimension, and changes of direction
• The document as a fixed part of the table and an expansion that does not contract
• The interpretive principles as infinitely many applications of one algorithm

Summary

General overview

The lecture moves from the challenge to the small common denominator to the construction of a large common denominator in order to learn that chuppah effects betrothal, by adding a document as a source that removes the challenge of their benefit is greater but creates a new challenge because it can also effect divorce. The lecturer translates the moves of the Talmud into binary tables and graphs, and uses “topological” criteria such as dimension (the minimum number of parameters), connectivity, and the number of changes of direction in order to decide between filling in a “1” and filling in a “0.” Later he presents another challenge to the large common denominator from the fact that “they exist even against one’s will,” and Rav Huna’s rescue, which corrects a datum in the table (money is not “against one’s will” in the realm of marriage), from which a very high sensitivity of the result to the change of a single value emerges. At the end he argues that the multiplicity of interpretive principles in the passage is the product of intuitive, step-by-step thinking, whereas under one formalization of “filling in a table and deciding,” what you really get is a unifying method that contains infinitely many “principles,” corresponding to the number of possible tables.

The course of the passage up to the challenge to the small common denominator

The lecturer sets up a table of properties for money, intercourse, and chuppah in relation to betrothal / kiddushin, marriage, redemption, a yevamah, and benefit, and places the discussion at the stage of a common denominator from intercourse and money to chuppah. He notes that the argument had gone from an a fortiori argument from money to chuppah, to a challenge to that a fortiori argument, to an inductive paradigm from intercourse to chuppah, to a challenge to that inductive paradigm, and then to a common denominator from money and intercourse to chuppah. He states that the challenge to the common denominator is “for their benefit is greater,” because money and intercourse involve benefit, whereas chuppah does not.

“A document proves it” and the expansion of the common denominator

The lecturer explains that a document is added in order to neutralize the challenge from benefit, because a document does not involve benefit, and therefore the common denominator is no longer subject to the argument “for their benefit is greater.” He formulates the role of “a document proves it” as adding a row to the table so that the desired filling once again becomes “1.” He presents the challenge to the document from the fact that “it can dissolve marriage with a Jewish woman,” adds a divorce column, and establishes that a document alone can divorce, whereas money, chuppah, and intercourse cannot. He describes the Talmud’s answer, “money or intercourse prove it,” as creating a broader common denominator which on one side includes money and intercourse as a single unit, and on the other side includes the document, in order to teach about chuppah.

Building graphs, dimension, and changes of direction as the deciding factor

The lecturer skips intermediate stages and transfers the decision to a graph model in which “1” and “0” are two possible fillings of the same expanded table. He builds inclusion / dependence diagrams for each filling, tries to assign alpha, beta, and gamma parameters, and concludes that in both fillings dimension 3 is required, so dimension alone does not decide the issue. He determines that the decision is made through changes of direction in the paths of the graph, and identifies the decisive difference as the disappearance of the edge N→A between the zero filling and the one filling. He demonstrates that moving from G to N requires more changes of direction in the zero filling than in the one filling, and therefore the change-of-direction criterion decides in favor of the one filling.

Interpreting the parameters and identifying beta with benefit

The lecturer calculates the “microscopic composition” of money, chuppah, intercourse, and document in terms of alpha / beta / gamma on the basis of the graph chosen as the correct filling. He identifies the parameter shared by money and intercourse that is not found in chuppah and document as beta, and interprets it as the property of “benefit,” which matches the challenge “for their benefit is greater.” He explains that the alpha parameter functions as the shared component that makes it possible to derive betrothal through a common denominator, and that anything not needed for the application to betrothal gets pushed by the model into other parameters that are not decisive for this question.

A challenge to the large common denominator: they exist even against one’s will

The lecturer moves to stage 12 and defines the challenge to the large common denominator as “for they exist even against one’s will.” He lists that the rule of money acquiring a Hebrew maidservant is against her will, intercourse with a yevamah is against her will, and a document in divorce is against her will, whereas chuppah is not against one’s will. He presents this as structurally identical to the challenge to the small common denominator, except that here there are three source cases sharing a property that the target case does not have, and therefore the ability to infer about chuppah collapses.

Rav Huna and correcting a datum in the table as a “rescue”

The lecturer presents stage 13 as Rav Huna weakening the challenge by saying that with money, “in the sphere of marriage we do not find against one’s will,” and therefore the value of money in the “against one’s will” column in the realm of marriage should be zero rather than one. He describes this as a different kind of challenge, one that does not add a row or a column but rather corrects a cell in the table, and he emphasizes that the result is extremely sensitive to the replacement of a single value. He recounts that in their model they got stuck because the result came out as “not a challenge,” until it became clear that they had copied an incorrect value, and the fact that they were unable to “force” reasonable criteria to produce a different result strengthened, in his view, the reliability of the technique. He explains intuitively that when the money entry is corrected to zero, the property “against one’s will” no longer characterizes all the source cases as against the target case, and so the challenge stops working and the common denominator remains valid.

Returning to the model: equivalence, a fourth dimension, and changes of direction

The lecturer again constructs diagrams for the one filling and the zero filling for the challenge, and shows that at a certain stage I and K either merge or separate depending on the filling, which changes the number of points and edges. He tries to fill the diagram with parameters and arrives at the conclusion that in the corrected case dimension 4 is sometimes required, so once again dimension does not decide between the fillings. He returns to the claim that changes of direction become the dominant criterion as the graph becomes more complicated, because they measure the hierarchical consistency of the parameters along the connections. He demonstrates that in one filling no more than one change of direction is needed between points, whereas in the other filling there are paths that require two changes of direction, and from this he concludes that Rav Huna “rescues” the a fortiori argument and the move is completed.

The document as a fixed part of the table and an expansion that does not contract

The lecturer explains that the table “only grows” over the course of the passage, because adding the document is a necessary part of the large common denominator and not a stage that can be erased when returning to money and intercourse. He formulates it so that money and intercourse as a group stand opposite the document, and the common denominator between the groups is the mechanism that makes it possible to derive chuppah, while “money and intercourse prove it” functions as the repair for the divorce challenge posed by the document. He argues that the technique in principle makes it possible to dispense with going “step by step” through the passage and obtain a result if all the correct data are supplied in advance in the table, whereas the Talmudic process is needed when the data are unclear or when they change over the course of the rescue.

The interpretive principles as infinitely many applications of one algorithm

The lecturer notes that the passage contains an enormous multiplicity of combined interpretive principles, including a fortiori argument, inductive paradigm, common denominator of various kinds and their challenges, and even a “large common denominator” that generates many combinations. He argues that there is no point in counting this as hundreds of interpretive principles, because everything converges into one principle: “give me a table, draw a graph, build a model, and decide according to topological and dimensional criteria.” He concludes that the number of “interpretive principles” in the formal sense is equal to the number of possible tables and is therefore infinite, and most structures are not even framed in the language of the Sages. He argues that every table has one of three possible outcomes: a preference for 1, a preference for 0, or no decision, and he compares this to an analogy from Zermelo’s theorem in game theory, according to which every game of a certain type has one definite outcome out of three, even though that is not trivial and requires proof.

Okay. We’re basically at the stage of the common denominator. I think last time I finished with an analysis of a refutation of a common denominator. A refutation of a common denominator—let me just remind you of the relevant table. Okay. Intercourse, chuppah, money. This is marriage, betrothal, or kiddushin—it doesn’t matter, it’s the same thing. Redemption, yevamah, and benefit. Okay. I’m just reminding you so we can move on from here. Basically, we’re in the Talmudic passage—you have the page with the passage; whoever doesn’t have it, I still have a few more here. Basically, in the first stage we went through what was in practice an a fortiori argument from money to chuppah. The second stage was a refutation of the a fortiori argument. The third stage was a paradigm case from intercourse to chuppah. The fourth stage was a refutation of the paradigm case. The fifth and sixth stages were a common denominator from intercourse and money to chuppah, right? Which is basically the table up to here. Up to here, this is exactly a common denominator, where the common denominator here is an a fortiori argument from money and here it is a paradigm case from intercourse. Okay? And in stage seven there is a refutation of the common denominator: that both of them involve greater benefit. Yes, money and intercourse involve benefit; chuppah does not involve benefit. Okay? That was the refutation. I think we did that too. I think that’s where we got to, and we saw that in fact in that situation the two fillings are equivalent. Okay. Now I move to stage eight. In stage eight: “a document will prove it.” Notice, this is a slightly more complex passage than usual in the Talmud, which is why I chose it. What does it mean, “a document will prove it”? What is the significance of this, actually? What is “a document will prove it”? What are we doing here? What is the Gemara doing here? What do you say? It adds another row. There—what is the document supposed to prove? That it has in it—that it effects marriage, it effects betrothal. No, what it does we’ll see in a moment, but what is its role? What are we trying to learn from it? What do you mean, you add another row below and try to add…? And I’m trying to fill in a one here. Yes. Right? I refuted the common denominator, right? At this stage. And okay, so let’s add one more row, where in that row there’ll be a document, and with that row the filling here will indeed be one. And isn’t that a refutation of the refutation? What? Isn’t that a refutation of the refutation? No—why a refutation of the refutation? Since the previous refutation was: what is common to money and intercourse is that both involve greater benefit, so that was the common denominator of money and intercourse. Correct. And then it refutes that because a document will prove it. Exactly. What does “a document will prove it” mean? That a document involves no benefit, and therefore that refutation won’t apply to it. It will succeed in filling in a one here and won’t be subject to that refutation of greater benefit. Okay? And then we say: the refutation is, what is unique about a document? That it dissolves marriage with a Jewish woman. Okay? The document basically divorces, and then we need to add another column here for divorce, and we see that money does not divorce, chuppah does not divorce, intercourse does not divorce, a document does. That will be one here, zero, zero, zero. Since this refutation applies only to a document, it’s quite clear that it will knock out this one. Okay? Then the Gemara says: money or intercourse will prove it. We return here, where money or intercourse will prove what? After all, they have a refutation on them. What will they prove? That they are not part of divorce. That they do not dissolve marriage with a Jewish woman, exactly. That they cannot dissolve it. And then what we’re doing here is a broader common denominator. So I’m already skipping all the intermediate stages and analyzing the final row directly. Okay? Basically what we’re doing here is adding another row here for document. Why did I mark it? W, I think, yes. Okay. And here: divorce. And a document effects divorce and all the rest do not. Now how do we fill this in here? A document does not effect marriage. Why not? It does. How can one acquire with a document? Kiddushin? Kiddushin. Not marriage. A document effects betrothal, not marriage. Redemption: a document does not do that. Yevamah: a document does not acquire, only intercourse. And benefit—unless someone has a borderline personality, he doesn’t get benefit from a document. Okay. Why redemption? Because it doesn’t redeem second tithe, which money does. Money has the power to perform halakhic acts that a document cannot perform. Okay. Now I’m skipping all the stages, because in the bottom line, instead of—how would we analyze this otherwise? We would say: basically we are making some inference from document to chuppah. I would again make a two-by-two table, refute it because a document divorces, make that a refutation of this inference, and then I’d say: let’s make a common denominator together with money and chuppah—well, money and intercourse—that also have a refutation on them, and together it will do the job. So I’m going straight to the whole thing together. Okay? If it does the job, I don’t care about the intermediate stages. The intermediate stages definitely work because it’s just a paradigm case and a refutation of a paradigm case, and all those stages we already checked. Meaning, the tables work. So all that remains to check now is the following table. Basically I enlarged it by one row and one column, and the claim is: where am I standing now? Right now I’m basically making a bigger common denominator. I had a refutation on the small common denominator, and what I’m doing now is making a larger common denominator whose two sides—you remember I said there are three kinds of common denominator. Either both sides involve an a fortiori argument, or both sides have a zero—that is, sorry, that’s an a fortiori argument—both sides have one side that is a paradigm case, or it’s an a fortiori argument on one side and a paradigm case on the other. That’s a small common denominator. But now we have a large common denominator. And a large common denominator means that one of its sides is itself a common denominator. That is, there is a common denominator here from money and intercourse. That’s one direction, yes. Money and intercourse. That’s one direction. And document. And they teach by a common denominator toward chuppah. This thing is itself also a common denominator, right? Just a common denominator that has some refutation against it. The structure is the same structure. It’s clear that the intuitive logic is the same intuitive logic. The question is whether it works here too with our formal tools. Okay. So let’s see whether it works or not. So. We’ll call this filling zero and this filling one. Okay? So let’s start with filling one. Who is the highest? Betrothal, of course. Usually that will be the case. A is the highest. Who comes next? N and H, right? N and H. They don’t enter one into the other, so there is no dependence between them. Who else is left? P. P enters H. Right? And G. G only enters A. That is the table. Okay? That’s in filling one. What happens in filling zero? So again the highest, let’s say, is A. And now Y. What? Didn’t we do Y? Right. Y enters through N and also H. Right? Y enters through N and H. Let’s do it like this. Okay? G enters from here and Y enters N and H. Okay. What happens here? So like this: A is the highest, but N does not enter into it. Right? N does not enter into it. P, Y, H—P, Y, and H do enter into it, and also G, right? So H enters A. H enters A. P, G, Y—each of them is one. So let’s see. P enters H. P enters H. Y enters N. And also A. And who is left? G. G only enters A, right? That is the diagram of filling zero. Okay? Now the situation, as you can see, is already starting to get a bit complicated. Now we have to—let’s say if this is alpha, then let’s try a few things. At worst, if it doesn’t work I’ll look in the book. Let’s try a few things. So here we have this: alpha. Here let’s say alpha is also beta, and here two alphas. Okay? That is the exit from here. Wait—G, what comes before G? It will probably have to be third already. G will probably have to be alpha and gamma, right? Because otherwise in any case it enters one of those two. Okay? Y—now here this is three alphas, sorry, two alphas and beta, right? And this—what do you say about it? Here there’s a problem. Two… We would make this alpha and two betas, but you’re not allowed to increase in two parameters, right? Or else it would have to be three alphas and beta and then there would have to be an arrow like this here, right? There is no choice but to add another parameter here, okay? But if I add another parameter, maybe I can use gamma. For example, beta and gamma… If we write here… it needs to be gamma… alpha beta, it has to enter and also gamma, but then we need to take care of this, right? So we need to take care of that, so I’ll make this two alphas. But then it will have a relation with N, no? With whom? It will have a relation with N. With whom? With G. Ah, right. There’s a problem here. Wait, so if so I have no choice but to raise this to three alphas and this too to three alphas. Now this seems okay to me, right? Okay. Not that Q—we said Q doesn’t play a role—I just need to make sure three parameters are enough for the dimension. Right? Three parameters—alpha, beta, and gamma—solve the problem. Let’s see whether I did something similar here. That was in filling one, right? In filling one we have A as alpha, G as alpha and gamma. Maybe I started with G—it doesn’t matter, it’s the same principle, yes. Here there is a slightly different solution, but exactly with the same dimension. There’s a whole family of solutions that have the same property, minimal solutions. Okay. What happens here? What happens here… What are the minimal solutions? Is there some way to find… What? Can you prove that you can’t do it with two parameters—is that what you mean? Or prove what the number of parameters is… what the minimal solution is, in short? There isn’t one minimal solution; there’s a family. After all, you can manipulate things. I could have put alpha and beta here, etc. Sure, sure—but with fewer than three parameters you won’t manage to fill this. Yes. I don’t have a theorem proving that; that’s one of the things where mathematical work has to be done here, to prove for every type of diagram what the minimal number of parameters is. I don’t think it’s such a complicated problem, but still it needs work. But intuitively it’s clear that with fewer than three parameters you won’t manage here. Heuristically I know how to do it. That is, heuristically you move backwards like this. You put alpha here, proceed to these two, and just build it backward. There are almost no alternatives once it gets complicated. And it gets more and more complicated, right? With complicated connectivity in many directions, definitely theorems will be needed here. Or some computer algorithm, whatever. Okay. What happens here? So if I make this alpha, this has to be beta, right? The extremities are always the start. There is no connection between them. Do you see here that Y enters H? Huh? Do you see Y here? Yes. Where? Y. Ah yes, right, Y enters H. Okay. So this is alpha, and here it will be beta. Y is alpha and beta. H1 and H2. So this will be two alphas and this two alphas beta. Right? And here too let’s say three alphas. And now I’ll have to do this with gamma, won’t I? I have no choice. Let’s see if I have no choice. There has to be gamma here. Yes, there has to be gamma here. Let’s take this as gamma. Because otherwise there are three exits here, so three exits requires gamma. Okay. This is exactly the form in which the gamma theorem will work, if you ask how many exits there are from each vertex. Okay. So we see that in terms of dimension it’s similar. Yes, here too there are three parameters. Fine, so in terms of dimension there’s nothing to be done—the details aren’t important, there is dimension three here and dimension three here. Valence, we said, doesn’t matter so much. What else remains? Change of direction. Connectivity is the same in both, the number of independent points is the same in both, there are no points that merge here, right? What remains to check? Only changes of direction. They can save us here, otherwise we’re in trouble, right? This is supposed to be filling one. Okay? So only change of direction can save us here. So let’s see what exactly the difference is. From Y to A—you see that arrow isn’t there, but that arrow from Y to A also exists in filling one. Where was Y? Ah yes, okay, it exists, okay. Fine. From Y to N it apparently gets canceled. Not from Y to N. Again, this zero becomes one. What does that mean? That N is no longer—from N to A. From N to A is the change between the graphs, right? Here N is connected to A, yes. And here N is not connected to A. That arrow disappears. Okay, that is basically the difference. What does that mean? It’s pretty clear that where an arrow disappears, problems of change of direction will start appearing. Right? Because an arrow always helps us with change of direction. When I have an arrow from N to A, that means the route from here to here doesn’t have to make all these transitions—it can go straight. So let’s see if this works. For example, when I have from P to Y. How many direction changes are there? One. Let’s see. One here and that’s it, just one, right? But if that arrow weren’t there… Why from H to Y? Why from H to Y? Why not P-H-Y? What? P-H-Y, you don’t need the whole detour. Ah, sorry, okay, still… right. No, I’m trying to see where it will affect things, probably from here to here. So let’s see from here to here. Right? I’m looking for where this will affect things. If this arrow is missing, that means that something on the route now will interfere and force me to make another direction change. So I’m looking where the maximum will be. I assume it will be these two. Let’s try. So in this case I have from G to Y. That’s one direction change. Right? From here? Wait, isn’t it two though? Why? One. Here I switch direction, and here it’s the same direction. Don’t you count it again? No, because it’s the same direction. The N-Y is always against the flow direction, so nothing changed. You choose a straight direction and then it goes with your route direction. Fine. Here too I went against the direction, so here too I’m going against the direction; that’s not a change. So here you see, from G to Y there is one direction change. Right? Here what happens from G to Y? One. No, here too one direction change. Wait. So then from G to N? Here there will be two, right? This is one and this is two. Agreed? What happens here from G to N? One. So there—that’s it. Okay? I was simply looking for where this arrow would affect things, where this arrow would cause trouble. When this arrow doesn’t exist, the simple path from G to N doesn’t exist. I need to take a complicated path, and therefore there are more direction changes. Okay? So the change of direction is decisive, and this is indeed filling one. Okay? Again, we see that… There are also changes of direction between A and N—there are two. Where? In blue, in filling zero. Ah. Fine, that’s not important, because you don’t need more than two between any two points. I take the maximal one required. There is no pair of points between which the transition requires more than two direction changes, and here there is no pair of points that cannot be reached with one direction change. So we proved the relation between N and A, let’s call it that, or between the lenient ladder—what does it have to resemble? Again. We proved the relation between N and A, basically. That is, why? Since, right, we said from G to N it’s simpler this way? From G to N it’s simpler because I can go through A, and here I cannot go through A. Okay? So that means the change of direction decides the table. Of course, what would happen if I did—let’s not say if, let’s now check just as an exercise. We’ve already solved these, right? Meaning, what are money, chuppah, and intercourse? We concluded that this is the correct graph, the other one was disqualified. Filling one is the correct, simpler filling. So now let’s see what the microscopic composition is of money, chuppah, intercourse, and document in terms of alpha, beta, gamma, and so on. So what is there in money? Money contains A, P, and H, right? It contains A, so it needs alpha; for H it needs alpha and beta as well, and here it needs all three. So money is one one one, right? Alpha, beta, and gamma in that order—I just did it simply. Chuppah contains N and A. N and A is three. Three zero zero is chuppah. Intercourse contains everything except P and G. That means it doesn’t have gamma, and it has three alpha and one beta—three one zero. And document contains A and G. A and G means it has two zero one. Why am I doing this? Because now let’s check our intuitions. After all, what we are making here is basically a common denominator between document and money and intercourse, right? There is something unique in money and intercourse, in money and intercourse that the others don’t have—that one, you see? That’s the small common denominator, right? The small common denominator is that one, and that one is the parameter responsible for the small common denominator, which in fact we already know what it is. What is it? Benefit. What is shared by money and intercourse? Benefit, right. So it’s pretty clear that this beta is connected to benefit. Let’s check that we’re right. Here, we are right, right? This beta is the benefit. We know that money and intercourse have benefit, unlike the others—that was exactly our problem. So I’m checking the consistency of the solution. So clearly this is okay. This beta basically reflects the benefit; I’m already giving it some interpretation. It’s like the chemical component, as it were, in money and intercourse that causes there to be benefit here. Chemical, as it were, meaning something—there is some property in money and intercourse because of which we derive benefit from those things, while in document and chuppah that component is absent, and therefore there is no benefit in them. What happens now with this parameter that is shared by money and document? Money and document have—no, so that’s not—sorry. The first parameter is the parameter of the common denominator, right? Why? Because in order to apply betrothal, what do we need? Alpha, right? And alpha exists in this one and also in those two, right? So if that is so, then one can learn from them by a common denominator to it—where of course it has the strongest version of all. Whatever all of them know how to do, it surely knows how to do as well. Therefore, if they have a force of up to three alpha, it has three alpha too, because if it didn’t have three alpha then you couldn’t make a common denominator, because maybe they do it because they have some higher alpha or something, while it doesn’t have enough alpha for that. So this alpha here must be the highest there is among all the others. Okay? That is exactly the idea of a common denominator. This is how the intuition of common denominator is expressed here. Why should that be specifically in chuppah and not in intercourse, for example? What? Why should that be specifically in chuppah? Because if the assumption is that I can learn chuppah—that it’s one—what does that mean? That the component that causes money, intercourse, and document to contain betrothal is apparently also present in chuppah, and that is exactly why I conclude that chuppah too succeeds in containing betrothal. But there’s no… Okay, but there is no significance to the three at the end? What do you mean, the three at the end? The hundred three? The one plus three in the hundred three? In that—three one zero, what is that? The three is at the beginning, not the end. That is alpha. Right, so there’s no significance to the fact that after that it’s zero and then one. No, no, because what matters—you see here—what matters for applying betrothal is only alpha. Right? That’s all. Whoever has alpha will succeed in doing the job. All the rest matter for other things. But that is exactly what the table does, this analysis. This analysis basically separates out all the things that don’t interest me—put them aside. What matters to me basically is whether you have alpha. That’s what matters. And this analysis shows that. Beta and gamma matter for various other things, but if I’m interested in whether chuppah effects betrothal, then what matters for me is whether chuppah has alpha or not. I don’t care what else it has. But the measure makes it the strongest. What? The measure? I’m saying it’s reasonable that if there is a common denominator that teaches from those three to it, then it too will have the highest alpha value found among the other three. Because if it had less, you couldn’t learn the common denominator from it; maybe they do the job because they have three alpha. Theoretically that could happen, yes. Okay, so this is just a translation in order to see the intuitive understanding of common denominator as it appears here, and in fact it does appear. Okay? Good. So that is the large common denominator. What is the next stage? We are now at stage twelve. “What is common to them is that they exist against her will.” What does “they exist against her will” mean? It’s a refutation, right? Money acquires a Hebrew maidservant against her will, chuppah is not against her will—that is exactly the problem. Intercourse with a yevamah is against her will, and a document in divorce is against her will. Okay? So basically we have one, one, one, zero here. And that is exactly the same idea as a refutation of a small common denominator. A refutation of a small common denominator—you see? It’s this sub-table here—that there is a property in the two teaching cases that is absent from the taught case. Right? The two teaching cases have some property, benefit in this case, that the taught case does not have. Here too it is the same thing, except that here there are three teaching cases. All three teaching cases have this property that the taught case lacks, and therefore this constitutes a refutation of the large common denominator. Now let’s see that it really works. Okay? So again, now let’s check filling one. But how does money acquire against her will? What? In what does money acquire against her will? A Hebrew maidservant. Yes. Now come on, let’s not waste our time. Let’s move straight to the next stage and analyze the biggest table, and then it will be… Twelve is: “What is common to them is that they exist against her will.” Or okay, we can still do it, let’s do it anyway. Because what is the next stage? Notice. What happens in thirteen? This is Rav Huna. After all, Rav Huna did use an a fortiori argument. They remove the idea that “against her will” could be something relevant to our issue in thirteen. Why? “As for money, in marriage we do not find against her will.” What does that mean? That this has to be changed to zero. Money is not one, because where it works against her will is not in the realm of marriage—it is in the case of a Hebrew maidservant. So don’t count that as one; it is really zero. Money does not work against her will in the field of marriage, and that is the important field for our purposes. So once again, notice, this is a very interesting refutation. It’s a refutation that does not add a row or a column, but claims that one of the data points I used in the inference is inaccurate. And if we change it from one to zero, suddenly we will see that the result here changes. In other words, this result is sensitive to the value of the variable here. If there is a one here, the result should be zero; and if there is a zero here, then here the result should be one. Okay? That is basically the claim. Okay? There is some sensitivity here, and that too is a type of theorem one has to look for: how parameters are related, how do I know which two cells are sensitive to each other, whether changing the value in cell i-j affects the value in cell k-l. And does a document really have an “against-her-will” dominance in matters of marriage? What? In the case of a document? Yes, in divorce. Yes. Okay. So what exactly are we doing here? By the way, this is the interesting point. With this table we got stuck. When—as I told you. With this table, when we built the model, we got stuck. It didn’t work. It simply didn’t work with all the criteria. There was supposed to be a refutation here, and it came out for us that it was not a refutation. And what turned out? It turned out that one of the values—we had simply copied it incorrectly. We were just analyzing the wrong table. And in that sense it was a great reinforcement for this technique, because one of the problems with a technique like this is that when you build criteria ad hoc, you can always build criteria so that whatever you want will work. So one has to make sure that the criteria also make intuitive sense. But still, there is the concern that you are forcing it, okay? We tried to force it with all our might—don’t tell anyone. We tried to force it with all our might, and it didn’t work. It didn’t work; you just couldn’t get to a result of one here, and that was very nice. Because the result zero is in fact a real result—you simply couldn’t change it by any forcing at all. That means this technique is good. That is, it’s not just some ad hoc thing. And after two days our heads were practically smoking. Then after two days we suddenly saw that in one value in this table we had switched something—we were simply working with an incorrect value. And that was very encouraging, even though it was frustrating for two days. But on the third day it was actually very encouraging. Because that basically means it’s not true that by force you can do anything. “The Merciful One exempts one who acts under duress.” Meaning, there are things we won’t manage to do with any reasonable criterion, and we looked only for reasonable criteria. You can always assume all sorts of ad hoc things. You’re always right, brother. Yes, exactly. So if that’s what’s right, then that’s the algorithm. But if you really try to work with reasonable criteria, and you genuinely can’t make it work, that means it really is a reliable technique. Okay? And here we see a similar phenomenon. If you switch this one—look, this is quite a large table overall, relatively speaking, yes? There are two, four, six, seven—twenty-eight values. Change this value from one to zero and you will never in your life be able to get a zero here; it will be one. In other words, these tables sometimes have very strong sensitivity: there are two cells—if we understand the intuitive idea behind it, you can also understand it overall. Because what is the idea? The idea is that now this refutation is no longer a refutation, because it no longer characterizes all three teaching cases in contrast to the taught case. Here, if this turns into zero, then it characterizes only two of the teaching cases but not the third, and so it ceases to be a refutation. Therefore the common denominator clearly remains valid. So the intuition, of course, gives that. But when we make the machinery, it’s hard to see why this particular cell is sensitive to the value in that particular cell. Okay. Clearly there is some mathematical theorem that would give this, but intuitively we can see it, and so I can skip the mathematics for now. I can work even without yet proving all the theorems needed along the way. But didn’t you say this is also supposed to develop something in our intuition? What? You said the goal isn’t only to formalize? To see it intuitively. No, but what I’m saying is: once I have this, something in my intuition should become sharper. Why? That’s what you said four or five lessons ago, something like that—that these algorithms would sharpen intuition? I don’t think so. I’m talking about something else; maybe you mean something else. In another moment I’ll get to that, okay? Remind me after I finish analyzing this, I just don’t want us to stop in the middle—let’s finish this chapter. Okay, so look, let’s begin with this. So I start with one at this stage; this is supposed to be a refutation. Keep in mind: the correct result is supposed to be zero. No—sorry, the correct result is supposed to be that it’s equivalent. Both one and zero are equally valid fillings; there’s no way to decide. Okay, so one—we say one is the maximum; N enters into it. Who else has twos? H also enters into it independently. K is actually larger. K contains H, right? H—is there a relation between N and K? No, there isn’t. Fine, so it’s like this, right? And what else is missing? Now all those with one. So we have P. P enters H and K, so P enters here. Right? And Y. Y enters N and H and also K. What? It enters H and enters K. Yes, everyone who enters H enters K. So it goes like this, to N and H, and who else remains? G. G enters K. Okay. And in zero we have A and K. In zero the connection between N and A is severed, exactly as before, right? So A and K are the same point. N—wait, what are you saying? A and K are the same point. A and K, right. A and K merge, and notice what happens here. A and K merge, this arrow disappears because they merge, and this arrow is deleted. That is basically the change. It’s not just deleting an arrow as in the previous case. We can already see what can do the job here, right? Number of points. At least in one direction. But the result, if you remember, is supposed to be that it’s equivalent, so here too there will be some advantage—maybe a change of direction as before. We’ll have to see. Okay? Is there a hierarchy among the properties? What? Is there a hierarchy among the properties? No, there isn’t. I said that once there is an advantage in each direction, then that too is intuitively correct. After all, intuitively, when you make a refutation, you don’t determine that the refuting property is stronger than the teaching property. It’s enough that there is a property showing the opposite hierarchy in order to say that you can’t draw conclusions. You just can’t know. Therefore the exact same idea is present here. Okay? So A and K merge, N is independent of them, all the rest is the same, right? Where does Y enter? Y also enters H, H enters A and K, right? And who else is missing here? P enters H, and that’s it—G. And G enters K. Okay? That is basically the diagram. This is for what? Filling… What? This is filling zero. It says so here, you see: blue is zero and black is one. Okay? Now how do you fill this monster? So again, here we have alpha, here two alphas, three alphas, alpha and beta, two alphas and beta—but now we need to take care of this. Right? Two betas, no? Let’s say this is three betas. You can’t do three alpha and two beta. And this is four. Okay. So far this works, right? And what happens here? Three alpha and beta. Right. No, but then here too it will need beta, right? So there will be an arrow like this. So here let’s say two alpha—we don’t need to, let’s raise the minimum necessary. Two alpha and beta—no, but two alpha and beta is no good because then there is an arrow like this. No, because then there is also an arrow from G to it. And then there will also be an arrow from G to it. No, from Y to G. From G—at the moment you have an arrow from G to H. Right. We somehow need to introduce a third parameter. In G maybe you need to do alpha and gamma, three alpha and gamma. Why should… wait, and if I make alpha then that’s no good either, because then there will be the reverse arrow. Yes. So in short, we need another parameter—that’s the principle. Okay? Let’s take the result from here. Yes, so filling one is like this. Y is three alpha, it says here; let’s see if that works. N is alpha and beta. K is two alpha. G is two alpha and gamma. H is three alpha. And this is four alpha. Actually it’s pretty intuitive: where the chain is longest, it’s worth running with the alphas. Okay. Good, so we see that we manage here with three parameters, right? What happens here? So A and K are alpha, G is alpha and gamma. It’s always good to keep similar parameters in both places, because then you can see the relation between them in the solutions. H is two alpha, P is beta, N is two alpha and beta, and this is three alpha. Okay. Fine. So in terms of dimension, both are three-dimensional; again, it’s not decided. The number of points favors that one, so that means here there will be changes of direction, because the connectivity is the same on both sides. So obviously there will be a problem of changes of direction here. We already know how to guess where the problem will be, right? If we go from N to A there, for example—from N to A there, that’s two. Right? Here there will be nothing at all that requires two, nothing. There we have one point that is internal; from G to Y, the furthest, it simply takes one. Right? From G according to this also one. Or from G to Y according to this is three? Huh? From G according to this is three. From G to N, sorry—let’s see: one, two, three. No, it’s not three, it’s only two. This is one, two. Here there’s no change. You’re going against the direction, and this too is against the direction. Here the direction doesn’t switch. It’s only two. Here there are two direction changes, a disadvantage, but the number of points here is better, and therefore it is a refutation. Okay. What happens now when this becomes zero? That’s stage 13, the last in the passage. It becomes zero. What changes is that K now functions differently. Everything else is the same, so let’s use these diagrams. Okay? K looks different. How does K look? Now K—G enters K, the thing from filling one. G enters K, so everything is normal. H—no, H does not enter K. We deleted it, that’s all. That is what changes, right? Only H does not enter K. Other than that, what could there be? K still enters A, everything as usual. So the line between H and K. But Y enters K. What? Y enters K. Where? Here? No, it’s supposed to enter K and it doesn’t enter K in the diagram. Where? Y. Let’s see. Y is supposed to enter K; it doesn’t enter K in the diagram. So wait, did we forget a line before, or…? Before there was a line between H and K, so that was okay. This line. Okay. This line? That line was here before—I don’t know, if it wasn’t here before, it needs to be put in, because otherwise we simply have a mistake. Three alpha and beta is stronger than two. It wasn’t there before because before it just passed through A. Ah, so you deleted the line between H and K. I see—right, exactly. Right. In any case that’s fine, and this line is deleted. Okay, so that is basically what happens to this diagram. What happens to this diagram? In this diagram H does not enter K, but K and A are separate. So we need to put a separate K here that H will not enter. Who will enter it? G probably will enter it, right? Or not? Yes, G enters it. Let’s put it above. G enters it, and who else? P enters A. That’s it, right? And P enters A. Ah, K enters A, of course. Yes, of course. There are supposed to be more changes in the first diagram. Why? P is supposed to enter A, for example. Where? In the first diagram P is supposed to enter A according to the diagram now. After we deleted the connection between A and K, there are a few things that… Ah, right, changed. So P enters H, and H now also has to enter A. Ah, of course. Because once we deleted that line, we need to connect it here now. Okay, of course. Fine. In short, this shortcut isn’t all that useful. We should have redrawn it from scratch; that would have been better. Okay. This is basically some sort of arrow like this. In short, I thought I could use the existing drawings and just make the changes, but it’s fairly complicated. What happens here is that we actually reach the fourth-dimensional domain. Meaning, now we already have four parameters. You can’t fill these tables with three; only with four parameters. So if you insist very much, I’ll do it. This is filling one. This is alpha. K is alpha and beta, and G is alpha beta delta. Okay? N is alpha and gamma. Y is two alpha beta gamma. And G, we said—what is H? H is two alpha. And P is three alpha. Okay. You see that there is a fourth parameter here. And what happens there—remember, the result is supposed to favor this. Right, now this is actually how Rav Huna presents his common denominator. So how… how does it work here? So alpha—what alpha is this? K is alpha and beta. K is alpha and beta. G is alpha, beta, gamma. Alpha, beta, delta. N is gamma. Why not two alpha, beta? Why shouldn’t G be two alpha, beta? It’s simply because later there would have to be additional arrows. G also includes H. Wait, let’s see. G would make two alpha, beta and also H. H is two alpha, as before. P is three alpha and Y is two alpha, beta, gamma. That’s it. Okay. In both cases I need four dimensions. In both cases I need four dimensions, and therefore how will it be decided? Number of points, connectivity, everything is the same, again it will simply be change of direction. Incidentally, the more complicated the graph, the more likely it is to be connected. Because the more complicated the graph, the lower the chance that it separates into two different regions, right? Because it can be connected to every point to create connectivity. And therefore it’s clear that change of direction becomes more and more important as the problem grows more complicated. What’s the idea behind that? The idea is that if the problem is complicated, then what determines whether one can draw a conclusion here or not is whether there is a consistent hierarchical relation among all the parameters in play. Right? Are they all pointing in the same direction? And therefore change of direction becomes dominant. So now what happens here—again, change of direction. Let’s see that it really works. So here, in terms of changes of direction, do we have more than one? There isn’t more than one. Wherever you want, there isn’t more than one here. I go from here to here, from here to here, from here to here, from here to here—in short, there is never more than one. From G to P also not. Where? From G to P, that’s one. Or this way—if not, no, not this way, this way. Fine. What happens here? For example, from P to K you need two, right? One here and the second here. So the change of direction decides in favor of this, and therefore Rav Huna succeeded in saving his a fortiori argument. And that’s the end of the metal book. Why? Why are we even using—taking the document into account as part of the parameters if now we’re proving only from money and intercourse? Because we are not proving from money and intercourse; that’s exactly the point. Money and intercourse join a larger common denominator. Money and intercourse together go along with document. Why do money and intercourse prove it? Isn’t “this case is not like that case, and that case is not like this case; the common denominator between them” just between money and intercourse, with no connection to document? No, no, no. Money and intercourse as a group. Money and intercourse as one unit—that’s what I drew here: money and intercourse on one side, and document, and the common denominator between them is that in both there is benefit, and in a document there is no benefit, so you need the document, otherwise you can’t learn from money and intercourse—you must have the document. Okay? And therefore the table is constantly growing; it never shrinks. So when you follow the passage, what you really see is that the meaning is that the table keeps growing. In fact this technique basically tells us that we don’t need to follow the passage at all. Just give me all the data, give it to me straight away—this is the whole table, the full list of all the data. Why go stage by stage now? How did the passage go? We started from this, made an a fortiori argument, right? Then a refutation came out, right? Then we said okay, so a common denominator—not a common denominator, sorry—a paradigm case. A refutation of a paradigm case. And then I say common denominator, this table, right? Then I say a refutation of the common denominator, from those two, right? Then I say a larger common denominator, all this, right? Up to here. And then a refutation of the larger common denominator, one one one, right? And what we ended up with in the end was replacing this one with zero. Okay? So basically the point is that when you work intuitively, you can’t know the result right away. You just can’t know it. You have to build it step by step: a fortiori argument, refutation, common denominator, refutation, large common denominator, refutation, and rescue of the refutation. Right? But with this technique—forget it, just give me all the data. Why do I care now about all those intermediate stages? Tell me what kiddushin is, what chuppah is, what inheritance is, what document is, write down all the data here and I’ll tell you the result. No need to go through all the stages brought in the passage. Except for the rescue. Even the rescue—just give me the correct data. The rescue is really just a change in a datum, from its perspective. Right, so give me the correct data. The stage before the rescue I would do like this, and then someone would come and say: not true, the datum there is wrong; it’s zero instead of one. But no—as long as that isn’t given to us, we basically need someone who saw the process in order to know—no, no, no, there doesn’t need to be any process. He is saying in principle: give me the correct data and I’ll tell you the result. You can argue maybe over what the correct data are—you think it’s one, he thinks it’s zero, doesn’t matter. Each person can say what he wants. So this isn’t the thirteen hermeneutic principles at all. An a fortiori argument, a paradigm case, and a common denominator in its three varieties, and the various refutations of them, and the large common denominator. We’ve done a huge number of hermeneutic principles here. That is, there is an a fortiori argument, and there is a refutation of an a fortiori argument. There is a paradigm case from one verse and a refutation of a paradigm case. Right? There are three common denominators, two paradigm cases, two a fortiori arguments and a paradigm case and an a fortiori argument—that’s already seven principles. Right? Refutations of each of the common denominators—that’s ten principles. Right? And now I don’t even know how many principles there are—large common denominator. A large common denominator can be an inference where each of the first ten is attached to one of the other first ten—that’s another hundred maybe. A large common denominator can already be a huge number of combinations, right? Refutation of a large common denominator, and so on and so on. So there are hundreds of hermeneutic principles here. Right? Now there’s no point in treating them as hundreds of hermeneutic principles. The whole distinction among the various hermeneutic principles exists only because we think of them intuitively. But if I understand that in fact everything works by one principle, I have some algorithm: fill in the table, tell me what’s preferable to what according to our criteria—three topological criteria and dimension—and that’s all, then this is one hermeneutic principle. All the rest are just different applications of it. Okay? So in fact, all the separation between the hermeneutic principles stems from the fact that we think about them intuitively and not in this way. If you think about them in this way, then really there is one form of reasoning here, albeit a somewhat complex one. That is: give me a table, draw the graph, build the model; the criterion lies in the question, the three topological indices and dimension. That’s all. And if that is the criterion, then all the hermeneutic principles—there are hundreds of them here. Hundreds? You could keep going. In fact the number of hermeneutic principles is the number of possible tables. Right? How many tables are there? Infinitely many, of course. And that means infinitely many hermeneutic principles. That is, there are as many tables as you like, of whatever size you like, with whatever filling patterns you like. Just look at this size: there are many possible filling patterns here. Most of the filling patterns here don’t even have names in the language of the Sages. Because only these structures—a fortiori argument, paradigm case, common denominator, refutation—those are clear structures. But they are very few out of the possible structures. What happens if I have some other inference? There will be dimensions that won’t work. No—what does it mean they won’t work? The result will either be one, or zero, or unknown. Not that you can infer nothing from them. If the whole first row is zeros, the whole second row is zeros, all the rows are full of zeros. Doesn’t matter, that’s a special case, but mathematically that too is a hermeneutic principle—it doesn’t matter. In the end the point is that every such table has some result. The result is either one is preferable, or zero is preferable, or there is no preference. Doesn’t matter—each of those is a hermeneutic principle. Either it is a principle of refutation, or it is a principle that proves something. To prove one is a proof; to prove zero is also a proof. A refutation means that one or zero doesn’t matter. Those are the three outcomes for every table. And therefore every table in fact has some solution—though again, that is a theorem that still needs to be proved. That every such table has one result out of the three: either one is correct, or zero is correct, or there is no way to decide. I’m not entirely sure that’s trivial. That is, that for every table one can indeed reach one of those three conclusions. This reminds me of Zermelo’s theorem, for those who know game theory. Zermelo’s theorem says that every game equivalent in some way to chess—it doesn’t matter—has one of three outcomes. Either White wins, or the first player wins, or the second player wins, or it’s a draw. It sounds like a trivial theorem, right? But it is not trivial at all, and it has a proof that is not so simple; it is proved by induction. Why is this theorem not trivial? Because it doesn’t say that every game will have one of three outcomes. Rather, it says that every game must have exactly one of the three. That is, for example in chess, exactly one of the three. Whoever plays optimally can force a win—either White or Black—or force a draw. Doesn’t matter. But chess has one correct outcome. I don’t know which one. So I say: it is either a win for White, or a win for Black, or a draw. In that wording it sounds trivial, but it isn’t trivial, because the theorem says there is one clear outcome. I don’t know who. The “optimal” in question is the result. When each side plays optimally, of course—that is the assumption. Each side plays optimally, meaning either White can force a win. You mean two players who keep playing forever? Not necessarily. Why? If White plays… Why? In chess White gets an advantage because he… “Advantage” is intuition. Here we’re doing mathematics. No, even in terms of scoring, that’s what. When White wins he gets fewer points. Fine, that’s again an intuitive claim because you’re playing non-optimally; now you can measure why.