2019-04-22 – Between Midrash and Logic – Lesson 13

Table of Contents

• [0:00] A complex hierarchy between the columns
• [4:07] Analysis of zero-filling – structure A, B, C
• [6:13] Comparing zero-filling and one-filling
• [7:40] The zero-filling graph – alpha and beta orders
• [14:33] The third table with two zeros
• [19:47] The three types of common denominator and the proof
• [22:09] Explanation of kal va-chomer and binyan av
• [27:50] The passage in Kiddushin and the connection to stringency

Summary

General Overview

The hierarchical relations between columns are not always a simple one-dimensional ladder. Sometimes you need a complicated internal structure in which one thing is “smaller than that,” but there is no relation between other pairs. The test is carried out by solving the “filling,” which assigns parameters like alpha and beta to points in the diagram, while preserving constraints such as the prohibition against increasing value on two axes at once and dependence or independence between parameters. From these solutions emerges a distinction between dimension-parameters and three decisive topological parameters—changes of direction, number of points, and connectedness—and on that basis a proof is constructed that each of them is necessary in order to explain the three types of “common denominator.” On that basis as well, a relative strength is established between kal va-chomer, binyan av, and common-denominator inferences, and then the tools are applied to the passage in Kiddushin concerning kal va-chomer from chuppah, its refutations, and the move to the common denominator and the refutation “since their benefit is greater.”

Simple Hierarchy versus Complicated Hierarchy, and Testing Dimension

A simple hierarchical relation is obtained when everything is arranged on one line of the ladder, and then the left-hand side is “stronger” because it has fewer changes of direction, but the decision also depends on dimension. Dimension is tested through an actual solution of the filling and not merely through graphic intuition, because it may be that dimension gives “the opposite side,” and then you remain in a state of refutation. The solution of the filling begins from the higher places as minimal indices and requires all the rest to keep increasing.

One-Filling: Parameter Possibilities, Non-Relation, and the Example of Rising Alpha with Beta Crossing It

In one-filling, two basic possibilities are presented for increasing beyond alpha: “two alpha” or “alpha and beta.” The attempt to place “two alpha and beta” or “alpha and two beta” is rejected because of connection constraints and the prohibition against raising on two axes. One possible solution is obtained by placing “alpha and also beta” at one point and “two alpha” at another, so that at an additional point you get “three alpha.” Another possible solution places “three alpha” opposite “two alpha and beta,” and creates a situation in which something is “smaller than that in terms of the alphas and bigger than that in terms of the betas,” and therefore “there is no relation between them.” Assuming alpha and beta are independent, it is determined that there is no relation, and a diagram is demonstrated in which “the alpha rises and the beta crosses it,” where beta marks what becomes stronger, but the quantitative measure of how strong it is is alpha.

Comparing Diagrams by Changes of Direction, Dimension, and the Conclusion that One-Filling is Preferable

In comparing the two diagrams, it is determined that both are two-dimensional and both require alpha and beta, and that valence is “not all that important,” so it does not change the picture. It is determined that one of them is “less good because it has more changes of direction,” and from that the conclusion follows that “one-filling is preferable.” Later they return to intuition through breaking down the forces of different actions and checking what each action “succeeds in doing” in relations of inclusion.

Calculating Forces in Zero-Filling and One-Filling, and the Intuition of a Common Factor

In zero-filling, it is determined that A contains B and C and therefore “it has two alpha,” that B contains only A and therefore “it has beta,” and that C contains everyone except C, so “it has alpha and beta,” but “that is not enough because C needs two alpha.” In one-filling, it is emphasized that you have to take into account the solution of the specific diagram, and then it turns out that B and C have “three alpha,” and that C gets “two alpha and beta.” The intuition is formulated as follows: in zero-filling there is a common factor for A and C, namely alpha, and it does not exist in B, so that is “less good,” whereas in the blue possibility in one-filling the common factor also exists in B, and therefore “that’s our zad, alpha is basically our zad,” and so it is preferable.

A Second Example: Deciding by Number of Points and Remaining with a Refutation until Throwing Out Indices

In another example, a graph is built in zero-filling in which A is the largest, B enters into A, and two components each separately enter into B. In one-filling, A and B collapse into a single point, and so “this line basically shrinks into a single point.” In calculating the parameters, one is forced to choose “two alpha” and not “alpha and beta” in order not to get stuck in the constraint that “we’re not interested in two” and “you can’t work in two indices,” but it is still determined that “not yet,” because if the dimension reverses, it will remain a refutation, and one must “throw out all the indices” unless two contradictory ones have already been found. In the end, it is determined that in terms of dimension and changes of direction there is no difference; the decision is determined “by the number of points,” and it is said explicitly that “this table is decided by the number of points,” whereas the previous one was decided by change of direction.

A Third Table with Two Zeros: Connectedness as a Third Parameter and the Need for an Additional Parameter

In the third table, where “there are two zeros,” zero-filling yields a split graph in which “A is not connected to the graph,” and therefore “the graph is split into two parts,” and it is said that this is “the third parameter.” In one-filling, everything enters equally into B, and then the number of points and changes of direction do not decide the issue, while connectedness does, after checking that in terms of dimension “nothing gets messed up for us.” In the attempt to assign only alpha and beta, the need for an additional parameter emerges, and it is agreed that the solution uses “alpha and gamma,” with the determination that the larger the table is, “the diagram gets more complicated and there are more microscopic parameters.”

Proof of the Necessity of the Three Topological Parameters and Convergence from Two Directions

It is determined that the three types of common denominator are decided in such a way that each one is decided by a different topological parameter out of the three, and this is presented as “a proof of the necessity of these topological parameters in the criterion.” It is argued that if one were to define a criterion without connectedness, the third common denominator would not be explained; if without changes of direction, the first would not be explained; and if without numbers of points, the second would not be explained. He says the beauty of it is that the three were “brought out even before we got to the common denominator,” from the analysis of kal va-chomer and refutations and from the understanding that “dimension and cleanliness are not enough,” and then it turns out afterward that each of the three really does decide a different common denominator, which is considered stronger than an ad hoc move.

Relative Strength: Kal Va-chomer versus Binyan Av, and the Status of the Common Denominator

In kal va-chomer, a preference is established for one-filling “because of connectedness” and also because of “dimension,” and therefore it is “a very strong preference.” In binyan av, it turns out that in one-filling the two columns are identical as one point, while in zero-filling you get “a reverse kal va-chomer,” and the preference remains only “in terms of number of points,” without any dimensional preference. From here an explanation is proposed that kal va-chomer is stronger than binyan av either because a dimensional preference is stronger than a topological preference, or because “two preferences are better than one preference.” A practical difference is then brought: “binyan av… can be refuted by any slight refutation, whereas kal va-chomer is not refuted by any slight refutation,” attributed to “the Talmud in Chullin 115.” Later it is determined that in generalizing inferences the preferences are only topological, and therefore kal va-chomer is stronger than they are. From that, it is explained why “the בעלי הכללים” claim that the common denominator has “the status of binyan av” and that one can “refute the common denominator with any slight refutation.”

Moving to the Passage in Kiddushin: Rav Huna, Kal Va-chomer from Chuppah, and Its Rejection

The passage opens with the statement of Rav Huna: “Chuppah acquires by kal va-chomer,” and it is clarified that the law is that chuppah does not create betrothal but only marriage, and the first kal va-chomer relies on terumah. The Talmud rejects this through Ulla, who establishes that “by Torah law, a betrothed daughter of Israel may eat terumah” from the verse “And if a priest acquires a soul, the acquisition of his money,” and explains that the reason she does not eat is “a decree lest they mix her a cup in her father’s house and she give it to drink to her brothers and sisters,” and therefore the first kal va-chomer falls away.

Building the Table for the Passage: Refutation, Binyan Av, Refutation, Common Denominator, and the Refutation of Benefit

The kal va-chomer that begins, “Just as money, which does not complete, nevertheless acquires—then chuppah, which does complete, should it not all the more so acquire?” is presented as a template of money versus chuppah in relation to betrothal and marriage, and it is said that the filling is one. The refutation, “What is unique about money? It redeems consecrated property and second tithe,” is described as adding a column to the previous table, and it is determined that here “zero-filling and one-filling are equivalent,” and therefore this is a refutation. “Intercourse will prove it” is built as another table in which intercourse effects both betrothal and marriage, while chuppah effects marriage only, and it is determined that this is “binyan av,” and that in binyan av “one-filling is better.” Then another refutation comes: “What is unique about intercourse? It acquires in the case of a yevamah,” as the addition of the yevamah column. Afterward, “money will prove it” joins with “and the law returns… the common denominator,” and a full table is built with all the collected columns, including the determination that money regarding yevamah is zero and intercourse regarding redemption is zero, so that the common denominator will work and not turn into a refutation. Finally, a refutation is brought against the common denominator: “What is unique about their common denominator? Their benefit is greater,” and this is translated into adding a benefit column in which money and intercourse have it and chuppah does not. The graphic analysis shows an advantage in each direction through changes of direction versus number of points, and so a refutation is created.

Methodological Notes: Independence as Vectors, Problems of Continuity, and the Further Complication of the Model

He says there is a desire to translate the structure into the language of vectors and independence in order to base algorithms on theorems of “Hilbert spaces,” but “for now I don’t know how to do that,” and “we need to work on the matrices.” Another point is raised about problems of continuity, where fillings are not just zero or one but continuous values, and then “you can no longer use these rules in a simple way.” In conclusion, he presents an exercise in assigning alpha and beta in graphs and infers that in the case at hand, “in terms of dimension they are both the same, and there is one preference in each direction, so basically that’s a pirkha, it works.”

The hierarchy relation between the columns is less simple. Here the hierarchy relation is simple: this is smaller than this, this is smaller than this, and this is smaller than this. Here there’s a terribly complicated relation: this is smaller than this, but this has no relation at all to that, so the internal structure is more complicated. When there’s a simple hierarchy—say, if they were all on one line—that would be the simplest possible case; that’s the simplest hierarchy, each one arranged on a simple scale, okay? So the one on the left is stronger? Exactly, so the one on the left is stronger because it has fewer direction changes, okay? But that’s still not certain; we also have to check the dimension. If the dimension gives us the opposite side, then we remain in a state of refutation, right? So how do we check the dimension? We solve it. What do we do? So let’s start with filling, filling one. In filling one, I say it like this: let’s say this is alpha, here I have two alpha, just for simplicity’s sake—what will be here? Alpha and beta, right? Those are always the two possibilities: one is two alpha, one is alpha and beta. Those are the two possibilities that were larger than alpha, the two possibilities that fit into alpha. What do I put here? Two alpha and beta? Two alpha and beta won’t work, because then there would have to be a connection to here. Alpha and two beta? Alpha and two beta won’t work because I’m not allowed to raise the valence on two axes, right? Alpha and beta? Alpha and beta won’t work because it would collapse into this, so let’s add another parameter. Something here doesn’t work out, right? So look, the simplest option is to put alpha and beta here too, and the second alpha here—it doesn’t really matter where they go. Then here you have three alpha, and that’s perfectly fine, right? A slightly more complicated option, but also possible, is to write three alpha here and two alpha plus beta here. If there were two alpha here, it would be less, but there are three alpha here, so that’s not it; it’s smaller than this in terms of the alphas and greater than this in terms of the betas, so there’s no relation between them. Clear? So then what is the relation? There is one—we just don’t know what it is. No, there isn’t a relation. Why, who said there is? You don’t know what beta represents. Doesn’t matter—assuming alpha and beta are independent, there’s no relation between them, okay? So now that’s a possible solution, and you can see here an example of the diagram I did last time: alpha rises, the alpha rises and beta crosses it, right? That’s exactly the example we saw last time, where beta only tells us what keeps getting stronger, but the quantitative measure of how strong it is is alpha. Okay? So that’s one possibility, or the other possibility—it’s not important. What happens here? So let’s start again. This is alpha and this is beta. I always start with the highest places—those are the minimal indices. And here too I always started with this, and then the rest have to keep growing. So here, for example, what would you put here and here? Above, two alpha, and there alpha and beta. Exactly. Here it has to be alpha and beta, because if I put two alpha here, that would be fine with this—but what about this? I’d have to add two alpha plus beta here. But then I’d get stuck here. Therefore here you have to put alpha and beta, and here two alpha—excellent, okay? What does that say about the relation between the diagrams? We see that both are two-dimensional, both require alpha and beta. The valence, we said, isn’t all that important; this is a bit worse in terms of valence, but we said valence doesn’t change anything, so it doesn’t change the picture. The previous picture remains as it was: this thing is less good because it has more direction changes. What does that mean? That filling one is preferable, okay?

Now, to see the intuition I presented earlier, let’s go back and see what this says about those. After all, what I found here is what the components are that make up each of the results. But now I ask: what are the powers or capacities that each of the actions has—canopy, intercourse, money, and so on, okay? How do I check that? So let’s start with filling zero, okay? In filling zero, I say this: what is A? A contains B and C, right? What does that mean? What does A have? A manages to make one for B and C. B needs alpha in order to be one. In order to contain C you need the power of two alpha. So it has two alpha. Agreed? So in fact it has two alpha. What does B have? B manages to contain only A. What does that mean? That it has beta, right? That’s it. What about C? C manages to contain everyone except C. That means it has alpha and beta. Agreed? For C that isn’t enough, because C needs two alpha, and it doesn’t have two alpha—it has only one alpha. Okay? Moving on, that was filling zero. What about filling one? In filling one… in filling one we have here A—what is A? In filling one it’s the same thing. Right? In filling zero, B has A, B, so alpha and beta. Right? And C also the same thing. Wait, no no, we have to be careful, it’s not exactly—it doesn’t have to be the same thing because the solution is not the same. No, we have to be careful, friends, we have to compare, enter this diagram. So look, A contains B and C. Let’s go back to B and C. For B and C that’s three alpha. Right? You have to take account of the solution that’s here, not the solution that’s here. Okay? So three alpha. B contains A and B. What are A and B? That’s alpha and beta. So that’s fine. And C—it also contains D, doesn’t it? What? Who? B? No. B doesn’t contain—that’s perfectly fine, it really doesn’t contain it. Okay? And C contains everyone except C, which means it has two alpha and beta. Right? Now look what that means intuitively. In filling zero, in filling zero what happens? Filling zero is the black one. You see what happens? There’s a common factor to A and C—what is it? Alpha. And it doesn’t exist in B. This is the zed—not our zed, this is the x and y. In each of them there is a common factor, one factor that doesn’t exist in lamed, right? Therefore it’s less good. What is the better alternative? This one. Why? Because here the common factor they both have also exists in B. So if alpha is—and notice, also in the solution—who contains B? After all, that’s what I was looking for, I’m asking who contains B. Alpha, right? We see it from here. And that is exactly the factor that exists both here and here, and after all it exists here too. So therefore that is our zed—alpha is basically our zed. Right? And therefore it’s preferable, the blue one. In filling one. So that fits the intuition I described earlier on the board. That’s one possibility, that’s the first example.

What’s the next example? The next example—let’s now do… this one was decided, remember, by direction changes. Right? What happens now here? Okay, what happens here now? I’ll start with filling zero this time, okay? In filling zero, A is here, it’s the biggest, right? B goes into A. B goes into A. And C and D go into B. Right? They both go into B. Agreed? Rabbi, can you remind us how you see the… I’m checking the relation between the columns. Column A is the highest, one one one, the highest possible. So I put it here. Okay? Now I look at who comes next, who goes into it. I say B is the second strongest. So B goes into A. Now I say these two each go separately into B, but not into each other—rather each one separately. So therefore this goes in from here and this goes in from here. Okay? So that’s the graph for filling zero. What happens in filling one? In filling one A and B are the same point, right? A and B are the same point. You can already see what the line here will be, the number of points, right? And C and D both go in here. Which means this line essentially shrinks to a single point. That’s what happened. Okay? C and D go in here. Now let’s start doing the calculation. So here, of course, this is the regular solution we have, okay? What happens here? Two alpha beta in C. Alpha. What happens here? Alpha beta. I could do either two alpha, right? But you need something that will also rank it separately from C and from D so that they won’t be identical to one another. So let’s see. If I do two alpha here, then here three alpha and here two alpha plus beta. Right? If I did alpha plus beta here, I’d be in trouble. Why? Because if you did alpha plus beta here, then here you’d have to do two alpha plus beta, and what would you do here? You’d have to do two alpha and two beta, and we said we’re not interested in doubles. But you can’t work with two indices, so I have to put two alpha here and not alpha plus beta. Right? But that already won automatically, didn’t it? Not yet, because if the dimension reverses us, then the dimension would make it like this. It would remain a refutation, it would remain open. I have to throw in all the indices. Unless I’ve already found two contradictory ones. If I’ve already found an advantage in one index and an advantage… then it no longer matters, because if it’s two against one then that isn’t important. What matters is that there’s an advantage in each direction, regardless of intensity.

Okay, now what happens here? So we see again: in terms of dimension, both are two-dimensional, and q, we said, doesn’t matter. Direction changes—what happens here? This is one direction change, and this is also one direction change. Right? So there’s no difference in direction changes. No difference in the number of points—they’re both connected. Right? Sorry, there is a difference in number of points. They’re both connected, no difference there, and this is decided by the number of points. Meaning this is better than what? Filling one is better because of the number of points. Let’s now do this calculation and see whether it really fits the intuition. So let’s start by solving A. So with A we’re now talking about filling zero. A contains A, B, and C. That means it contains everyone except D. That means it has three alpha. Right? This is what succeeds in containing everyone except D. What does B have? It contains only A, so it has alpha. And what does C have? C contains everyone except C, so it has two alpha and beta. Right? Wait, this is filling… two alpha and beta. Wait, which filling is this now? Wait, no—C doesn’t manage to contain C. It should be two alpha and beta. Why? Because if it also had B, then it would contain C. It doesn’t manage to contain C, but it does contain D because it has two alpha. So it has two alpha. We’re talking about the upper one. Ah, wait, that’s it, I saw something here wasn’t fitting. Sorry, I looked at the lower one. Wait, so once again, are these right? These are right. B is only A, so that’s alpha, yes. And C? C contains everyone except C, so it has two alpha and beta. Okay? That’s in filling zero. What happens in filling one? In filling one, then A contains everything except D, so alpha plus beta. B contains A and B, which is alpha. And C contains everything except C, which is two alpha. Right? Okay, now here you see—just one second—in order to contain A you need alpha. Something here doesn’t fit. I did this wrong. It can’t be that it has alpha. Yes, it comes out the same height. Not only that—it comes out the same, the intuition… what tells me I missed something here? That the intuition says there’s a common factor, there is a common factor both here and here, so it can’t be. Clearly I missed something here. And clearly I missed something here because if it had alpha, it could contain B. Wait, who am I asking about? I’m asking about B. Ah, wait, no, sorry, then this is fine. B… Ah, no, no, fine, this is fine. It’s alpha. It’s alpha. A is fine. Why is this still fine? It’s still fine because I’m asking who contains B. Right? After all, my question mark was here. What do you need for B? Two alpha, not alpha. Let’s now look here. Here both sides have two alpha, and this one doesn’t have two alpha. Right? It has only one alpha. So for us that’s as though it doesn’t have the relevant parameter. Okay? Here, by contrast, in this solution, alpha is enough in order to contain B. So here both sides have the alpha parameter, and B has alpha too, so therefore this is fine. Okay. Good. So this table is decided by the number of points. The previous one was by direction changes, and this one by the number of points.

Now I move to the third table. The third table is when there are two zeros. Right? And there’s no longer any further equal side—you can’t have any more equal sides in the whole Talmud, aside from these three types. So this completely exhausts what we’re doing here. What happens here? So we start again with filling zero. Filling zero: who is the strongest? B is the strongest, right? Let’s put B first. A doesn’t depend on it at all, and C and D go in. A binyan av from two verses. Fine? We’re now going to see what will decide this table, right? What will decide this table? Connectedness. Connectedness. You see? A is not connected to the graph—there are two parts here. The graph is split into two parts. That’s the third parameter. Okay? That’s in filling zero. What happens in filling one? So B is the strongest, and all of them go into it equally, right? C, A, and D. That’s the graph, right? Each of them goes into it separately, not through the other. Now we can already see what will decide the graph. Number of points: four and four. Connectedness is the same thing. Here too there are direction changes, sorry. Here there’s one and here too there’s one. There isn’t more than one, right? But connectedness tips the balance. Exactly the third topological parameter—we just need to check that in terms of dimension nothing gets messed up. So let’s begin here. Let’s say this is alpha, and this is two alpha, and this is alpha plus beta. What do you do here? Gamma. Or anything else that doesn’t connect it at all to the other two. I think you can do this with two. Let’s do two beta. Why beta? No, two beta. Beta—but why beta? I have two alpha here. But wouldn’t one beta work? One beta wouldn’t work, because then there’d be an arrow from D to A. Yes. But why not two beta? Why not two beta? Because then there’d be an arrow from there. No, I’m not allowed to go up with two beta, I’m not allowed to go up once… because it’s connected because its parameter went up. No, no, there’s a constraint: I’m not allowed to go up in two parameters in this graph. Because it’s in a different graph, then we can? No, it’s the same graph, it’s an unconnected graph. It’s the same graph. So wait, but there is a solution here—just a second. Let’s look here for a second. Just to check. Let’s look here for a second. What happens here? Here we have alpha, and this is two alpha, this is alpha plus beta, and this is just beta. No, just beta? So here too, then something goes wrong here too. So apparently in both we need to introduce gamma. And here too gamma. In a second we’ll see whether I didn’t find a better solution here. For some reason I don’t remember having used another parameter here. Ah, yes. Yes, I used another parameter here—of course not gamma but alpha and gamma. Right? Because there has to be an arrow in relation to alpha. So alpha and gamma. The A of the upper graph is also the separate one. It’s also the separate one because it has no arrow. Because… yes, so here indeed you need—it shows you, we’re already getting to… the bigger the table, the more complicated the diagram becomes, and the more microscopic parameters are needed to explain it. There are more properties at play in the field. Okay? So here we really get to three properties. And now again: both are three-dimensional, the direction changes are the same, the numbers of points are the same, and connectedness decides the issue. Okay?

Let’s check our intuitions again. So we have the following: I start with filling zero. A contains B and C. That means it has two alpha. Right? B contains only A, which means it has gamma. And C contains D and B. Alpha and beta. Both B and D—that’s alpha and beta. Okay? What happens in filling one? In filling one, A contains B and C. Two alpha. B contains A and B. Alpha and beta. And this… alpha and gamma. B and D? Alpha and gamma. Okay? Again, we check our intuitions. What… what do you need in order to contain B? Look, here you need alpha in order to contain B. So these two have alpha, but here there is no alpha. So there again that same intuition comes back, right? They have a common factor, which is the relevant factor that contains what we’re looking for, and it doesn’t have that factor. By contrast, what happens here? Here, who contains the… B? Alpha. These two have alpha, and this one has alpha too. So therefore this is indeed preferable. Okay?

So look at the interesting thing we got, actually: the three kinds of common-denominator argument that exist—one of them is decided… each one is decided by a different topological parameter out of the three. This is actually a proof of the necessity of these topological parameters in the criterion, right? Because if we had defined a criterion without connectedness, for example, then the third common-denominator argument our model wouldn’t explain, right? If connectedness were not needed in order to prove superiority, I wouldn’t put connectedness into the criterion, so that last common-denominator case couldn’t be explained by the model. If I hadn’t included direction changes, then the first one couldn’t be explained. And if I hadn’t included the number of points, then the second one couldn’t be explained. That means there is a proof here that these three topological criteria are required in order to have a good model describing these inferences. The nice thing is, as I probably already told you, that we derived these three parameters before we ever got to the common-denominator argument. Meaning: we thought to ourselves—we reached the conclusion, as I explained at the end of last semester, we reached the conclusion from the simple elements themselves, the issues of a fortiori reasoning and refutations, and we immediately saw that dimension and cleanliness are not enough. You need to add topological parameters. Then we asked ourselves, in graph theory, which topological parameters sound reasonable. We wrote down three topological parameters, and now it turns out—we kept moving forward, we did the common-denominator argument—and it turns out that exactly each of those three decides a different common-denominator case. Meaning that this is a proof that exactly those three are required. So that’s nice because it converges from both directions. If I had done it ad hoc, it would have been less strong. Still fine, but less strong. Here I’m saying there was some logic to it, and we wrote it down even before we saw that this was really what works. And after checking, we see exactly that those three are needed. Okay?

Now another point: there really is room now to discuss what the parameter is, what the stronger inference is. Which is the stronger inference? So I’ll start first with a fortiori reasoning and binyan av. A fortiori reasoning and binyan av—what was the situation in a fortiori reasoning? In a fortiori reasoning the table was this one, right? And the table was A B or A B, right? That was filling one and filling zero. That was a fortiori reasoning. Why is filling one preferable? Because of connectedness. Both connectedness and dimension, because here it’s alpha and two alpha, and here it’s alpha and beta, right? So both connectedness and dimension, right? So the superiority of a fortiori reasoning is a very strong superiority. Okay? By contrast, in binyan av the situation is like this. So in one table we basically have this in filling one, and in filling zero we have this, right? Exactly like a fortiori reasoning. Agreed? In filling one, basically the two columns are identical, so we have only one point, right? And in filling zero we simply got the reverse a fortiori. So overall, what is the superiority of this over that? In the number of points? No, in dimension it’s the same dimension, because this is alpha and this is two alpha and this too is alpha. So in terms of dimension there is no superiority; there is only superiority in terms of number of points, and that’s it. So now that we compare these two inferences, binyan av and a fortiori reasoning, we have an explanation why a fortiori reasoning is considered a stronger inference than binyan av. It is stronger than binyan av because it has two advantages for filling one. And binyan av has only one advantage. There was also another possible explanation: to say that superiority in dimension is stronger than topological superiority, because after all our initial intuition about why one is preferable to another was dimension—how many microscopic parameters you need. That’s simpler. What? Because again it comes back to this—that it’s simpler. What’s simpler? When there is only one dimension, that simplifies it much more than with… Yes, exactly. Meaning our initial intuition, when we asked ourselves which diagram would be simpler, we immediately said dimension, right? What has fewer parameters. We only added topology afterward because of constraints. So there are two possible ways to explain in this model why a fortiori reasoning is stronger than binyan av. One explanation is that superiority of dimension is stronger than topological superiority. A second explanation is that two advantages are better than one advantage. Therefore this is an inference—again, they are both good inferences. In both of them filling one is the correct filling. But still, a fortiori reasoning is a stronger inference than binyan av. There’s a practical consequence to that: a binyan av… and here one can refute with any slight refutation, while a fortiori reasoning is not refuted by just any slight refutation. That’s a Talmudic text in Hullin 115. Meaning the Talmudic text itself says that a fortiori reasoning is a stronger inference. Okay?

Now those are the two explanations. Now let’s move for a moment to the generalization inferences, and according to these two explanations we’ll understand something else. Quite a few later authorities—maybe it already starts among the medieval authorities (Rishonim)—claim that the common-denominator argument has the status of a binyan av. Even though a common-denominator argument can begin with a fortiori reasoning. Right? You begin with a fortiori reasoning, it doesn’t work, you add another one; that too can sometimes be a fortiori reasoning, or a binyan av. But to say that both are a fortiori reasoning. And when I do the common-denominator argument, when I finish the process, what I get is a binyan av and not a fortiori reasoning, even though it is built on two a fortiori reasonings. Many later authorities say this; it appears a lot in the books of rules. The practical implication is whether you can refute it with any slight refutation. There is a Talmudic text saying that one can refute a common-denominator argument with any slight refutation. What does that really mean? Why indeed is that so? Now I can look for an explanation. Because if I understand why a fortiori reasoning is stronger than binyan av—and we explained that earlier—let’s see whether a fortiori reasoning is also stronger than a common-denominator argument in the same sense. So we had two explanations for why a fortiori reasoning was stronger than binyan av. One explanation: because in a fortiori reasoning the superiority was superiority of dimension and not topological superiority. And in binyan av it is superiority in number of points. Okay? What happens in the generalization inferences? There we saw that the superiorities are all topological—each of the three topological indices, right? So therefore it is obvious that a fortiori reasoning is stronger than them. Since the superiority in a fortiori reasoning is superiority of dimension, not topological superiority. Or: because in a fortiori reasoning there are two superiorities—both dimension and connectedness, also a topological matter—whereas in all the generalization inferences there is only one topological index that determines the superiority. Therefore, just as a fortiori reasoning is better than binyan av, a fortiori reasoning is also better than the three common-denominator inferences. So suddenly we understand very well why the authors of these rules say that the common-denominator argument has the status of a binyan av, meaning that it is weaker than a fortiori reasoning. Okay? Good, so that’s the explanation regarding the relative strengths.

All right, before I… look, there are pages here, I’ll hand you these pages, maybe pass them back. This is the topic in Kiddushin, and I want to go through it from now on, because there we’ll be able, first, to use the tools we learned, and also to continue one more step forward because it gets a bit more complicated. Okay, so let’s look at the Talmudic text. Do you have it? Rav Huna said: a canopy effects acquisition by means of a fortiori reasoning. The law is that a canopy of course does not effect acquisition. This is really a difficulty. Yes? A canopy effects acquisition by means of a fortiori reasoning. “Effects acquisition” means it creates betrothal with a woman, as distinct from marriage; a canopy creates marriage but does not create betrothal. “And if money, which does not enable eating of terumah, effects acquisition, then canopy, which does enable eating of terumah, is it not all the more so that it should effect acquisition?” So there is a fortiori reasoning here. I’m not treating it, as you see, as starting the numbering, because it is immediately rejected. “And does money not enable eating? But Ulla said: by Torah law, a betrothed daughter of Israel may eat terumah.” So we see that if I acquired her with money she becomes my betrothed, and that is enough for her to eat terumah, so why are you saying money does not enable eating of terumah? As it is stated: “And if a priest acquires a person, the acquisition of his money, he may eat of it,” and this is indeed “the acquisition of his money.” “And why did they say she may not eat? A decree lest they mix her a cup in her father’s house and she give to drink to her brothers and sisters.” Fine? So in truth a betrothed woman does not eat terumah, but that is a rabbinic law. On the Torah level she does eat terumah, and therefore this whole a fortiori reasoning collapses.

Now we begin the topic itself. “Rather, one may challenge as follows.” Why is it called “challenge”? It’s called “challenge” because Rav Huna’s whole a fortiori reasoning is really a kind of difficulty, right? Since canopy does not actually effect acquisition; Rav Huna is only raising here an a fortiori reasoning that could seemingly yield the result that canopy does effect acquisition. So he begins like this: “And if money, which does not complete, effects acquisition, then canopy, which does complete, is it not all the more so that it should effect acquisition?” So we begin the topic in the following way. We have a fortiori reasoning. We have money—this is the teacher—and this is canopy. Here we have marriage and here betrothal. Let’s call this A for convenience, marriage. If, of course—how could it be otherwise?—money is money and canopy. So what happens? Money does not manage to impose marriage, yet it does impose betrothal, right? Canopy, which imposes marriage—is it not all the more so that it should impose betrothal as well? That is basically the a fortiori reasoning, right? We already know how to solve it, I won’t do it again. Okay, we know that the filling is one, we have the diagrams. I’m just beginning to construct it so you can see how one analyzes a topic using these tools.

Step two: “What about money, for it redeems consecrated items and second tithe?” What is that? A refutation, right? It’s a refutation of the a fortiori reasoning. What kind of refutation? Money has another parameter. Meaning, here—redemption, right? Another column is added. One, zero. Money redeems and canopy does not redeem, right? Notice, the refutation is not a new argument; I attach it to the previous table. What do I do now in order to see the situation? I analyze the complete table. We already did this too, and we already saw that here it really remains open. Filling zero and filling one are equivalent. Okay, so that is a refutation.

Step three: “Intercourse will prove it.” What is “intercourse will prove it”? Here you have to be a little careful. In principle we would now have to add another intercourse here, another row. But this is the stage where we arrive at the common-denominator argument; that will happen in stages five and six, sorry. Basically what intercourse does is the following. Another table. A, N, money, canopy—sorry, intercourse, canopy. Let’s call these B and H. One, one, one, question mark. Intercourse, after all, creates both betrothal and marriage, right? And canopy creates marriage but not betrothal. And now what do we do? We already see from the data—that is, the Talmudic text doesn’t say anything here, but it’s clear to us from the data—that this is a binyan av, right? Not a fortiori reasoning, but a binyan av. We analyzed that too already. And we know that in binyan av filling one is better. You can already see it, because this is one point, and if you fill one both are identical. If you fill zero here, there is an arrow between two points. That’s what we did just a moment ago. Okay, so the binyan av works.

But, next stage: “What about intercourse, for it effects acquisition with a yevama?” So what does that mean? Again, yevama—intercourse effects acquisition with a yevama, canopy does not effect acquisition with a yevama, right? Exactly the same thing; we also did that already. It’s a refutation of the binyan av. Okay. Yevama. That is section five: “Money will prove it.” Now “money will prove it”—what does that mean? They don’t just go back again to the first one. But what does it mean to go back again? They aren’t really going back, because the first one didn’t work; rather, it is clear that stage five comes together with six. Stage five comes precisely together with six: “And the law returns. This is not like that, and that is not like this; rather, the common denominator between them is that they effect acquisition generally and effect acquisition here, so I too will bring canopy, which effects acquisition generally, and let it effect acquisition here.” What does that really mean? I am now making a complete table. And I say as follows: how many columns will there be in the table? Four, right? Notice, because we gather all the… wait, that’s one column too many. We gather all the columns we have among all of them. Notice—that’s the method. At each step as we progress, the table grows. We do not make a different table. The table grows. That is a methodological stage on the way to constructing this table. So we have, just for names, we have A, N, P, and Y, right? And here we have money, canopy, and intercourse. Okay, now notice: what we have actually obtained here is exactly the common-denominator table that we analyzed at the beginning of the lesson, right? This is the end. But first let’s see how to build it; let’s try to do it by hand. So look, money is zero, one, one—but what about money… with a yevama? We don’t have that datum here. We have no idea, we don’t have the datum—but clearly it’s zero. We know that money does not effect acquisition with a yevama. We know that from elsewhere. Let’s see here. Right. If it weren’t so, there could be no common denominator. From that too we can know it, as I said earlier. If there were a one here, one one one would refute the whole common-denominator argument. Therefore in practice, if the Talmudic text makes a common-denominator argument, you can already put a zero there. But we also know that that is the law. Okay. What happens with canopy? So look, betrothal is one, here it’s a question mark. Redemption is zero, and yevama is zero. All that data we have from here; no need to add anything from outside. What about intercourse? One one, so we have one one, yevama we already know is one, and zero, and here again—intercourse does not effect redemption, we know that from elsewhere. Yes? So that’s zero. But from the very fact that this is a common-denominator argument, I said, this and this have to be zero. So that they remain independent. Right. Okay. Or so that this won’t be a refutation, really. Not so that… in the graph you’ll see it through independence, but in terms of what matters to us: if there were one and one here, that would be a refutation. You can see that intuitively too.

Could there be a situation where only if we know it from somewhere else, and it wouldn’t be as you said, that we know it from both sides? Because there wouldn’t be a refutation if it were money and yevama? You said we know it doesn’t happen… doesn’t acquire… In any case it’s zero. You said we also learn it from the table. Right. No, I learn it from the fact that the Talmudic text wants to make a common-denominator argument. If the true law here were one, the Talmudic text would not be able to make a common-denominator argument. Because if you analyzed the table, you would see that it doesn’t work. And it is also intuitively clear to us that it doesn’t work. Why is it intuitively clear that it doesn’t work? Because if there were a one here, it would come out that there is some parameter for money and intercourse that contains yevama, right? And canopy doesn’t have that parameter. So consequently betrothal also couldn’t be learned through that same parameter. So there is no reason to assume that it depends on the parameter common to all three of them. Therefore logic too says that if there were a one here, we couldn’t make a common-denominator argument. And if you checked the table, you’d see that you can’t make a common-denominator argument. It wouldn’t work. Okay, why again? Once again. If there were a one here, or here, it doesn’t matter, then there would effectively be a column here with a one here and a one here and a zero here. Right? That means that money and intercourse have some property that succeeds in containing yevama, and canopy doesn’t have that property. But if money and intercourse have a common property that does not exist in canopy, that is a refutation of the common denominator. Because then you can say that money and intercourse have their special power in betrothal because of that special parameter—but canopy doesn’t have that parameter, so it also won’t succeed in containing betrothal. That is exactly how a refutation of a common-denominator argument is made. How do you make a refutation of a common-denominator argument? You add another column here, and here you have one and one. Understood? That’s how a refutation of a common-denominator argument is made. Now if it were one in one of these two, then this whole thing would have moved further along—you already would have acquired a refutation of the common denominator. There in section seven this is exactly such a refutation. Right. Exactly. So you understand—this is the table we analyzed earlier, so I also don’t need to do it. But you see how this whole business is constructed.

And now I move one stage further. And that’s like constructing… complicating the table with another column and another row in the next stage. Exactly. And then the Talmudic text says: “What about the common denominator between them? For their benefit is great.” We’re in stage seven. What is “the common denominator between them”? Who are “them”? Money and intercourse. Money and intercourse are the teachers, and canopy is the lamed. Right? There is one zero one there. So now we basically need to add this thing; let’s call it benefit. This one and this one have benefit, and this one doesn’t. Right? And we already understand on our own that this is a logical refutation. Okay. If we do the analysis we’ll see it in the tables themselves as well. Okay, so let’s do that. Is there a relation here to independence—like if we take this as vectors then really… It looks like a matrix… Not in a simple way. Obviously simple independence, one zero zero one zero one, is good because you know that creates for you two parameters, alpha and beta. But more complicated independence of zero one one and zero one one is already more complicated. It’s not independence in the… I’m very much looking for a way to translate this into vectors and vector independence. Because then I can base all the algorithms on theorems of Hilbert spaces and it becomes much simpler. But for now I don’t know how to do that yet. We need to work on the matrices. If someone wants to do a master’s degree someday, he’s welcome—we have several suggestions for him.

Anyway, how do we do this thing? So let’s see. We start with filling zero and filling one. But let’s make sure there really is a refutation here—even though intuitively it’s clear to us there’s a refutation, let’s see it. I just want to see that the model works. Okay? So let’s do it. Here there’s filling one. I’ll start with filling one. Filling one—you see, A is the strongest. Right? So it is first. Who comes next? N. Okay. After that? H. H, right? H goes into A as well, but doesn’t go into N. Therefore it has to be like this. And now Y and P. So P goes into H, right? P goes into H, and Y goes both into H and into N. Right? Y goes both into H and into N. And we have no idea what the relation is between it and P. What? There is no relation between them. Fine? That’s the table of filling one. What happens in filling zero? In filling zero A and H collapse together, notice? Right, in filling zero. Notice, this line simply disappears and they collapse together. If we did this with coordinate letters for those points? What? Meaning, I see one, zero, one. Can we treat it like coordinates? You can treat it like coordinates. What? In space… It’s the same question as before. If it were like that, then they’d be vectors. It isn’t. Because for example there are more complicated problems. For example, if the fillings here can’t be just zero or one but can be continuous. For this they pay a hundred shekels, for that they pay eighty-two shekels… then you have a continuous problem. In a problem where you have a continuous problem, you can no longer use these rules in a simple way; you have to work with vectors. Therefore it’s very important for us also to find some translation into the language of vectors.

Okay, so A and H collapse together. Right? Y and P… sorry. N first of all. N is independent, agreed? Independent. And H… no, Y and P. What remains for us? Y and P. So Y goes into N and A, and P goes into H, right? Since they are the same point, then how is there a problem with H and A being the same point? P goes only into A and not into H, but A and H are the same point? Right, so A and H are together, it’s the same point. What does that mean? If it goes into this, it goes into that—they are the same values. Okay? So now let’s first see what happens here. What are we supposed to get here. Notice, this is a refutation, remember. We had zero before; it’s clear to us that here there are fewer direction changes. What? No, the question is compared to what—you have to compare it to something. Here there are two direction changes. How many direction changes are there here? One, two, three at least. At least three. What do you say? One. One? Yes. Choose any two points you want, any two points at all, and I get from one to the other with one direction change. For example this one, right? You think there are two changes. But that’s not right—you can go like this and with a direction change like this. This connection does the job. Once there is a connection like this, it reduces the direction changes. Here the direction change is one, here there are two direction changes. Okay. So this has an advantage in terms of direction changes, right? Let’s write down an advantage in direction changes. Number of points—this has the advantage. And you can already see this is a refutation. Now we don’t even need to keep checking. We don’t need to check—why? Because whatever happens later, once there is an advantage for each of the two directions, I don’t care how many advantages there are in each direction. Once there is an advantage in both directions, that’s a refutation. But we don’t know which advantage is preferable. Yes, exactly. Except for dimension. In dimension that maybe—this is one possibility—but there too, it’s preferable in terms of the inference being stronger, but if in the same diagram there were superiority in dimension for filling one and superiority in something else for the other side, that would be a refutation. But disconnectedness—why disconnectedness… and disconnectedness is something… why isn’t that the criterion that, let’s say, takes over? Meaning, if we have one parameter that isn’t connected to anything at all, then it’s hard to see how… Who said it determines the relation in the parameter that interests us? I’m interested in what causes betrothal, so suppose H isn’t connected—fine, it isn’t connected; let’s instead relate to the others. It’s maybe even simpler in a certain sense. What? Yes, we also remove a property, exactly. So we use that property. Preliminary point, because in that place it matters less. By the way, this is part of the theorems that also need to be proved here. Meaning, when we can remove part of the table and do the… do the calculation again, is it possible, can one ignore the effect of the… of the disconnected parts?

Okay, so now here, just for the exercise, how do we do the model here? Let’s say this is alpha, this is alpha beta, this is… one second, alpha beta, two alpha and beta, this is two alpha, no, two alpha won’t work here, right? Because then there would have to be such an arrow. Three alpha. Three alpha is fine. And four alpha and beta. But it has to have a connection to here, right? Why? Doesn’t alpha connect it to this? No, it needs both of them. There can’t be something here that isn’t also here. But doesn’t that connect Y and P? Ah, but that’s not good, right, that connects Y and P. Why? Huh? Why does it connect Y and P? Because there is such a relation here. But there is either more… ah, okay, greater than it. Right? So wait, then let’s switch. Right. Let’s switch, wait. Why not call it… so this is two alpha, and this is… and we can leave this, turn Y into three alpha. Wait. Right. No, I can’t do three alpha here. Three alpha beta. Three alpha beta, and keep the three alpha. That one, two alpha beta. There’s no point keeping three alpha, this one below is four alpha beta. No, four alpha beta won’t work because then there is an arrow between them. So it needs to be two alpha beta. Then between them too that won’t work, so let’s do three. I think that’s okay, no? No, but you still have the arrow between Y and P. Between Y and P. Ah, right, there is an arrow between Y and P. Wait, so something is off here. So let’s start differently. Maybe let’s start with beta in… it doesn’t matter. Wait. Let’s see if this works; we got a bit tangled here. Yes, here I did it backwards. This is two alpha, this is alpha beta, and this is two alpha beta. It has to be, and that one is three alpha, three alpha. Fine? Now it’s okay, right? Yes. Now it’s okay. And in this table, then this is beta, this is alpha, alpha beta—we already did this table. Two alpha and alpha beta, and then two alpha. Okay? No, the other way around, the other way around. What? Beta, two alpha—you reversed it there. Ah, okay, sorry. Fine? We did that earlier too. And basically we see that in terms of dimension they are the same, and there is one advantage in each direction, and therefore this is a Pareto situation—it works. Okay.