 Hi, welcome back to Fill 320, Deductive Logic. I'm Professor Matthew Brown, and today we're going to be talking about proofs in SL again. And specifically, we're going to be focusing on rules of indirect proof. So let me remind you of where we were at the end of last time, which is this distinction between direct and indirect proof. Remember that direct proofs of the kind we looked at in our last lecture derive their conclusion from the assumptions using introduction and elimination rules and nothing else. Every line in the proof, every intermediate step in the proof derives from the application of the rules to the premises and the prior intermediate steps. Indibreck proofs, by contrast, in one way or another introduce assumptions that are not there in the premises. Introduce additional assumptions which are not directly justified by a rule. Let me show you what I mean in this contrast. So with the direct proof, and here's a simple one, you have your premises on line 1 and 2. We know we want to prove A and B here. And lines 3 and 4 are each justified by the application of a rule. Line 3 is justified by the Biconditional Elimination Rule, applied to line 1 and 2. And line 4 is justified by the Conjunction Introduction Rule, which is acting on lines 2 and 3. And that's all we need to prove it. Quaterrat demonstratum, QED, as we sometimes say when we've proved something. An indirect proof, by contrast, involves some kind of sub-proof. That's what the second vertical and horizontal line in this proof mean, where additional assumptions have been introduced besides those that are in the premises. When the second vertical line ends there after line 3, the sub-proof is closed. Closing a sub-proof is called discharging the assumptions of that sub-proof. And you can discharge a sub-proof using the indirect proof rules. Crucially, you cannot use any of the lines in the sub-proof once it is closed anywhere outside the sub-proof itself, except in that line that closes it. I can no longer refer to A or B in lines 4 and afterwards through, say, the reiteration rule or other applications of other rules. This is this specific kind of indirect proof we might call a conditional proof. And it's the primary way we introduce conditionals and also by-conditionals. Here we assume A, and we derive B, and that gives us this conditional at the end. So I'm going to now go through the additional rules of our natural deduction system that are indirect proof rules. So there are several. The first we've just looked at an instance of, it's conditional introduction. How does conditional introduction work? You have a sub-proof with A as an assumption. A in script A, remember, is a meta variable for any arbitrary sentence. We are able to derive B. There may be intermediate steps before we get to B. And if we do this, then we can introduce the conditional if A then B. And we number the lines that the sub-proof starts with and the sub-proof ends with, and those are the ones we refer to. You notice here it's M-N. You refer to the whole range of lines that constitute the sub-proof, not just two lines. It's not M, N. It's M-N. That's an important distinction. By-conditional introduction is also a type of indirect proof. And it works very similar to conditional introduction, but you have to do it in both directions. So again here, you've got the sub-proof from A to B, and the sub-proof from B to A. And we number those sub-proofs M through N and P through Q. And again, there may be many intermediate steps between M and N and P and Q in these sub-proofs. And once we have got those two sub-proofs, we can discharge them and introduce the by-conditional. And that's how we write it. Interestingly, both of our negation rules are indirect proof rules. So if we look at negation introduction, again here with negation introduction, we assume some A, some arbitrary sentence A. And what we are trying to derive for negation introduction is a contradiction. That is, two lines which have B and not B on subsequent lines. This allows us to introduce the negation of A, some arbitrary sentence. And the reason is, we know that both B and not B cannot both be right. So if on the basis of this sub-proof, if assuming A generates a contradiction, generates two logically incompatible sentences B and not B, then we know that A must have been wrong. It must, in fact, be the opposite. Not A must be the case. This sort of indirect proof is often also known as proof by contradiction. Negation elimination is also a type of proof by contradiction, where instead of starting with A, we start with not A. We generate a contradiction, and this allows us to derive A. So similar deal, but here we are actually starting with the negation and getting the non-negated version of that same arbitrary sentence. Okay? So those are indirect proof rules. And in some cases, they're going to be the kind of rule you need to adopt in order to get something that you can't get through a direct proof. I wanna turn to the first of the problems in your homework, in your practice exercise, 6A, right? What it does is it gives you a proof, already complete, right? And it asks you to fill in the rules that justify each of the steps between the premises and the conclusion. This is what your first practice exercise problems are gonna look like. Now above the line, the premises need no rule. There's no rule required. Often we write at the end of all the premises that we want K, K is what we're trying to prove. In this case, we want, and then the sentence that we're trying to prove. That's actually optional in this system. When we're doing proofs in Carnap, however, after each premise, we are gonna use a rule or a rule called PR just to indicate that it's a premise. That's just a sort of feature of that system, right? When you're writing proofs by hand, you wouldn't need to do that. You could just write it this way or even leave the want K out of it. So let's look at our proof. So we've got line four is W. How can we get W out of our premises? We see, number two, there's that conjunction A and W so we can actually apply our conjunction elimination rule to get W. That's good, that's easy. Now that we've got W and we have this conditional in line one, if W, then not B. So we can use that to get not B using the conditional elimination rule, right? And then we have our line six, J and K. This is something we can do through the disjunction elimination. Line three is our disjunction and line five is the negation of the first disjunct, right? So we can do that to get J and K. And then another conjunction elimination will get us K, right? Applied to line six, okay? The other homework problems will work much the same, right? So you'll have an opportunity to practice those after this lecture is concluded. This is a case of completely direct proof but there will also be cases of indirect proof in the practice exercises. Let's look at doing some additional example proofs. What I want you to do is I want you to pause the video here and look at these different arguments. Pause the video and try on your own to write down the complete proofs for each one. And then when you're done, unpause the video and we'll go through it together via Carnap. I'll show you how I would do the proofs. Okay, are you finished? Let's try it together. So here I've already loaded all of these problems into Carnap and let's see how I would do these. So we've got the premises loaded in here for our first problem. And what we wanna get to is not A, okay? Well, we've got not C here. So we can go ahead and use our conditional elimination rule to get if a then C, right? That's conditional elimination on line one and two, right? If we've got if a then C, can we get not A out of that? Well, that's not totally clear how we can do that directly. So let's try an indirect proof. To start a sub proof, I do a space here that creates a new sub proof line and I'm gonna assume A, right? I'm gonna try the negation elimination route. I do colon AS to show that's an assumption, right? Now I've got here on line three a conditional and on line four I've got the antecedent of the conditional. So this will allow me to derive C through the conditional elimination rule. Line three is my conditional, line four is my antecedent. That's great, but not super surprising. How can I use negation elimination? Well, I need not C, right? That gets me the contradiction that I want and I can do that easily through the reiteration rule on line two, remember reiteration, right? We can bring things from above the sub proof, outside of the sub proof into the sub proof. We can't do it the other way once the sub proof has been discharged though. Okay, so that gets me my contradiction. That means I can conclude not A, right? And the reason I can conclude it is negation, introduction and my sub proof is lines four through six, right? And that's it, I've done it, right? That checks out according to Carnot. Let's look at line two. More premises in this proof. We've got four premises here. What I'm trying to get to remember is if G, then E. That's what I'm trying to get to. I've got G and T and T and E. Not clear how I'm gonna totally get that, but let's start working on it. I've got conditionals and antecedents in my premises. So I'm gonna start there. I can get G implies T from my conditional elimination one and three, right? That checks out. I can get similarly if T, then E through conditional elimination lines two and four because four is the antecedent of two. All right, now I've got G, then T, T, then E. Where am I gonna go next? Well, I'm trying to prove a conditional. So let's try a conditional proof. Let's try conditional introduction. So I wanna start with G, right? That's the assumption that's gonna be the antecedent of the conditional I'm trying to prove, right? Now, if I have G, I can get T, right? That's just a simple conditional elimination lines five and seven. And if I have T, hey, I can get E. That's another conditional elimination on the conditional from line six and the antecedent from line eight, right? That gets me where I want. I've got G on the first line. I've got E on the last line. So I can conclude if G, then E through conditional introduction, the subproof numbers are line seven through nine, not comma, but a dash, right? And that gets me there. That is lights up green. That means that's a good proof, valid proof of our conditional. Let's look at our third example together. Here I'm trying to get from these two premises to another conditional. I'm gonna guess here again, we need a conditional proof. So I start by assuming the antecedent N. Now I can get O through conditional elimination on one and three, right? And I can get P through conditional elimination on two and three, right? So far, so good. I've got O and P. So I can do a conjunction introduction, right? That's conjunction introduction. Lines four and five are the two conjuncts. That checks out. Oh, no, that doesn't check out what I do wrong. Conjunction introduction, let's see. Can't match these premises with this conclusion. Oh, as I see what it is, I used E not P. That was a little typo. Okay, now it checks out. Now I've got the plus, okay. Notice what I did here when I had it wrong, I got the X and I can actually hover over the X and it gives me some information that can maybe help me figure out what I did wrong. That's one of the nice things about Carnab. That'll be true in your practice exercises. When we come to the exam, of course you won't have access to that automatic hints. So I think that gets me the whole way because now I can introduce if in, I can discharge the subproof and introduce if in, then O and P, right? That is our conditional introduction rule. And the subproof starts on line three and goes through line six. It checks out, done. Okay, now this is a kind of funny one. We've got only one premise and we're using it to conclude not S, right? Now let's see. It's not all that obvious how I'm gonna get there. I don't see any direct proof rules I can use because I just have the one premise, the conditional. So let me try a proof by contradiction, right? Let's try a negation introduction since we've got a not S. So we start by assuming S. We wanna prove the opposite is true, right? If I have S, I can get S or T because I can use disjunction introduction on line two. Remember if I've got S, I can introduce anything as the second disjunct and that's legitimate on the rule of disjunction introduction, okay? Well, S or T is the antecedent of one. That's why I picked T, right? And so I can use now the conditional elimination, right? I can get not S by using conditional elimination. Number one is our conditional and number three is our antecedent. Again, that checks out. I've got not S, I also need S, right? Because that's gonna create our contradiction right there. And I'm just gonna use reiteration on line two to bring that down. And that's all I need, I've got my contradiction there. So I can conclude not S on the basis of negation introduction this time and our line numbers are two through five. Did that work? Yes, it did. So that all checks out. So those are four examples of indirect proof. In the homeworks though, some of them will be direct proofs too. So you'll get a chance to practice both. That's all I wanted to share with you today for our lecture on indirect proof rules. Next time we're gonna learn the last set of rules for SL that we're gonna talk about, which are derived rules and our rules of replacement. Now, strictly speaking, those are rules you don't need to do proofs in SL. You certainly won't need them for the first set of homework problems, but give it a go and I'll share those rules with you next time. Bye.