 Hello, and welcome back to Fill 320, Deductive Logic. I'm Professor Matthew Brown, and today we're talking about proof strategy and proof theoretic concepts. This will be our last lecture for the unit on proofs in an SL. Let's get into it. So I wanna start by talking about proof strategy. I'm gonna talk at the beginning of this lecture in general terms about proof strategy, and then at the end of the lecture, we'll work through some proofs together, and you'll see how these strategies are applied in practice. The first strategy I wanna talk about is working backwards from what you want, right? So when you have a proof, you know the statement you're trying to prove. You need to ask yourself, what rules could get me what I need? For example, if you're trying to prove a conditional, if the main connective in the sentence you're trying to prove is a conditional, you might need to use the conditional introduction rule at the end of the proof. So how can you get there, right? How can you get what you want using the appropriate rules? So the next strategy is to work forward from what you have, right? You ask yourself, what rules can be applied to your premises in order to derive other things that might be helpful in moving your proof forward? Another strategy is to change what you're looking at. Think about your replacement rules. They can often make your life easier, right? If a proof seems impossible, try some different substitutions to change what you're looking at, and that may give you some idea of how to move forward. Definitely do not forget indirect proof. The conditional introduction rule, the negation introduction and elimination rules, these are your friends, right? They can help you prove things that direct proof rules cannot help you with, right? And then repeat as necessary. Keep working forwards and backwards until you figure out how to connect your premises to your conclusion with a valid derivation and persist, right? I mean, this is perhaps the most important strategy. There's a lot of trial and error in coming up with valid proofs and it's not always obvious. You might go down some dead ends, but just keep at it, right? And you should be able to find a proof eventually that will work. So those are some ideas for proof strategies. In the last part of the lecture, we'll practice some of these strategies as we work together on some proofs. Next I want to talk about some proof theoretic concepts, right? These are concepts that help us think about what we're doing when we're deriving proofs, right? The first is this symbol, right? We'll use this symbol here called the single turn style or just the turn style to indicate that a proof is possible, right? We call it a single turn style to underscore the fact that it is not the double turn style symbol that we'll use for a different but related concept in the next unit in chapter five, right? When we're talking about models, right? You've seen it actually in some of the truth table problems that we did in the last unit. Okay, so what do I mean that this means a proof is possible? So suppose you have a turn style B, right? I have some sentence A, some sentence B. This means that we can prove B on the basis of A, right? Or we might have a whole set of sentences A1, A2, A3, et cetera, and on that basis we can derive B, right? So the turn style here means that B is provable assuming A or on the basis of premises A or A1, A1, A2, et cetera, right? Or we say it's derivable from A, right? B is derivable from A, right? To say it's derivable is to say we've provided a proof. Sometimes you see the turn style here with a sentence on the right, but nothing on the left, right? This means that C is a theorem of the proof system we're using. So for our purposes right now, it means that C is a theorem of SL. A theorem is like a tautology. You can prove it without any premises. That means it's always true no matter what premises you assume. Finally, if we have both, if we have a proof from A to B and a proof from B to A, if we can derive B from A and we can derive A from B, we say that A and B are provably equivalent, right? This is like logically equivalent that we saw before, right? But instead of using truth tables and instead of using a semantic style to show that they're equivalent, here we use proofs, that we can derive them from each other. That means they're equivalent. So those are some of the proof theoretic concepts that I want you to know that help us think about what we're doing when we're doing some proofs. Finally, today I want us to go through and try some proofs. So take a moment and pause the video and try to write down a proof on your own for each of these four arguments. And when you're done, come back, unpause the video, and we'll work through them together in Carnap. Okay, welcome back. I've got Carnap open here and I'm gonna try to work through these proofs. I'm gonna start with the first one here. And first thing I need to do is I need to put in my premises, right? So I've got four premises here. P is one of my premises. What else do we have? We have if P then Q, sub-premise, if Q then R is a premise. Oops, made an error there. Let's see, there we go, Q then R. And then if R then S is a premise. And I'm trying to prove that S, okay? Let's see. So there's some different options I might pursue here, but I'm just gonna start by thinking about what can I do with what I have, right? I've got a conditional P then Q and a P, so I know I can use conditional elimination to get Q. That's on lines two is the conditional and one is the antecedent, right? Well, now I can, okay, I think I see how this goes. I can get R the same way, conditional elimination. On three is the conditional and line five is the antecedent. And then I can get S conditional elimination through line four is my conditional and line six is the antecedent. And that gets me my proof. Great, is that the way you did it? Another way we could do it is to use one of our derived rules that we talked about last time, right? Which is the hypothetical syllogism. Just to remind you how the hypothetical syllogism goes, if I have two conditionals A then B, B then C, I can derive if A then C, right? So let's go back and try it that way. I have P then Q, Q then R, so I can get the P then R, right? Through hypothetical syllogism lines two and three. There we go, okay. So I got P then R and I can get P then S the same way, hypothetical syllogism on line five and four. And now I have P then S and I have P so I can get S through conditional elimination on line six and one. Another valid proof. Let's look at the second one. I already got my premises in here, so I just need to figure out how to get what I want, which is Q and S. Okay, let's see. So I have P and Q, that means I can get P through conjunction elimination on line one. And then I can use conditional elimination to get R and S through conditional elimination on line two and three, right? So I'm just, right now I'm just again, transforming what I have. I want Q and S. Okay, well I can get Q through conditional elimination, right? I'm gonna need to also get S on its own because to get a conditional introduction, I'm gonna need S and Q, right? Okay, I can get S on its own too through conjunction elimination on line four. Did I say conditional elimination up here? I meant conjunction elimination on line five. Okay, so I've got Q, I've got S, that means I can get Q and S through conditional introduction on lines five and six. And that gets me a valid proof. That's done. Can you even tidy it up if I like? I'm doing the rules out of those at the right button. I'm done. Let's look at the next one. It looks a little tough because I only have one line and it's a conditional. So I know there's nothing I can do with that in terms of a direct proof. My conclusion that I want is also a conditional. So that suggests I'm gonna need to do an indirect proof. That suggests I'm gonna need to use conditional introduction. The way that works, remember, is I start with the antecedent of the conditional as an assumption, use AS, to stand for assumption in Carnap. And I work from there. If I have Q, I can introduce P or Q. That's how disjunction introduction works. If I have P or Q, that's the antecedent of our conditional in line one. So I can get R that way through conditional elimination on lines one and three. And R is what I'm looking for in the consequent of my conditional Q, if Q, then R. So I can close out my subproof. I can discharge it. And I can use the conditional introduction rule to get if Q, then R using conditional introduction on lines to through four. Remember, with conditional introduction and all indirect proofs, you give the full range of line numbers for the subproof as your lines here. So there's my valid proof. Did you get it or did you get it a different way? You can let us know in the comments or on Discord if you tried it a different way and it worked or you think it did, you can check it. All right, here is another case where we've only got one premise. We're not gonna be able to simply use a direct proof here. We're gonna have to use some kind of indirect proof. And again, I have a conditional. So I think that is going to weigh in favor of conditional introduction. Let's try it and see if it works. P and Q is the antecedent of the conditional in my conclusion, right? So that's a start. That's the assumption of my subproof. On that basis, I can get P through conjunction elimination on line two. And then through conditional elimination, that was conjunction elimination, through conditional elimination, I can get not Q and R. That's conditional elimination on lines one and three. Yes, good. Now, what can I do with not Q and R? Let's see. I think what I wanna do here is transform it into something else. I'm gonna go look at my reference sheet from the back of the book and my replacement rules. And I'm gonna look here at the DeMorgan's rules. And I see if I've got not A and B, I can exchange that for not A or not B. So that's what I'm gonna do. Not Q or not R through DeMorgan's on line four. Yes, successful, okay. If I've got not Q or R, how can I get not R? That is what I want in the antecedent. I'm close. What I need is, let's go back and look at our rules again. If I have, by the way, these are at the back of the book. So if you need to quick reference, you can always look at these rules. I want to do disjunction elimination to get the not R. I need the negation of the other side. So I need the negation of not Q, which is Q, which I can get through conjunction elimination on line two. Then I can get not R through disjunction elimination on lines five and six. That says no, ah, I know why. It's because I need to transform this to not not Q using the double negation rule. And then I can use my disjunction elimination on lines five and six, on five and seven. Yes, the rule says, if we look at it here again, if I've got A or B and I have not A, I need, I can get B, right? Well, we look here, I have in the disjunction not Q or not R. So what I need is the negation of this, which is not not Q. That's why that's how that works. All right, that is all I need to discharge my subproof. So I have P and Q, then not R, and that is through conditional introduction lines two through eight, oops, two through eight. And that's it, I'm done. So that's all we have for today. I hope that you had good luck with those proofs. You understood what we were doing. If not, if you have any questions, feel free to drop them in Discord, send me an email or get in touch during office hours, or leave a comment on the video. That's all we have for unit four on proofs. Please go ahead and work on the practice problems and give the exam a try when you feel ready. And I'll see you in the next unit. Bye for now.