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BSIC Chapter 3, Section 3.6

BSIC Chapter 3, Section 3.6


Hello and welcome to video number seven, Chapter 3. This is the last video about the content of the chapter. The last video, the eighth of eight, will just be me working through the practice quiz for chapter three. And in fact in this video I’m not going to be introducing any new material. What I’m going to be doing is talking about strategies for looking at proofs, and in particular what what we can call “sub-proof frames.” And this isn’t anything new, but this is just a way to think about what you’re doing when you’re using sub proofs. So let’s look at this proof. With this proof the conclusion is a conditional. I might think hey I’m going to use conditional proof to derive the conclusion of the argument. Once I’ve made that decision notice that kind of sets up the structure of my proof, right. And schematically it means my proof is going to look like this. I want this conditional. My plan is to get if B then D, from conditional proof. That tells me what my ACP is going to be. It’s gonna be the antecedent. It also tells me where when I’m going to close the sub proof once I have derived D and then once I do that I’m gonna get this conditional out so I I don’t know all the details but I do know that on line five I’m going to assume B assumption for conditional proof at some point unless things go haywire I’m going to derive D I don’t know what line number that is just call it X but once I get D I’m going to close this sub proof off and my next line y I’ll get if B then D and the justification will be 5 through X whatever that line is conditional proof so once I’ve decided to use conditional proof that sets up the whole structure of my proof for me and then it’s just a matter of when I go through doing line five like this and then just figuring out what this stuff in the middle is how I’m actually going to d that’s the only part I don’t know and then once I get D this I will just become that so let’s go ahead and do the proof assume B and remember my goal is to get D how can I get it well there’s a actually first off let’s just do it and then I’m gonna I’ll come back and I want to point something out about this proof so there’s a lot of ways to get I can go a B or D so I can just go di on B to get B or D then I can get D look I’ve got the negation of this disjunct so I can derive the other disjunct by D s on 2 & 6 so there’s d s– and i can close this subproof off and the next line is if b then d okay so if you look back this is exactly like my frame was and said now I know what X X is X is 7 and Y is 8 and I know how this middle part played out okay but so basic I was just kind of filling in the blanks of that frame that I’d constructed okay now there’s something else that’s kind of unrelated to subproof frames or conditional proof that i want to sort of point out it’s helpful notice that at this point in the proof I had a contradiction I had a statement and it’s negation okay why did that happen well sometimes you might have it just because your premises are inconsistent and you’ll be able to derive a contradiction if you get that in this case it’s because my one of the things that was contradictory was an assumption I just made but doesn’t matter for purposes of the point I’m trying to make now at some point you might in doing a proof see that once you see that you have that contradiction like this a statement at its own negation you can derive anything in two steps in just the way I did it you use di on one of them and then you just disjoin anything you want whatever you want to derive you can just dis join with di I wanted D so I just joined that and then using disjunctive syllogism because I have the negation of this I can take this thing I wanted and get it by itself once you’ve got a statement its own negation you could use this little recipe to derive anything I could have derived Z here write the B or Z and then get Z by D S on 6 + 2 whatever I added here whatever I disjoined I could then immediately get on a line by itself by using D S with the other the negated version of that so this is just a recipe you can use whenever you see that happen ok let’s do another example another proof so looking at this look at the conclusion that we’re trying to derive here the conclusion is a conjunction and each conjunct is a conditional well what this suggests is that one way you could do this proof is to derive each conditional each conjunct by itself and then can join them using the rule of conjunction okay which means that strategically the proof might look like this I might get this conjunct the first one somehow who knows I’m leaving that blank then somehow get the other conjunct I don’t have how that’s gonna happen I don’t know yet but once I do that if I get this and this then I can conjoin them with the rule of conjunction okay I don’t know what these lines are gonna be that’s why I just put an X there well now I can think how can I get these well notice this is a conditional one thing I might do is use conditional proof and if I do that I’ll get this conditional by assuming R and deriving a now that’s how I’ll get that and then I can also do the same thing with this conditional I can assume s and derive B notice here I’ve got the whole frame for my proof I don’t know everything I’ve got blanks that need to be filled in I don’t know what’s gonna go on in here I don’t know what’s gonna go on in there but the basic structure of my proof is now set so I’m just gonna go ahead and do it now remember first get if R then a with conditional proof that I’m gonna get FS then B with conditional proof then I’m gonna conjoin them okay first assume R remember my goal is now to derive a and you can see that’s gonna be pretty easy I’ve got R which is this antecedent I can get DNA and then simplify a off so that’s what I’m gonna do DNA by 1/4 modus ponens a from line 5 and simplification now I can end my sub proof because this is the consequent I want my conditional to have so I’m going to close that subproof off if R then a lines 4 through 6 conditional proof well now I’m not done obviously because this isn’t this is just part of what I need next I need this conditional if S then B remember the frame I had a strategy for this I’m going to assume s try to derive be super easy look I’ve got s I can just do disjunctive syllogism on 3 & 8 because this is the negation of that disjunct so I can drive the other disjunct by D s so 3 8 DS I’ve got B that was fast oh I took one step I can close that and I get if S then B now I have both of the con junks I have this is the first conjunct if R than a and then the second conjunct if S then B is right here on line 10 so I just can join those and I’m done with the proof so you can see how you can especially if you’re using sub proofs but even if you’re not using sub proofs you can sort of plan out the structure of your proof beforehand and if you’re using a sub proof that actually sets up a lot of the structure for you okay well that’s it next and last video for chapter 3 I’ll be working through the practice quiz

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