Dr Greg Bahnsen - Critical Thinking: Deductive Proof Part 1 (Part 16)

Let's pray. Lord, we come to you today and ask that you give us clear minds and strong bodies that we might be able to do the work that is set before us. And above all, we pray that you would give us hearts, hit us out and glorifying you. We do pray that you would be with us, forgive our sins and strengthen us by your spirit, that everything that we do or say would be for the sake of the kingdom of our Lord and Savior, Jesus Christ, and learning to love him with all of our minds. We pray these things in his most blessed name.

Amen.

All right, as we come to chapter nine in Kofi, the Method of Deduction, Basically, we're going to be looking at how to prove the validity of a symbolic argument or an argument that's put in symbolic notation and to prove it without making reference to the truth tables. We've already seen that we can test the validity of any argument by writing out a truth table. But as Kofi says at the beginning of this chapter, truth tables become rather unwieldy as the number of component statements increases in an argument. He says a more efficient method of establishing the validity of an extended argument is to deduce its conclusion from its premises by a sequence of elementary arguments, each of which is known to be valid. And so he offers us rules of inference eventually, on pages 297 and 298, and by means of these initially nine rules, we'll expand them in a little while in terms of these nine rules. We can take the first premise, or some premise that is a given in an argument and use it to deduce another step which is legitimate by the rule. The rule allows you to take this step, and then whether it's 2, 3, 4, 10 steps or more. The point is that every one of the steps in the argument is made legitimate by these rules, and you can get to the conclusion using those steps. Then you have deductively, you have proven the validity of this deductive argument. You've shown it to be deductively valid. This is a little bit analogous to what we did with respect to categorical syllogisms. In terms of categorical syllogisms, we found out that we could use logical analogy. We could come up with a counterexample in the same form. But then that becomes kind of difficult sometimes. And if we're not clever enough, we're not sure whether we really have got to accept the validity or invalidity of the argument in front of us. And so we came up with rules. Eventually, rules for Categorical syllogisms, there were six, and those rules tested the syllogisms. Now, in the same way, we have found a way in terms of symbolic logic or propositional logic, using compound statements, we have a way to test any argument that's given to us by the truth tables. But then who wants to get up to 32, 64 and 128 lines on a truth table? Once you get so many components, it's just unwieldy. So what we're going to do is we're going to formulate rules that can be used, and these rules will allow us to go step by step by step from the premises, spanning the gulf, if you will, from the premises to. To the conclusion. We're going to build a bridge from the premises that are given to the conclusion. And that bridge has to be built according to certain rules. And those are the rules of inference given to us on page 297. In order to make use of the rules of inference, we have to understand the concept of an argument form, an elementary valid argument form. So in order to explain that, I'm going to try to have you visualize it on the board here. Let's say that I run into an argument in ordinary language and I have schematized that argument. I've assigned capital letters for the major propositions in the argument. And I found the relationship between them. There's the or relationship, the and relationship, the if, then relationship. I've noted the negations in there and I've got it all schematized so that. Now these capital letters stand for. They're short for an English expression or a proposition. If I have schematized my argument, if B, then L, but not L. The question is, what can I deduce from this by means of the rules that are given to me by copy? Now, in your book, please look at rule of inference number two, called modus tollens. Now, you already understand modus tollens in a kind of intuitive or ordinary way, because we studied that in the ordinary language aspect of deductive logic. Modus tollens. As if I say if this, then that, but this, that is not true, then I know that this has got to be untrue as well. Anyway, modus tollens is written down as a rule of inference. And now notice what I'm doing here. I'm going to write down this form. When we give an argument forms, we use small letters just to show that anything can go in here. And I'm going to put down the first line of the rule is if P then Q. The second line of the rule is not Q. And now if we just look at these two, hopefully it's obvious that they are exactly parallel to each other. I have one form here, if then, and another form here if then. And I have another form here, not L, and a form very similar, not Q. But is everyone with me so far? You see how they look the same. Now here's the difference though. On this side over here, what we call a rule of inference, another line is given. Therefore not P. We don't have not P on this side. All we have are the two premises. If B, then L, not L. Our argument has only provided us two premises. Now what can we do with those two premises? The rule of inference says you can draw down the negation of B. And so if you impose this argument on the left hand side of the board, if P, then L, not L, on top of if P, then Q, not Q, you notice that they fit together. And this bottom line can now be filled in from over here. You can draw this down so that it's exactly parallel to this form of argument here. If I say any step like this will count in my game of logic. Anytime you're given this much, you can use this as a legitimate premise as well. Then what we have here is a way of constructing a bridge to go further than what you have. You only have these two premises. Now you can get a third premise which is legitimate just by following this pattern here. So in the place of this P and Q in the rule is a place marker. It's a place marker for any proposition or any compound proposition, any simple or compound proposition. And I'm going to come to this point that a compound proposition can be filled in there in just a moment. But I want to make sure this first step is understood. I'm going to restate what we've just talked about so that I can go on to take another step to show how this develops with the use of compound propositions in the argument itself. But let's forget about the argument in ordinary language that gets schematized. Let's just look at the rule of inference. The one that we took as our example was if P, then Q. And I'm going to write the premises across the top for a reason here. The second premise is not Q and the conclusion, therefore not P. Now, if we're going to follow this pattern and we are testing an argument that said if B then L, not L, you'll notice how we can go through and as it were in the Place of each one of these variables. The P, the Q, the not q, and the not p are all variables, or if we will place markers, they're just holding positions for a proposition. And so in the place of P, I put B In the place of Q, I put L in the place of not Q, I put not l, and in the place of therefore P, I put therefore not P, therefore not. As we put these across the board this way we see how each one of these things fills a place in the argument. And the reason why I could fill in this one, which was originally blank, is because this gives me a rule to follow. The top line that has the variables gives me a rule to follow. I can fill in these two spots and. And the rule says, therefore it's legitimate to fill in this one according to the pattern seen up here. Now, it will help, I think, if we realize that since this is a place marker, the thing that can fill the place marker can be complex. It can be simple or complex. There's no reason why it has to be just one proposition as long as you treat the compound as a unit in itself. So now, underneath P instead of B, I'm going to put B and G and put it in parentheses. Now, by bracketing it or putting it in parentheses, what I'm saying is, for purposes of logic, you can treat that as though it's one unit that you know. And later you're going to finally deal with this fact. You know, there's two propositions in there, but for working the operation upon it, you can treat it as though it's a unit all by itself. Likewise, in the top line, I have the variable Q and the second line underneath it. For that place one marker, I used L. But as I just said, we could use something else, as though we could fill in something else to that place for that place marker that was compound. So here, let me put down. L or r. Now, in the third line here, where I have B and G filling in the place marker of P, I'm saying if B and G, then whatever fills in the place marker for Q, L, or R, the next step is the denial of what was the second place marker. What is the second place marker? Okay, so if I put down not L or R, remember, this is in parentheses to remind you this is a unit. It goes into one of the slots all by itself, not L or R. Now I can conclude. Let's see if we can follow the pattern J up here on the variable. This is the rule line. It shows me how I can proceed. I'VE got these two steps. I don't allowed this step on this one where I'm using simple statements. If B, then L, not L. I was able to fill in what not B because that's the same pattern as. Okay, and then so not B. So this is all a matter of just seeing this fits the slot. It's the very same pattern. So now let's see if you can tell me what am I going to negate. I've already negated this second compound unit treated as a Q all by itself. So now I'm going to negate what? That's right, the first one, whatever was, whatever took the place of P is now going to be negated. So down here I put not parentheses, B and G. So I think the crucial thing is to note that the variable in the rule can stand for a simple statement or a compound statement which is bracketed and treated as though it were just one unit by itself. And then what will happen, though you don't have to understand this yet, is when we get to the next step, we're going to see there's a way to replace things within the parentheses with their logical equivalence. But right now we're just treating the variable as a simple or a compound that is bracketed. And so anything you do to the P and the Q and you get down to the conclusion, you can do to whatever stood in the place of the P or so this is a rule of inference by which we can develop step by step, valid, new, valid premises, which, if we're clever enough, will get us to the conclusion we're trying to establish. Now, in a deductive argument. Let me finish this and then we'll do a few of them. In a deductive argument, you have a certain number of premises. The number of premises can't be determined ahead of time. That all depends on what you're studying and what comes out. In most logic books, it doesn't get to be more than four or five or six, and early on it's one or two or even one. But the point is, whatever the number of premises, Let me call them P1, P2, P, N, or whatever the number turns out to be, whatever the number of premises at the end of this, there's going to be a C, a conclusion. And the whole point here is to show that between. Between the end of the premises and the conclusion. This gap that I have symbolized on the board here, we want to be able to use the rules to build a step, another step, another step, or however many are necessary. We want to get steps which are legitimate by the rules that will eventually give us the conclusion. So if we can go by the rules, rules from these premises by steps to the conclusion, then we know that the conclusion is validly drawn from the premises. And this is it's filling in these steps right here that is the challenge in a deductive proof. And how are we going to fill in these steps? By using those rules of inference. That is, we're going to follow, we're going to get the pattern, and then we're going to try to imitate that pattern using what we have in our premises. So if I know this pattern is legitimate and I can take one of my premises and put it in here and then get something else that becomes then the step. And I can use that step as much as I can use these premises. I can go back to the premises or use this premise in this new one. So any number of combinations, I would try using the rules of inference to bring the conclusion down. On page 298 we have some examples, some exercises, and in Roman numeral one, let's begin with number one. We have as a premise A or B bracketed, then C. And we're supposed to show that from that we can get to the conclusion A or B. Then bracket parentheses A and B and C end of bracket. Jay, would you take that one?

Okay.

Number one then is by absorption.

Let's all look Back to page 297 in COPI. Rule 6 at the bottom, called the rule of absorption, says that if you have a P then Q, you can legitimately take this step P then pnq. So whatever stands for P can be repeated in the parentheses at the end of that. If P then Q means if P then P and Q. I think you can see that's intuitively obvious. If you're asserting P, then of course you can assert P again along with whatever implies P. Does that make sense to everyone? So in this particular case, the A and B in parentheses stand for the P. If A and B then P. So that means if A and P then A and P and C. Oh, I said, Did I say it right? P. I don't know where I got the P should have been. If A and B then C. Therefore, if A and B then A and B and C. By the rule of absorption, we just follow the very same pattern that we find in the rule making the compound A and B stand in the place of the variable P. And then every time P appears in the inference rule, we write A and B in its place. Okay, number two, I'll just assume for the sake of time and those listening by tape that you have the book in front of you. I'm not going to try to read them all. We have one line here. And how can we get from that line line to the one below it?

By simplification.

That's right. Simplification says if you have two things conjoined, P and Q, then you can infer one or the other. Of course, both of them are already there, but the point is you can drop one off, since they both have to be true. So simplification is just a matter of saying once you know P and Q, you know P as well as you know Q. So you can simplify either way. Now, in the case of number three, we have the proposition if H then I, but the conclusion says if H then I or if H then not I. Now, how could we possibly go from if H then I to that more complex treat if H then I as a unit, all by itself, that's P. Remember, it's bracketed. And so it can be treated. It can be put in the place of one of those place markers, one of those variables all by itself. And then if we go up, actually, it's right on this page, 298 to rule number nine, the rule of addition. And if you look at the rule of addition, it says to any true proposition, you can add any other proposition you want conjoined with the disjunction, or that is, if P is true, as long as you know that P is true, then you can add anything you want to it. The important point here is in Rule 9, the premise line doesn't contain Q at all. Where does Q come from? It comes from out of the air. It's totally arbitrary. All you know is P. But if you know P is true by the rules of logic, you can add anything to it using the disjunction. Or that's because as long as P is true and you say or, or, or, or anything else after can be false, and it will still be as a compound true. And therefore the rule of addition allows you by to put through disjunction any other proposition along with P. One more time. Since Q is not found in the premise line, you notice Q is supplied by U, meaning U I, and that means this is the easiest of all. Whenever you have an argument form, last statement form, be it simple or complex, you can add anything to it using the disjunction. So in number three, that's what we have. If H, then I is A statement form that you treat as one variable. And the rule says you can add anything you want to your variable using the disjunction. And in the conclusion, that's what Copie has done. He's added to the if H, now the or. And then he could put anything. He could have put if R, then not S, or if A, then B. He could have put anything in there. But since you can put anything in there, then the rule of addition justifies the step that happens to be there. It might be helpful before we go any further, if we just look at these rules of inference to see if they aren't somewhat intuitive, intuitively obvious to us. If we understand why the rules work, then it'd be easier to choose one to use. Improving our exercises. Rule number one is modus ponens, and we've already studied that. That should be real obvious to us. If P, then Q and P is asserted, then the next step can be added. Q must be true as well. So whenever we see the antecedent affirmed after a conditional that has the antecedent, then the consequent, we know that the consequent can be affirmed as well. If P, then Q. P, therefore Q. Modus pollens, remember, is the method of denial. If you say if P then Q, and you deny the consequent, not Q, then you can deny the antecedent as well. That should be fairly obvious. From the exercises we did before using modus tollens, we've also used hypothetical syllogism number three. If P, then Q. If Q then R. Now if you have those two as premises, you can draw the conclusion. If P then R. This is just like transition in geometry. You know that P leads to Q. You know that Q leads to a R. Therefore P must lead to R as well. That's called hypothetical syllogism. Disjunctive syllogism we found in our previous assignment. Someone says P or Q and then denies one of the disjuncts, in this case P, P or Q, then not P. Therefore, since one of them has to be true and P isn't, Q must be asserted. That's what's called disjunctive syllogism. Constructive dilemma is one that you probably don't think about as quickly when you're doing the assignments, although you have to to get one of them right. I know from last night. But if you have two conditional statements joined with the conjunction the dot, and then the next line asserts that the first are the antecedent of the first conditional or the antecedent of the second conditional. So we have if P then Q and if R then S for the first line. And the second line then takes the P and the R and puts them together with the disjunct P or R. Therefore, you can draw the conclusion Q or S. And hopefully you can see that, because what you're doing here is just following modus ponens twice and then taking the conclusion of the two arguments using modus ponens and joining them together with the disjunction. Just like the first premise, the two conditional statements at a disjunction that's called constructive dilemma. Rule 6 is absorption, and it says if P then Q. And from that you can always conclude if P then P and Q. And that's just to say if you assert P, obviously you can assert it again and you can add to it anything that P implies. So if Q is implied by P, then P implies itself and Q together. That's called the rule of absorption. Rule 7, on page 298, the rule of simplification simply says if we conjoin two statements, P and Q, then we can deduce from that either one of them separately P. And then you could also put Q there. Conjunction is the other way around. Rule 8 says if you have one premise that asserts P and another premise asserts Q, then you can have as a third premise a concluding premise, P and Q. It's just conjoining those two which are given in your premises, so that now in the third premise they're just reformulated on the same line. And then finally addition has been explained already. If you have any premise that is taken as true, you can add using the word or any other premise, be it true or false, and you can choose whatever you want. You can add the Q and make Q anything you want it to be, provided you know, P to be true. Okay, back to our exercises. Let's look at number four now and again. I won't bother to read it because that just gets too complicated. We're all looking at the same thing. How do we get that conclusion from the premise that stands above it? Now let me ask you, Pat. When you look at that premise standing above the conclusion, do you. Do you see anything that looks like the conclusion as a unit? Yeah.

Not J. Not J. Ok, Right.

In fact, if you treat that as a unit standing by itself, that is the same as the conclusion itself. Right. So you see that the conclusion has. In the premise, in the premise line, what we call the conclusion has been asserted and has been conjoined with something else as well that is asserted to be true. Now, whenever you have a P and a Q together, we know by the rules of inference that we can take one of them. And what rule is that, Pat?

Simplification.

That's right. Very good. And then number five, J. Would you do that for us?

Constructive dilemma.

Hopefully you can see by looking at the order of mention of these letters that it's in the very same form as what is Constructive dilemma on page 297, rule number five, except that you have to treat bracketed statements as a unit. In this case, the parentheses onp stands for one of the variables in the rule. Can you see that visually, O and P stand together as a variable. And once we do that, O and R likewise stand together as a variable. Once we catch on to that, then obviously the form is identical to the rule, and the rule allows you to draw the conclusion that's in the form given there with the disjunct at the end. Okay, number two, Copi says in Roman numeral two. Each of the following is a formal proof of validity for the indicated argument. State the justification for each line. That is not a premise. And so, Pete, let's have you take number one. Would you state the justification for line three?

Line three is a simplification of line one.

Correct.

Line four is an addition to line two. No, three.

Excuse me, three.

And then we get line five from line two and line four. That would be modus ponens.

Correct.

And then from line three and line five, we have conjunction, which leaves us with the conjunction of six.

That gives us the conclusion, which is. That's correct. Does anyone have any questions about how that one was done? Okay, number two.

Line four is simplification from line one.

Correct.

Line five is constructive dilemma from lines two and four.

Wait a minute. You say line five is constructive dilemma. Why don't. Why isn't line five just the simplification of line one? Well, it could be, I guess.

Well, no, the way I understood this with the simplification is you have to take the. Have the statement be on the same on the right hand side. Well, because.

No, if you can simplify either one of them. Well, it works.

It works as constructive dilemma as well. From line two and four.

Well, that makes it a little bit more difficult. What's your justification for line 6?

Line 6 was from line 6, 5.

And 3 by.

Disjunctive syllogism.

Right. You might want to notice something here if you do it the way that I had proposed. Line five being a simplification of line one. Line four is actually unnecessary to the proof.

Line two would be unnecessary. Two also.

Well, two's given us a premise, and so you don't talk about that. Given these premises, what steps would be necessary to get to the conclusion? Okay, Patla said you do number three. How do we justify line five?

Line five is a hypothetical syllogism from lines one, lines one in line two.

Correct. Then line six. How do you justify that conjunction of line.

One and line three?

Okay, and then line seven. Wait a minute. You said line six is a conjunction of line lines one and line three. Well, line one says if I then. Wait a minute. Let me just get this clear here. Line one says if I then J, and if I then J doesn't appear anywhere on line six, and so that can't be correct.

I then kick I, then K, L, then M. And so we make a conjunction of those two statements. Number six.

Yeah, three and five conjoined. Okay. And then how do you get number seven? Addition allows you to add anything you want to some individual proposition or combination that you know to be true. But you don't have anything like that in the previous premises to add to K, so it can't be addition. You notice that you don't have K or M anywhere above line seven to add to. You'd have to have K or M standing alone. Then you could, by the rule of addition, add anything by disjunction to it, but you don't have that.

The constructive dilemma from line 6. Line 5.

Well, 6 and 4, actually, Yeah. Six gives us our two conditionals joined together with an. And six gives. Excuse me, Line four gives us the disjunction irl, which is the disjunction of the two antecedents. And so we can conclude that disjunction of the two consequences, K and M. Right. Constructive dilemma number four. Jane, Line.

Four is absorption from line one.

Correct.

Line five is hypothetical syllogism from lines four and two.

Hypothetical syllogism from lines two and four. Okay.

And then six is another absorption from line five.

Correct.

And then seven is modus tollens from lines three and six.

Correct. Modus tollens from lines three and six. Does anyone have any questions or problems with number four? Okay, Number five gives us four premises. And then the conclusion, R or U. And in order to get to R or U, Copi supplies the proof. He begins first of all by saying if T, then U. And if I look above, I notice that if T, then you notice line two in the premises. Pardon me, line two in the premises has if T then u as the consequent of a conditional if not S, then if T then u. Treating the if T then U as a unit, that unit is the very thing that I find on line five. And so in order to get to it, all I would have to do is use modus ponens, if P, then Q, P, therefore Q I could use modus ponens in if I could find that not S is true somewhere. Oh, and lo and behold, on line 4, not S is found. So from lines 2 and 4, by modus ponens, I can get line 5. Okay, now let's look at line 6 of the proof. Line 6 has if Q then r conjoined with if T, T then u. And as I look above, I notice that line one gives me if Q then r, and line five, which we just proved gave me if T then u. So I can conjoin those two lines together. With me so far, Line 7 now has to be proved. It says Q or T, Q or T. Now that Q or T as a unit turns out to be found in line three, where we have S or Q or t. And so if I could deny S, I would be able to use disjunctive syllogism and assert Q or t. Now can I deny S? Yes. Line four gives me the denial of S. And so now I can, by disjunctive syllogism from lines 3 and 4, justify line 7, which is Q or t. And now how can I get R or U from all of the lines that have been justified or have been given as premises? Well, it turns out that R is the consequent of the if Q then r in line 6, and u is the consequent of the if T then u, and they're conjoined with the dot. So if I could find Q or T to be true, then by constructive dilemma, I could include R or u. And lo and behold, line 7 is Q or T, which has already been proven. So we have constructive dilemma on lines six and seven to prove line eight. They're all pretty clear. Now, I think the key is in recognizing that when you put something in parentheses, you treat that whole thing as one unit in comparing it to the rules of inference. Okay, Roman numeral 3. Now read to the following. Adding just two statements to the premises will produce a formal proof of validity. Construct a formal proof of validity for each of the following arguments. Okay, Argument number one gives us a premise. A gives us a premise B and a conclusion A or C and B J. How would you prove that?

In two steps I would take line 1A and add to it by addition C to get A and C, A.

And C A or C. It's very important that though that's called addition, we don't use the and it's always or A or C. Good. Then the next step, and then I.

Would take that and add to that. Am I using the right word?

Conjoined.

Conjoin. Thank you. Conjoined to it B and therefore get the quantity A or C and B.

Okay, so line three in your proof would be addition to line one, and then line four in your proof would be conjunction of line three and two. Very good. Is that clear how that was done? Okay, Pat, what said you do number two? The first premise is if D then E. The second premise is D and F. And in two steps we're supposed to get get from those two givens to the conclusion E. What would be your third step?

Take d from line 2 by simplification.

Line 2. Simplification gives you D. Very good.

Then to take.

Take E. You don't take E. You're going to assert E on the basis of lines 1.

Line.

3, line 1, and 3. By what rule? Modus ponens. Very good. Now this one would be a good example for figuring out a procedure in constructing a proof. You notice what you want to prove here is E. Now you go above and look at the premises. There's only two of them. You say, now how could I get to E given what I know here, by the rules that I'm familiar with. Well, modus ponens would make it real easy because I've got if D then E. So all I have to do is get D and I've got the step to E. Now how could I get D? So now you start ranging through the premises to say, how could I prove D? Well, if you have dnf, there's a rule that says you can simplify and have just one of them. And that would be D. And so now you're able to prove it. But the point is we work backwards. You first have to look at what you want to prove, and then you go find a premise, or a couple maybe, but a premise that looks helpful in getting to that conclusion. Then start manipulating the other premises to provide the missing steps. I'll do number three for you. We have the premise G and the premise H and the conclusion G and H. Or I please notice something here that there's a typo in this problem. Do you see the typo I hope that didn't throw you off. I worried about that when I was doing them last night. Here you have is the conclusion parin G dot H parin paren or I. That double paren is unnecessary. That's not a compound within a compound H or I. Well, I see part of the conclusion by looking at the premises, because my premises are G and H, so all I have to do is conjoin them. That's step three. I conjoin one and two. And therefore I assert on line three gnh. Now, I know that I can add anything to a true proposition with disjunction. And so by addition to line three, I put disjoin. I follow that. Okay. Number four, Pete.

Number four.

We get.

Line three from, which would be K. From bonus from line one two, which would be modus ponens.

Yeah. Let's try to express this clearly for those listening. Line 3 asserts what k on the.

Basis of 1 and 2 by the rule of modus ponens.

Very good. Then line four.

Line four would be K or L. It would assert K or L, which is the conclusion from line three by the rule of the addition.

From line three. Very good. Let's move ahead. We were to do some in Roman numeral 5. This is where it gets a little trickier because now we are to construct a formal proof of validity for each of the following arguments. Jay, let's have you do number one.

Okay. Line four of this proof we would assert from simplification of line three.

Let's make clear before we go any further, there are three lines given as premises. That's the raw material we're going to start with. And we want to end up with C as our conclusion. Okay, so then line four, the first step of the proof, we would assert.

By simplification of line three, not B.

Very good.

And then taking lines one, which is if A, then B, and line four, which was not B by modus tollens, we assert not A.

That's correct. Modus tollens. Excuse me.

And then on line 6, taking line.

2, line 6 says not a.

Correct.

Line 5 says not a.

Okay. Line 5 says not a.

So on line 6, we're going to take line 2 and line 5. And by disjunctive syllogism, we will. We will assert C and D correct. And then on line seven, we will, by simplification, assert C. Very good.

Let me try number two for you. We have another typo here in number two. Do you notice that in the first premise, H has been put with parentheses all by itself. You don't need to do that. Obviously, that's supposed to be H then I in one compound. So with that typographical error eliminated, we have three premises as givens. By the way, that typo is repeated in line three. So at least they're consistent in their mistakes. We have to give them credit for being logically consistent, I guess. Okay, three lines as premises. The fourth line, which will be the first step of proof, would be to assert if F, then G. And I can do that by simplifying line one. Okay. Line five is going to be F or J. And I get F or J by simplifying line three. And then line six will be. This is going to be fairly long one compound. If F, then G conjoined to another compound. And if J, then K. And I was able to get line six by conjoining lines two and four. Line two gave me the if J, J then k part. Line 4 gave me the if F, then G part. And so I can put those two together now. And then line seven, which is the conclusion G or K follows by constructive dilemma from line 6. And line five. Okay, very good. Let's have Pete do number three.

Number three. Okay, we have three lines of premises here. So we'll start from line four. Line four we get. I might have done this wrong. Line 4 asserts not n from number lines 2 and 3 by the rule of modus tollen.

Correct.

Line 5 asserts not m and not n by conjoining lines 3 and 4.

Correct.

Line 6 asserts f u then N from U or F O then N.

If O, then N based on modus.

Ponens of from lines 1 and 5.

Very good.

And then line 7 asserts not O, which is our conclusion from lines 4 and 6 by the rule of modus tollens. That's right.

Then we'll look at Roman numeral 6, which is given to us in a natural language. And he provides the letters the teacher stand for the proposition. So you have to construct what the premises are. And then you have to come up with the conclusion. And so I'll do that for you on the board.

Here.

We read, if Adams joins, then the club's social prestige will rise. Letter A will stand for Adams joins. And letter S for the club, social prestige will rise. So we have if A then S. The next line says, if Baker joins, then the club's financial position will be more secure. And given what Copie provides, Baker joins as B. So if B, then F, financial position is more secure. And then the next thing told to Us, either Adams or Baker will join. So we have A or B. And the next line tells us if the club social prestige rises, which is S, then Baker will join. So if S then B. And then if the club's financial position becomes more secure, which is F, then Wilson will join. If F, then W. And those are all of the premises that are given to us. The conclusion is, therefore either Baker or Wilson will join, that is either B or W. So what we want to be able to prove is B or W. We have five premises. How can I get to B or W from these five premises? Well, by looking at the conclusion. By looking at the conclusion, I see how my work is cut out for me. I have to get a B or a W or them together. If. I look at lines 1 and 4, line 1 tells me if A then S. 4 tells me if S then B, then I know if A then B. This is from lines 1 and 4. By. What rule I go if A then S, if S then B. Therefore, if A then B. What rule have I followed? No, that's not a dilemma at all. Let's look back at the rules. This is the pattern of hypothetical syllogism number three. It starts if P then Q. And then you cover the Q. If Q then R. Then you can take the extremes if P then R. Okay, now line seven. Is if b then w. Pat, can you tell me how I would get if B then W from these premises here or that one? You see any B here that would start out an argument and then do you find any given in line five? Okay, so two and five together in the same way will justify line seven. Okay, now line eight. We'll say if A then B has one unit conjoined with if B then W. And how can I justify this long line right here? Have I done any of that work previously?

6 and 7.

6 and 7 give me the component parts. So all I have to do is conjoin six and seven. Then line nine will be my conclusion, B or W. And this by constructive dilemma. Looking at Line 8, it says if A then B. And if B B than W. And line three had given me A or B. So if A is true, the B will follow. If B is true, the W will follow. So since these are conjoined together, we know it's going to be one or the other, or maybe one. Both. Understand how constructive dilemma fits in here? Here's one unit conditional statement, another unit conditional statement, and the antecedents for both are provided in a disjunction in line Three.

Could you have joined A if A then S, and if B, then F? Because in the way it's stated in here, says if Adams joins, then the social club's social, then the club social prestige will rise. And if Baker joins, then the club's financial position.

Well, I thought about that last night because, you see, I have this step here, eight, where I bring those. I bring the two things together. @ some point you're going to have to conjoin them if you don't do it at the very beginning. And I thought it wasn't legitimate to do it at the beginning because there was a semicolon there. And I thought it should be treated as a separate premise. But, you know, that's. Again, maybe he meant it to be taken as one big long premise, in which case the proof would be a little bit different, but the steps would eventually have.

To nine steps. It takes.

Five. All right, we move ahead then to section 9.2 in the book where Copi discusses rules of replacement. He tells us in the middle of page three of four, the rule of replacement permits us to infer from any statement the result of replacing any component of that statement by any other statement logically equivalent to the component replaced. That's very tight, nicely worded English, but sometimes we get a little lost. And it would be helpful, I think, if we go over this slowly, the rule of replacement, for mit us to infer from any statement. Now, he's talking about a compound statement, so I'm just going to give you a compound statement to use as an illustration. P and Q. It could be P if P then Q, P or Q, not P and Q whatever. But it's a compound statement. Let's make it more compound than that. Let's say it's P and Q or P then Q. Now we've got a real compound statement, don't we? Compound here and compound there, and it's joined together with a disjunction. Now, what he says is these rules of replacement permit us to infer from any statement, any compound statement. That's what I have up here on the board. We can infer from this the result. I'm going to put down here a blank line. I can now put on the next line of my proof the result of replacing any component of this statement. I can replace any component of this statement with any other statement that's logically equivalent to it, so that if I knew something was logically equivalent to P then Q, I could take whatever is logically equivalent and put it in the place of The P then Q. Now let's jump ahead and look at the rules of replacement that are offered to us. A rule of replacement is stated in the form of a logical equivalence, that is to say, a material equivalence where all of the lines in the truth table would be true when we state that these two expressions are equivalent, logically equivalent to each other, and that equivalence is written in the form of the three line equivalent. So now, whatever is on this side of the three lines and whatever is on this side of the three lines, whatever's on the left side, whatever's on the right side, are logically equivalent to each other. Consequently, any time if you have this rule of replacement, and I'm just using one and two to make it clear, one equals two is logically equivalent to two, anytime you find a one, you can put two in its place. And anytime you find a two, you can put one in its place. These are substitutable. So now what? We have our substitute substitutable parts. Let's go back to my compound statement. I said we have PNQ disjoined with if P then Q. Now I can take any one of these components and find something logically equivalent to it. That is where we have the three lines. And if I find that component on the one side of those three lines in a rule, I can take what's on the other side of the three lines and put in its place. In my illustration, I want you to look at rule number 16, material implication. Notice, material implication has one compound on the left hand side that says if P then Q, it's bracketed to remind you that it stands together as a unit and that is logically equivalent. That's the same as saying bracketed, not P or Q. It may not be immediately apparent to you that we went through that earlier in the class and explained why that is, those are really logically equivalent statements. But since they are logically equivalent, any place I find if P then Q, if I wish to, I could rewrite it as not P or Q. And that's all I'm doing is rewriting it. I'm just translating it into another form. Now notice in my illustration. Now here's this component part of this long statement. Since if P then Q is logically equivalent to not P or Q, I can write that in my next line in its place. So now my compound statement originally said P and Q or if P then Q. Now I have P and Q or not P or Q, because I've taken one command component and replaced it with something logically equivalent to it. Everyone follow that. Once we catch on to the fact that you can take a part of a line as long as it's a component standing by itself, you can take a part of a line, a component, and replace it with something logically equivalent to it, then you understand what a rule of replacement is all about. We are given 10 such rules of replacement. It probably would be worthwhile to look at them to see to whatever degree you can. You catch the intuitive thrust of them. All of them are provable by the truth tables. That's where we come up with the people that just write down, say, okay, these are the rules we'll play today in the game of logic. These rules have been proven by the truth tables. And the point is that there's an assertion of an equivalence between something on the left side and the right side of that three line equivalent. Mark Demorgan's theorems tell us that when you deny the conjunction of two things, that's the same as disjoining the denial of them both. That's the first line. Another way of doing it is to say if you deny the disjunction of two things, that's the same as conjoining the denial of them both. And if you look at that for a minute, I think you can almost see logically why that would be. If I say it's not the case P and Q, then that's the same as my saying it's not P or it's not Q. At least one of them is going to have to be not the case. Case if them join together is not true. And then if I say it is not the case that P or Q, then I'm saying neither one of them can be true. So that's the same as conjoining not P and not Q. Again, I hope the intuitive thrust is enough to help you remember how to use this rule. That's not the way you prove it. You don't cite an intuition and say, okay, we've got this logical rule, but in this case, most of us can probably see without the truth table, that's likely to be true. Number 11, commutation. This is very important. This just tells you you can write things in reverse order. P and Q could be rewriting Q or P. Now, I may have used the rule of commutation in one of the exercises previously and may not have happened, had the right to do that in terms of what Coby wanted. But it should be intuitively obvious that A or B or P or Q is the same as saying Q or p and the same with the and sign P and Q is the same as saying Q and P. The law of association is very helpful. This tells you that if you have all disjuncts or all conjuncts in a line that has brackets, you can rewrite the brackets two at a time. You see how that's the case. On the left hand side we have P or and then in parentheses Q or R. That is the same as saying parentheses P or Q and then or R. You can shift the brackets in twos. Remember compound statements, you can't have more than two in the bracket. Number 13. Distribution. Here you can distribute the effect of the and and the or as we use them in logic. If you have P and and then this parentheses Q or R, that's the same as saying P and Q or P and R. Met algebra. You can almost see that as an algebraic function, can't you? You factor both the things in the parentheses by what is outside the parentheses. So you end up with P and Q or P and R. It works the other way too. If you have P or Q and R, then you can say, well then it's P or Q and P or R distribution. Double negation requires no explanation. Transposition. Transposition, I hope you recognize, is a form of modus tollens. It's writing out modus tollens. If P then Q is the same as saying if not Q then not P. If you deny what is the consequent, you have to deny the antecedent. Number 16. Material implication was discussed in yesterday's class. It turns out to say if P then Q. Remember the truth table is the only way an if P then Q can be false is if you have the P be true and the Q be false. So if it's true that if P then Q, then either the P is false, in which case anything follows from it, or the Q is true, in which case anything implies it. So almost intuitively you can catch on to that one. Material equivalence. This just tells us that to state that P is equivalent to Q is the same as saying if P then Q and if Q then P. It's a biconditional. The horseshoe or the arrow of conditionality goes both ways. Exportation. This is an important rule because it allows you to regroup when you have a horseshoe or an implication. If the antecedent is P and Q and the consequent is R. That is, if you have an if P and Q then R statement that is the same as saying if P then Q then R It's kind of like proving it in steps. If I say P and Q, if P and Q, then R, you say okay, that's if P then Q. Okay, but now I'm going to assert P. And if I assert P, then I know that Q will lead to R, because P has been asserted already. So exportation is a matter of understanding what is involved in saying that two things are the antecedent of the truth of a third P and Q, then R is the same as saying, well, if P, then if Q Q than R. And tautology is obvious.

Enough.

You can repeat any true premise either with an or, or with an and. The next paragraph in Copie simply reminds us that we replace component parts. We replace components parts here, rather than working with whole lines. In the rules that we followed previously, we dealt with a whole line of proof at a time and looked at another whole line of proof and then came up with the third line or another step in our proof process. Here what we're going to do is replace a component part of a line with something that is logically equivalent to it. So in the very middle of the paragraph he says, but in moving from one statement to another by way of replacement, we can replace a component of the first only by our statement certified to be logically equivalent to that component. Or if you will, you can replace a component with something that is logically equivalent to it. He reminds us on the bottom of page 307. To skip ahead, the first nine rules can be applied only to whole lines of proof, and then page 308, a few lines down. He says any of the last 10 rules can be applied either to whole lines or to parts of line. By replacement. Logically, equivalent expressions can replace each other wherever they occur, even where they do not constitute whole lines of a proof. So that's the difference between the first nine rules and then the replacement rules. The first nine have got to be applied to a whole line. The the second nine can be applied to a whole line or to a component part. Now what I would like to do is to get into the exercises and do a few of them so we understand the rules. But then I want to go back and finish covering what Kopi says about these rules. He describes them from the middle of page 305 to the bottom of 307. And essentially what he says is that these rules are redundant, understand that they are in a certain sense deficient, and yet they are complete. They are mechanical, but they are not completely effective. So there are advantages and disadvantages, and we'll go back and look at that, just so we have an overview or an evaluation of what we have learned today in this. But right now, let's see if we can't do some exercises on page 308, for your information. He says, when you're constructing a formal proof, there's no purely mechanical way to do it. And so he gives a few hints as to how to do it. He says, begin deducing conclusions from the given premises. Just take the given premises and play with them for a while. Look at the rules and say, now here's one in that form. Here's one in the form of that rule over there. So I could rewrite this in the following way. I could rewrite that in the following way. Just kind of give yourself more material to work with. That's one guideline. Secondly, he says, try to eliminate statements that occur in the premises but not in the conclusion. So pay attention to what's in the conclusion. That's what you've got to get to. And eliminate stuff in the premises that appear superfluous. Another guideline, he says, introduce by means of addition a statement that occurs in the conclusion but not in any premise. Introduce by means of addition a statement that occurs in the conclusion but not in any premise. So if the conclusion is a compound that has A, B and C in there, You can add to any of those and see if that helps you. Then if you can get that one letter from your premises, then you've got another step to the conclusion. And then finally, he says another method is to work backward from the conclusion by looking for some statement or statements from which it can be deduced. I'm not sure why he waited to say that. I consider that the first step or the first guideline myself. So with these guidelines in mind, play around with the premises, see what you from them. Eliminate unnecessary premises, look at the conclusion, work backwards, and if necessary, introduce by means of addition a statement that occurs in the conclusion but not in any premise. Let's see if we can't work out the problems given to us. First of all, In Roman numeral 1, for each of the following arguments, state the rule of inference by which its conclusion follows from its premises. Okay, we're going to begin real easy here. He gives us the answer. We just have to know which rule justifies it. J number.

One. The rule of transportation. We have if C, then D in the premise, and in the conclusion we have if not D, then not.

C. Okay, that is transportation. Is that what you called it? Remember, that looks a lot like Modus tollens. You perform a modus tollens on that little bit there. If C then D, and you write it out. If not D then not C. And that component part replaces, because it's logically equivalent, that component part replaces the if C then D. Pat, in premise one, do you notice the second conjunction? The second part of that says if C then D. Now it turns out that if C then D can be rewritten according to the law of transportation as if not D then not C. Okay, so component part replaces component part with a logically equivalent expression. Number.

Two. Number two. Component is material implication of E if E than F. Very.

Good. That one component part, if E then F can be rewritten in a logically equivalent way as not E or F. Right. Pat, take number three, the exportation. Very good. Exportation. What does exportation.

Do? It says that we can have P and Q, therefore. Well, in the example P and Q therefore R is equivalent to P therefore Q therefore R. And that's the same form as we have in the first statement and number three and also in the first part of the.

Conclusion. Okay, now did. Did we work from the left to the right hand side of the equivalence or from the right to the left hand side of the equivalence in this example, Right to the left. And that's why that I think this is here is just to force you to realize it's not always the way you read English. Left or right. When it's an equivalence, it can go either way. It's a biconditional. Okay, number.

Four, that is a tautology. So therefore it only needs to be stated.

Once. Yeah, or you can state it twice if you wish. Again, it's biconditional. You can go either way. Okay, everyone see, it's a tautology. Good, let's move to Roman numeral 2. Each of the following is a formal proof of validity. For the indicated argument, state the justification for each line. That is not a premise. Let's begin with number two. And Pete, would you lead us.

Here? Okay, line three is a commute commutation of line.

One. That's right. Does everyone see that it's a commutation from line.

One then line.

Four. What basically does commutation do? Reverses the order of.

Mention. Okay, then line four is exportation of line three. Very.

Good. That shows us two things that are in the antecedent. It means it can be restated that if the first antecedent, the first thing mentioned is true, then it follows that if the second is true, what was originally the consequent would Be.

True. And then five, then five is a hypothetical syllogism from lines two and.

Four. Very good. Let's try number three then. It might be helpful to remember that a compound statement can be treated. You know, let's just call it a blob because it's one unit. And so if we have in this particular case in line two, the blob, if D, then F, we can say if the blob then G. And it turns out that line four is if E, then that blob, which we called if D then F. So line 4 says if E, then the blob. Line 2 says if the blob then gave. And we know the conclusion is if E then G mediated through the blob. Okay. I don't know why they don't use technical expressions like blob and so forth in the book. It would certainly help. Alright, whose turn is it? Number three. Okay.

Pat. Okay, line three, we have a commutation of volume.

One. A commutation. H or I. Yeah. You have to remember, if it's a commutation, you would have to write out the component part and finish the line. You can't drop the rest of the line just because you replace a component. Think of it this way, Pat. If I say I will buy a ball at the store and then I tell you the word ball can be replaced, you know, with another word if you wish. And then you can rewrite the sentence. When you rewrite the sentence, you have to say I will buy a dish at the store. You don't just say I will buy a dish or you don't rewrite it just as dish. Everything that's on the line has to be rewritten and just the component parts replaced. So line two is not a commutation of line one, because then you'd have to go on to write the rest of line one down. Since it wasn't written down, that isn't what the author was doing. Copie was not intending that to be a commutation of that long line or else the rest of the line would be found in the same way that the rest of the sentence I will buy a blank at the store would be found. You don't just write dish as the second.

Line. So the second part of the line wouldn't be excluded because I was affirmed in line.

Two. Well, now you're going through a reasoning process which is what these steps are intended to teach. But you can't do all the reasoning process in one step. You got to go step by step by Step. Every line, every step along the way has got to be justified. So how could we justify asserting I or H given the two premises that are there? This is very simple. I think I can give you a.

Hint. Primodus Pohens.

Ponens. We.

Have the hori is asserted or I is asserted in line.

Two. Line two is a given. You don't have to prove line two. Line two is one of the premises that you start out with. Remember our notation. We have premise, premise, and then finally a slash line to indicate what the conclusion is. And everything following that then must be a step in the argument. The conclusion is J and K. And the last premise given to us is I. That's line two. So let's start again. How could we get line three of this proof? How could we justify Copie? See, this is Copie's work. He's saying, I've done it for you. Now you tell me what justifies each one of these steps. Here's the step I'm going to take. I or H. How could you justify that from the premise to I? Or how could you justify from the premise 1H or I than J and K and.

L. By.

Addition? That's right, by.

Addition. From line.

Two. Exactly. If he gives you I, you can add anything to it with a disjunction. Very good. And then line four. How could we assert H or I given the premises that we've proven or that we're given in the.

Problem? It's a permutation of line.

Three. Very good. And then line five. How could we get J and then the blob K and L. Do you see that combination of symbols.

Anywhere? It's in line.

One. In line one as the big blob.

Right.

Okay. So by.

Association. No, not by association. Notice what you have in line one, Path. You have the big blob, which says J and K and L as being implied by something else. Now, if you can find that which implied it, then you could conclude that J and K and L can be written as well. So what would imply J, K and L? H or.

I? H or.

I. Do you have an H or I in any of your proof lines in line four? That's right. So that five is justified from line one and four. By what rule? Great. Now six gives us J and K in the parentheses and.

Out. That would be taken.

From. Does that look like anything that we see above? Similar to anything above it. Like right above it.

Maybe. Oh yeah. There'll be a.

Association. That's right. The law of association lets us Move the brackets. Remember though, all of the connections have to be the same kind. You can't move to the brackets when you have ors and ands in the line. But when they're all ands or all ors, you can move the brackets anywhere you want. Okay, and then seven is the conclusion. J and K. How do we get J and K from any of the above.

Lines? Simplification of line.

Six. Very good. Okay, let me do number four for you so we can move ahead quickly. We have two premises, 1 and 2. The conclusion we're looking for is not M. The first step suggested by Copie is to assert not O or not P. Well, since he gave me not O in line two, I'll just assert that on the basis of addition. Now then, how could I get not and then parentheses O and p? Well, Demorgan's theorems. You look at Demorgan's theorems and how they work, you have to check both sides of the equivalence. Demorgan's theorem helps us. If we have a double negation. Well, it's not really a double negation if we have a negation conjoined to a negation. You notice we have that in the second line of DeMorgan's theorem, a negation conjoined to a negation on the right hand side. So that means it's logically equivalent to write that as not and then put in parentheses P or Q. So that's what I've done. In line four, I've written not. Ok, I looked at the wrong line in Demorgan. Let me back up. The top line in the Demorgan's theorem says if you have two negations disjoined, then you can rewrite them as conjoined with the negation outside the parentheses. And that's what line four is. So DeMorgan applied to line three produces line four. Now, line five can be justified according to modus Tollens. Looking at line one and line.