CHAPTER 5 - DISJUNCTIONS

Transcription

1 1 CHAPTER 5 - DISJUNCTIONS Here, you ll learn: How to understand complex sentences by symbolizing disjunctions applying truth conditions for disjunctions How to assess arguments for validity / invalidity by establishing that arguments are valid using I, O, DM, COM, ASSOC, SD, CD, AR, NEG AR, and EXCL-MID rules Symbolizing We have only one more connector to discuss in our treatment of propositional logic, the connector or, which is symbolized as the wedge. Rule: P or Q will be symbolized as P Q A statement with as the main connector is called a disjunction and the statements connected with are called disjuncts. (If these terms sound familiar to you, there s good reason for that. They re closely related to the terms conjunction and conjuncts, which we learned when we studied &. ) As with all connectors, the propositions connected with may be simple or compound, so A B and (P Q) (C & D) are both disjunctions. Practice To get used to disjunctions, symbolize the following sentences. The answers are in the endnotes. 1. Either he ll read the book or he ll watch the movie. 1 R = He ll read the book. W = He ll watch the movie. 2. Either he ll read the book or he ll watch the movie or both. 2 R = He ll read the book. W = He ll watch the movie. 3. Either he ll read the book or he ll watch the movie but not both. 3 R = He ll read the book. W = He ll watch the movie.

2 2 Before we continue symbolizing disjunctions, let s compare the second sentence above, Either he ll read the book or he ll watch the movie or both, with the third sentence above, Either he ll read the book or he ll watch the movie but not both. Sentence 2, Either he ll read the book or he ll watch the movie or both, is an inclusive or because it includes the possibility that both disjuncts might be true. Sentence 3, Either he ll read the book or he ll watch the movie but not both, is an exclusive or because it excludes the possibility that both disjuncts might be true. In logic, we always read or inclusively and so A or B is equivalent to A or B or both. If we want an exclusive or, we need to specify that explicitly by saying A or B but not both. Knowing that is inclusive allows us to complete the truth table for as follows. (Can you see how the first row of this table indicates that is inclusive?) Okay, now back to symbolizing. P Q P Q T T T T F T F T T F F F 4. If he reads the book or watches the movie then he won t be left out of the conversation. R = He ll read the book. W = He ll watch the movie. L = He ll be left out of the conversation He ll be left out of the conversation, unless he reads the book or watches the movie. R = He ll read the book. W = He ll watch the movie. L = He ll be left out of the conversation Either watching the move is sufficient for being able to discuss the plot intelligently or reading the book is necessary for being able to discuss the plot intelligently. R = Someone reads the book. W = Someone watches the movie. D = That person can discuss the plot intelligently If he either reads the book or watches the movie then he ll be able to discuss the plot intelligently and he ll impress his friends. R = He reads the book. W = He watches the movie. D = He ll be able to discuss the plot intelligently. I = He ll impress his friends. 7

3 3 8. Provided that he s joined the book club, he ll read the book or watch the movie but not both. J = He s joined the book club. R = He reads the book. W = He watches the movie. 8 Primitive Inference Rule: O Our first inference rule for is delightfully intuitive. Suppose I tell you that my aunt Helen will be arriving on the 6:00 or 11:00 train, and then, after speaking to Helen on the phone, tell you that she won t be on the 6:00 train. What would you conclude? You d probably (and correctly) conclude that Helen will be on the 11:00 train. We reason this way all the time. If we know that a disjunction is true, and that one of the disjuncts is false, we conclude that the other disjunct is true. From P or Q and Not P (negating the first disjunct) we conclude Q (the second disjunct) and from A or B and Not B (negating the second disjunct) we conclude A (the first disjunct). This gives us the O rule. i O:, Opp -, Opp - Practice Prove the following arguments. The answers are in the end notes. 1. Either I m hallucinating or that cat is dancing. I m not hallucinating. Therefore, that cat is dancing. H = I m hallucinating. C = That cat is dancing Either there is no gas in the car or the battery needs a charge. There is gas in the car. Therefore, the battery needs a charge. G = There is gas in the car. B = The battery needs a charge. 10 i We ve verified O by appealing to our logical intuitions, but we can also verify it by constructing a truth table as follows: P Q P Q ~P Q T T T F T T F T F F F T T T T F F F T F The only row in which P Q and ~P are both true indicates that Q is also true. (I ve highlighted that row.) This means that whenever P Q and ~P are true, Q must be true as well, and that means that the argument P Q, ~P - Q is valid.

4 4 3. P & ~Q, R Q - R P Q, ~P ~Q - P (Hint: Do a proof by contradiction.) P Q, ~Q ~R, P Q - ~R (Hint: Do a proof by contradiction.) 13 Primitive Inference Rule: I To get a handle on the next primitive inference rule, let s consider the following situation: Sally recently turned five years old and she got a new puppy named Winston for her birthday. Sally s mother has a very strict rule to the effect that Winston is not allowed on the furniture and is especially not allowed on Sally s bed. (Winston sheds.) Sally, however, likes sleep with Winston and she will sometimes put him on the bed in order to dress him up in various outfits. (Winston is a placid little fellow and doesn t mind.) One rainy Saturday afternoon, while playing with Winston on her bed, Sally heard her mother call from the living room. Sally, where s Winston? Not wanting to lie to her mother, but not wanting to admit that Winston was on the bed, Sally answered, He s either on my bed or in the kitchen. or

5 5 Of course, Sally s response is unlikely to fool her mother, but the logically interesting feature of Sally s answer is that it wasn t technically a lie. Because Winston was on her bed (we can symbolize Winston is on the bed as B ) it s certainly true that Winston was either on her bed or in the kitchen (we can symbolize Winston is on the bed or in the kitchen as B K ). Sally certainly implied that she didn t know Winston s exact location, she never explicitly said this. What she said was He s either on the bed or in the kitchen, and that was true, strictly speaking. In short, given that B is true, B K is true as well. This means that B - B K is valid and that gives us our next primitive inference rule, I. ii I: - - It s important to note that can be anything at all regardless of whether or not it appeared in the proof before because, given that is true, both or (anything at all) and (anything at all) or are true. Sally, for instance, could have said Either Winston is on my bed or he s touring France. or That simply wouldn t have served her purposes. Note 1: The formula that s wedged in can be anything at all. ii We ve verified I by appealing to our logical intuitions, but we can also verify it by constructing a truth table as follows: P Q P P Q T T T T T F T T F T F T F F F F All of the rows where P is true have P Q true too. (I ve highlighted those rows.) This means that whenever P is true, P Q must be true as well, and that means that the argument P - P Q is valid.

6 6 It s also important to note that unlike &I, I is a one-line rule. This is because we aren t taking two formulas on different lines and connecting them a, as we take two formulas on different lines and connect them with &. Instead, we re taking a formula on one line and attaching to it with some other formula that we effectively pull from the sky. For example, consider the following argument: P&Q - P Q. Can you see what s wrong with this proof of that argument? 1. P&Q P 2. P 1 &O 3. Q 1 &O 4. P Q 2, 3 I The problem is that line 4 treats I as if it were a two line rule that connects lines 2 and 3 with an or. But I doesn t work that way because I is a one-line rule. We can repair this proof in two ways. First, we can think of P Q as being generated from exclusively from the P on line 2, like this: 1. P&Q P 2. P 1 &O 3. Q 1 &O 4. P Q 2, 3 I Second, we can think of P Q as being generated from exclusively from the Q on line 3, like this: 1. P&Q P 2. P 1 &O 3. Q 1 &O 4. P Q 2, 3 I Note 2: I is a one-line rule. Before we practice with I, it s good idea to see how I differs from the other in rules we ve seen: the other in rules can be used to indicate how we might derive the conclusion of an argument, but this is generally not the case with I. For example, if the conclusion of an argument is P & Q, there s a good chance that we ll use &I to derive it. If the conclusion of an argument is P Q, there s a good chance that we ll use I to derive it. If the conclusion of an argument is P Q, there s a good chance that we ll use I to derive it. And if the conclusion of an argument is ~P there s a good chance that

7 7 we ll use ~I to derive it. However, if the conclusion of an argument is P Q, there s a good chance that we won t use I to derive it. This is because in order to prove P Q with I, we would first need to prove P or we d first need to prove Q. But if either of these follows from the premises then the argument would probably take that as the ultimate conclusion rather than the disjunction, which contains roughly half as much information. Practice Prove the following arguments. The answers are in the end notes. 1. If either shirts or slacks are on sale then the store will be very busy. Therefore, if slacks are on sale then the store will be very busy. H = Shirts are on sale. L = Slacks are on sale. B = The store will be very busy E, (A C) (B & D), B ((E F) A) - A B 15 Derived Rules Now that we have O and I, we can get some very handy derived rules. DeMorgan s Law Our first derived rule is DeMorgan s Law, which deals with negated disjunctions, like ~(P Q), and negated conjunctions, like ~(P & Q). DeMorgan s Law is exceptionally useful because, by showing us how negated disjunctions and conjunctions should be handled, they help us to avoid some very common reasoning errors. Let s start by examining negated disjunctions. DeMorgan s Part I: Negated Disjunctions To start us thinking about negated disjunctions, suppose I tell you that Sandra is either a physics major or a chemistry major. Letting P be Sandra is a physics major, and C be Sandra is a chemistry major, we can symbolize my statement as: P C. Now suppose that you disagree with me, knowing that Sandra doesn t like science of any sort. No, you say, It isn t the case that Sandra is either a physics major or a chemistry major. We can symbolize your statement as ~(P C) Now, think about your claim that it isn t the case that it isn t the case that Sandra is either a physics major or a chemistry major, ~(P C). Which of the following statements is equivalent to what you said: the claim that either Sandra isn t a physics major or she isn t a chemistry major, ~P ~C, or the claim that Sandra isn t a physics major and she isn t a chemistry major, ~P & ~C?

8 8 If you said that It isn t the case that Sandra is either a physics major or a chemistry major, ~(P C), is equivalent to Sandra isn t a physics major and she isn t a chemistry major, ~P & ~C, give yourself a gold star! If we know that it isn t the case that Sandra is either a physics major or a chemistry major then we know she can t be physics major (because if she were then she d be a physics major or a chemistry major) and we know that she can t be a chemistry major (physics major or a chemistry major). To say Either Sandra isn t a physics major or she isn t a chemistry major, on the other hand, is too weak because it leaves open the possibility that she might me a physics major (while failing to be a chemistry major) or a chemistry major (while failing to be a physics major). In fact, we know that she can t have a major in either subject. All of this tells us that ~(P C) is equivalent to ~P & ~C. In general, we can transform a negated disjunction into the conjunction of the opposites of the disjuncts. (I think of this as distributing the negation over the or and changing the or to an and. ) In order to formally establish this derived rule, we ll need to prove each of the following four equivalences, which cover all possible cases of negated disjunctions: I) ( ) & II) ( ~ ) & III) (~ ) & IV) (~ ~ ) & I ll prove the first equivalence and let you prove the other three. (You can find my proofs of II, III, and IV in the endnotes.) I) ( ) & ( ) - & 1 1. ( ) P want & 2 2. PP want contradiction I 1,2 4. ( ) & ~( ) 1,3 &I 1 5. ~ 2-4 ~I 6 6. PP want contradiction I 1,6 8. ( ) & ~( ) 1,7 &I 1 9. ~ 6-9 ~I ~ & ~ 5, 9 &I

9 9 & - ( ) 1 1. & P want ( ) 2 2. PP want contradiction 1 3. ~ 1&O 1,2 4. 2, 3 O 1 5. ~ 1 &O 1,2 6. & ~ 4,5 &I 1 7. ( ) 2-6 ~I II) ( ~ ) & 16 III) (~ ) & 17 IV) (~ ~ ) & 18 DeMorgan s Part II: Negated Conjunctions To pump our intuitions about negated conjunctions, suppose that you re helping me to buy a car. This one looks good, I say, It s both new and under $10,000. Letting N be The car is new, and U be It s under $10,000, what I said can be symbolized as N & U. That s impossible, you respond, having noticed how expensive all of the new cars are. It can t be the case that the car is both new and under $10,000. We can symbolize your response as ~(N & U). Now, think about your claim that it can t be the case that the car is both new and under $10,000, ~(N & U). Which of the following statements is equivalent to what you said: the claim that the car isn t new and isn t under $10,000, ~N & ~U, or the claim that either the car isn t new or it isn t under $10,000, ~N ~U? If you said that It isn t the case that the car is both new and under $10,000, ~(N & U), is equivalent to Either the car isn t new or it isn t under $10,000, ~N ~U, you re exactly right! When you say that the car can t be both new and under $10,000, you aren t saying that it can t be new and you aren t saying that it can t be under $10,000. After all, it might be new, as long as it s at least $10,000, and it might be under $10,000 as long as it s not new. It just can t be both simultaneously; it s not new or it s not under $10,000, ~N ~U. This show us that ~(N & U) is equivalent to ~N ~U. In general, we can transform a negated conjunction into the disjunction of the opposites of the conjuncts. (I think of this as distributing the negation over the and and changing the or to an or. ) In order to formally establish this derived rule, we ll need to prove each of the following four equivalences, which cover all possible cases of negated conjunctions:

10 10 I) ~( & ) ~ ~ II) ( & ~ ) III) (~ & ) IV) (~ & ~ ) Once again, I ll prove the first equivalence and let you prove the other three. (You can find my proofs of II, III, and IV in the endnotes.) I) ~( & ) ~ ~ ~( & ) - ~ ~ 1 1.~( & ) P want ~ ~ 2 2. ~ (~ ~ ) PP want contradiction 2 3. & 2 DM 1,2 4. ( & ) & ~( & ) 1, 3 &I 1 5. ~ ~ 2-4 ~O You might be concerned about step 3 in the previous proof because it looks like we re using DeMorgan s Law to prove DeMorgan s Law. This inference is okay, however, because we re using Part I of DeMorgan s Law, which we ve already established, to prove Part II of DeMorgan s Law. That s just fine. After all, we could have given these two parts of DeMorgan s Law different names, if we wanted to, and treated them as two different inference rules altogether. ~ ~ - ~( & ) 1 1. ~ ~ P want ~( & ) 2 2. & PP want contradiction &O 1,2 4. ~ 1, 4 O &O 1,2 6. & ~ 4,5 &I 1 7. ~( & ) 2-6 ~I II) ( & ~ ) 19 III) (~ & ) 20 IV) (~ & ~ ) 21 We now have DeMorgan s Law, which is an incredibly useful equivalence rule. I ll indicate that DeMorgan s Law has two parts, but in the context of a proof we don t distinguish between them. Whenever we use DeMorgan s Law on a negated disjunction or a negated conjunction, we just call it DeMorgan s.

11 11 DM ( ) Opp & Opp ( & ) Opp Opp There are a couple of things we should note before we practice with DeMorgan s Law. Note 1: DeMorgan s Law is an equivalence rule and so it can be used on parts of lines. This means that not only may we use DeMorgan s Law on formulas like ~(A & B) and ~(C D), as follows: 1. ~(A & B) P 2. ~A ~B 1 DM 1. ~(C D) P 2. ~C & ~D 1 DM But we may also use DeMorgan s Law on formulas like ~(P & Q) ~(R T), as long as we only use one instance of DeMorgan s Law for each inference. The following proof, for example, uses DeMorgan s Law on the antecedent of the conditional, and it s fine: 1. ~(P & Q) ~(R T) P 2. (~P ~Q) ~(R T) 1 DM We can then use DeMorgan s Law on the consequent of line 2, like this: 1. ~(P & Q) ~(R T) P 2. (~P ~Q) ~(R T) 1 DM 3. (~P ~Q) (~R & ~T) 2 DM What we can t do is go directly from line 1 to line 3 in the proof above.

12 12 Note 2: DeMorgan s Law is an equivalence rule and so it can be used in both directions. In some sense, DeMorgan s Law is most naturally used to hop negations into conjunctions (transforming them into disjunctions) and to hop negations into disjunctions (transforming them into conjunctions). Like this: (A & B) Hop! (A & B) A B (C D) Hop! (C D) C & D

13 13 But because DeMorgan s Law goes in both directions, it can also be used to suck out negations from two disjuncts (changing the disjunction to a negated conjunction), and to suck out negations from two conjuncts (changing the conjunction to a negated disjunction). Like this: Suck L L M M Pop! (L & M) Suck P & P & Q Q Pop! (P Q) And the really nifty thing is that we can transform a conjunction into a negated disjunction even if the conjuncts aren t negated, and we can transform a disjunction into a negated conjunction even if the conjuncts aren t negated. This is because thanks to Double Negation we can always imagine an even number of invisible tildes in front of any formula! P = P

14 14 This allows us to perform the following inferential moves: R & S This is the same thing as ~ ~ R & ~ ~ S, so sucking out one of those tildes from each conjunct and changing the and to or gives us. ~(~R ~S) and U W This is the same thing as ~ ~ U ~ ~ W, so sucking out one of those tildes from each conjunct and changing the and to or gives us. ~(~U & ~W)

15 15 Of course, we can always double check this backward uses of DeMorgan s Law by seeing if we can use DeMorgan s Law on the new formula to get back to the original formula. ~(~R ~S) Does using DeMorgan s Law on that formula give us. R & S Yes! ~(~U & ~W) Does using DeMorgan s Law on that formula give us. U W Yes! Now that we ve seen some of the finer points of DeMorgan s Law, we can practice with it. Practice Answer the following questions and prove the following arguments. The answers are in the end notes. 1. What do you get when you use DeMorgan s Law on the following formulas? 22 a. ~(A & B) b. ~(A B) c. ~(~P & Q) d. ~(P ~Q) 2. What do you get when you use DeMorgan s Law on the following formulas? 23 a. ~A & ~B b. ~A ~B) c. ~P & Q) d. P ~Q

16 16 3. (P & Q) R, R - P Q P (A B), A & B - P (P Q) (A & B), ~A ~B - ~P & ~Q (P & Q) R, R S, S, Q - P 27 Or Commutativity In Chapter 2, we established the derived rule And Commutativity, or & COMM, which tells us that the formulas & and & are equivalent to each other. Intuitively, we should have a commutativity rule for disjunctions as well, because Either she s a physics major or a chemistry major and Either she s a chemistry major or she s a physics major, seem to mean the same thing. In order to establish the commutativity of disjunctions, we need to prove the following equivalence: And we ll do this by proving the following two arguments: I) - II) - I ll prove the first argument and let you prove the second. You can find my proof in the endnotes. I) - 1. P want 2. ~(~ & ~ ) 1 DM 3. ~(~ & ~ ) 2 & Com 4. 3 DM II) - 28 We have now established the commutativity of Or and earned the following derived equivalence rule: COM:

17 17 Or Associativity In Chapter 2, we established the derived rule & Associativity, which tells us that the formulas & ( & ) and ( & ) & are equivalent to each other. Disjunctions have this associative property as well. In order to establish the associativity of disjunctions, we need to prove the following equivalence: ( ) ( ) And we ll do this by proving the following two arguments: I) ( ) - ( ) II) ( ) - ( ) I ll prove the first argument and let you prove the second. You can find my proof in the endnotes. I) ( ) - ( ) 1. ( ) P - want ( ) 2. ~[~( ) & ~ ] 1 DM 3. ~[(~ & ~ ) & ~ ] 2 DM 4. ~[~ & (~ & ~ )] 3 & ASSOC 5. ~(~ & ~ )] 4 DM 6. ( ) 5 DM II) ( ) - ( ) 29 We have now established the associativity of Or and earned the following derived equivalence rule: ASSOC: ( ) ( ) The Law of the Excluded Middle In Chapter 4, we derived the Law of NonContradiction, which allows us to write any formula of the form ~( & ~ ) on any line of any proof. We can do this because ~( & ~ ) is a tautology, or a sentence that it always true in virtue of the logic of the connectors that it contains. Now we can prove another tautology, the Law of the Excluded Middle. The Law of the Excluded Middle says that ~ is always true, and when you think about it, you ll see that s right. Right now, you are either sitting down or you aren t. There is either intelligent life of other planets or there isn t. Portland is the capital of the United States, or it s not.

18 18 In order to establish the Law of the Excluded Middle, we need construct a proof for the following argument: - ~ Can you see how we might do that? Since we already have the Law of Noncontradiction, let s start there. 1. ~( & ~ ) NON-CON We can then complete the proof by using Commutativity and DeMorgan s Law as follows: 1. ~( & ~ ) NON-CON 2. ~(~ & ) 1 & Comm 3. ~ 2 DM EXCL-MID: ~ After we discuss the derived rules Simple and Complex Dilemma below, we ll see how the Law of the Excluded Middle sometimes appears in an argument. Simple Dilemma Suppose that we both have been invited to a potluck, and I m wondering if I want to go. (I m not much of a potluck person.) Look, you say, knowing my fondness for baked goods, George is either going to bring his famous chocolate cake or he s going to bring his equally famous apple pie. If he brings the chocolate cake, you should be there. And if he brings the apple pie, you should be there. Clearly, you want me to conclude that I should be at the potluck. And that s exactly what I should conclude because the following argument is valid.

19 19 George is going to bring chocolate cake or he s going to bring apple pie. If he brings chocolate cake then I should be there. If he brings the apple pie then I should be there. Therefore, I should be there. C = George is going to bring chocolate cake. P = George is going to bring apple pie. B = I should be there. C P, C B, P B - B This is a fairly common inference form so it s worth proving it in general, as a new derived rule. In order to establish this rule, we ll need to prove the following argument,, - And here is one way to construct this proof: 1 1. P 2 2. P 3 3. P want R 4 4. ~ PP want contradiction 2,4 5. ~ 2, 4 MT 1,2,4 6. 1, 5 O 3,4 7. ~ 3, 4 MT 1,2,3,4 8. & ~ 6, 7 &I 1,2, ~O We have now established the derived rule Simple Dilemma. SD:,, - This rule is called Simple Dilemma. It s a dilemma because it presents us with a unresolved choice between two possibilities. (It s either or.) It s simple because no matter which possibility happens to occur, we ll have the same outcome. (We re going to get no matter what happens.) Simple dilemma is the first of two three-lines rules we ll have, and with it in hand, we can complete your argument about the potluck very easily.

20 20 Practice George is going to bring chocolate cake or he s going to bring apple pie. If he brings chocolate cake then I should be there. If he brings the apple pie then I should be there. Therefore, I should be there. C = George is going to bring chocolate cake. P = George is going to bring apple pie. B = I should be there. C P, C B, P B - B 1. C P P 2. C B P 3. P B P - want B 4. B 1, 2, 3 SD Prove the following arguments. The answers are in the end notes. 1. If our character is determined by our genetic make-up then we aren t responsible for what we do. If our character is determined by our early-childhood conditioning then we aren t responsible for what we do. Our character is either determined by our genetic make-up or by our earlychildhood conditioning. Therefore, we aren t responsible for what we do. G = Our character is determined by our genetic make-up. R = We are responsible for what we do. C = Our character is determined by our early-childhood conditioning I have certain sense impressions. If I have sense impressions, they re either veridical or hallucinatory. If my sense impressions are veridical, then I must exist (because I m actually experiencing things external to myself). If my sense experiences are hallucinatory, then I must exist (because I m hallucinating). However, something exists, if I exist. Therefore, I can be certain that something exists. 31 S = I have certain sense impressions. V = My sense impressions are veridical. H = My sense impressions are hallucinatory. I = I exist. E = Something exists. 3. A B, P & Q, (A & P) (Q C), B (C D) - C D 32

21 21 Complex Dilemma Having been convinced by your argument that I should attend the potluck, I m wondering what I should bring. I don t have much time to prepare anything. How about this? you say, I m either going to bring lasagna or a tuna casserole. If I bring the lasagna then you bring a salad. If I bring a tuna casserole then you can bring some bread. From this I can conclude that I ll either bring a salad or bread (two relatively simple things) because the following argument is valid: You will bring lasagna or a tuna casserole. If you bring lasagna then I ll bring a salad. If you bring a tuna casserole then I ll bring some bread. Therefore, I will either bring a salad or some bread. L = You will bring lasagna. T = You will bring a tuna casserole. S = I will bring a salad. B = I will bring bread. L T, L S, P B - S B This is another fairly common inference form so it s worth proving it in general, as a new derived rule. In order to establish this rule, we ll need to prove the following argument,, -

22 22 Here s one way to construct the proof P 2 2. P 3 3. P want 4 4. ~( ) PP want contradiction 4 5. ~ & ~ 4 DM 4 6. ~ 5 &O 2,4 7. ~ 2, 6 MT 1,2,4 8. 1,7 O 1,2,3,4 9. 3, 9 O ~ 5 &O 1,2,3,4, 11. & ~ 9, 10 &I 1,2, ~O (We can also prove this by first proving P (R S) and Q (R S) and using SD. You can find this proof in the end-notes. 33 ) We have now established the derived rule Complex Dilemma. CD:,, - A Complex Dilemma is a dilemma for the same reason that a Simple Dilemma is a dilemma: it presents us with an unresolved choice between two possibilities. (It s either or.) Unlike a Simple Dilemma, however, a Complex Dilemma presents us with a situation in which the two possibilities carry different outcomes. (If happens, we ll get. But if happens, we ll get.) That s what makes it complex. Complex dilemma is the second of our two three-line rules, and it allows us to prove the second potluck argument very easily. You will bring lasagna or a tuna casserole. If you bring lasagna then I ll bring a salad. If you bring a tuna casserole then I ll bring some bread. Therefore, I will either bring a salad or some bread. L = You will bring lasagna. T = You will bring a tuna casserole. S = I will bring a salad. B = I will bring bread.

23 23 L T, L S, P B - S B 1. L T P 2. L S P 3. P B P - want S B 4. S B 1, 2, 3 CD Before we practice using Complex Dilemma, there s one more thing we should discuss. Suppose that while we re at the potluck, we engage in a conversation with someone who brags about having sold a wreck of a car to a stranger. Yeah, he says, I rolled back the odometer and really charmed the guy. I ended up getting about $7000 more for the car than it was worth. The poor sucker s going to end up putting at least that much into the car to cover his repair bills in the next few months. Say, while we re talking about cars, I can get you a great deal on tires. He s my card if you re interested. After we wander away from this person in search of more pleasant company, you say Can you believe he was trying to sell us tires? If that guy was telling us the truth about how he sold that car then we shouldn t trust him. And if he wasn t telling us the truth about how he sold that car then we shouldn t trust him. Obviously, we shouldn t trust him! In saying this, you are presenting me with the following (very good!) argument: If he was telling us the truth about how he sold that car then we shouldn t trust him. If he wasn t telling us the truth about how he sold that car then we shouldn t trust him. Therefore, we shouldn t trust him. T = He was telling us the truth about how he sold that car. S = We should trust him. T ~S, ~T ~S - ~S We can certainly prove this using as follows: 1 1. T ~S P 2 2. ~T ~S P - want ~S 3 3. S PP want contradiction 1,3 4. ~T 1, 3 MT 2,3 5. T 2, 3 MT 1,2,3 6. T & ~T 4, 5 &I 1,2 7. ~S 3 6 ~I

24 24 But that proof requires us to take a provisional premise and keep track of premise dependence. There s an easier way. Remember the Law of the Excluded Middle, which allows us to put a formula of the type ~ on any line of a proof? We can use that here, like this, to construct a much simpler proof: 1. T ~S P 2. ~T ~S P - want ~S 3. T ~T EXCL-MID 4. ~S 1, 2, 3 SD Now suppose that you and I are wandering around the living room and we hear someone say, If I get the job, I ll be moving to Seattle. If I don t get the job, I ll join the Peace Corps. From this, we can conclude that this person will either move to Seattle or join the Peace Corps because the following argument is valid. If she gets the job then she ll move to Seattle. If she doesn t get the job then she ll join the Peace Corps. Therefore, she ll either move to Seattle or she ll join the Peace Corps. J = She gets the job. S = She ll move to Seattle. P = She ll join the Peace Corps. J S, ~J P - S P We can construct a proof for the argument like this: 1 1. J S P 2 2. ~J P P want S P 3 3. ~(S P) PP want contradiction 3 4. ~S & ~P 3 DM 3 5. ~S 4 &O 1,3 6. ~J 1, 5 MT 3 7. ~P 4 &O 2,3 8. J 2, 7 MT 1,2,3 9. J & ~J 6, 8 &I 1,2 10. S P 3-9 ~O

25 25 However, thanks to the magic of the Law of the Excluded Middle, we can also construct the following, much simpler, proof: 1. J S P 2. ~J P P want S P 3. J ~J EXCL-MID 4. S P 1, 2, 3 CD The moral of these two arguments, of course, is that both Simple and Complex Dilemmas sometimes take the Law of the Excluded Middle as an unstated premise. By supplying this assumption, we can simplify the proof the argument. Practice Prove the following arguments. The answers are in the end notes. 1. One of my favorite quotations from Abraham Lincoln is If this is coffee, please bring me some tea; but if this is tea, please bring me some coffee. This quotation gives rise to the following argument: Either Lincoln is drinking coffee or he s drinking tea. If Lincoln is drinking coffee then he d like to have tea. If Lincoln is drinking tea then he d like to have coffee. Therefore either Lincoln would like to have tea or he d like to have coffee. C = Lincoln is drinking coffee. T = Lincoln is drinking tea. E = Lincoln would like to have tea. O = Lincoln would like to have coffee Either things are right and wrong because God says they are or God says that things are right and wrong because they really are right and wrong. If things are right and wrong because God says they are, then ethics are arbitrary. If God says that things are right and wrong because they really are right and wrong then God is not the source of moral value. Ethics are not arbitrary. Therefore, God is not the source of moral value. S = Things are right and wrong because God says they are. R = God says that things are right and wrong because they really are right and wrong. A = Ethics are arbitrary. G = God is the source of moral value T only if either R or S. R is sufficient for I. E is necessary for S. Therefore, if T then I or E A B, A (C & D), E B - D E Hint: Prove two conditionals, one with A as the antecedent and one with B as the antecedent, and then use CD. 37

26 26 5. A & (B C) (A & B) (A & C) Complete the following proofs by filling in the blanks A & (B C) P want (A & B) (A & C) 1 2. B C 1 &O 3 3. B PP want A 1 &O B. 3-5 I 7 7. C PP want A 1 &O C. 7-9 I 11. (A & B) (A & C) 2, 6, 10 CD 1 1. (A & B) (A & C) P - want A & (B C) 2 2. A & B PP want A & (B C) &O &O I 6.. 3, 5 &I 7. (A & B) (A & (B C)) A & C PP want A & (B C) &O &O I , 11 &I 13. (A & C) (A & (B C)). 14. A & (B C).

27 27 6. A (B & C) (A B) & (A C) Complete the following proofs by filling in the blanks A (B & C) P - want (A B) & (A C) 2.. PP want (A B) & (A C) 3. A B 2 I I 5.. 3, 4 &I 6.. ((A B) & (A C)) 2-5 I 7.. PP want (A B) & (A C) &O I C 7 &O I , 11 &I 13.. ((A B) & (A C)) 7-12 I (A B) & (A C) 1, 6, 13 SD (A B) & (A C) - A (B & C) 1 1. (A B) & (A C) P- want A (B & C) 2 2. ~(A (B & C)) PP want contradiction DM 1 4. A B 1 &O 2 5. ~A 2 &O 1,2 6. B 4,5 O 1 7. A C 1 & O 1,2 8. C. 1,2 9. B & C 6,8 & I ~(B & C) 3 &O 1,2 11. (B & C) & ~(B & C) 9, 10 &I A (B & C). In proving the two previous equivalences, we have established Distribution, which is often presented as a dervied rule. Distribution, or DIST, notes that from a line of the form (( & ( )) we may derive a line of the form (( & ) ( & )) and vice versa. And from a line of the form (( ( & )) you may derive a line of the form (( ) & ( )) and vice versa. However, I have never actually used Distribution in real life, so we won t be taking it as a derived rule here.

28 28 Arrow Now that we have DeMorgan s law, we are able to handle negated conjunctions, like ~(P & Q) and negated disjunctions, like ~(P Q). Negated conditionals, sentences like ~(P Q), still elude us, but they won t elude us for long because every conditional is equivalent to a disjunction and every disjunction is equivalent to a conditional. To establish, we need to prove the following two equivalences: I) II) ~ I ll prove the first equivalence and let you prove the second. You can find my proof in the endnotes. I) 1 1. P - want 2 2. ~( ) PP want contradiction 2 3. & ~ 2 DM &O 1,2 5. 1, 4 O 2 6. ~ 3 &O 1,2 7. & ~ 5, 6 &I ~O 1 1. P want 2 2. PP want 1,2 3. 1, 2 O I II) ~ 40 We have now established the Arrow rule. AR: Opp Opp

29 29 Basically, what AR tells us that we can transform a conditional into a disjunction by keeping the formula on the right the same, changing the to, and transforming the formula on the left to its opposite. Like this: OPP And we can also transform a disjunction into a conditional by keeping the formula on the right the same, changing the to, and transforming the formula on the left to its opposite. Like this: OPP In practice, this means that we can perform the following inferences: 1. P Q P 2. ~P Q 1 AR 1. ~A B P 2. A B 1 AR 1. P ~Q P 2. ~P ~Q 1 AR 1. ~A ~B P 2. A ~B 1 AR 1. ~P Q P 2. P Q 1 AR

30 30 1. A B P 2. ~A B 1 AR And because AR is an equivalence rule, and because equivalence rules can be used on parts of lines, we can also perform the following inferences: Practice 1. (P Q) A P 2. (~P Q) A 1 AR 1. C (~A ~B) P 2. C (A ~B) 1 AR Prove the following arguments. The answers are in the end notes. 1. P Q, Q R - ~P R P Q, Q R - P R P Q, P R, Q S, ~(~R S) (T V) - ~T V P Q, P Q - Q 44 Hint: Use Arrow, the Law of the Excluded Middle, and Simple Dilemma Negated Arrow With the Arrow rule in hand, we can now see how to handle negated conditionals like ~(P Q). Thanks to the fact that the Arrow rule can be used on parts of lines, we know that ~( ) is equivalent to ~(~ ). And thanks to DeMorgan s Law, we know that ~(~ ) is equivalent to & ~. In other words: ~( ) ~(~ ) & ~, and so ~( ) & ~ We can reason in a similar way to show that ~( ~ ) & You can find my explanation of this equivalence in the endnotes. 45 With these equivalences established, we now have the Negated Arrow rule: NEG AR: ~( ) & Opp

31 31 It s interesting to note that this rule makes perfect sense when we remember the truth conditions for. A sentence of the form ~( ) true when and only when is false, and is false when and only when the antecedent is true and the conclusion is false, or & ~. It follows that ~( ) true when and only when & ~. Neat, isn t it? In practice, Negated Arrow means that we can perform the following inferences: 1. ~(P Q) P 2. P & ~Q 1 NEG AR 1. ~(P ~Q) P 2. P & Q 1 NEG AR 1. ~(~P Q) P 2. ~P & ~Q 1 NEG AR 1. ~(~P ~Q) P 2. ~P & Q 1 NEG AR Negated Arrow is a very handy rule because sentences of the form ~( ) are actually relatively common. Remember that A is not sufficient for B is symbolized as ~(A B) and that C is not necessary for D is symbolized as ~(D C). Practice Prove the following arguments. The answers are in the end notes. 1. Being socially intelligent is not sufficient for being academically gifted. Being socially gifted is sufficient for being a good teacher. Being academically gifted is necessary for being a good researcher. Therefore, one can be a good teacher and not be a good researcher. S = Someone is socially intelligent. A = That person is academically gifted. T = That person is a good teacher. R = That person is a good researcher ~(P Q), R Q, R & P, (R & P) S - S ~(~A ~B), (B & ~A) (C D), C ~E, D F - E F ~(L ~M), O ~L, ~P ~M, R (~P O) - ~R 49

32 32 CHAPTER SUMMARY (new material Indicated with * ) UNDERSTANDING COMPLEX SENTENCES and = & & is true when and only when both and are true. If truth trickles down over a connector from the whole sentence to the parts of the sentence, the connector is symbolized as &." * or = is false when and only when both and are false. Or is inclusive by default. If then = is false if and only if can be true and can be false at the same time. Whatever follows if ( should and provided that ) becomes the antecedent. Whatever follows only if becomes the consequent. Whatever is being described as sufficient becomes the antecedent. Whatever is being described as necessary becomes the consequent. if and only if = is true when and only when and have the same truth value. Necessary and sufficient = iff = If and only if Not = ~ ~ is true when is false and ~ is false when is true. Niether nor = ~ & ~ is not sufficient for = ~( ) is not necessary for = ~( ) Unless, =, unless = ~ If the sentence has more than one connector, we identify the main connector by i) Identifying all of the connectors. ii) Seeing what each connector connects. iii) If there is nothing left over, that connector may be the main connector. If there is part of the sentence left over, that connector isn t the main connector. Once we ve identified the main connector, we symbolize around that main connector To see if two sentences say the same thing, symbolize the sentences and see if the symbolizations are the same or equivalent to each other (see the equivalence rules below). If so, then the sentences say the same thing. If not, then they don t.

33 33 ASSESSING ARGUMENTS FOR VALIDITY / INVALIDITY If we can proceed from the premises of an argument to its conclusion by a series of valid inference rules then the argument as a whole is valid. If we can proceed from the premises of an argument to its conclusion only by committing a formal fallacy then the argument as a whole is invalid. Valid Inference Rules & O & - & - & I, - &, - & & COM & & & ASSOC & ( & ) ( & ) & * O, Opp -, Opp - * I - - * COM * ASSOC ( ) ( ) O MT I CH CN, -, Opp - Opp Take as a PP and derive from -. [ depends upon all of the premises upon which depends less the premise that corresponds to.], - Opp Opp O - - I, -, - CH, - COM CN Opp Opp I / O Take Opp as a PP and derive a contradiction from Opp -. [A statement derived by I / O depends on all of the assumptions on which the contradiction depends, less the assumption corresponding to Opp.] DN *SD,, - *CD,, - EXP ( & ) ( ) *DM ( ) Opp & Opp ( & ) Opp Opp *AR Opp Opp *NEG AR ~( ) & Opp

34 34 NON-CON *EXCL-MID ~( & ~ ) ~ Formal Fallacies FAC, - FNC, ~ - ~

35 35 ANSWERS Symbolizing Practice 1 1. Either he ll read the book or he ll watch the movie. R = He ll read the book. W = He ll watch the movie. R W 2 2. Either he ll read the book or he ll watch the movie or both. R = He ll read the book. W = He ll watch the movie. (R W) (R & W) 3 3. Either he ll read the book or he ll watch the movie but not both. R = He ll read the book. W = He ll watch the movie. (R W) & ~(R & W) 4 4. If he reads the book or watches the movie then he won t be left out of the conversation. R = He ll read the book. W = He ll watch the movie. L = He ll be left out of the conversation. (R W) ~L 5 5. He ll be left out of the conversation, unless he reads the book or watches the movie. R = He ll read the book. W = He ll watch the movie. L = He ll be left out of the conversation. ~(R W) L 6 6. Either watching the move is sufficient for being able to discuss the plot intelligently or reading the book is necessary for being able to discuss the plot intelligently. R = Someone reads the book. W = Someone watches the movie. D = That person can discuss the plot intelligently.

36 36 (W D) (D R) 7 7. If he either reads the book or watches the movie then he ll be able to discuss the plot intelligent and he ll impress his friends. R = He reads the book. W = He watches the movie. D = He ll be able to discuss the plot intelligently. I = He ll impress his friends. (R W) (D & I) 8 8. Provided that he s joined the book club, he ll read the book or watch the movie but not both. J = He s joined the book club. R = He reads the book. W = He watches the movie. J ((R W) & ~(R & W)) Primitive Inference Rule: O Practice 9 1. Either I m hallucinating or that cat is dancing. I m not hallucinating. Therefore, that cat is dancing. H = I m hallucinating. C = That cat is dancing. H C, ~H - C 1. H C P 2. ~H P want C 3. C 1,2 O Either there is no gas in the car or the battery needs a charge. There is gas in the car. Therefore, the battery needs a charge. G = There is gas in the car. B = The battery needs a charge. ~G N, G - N 1. ~G N P 2. G P want N

37 37 3. N 1,2 O P & ~Q, R Q - R 1. P & ~Q P 2. R Q P want R 3. ~Q 1 &O 4. R 2, 3 O P Q, ~P ~Q - P (Hint: Do a proof by contradiction.) 1 1. P Q P 2 2. ~P ~Q P want P 3 3. ~P PP want contradiction 1,3 4. Q 1,2 O 2,3 5. ~Q 2,3 O 1,2,3 6. Q & ~Q 4,5 &I 1,2 7. P 3-6 ~O P Q, ~Q ~R, P Q - ~R (Hint: Do a proof by contradiction.) 1 1. P Q P 2 2. ~Q ~R P 3 3. P Q P want ~R 4 4. R PP want contradiction 2,4 5. ~Q 2,4 O 1,2,4 6. P 1,5 O 1,2,3,4 7. Q 3,6 O 1,2,3,4 8. Q & ~Q 5,7 &I 1,2,3 9. ~R 4-6 ~I Primitive Inference Rule: I Practice If either shirts or slacks are on sale then the store will be very busy. Therefore, if slacks are on sale then the store will be very busy. H = Shirts are on sale. L = Slacks are on sale. B = The store will be very busy. (H L) B - L B 1 1. (H L) B P - want L B

38 L PP want B 2 3. H L 2 I 1,2 4. B 1,2 O 1 5. C B 2-4 I E, (A C) (B & D), B ((E F) A) - A B 1 1. E P 2 2. (A C) (B & D) P 3 3. B ((E F) A) P - want A B 4 4. A PP - want B 4 5. A C 4 I 2,4 6. B & D 2,5 O 2,4 7. B 6 &O 2 8. A B 4-7 I 9 9. B PP - want A 3,9 10. (E F) A 3,9 O E F 1 I 1,3,9 12. A 10,11 O 1,3 13. B A 9-12 I 1,2,3 14. A B 8,13 I Derived Rules DeMorgan s Law DeMorgan s Part I: Negated Disjunctions 16 II) ( ~ ) & ( ~ ) - & 1 1. ( ~ ) P want & 2 2. PP want contradiction 2 3. ~ 2 I 1,2 4. ( ~ ) & ~( ~ ) 1,3 &I 1 5. ~ 2-4 ~I 6 6. ~ PP want contradiction 6 7. ~ 6 I 1,6 8. ( ~ ) & ~( ) 1,7 &I ~O ~ & 5, 9 &I & - ( ~ )

39 & P want ( ~ ) 2 2. ~ PP want contradiction 1 3. ~ 1&O 1,2 4. ~ 2, 3 O &O 1,2 6. & ~ 4,5 &I 1 7. ( ~ ) 2-6 ~I 17 III) (~ ) & (~ ) - & 1 1. (~ ) P want & 2 2. ~ PP want contradiction 2 3. ~ 2 I 1,2 4. (~ ) & ~(~ ) 1,3 &I ~I 6 6. PP want contradiction 6 7. ~ 6 I 1,6 8. (~ ) & ~(~ ) 1,7 &I 1 9. ~ 6-9 ~I & ~ 5, 9 &I & ~ - (~ ) 1 1. & P want (~ ) 2 2. ~ PP want contradiction &O 1,2 4. 2, 3 O 1 5. ~ 1 &O 1,2 6. & ~ 4,5 &I 1 7. (~ ) 2-6 ~I 18 IV) (~ ~ ) & (~ ~ ) - & 1 1. (~ ~ ) P want & 2 2. ~ PP want contradiction 2 3. ~ ~ 2 I 1,2 4. (~ ~ ) & ~(~ ~ ) 1,3 &I ~I

40 ~ PP want contradiction 6 7. ~ ~ 6 I 1,6 8. (~ ~ ) & ~(~ ~ ) 1,7 &I ~I & 5, 9 &I & - (~ ~ ) 1 1. & P want (~ ~ ) 2 2. ~ ~ PP want contradiction &O 1,2 4. ~ 2, 3 O &O 1,2 6. & ~ 4,5 &I 1 7. (~ ~ ) 2-6 ~I DeMorgan s Part II: Negated Conjunctions 19 II) ( & ~ ) ~( & ~ ) - ~ 1 1.~( & ~ ) P want ~ 2 2. ~ (~ ) PP want contradiction 2 3. & ~ 2 DM 1,2 4. ( & ~ ) & ~( & ~ ) 1, 3 &I 1 5. ~ 2-4 ~O ~ - ~( & ~ ) 1 1. ~ P want ~( & ~ ) 2 2. & ~ PP want contradiction &O 1,2 4. 1, 4 O 2 5. ~ 2 &O 1,2 6. & ~ 4,5 &I 1 7. ~( & ~ ) 2-6 ~I 20 III) (~ & ) ~(~ & ) - ~ 1 1.~(~ & ) P want ~

41 ~ ( ~ ) PP want contradiction 2 3. ~ & 2 DM 1,2 4. (~ & ) & ~(~ & ) 1, 3 &I 1 5. ~ 2-4 ~O ~ - ~(~ & ) 1 1. ~ P want ~(~ & ) 2 2. ~ & PP want contradiction 2 3. ~ 2 &O 1,2 4. ~ 1, 4 O &O 1,2 6. & ~ 4,5 &I 1 7. ~(~ & ) 2-6 ~I 21 IV) (~ & ~ ) Practice ~(~ & ~ ) ~(~ & ~ ) P want 2 2. ~ ( ) PP want contradiction 2 3. ~ & ~ 2 DM 1,2 4. (~ & ~ ) & ~(~ & ~ ) 1, 3 &I ~O - ~(~ & ~ ) 1 1. P want ~(~ & ~ ) 2 2. ~ & ~ PP want contradiction 2 3. ~ 2 &O 1,2 4. 1, 4 O 2 5. ~ 2 &O 1,2 6. & ~ 4,5 &I 1 7. ~(~ & ~ ) 2-6 ~I What do you get when you use DeMorgan s Law on the following formulas? a. ~(A & B) = ~A ~B b. ~(A B) = ~A & ~B c. ~(~P & Q) = P ~Q d. ~(P ~Q) = ~P & Q What do you get when you use DeMorgan s Law on the following formulas?

42 42 a. ~A & ~B = ~(A B) b. ~A ~B) = ~(A & B) c. ~P & Q) = ~(P ~Q) d. P ~Q = ~(~P & Q) (P & Q) R, R - P Q 1. (P & Q) R P 2. R P want P Q 3. ~(P & Q) 1, 2 MT 4. ~P ~Q 3 DM P (A B), A & B - P 1. P (A B) P 2. A & B P - want P 3. ~(A B) 2 DM 4. ~P 1, 3 MT (P Q) (A & B), ~A ~B - ~P & ~Q 1. (P Q) (A & B) P 2. ~A ~B P want ~P & ~Q 3. ~(A & B) 2 DM 4. ~(P Q) 1, 3 MT 5. ~P & ~Q 4 DM (P & Q) R, R S, S, Q - P 1. (P & Q) R P 2. R S P 3. S P 4. Q P want P 5. ~R 2, 3 O 6. ~(P & Q) 1, 5 MT 7. ~P ~Q 6 DM 8. ~P 4, 7 O Or Commutativity 28 II) - 1. P want 2. ~(~ & ~ ) 1 DM

43 43 3. ~(~ & ~ ) 2 & Com 4. 3 DM Or Associativity 29 II) ( ) - ( ) 1. ( ) P - want ( ) 2. ~[~ & ~( )] 1 DM 3. ~[~ & (~ & ~ )] 2 DM 4. ~[(~ & ~ ) & ~ ] 3 & ASSOC 5. ~(~ & ~ ) ] 4 DM 6. ( ) 5 DM Simple Dilemma Practice If our character is determined by our genetic make-up then we aren t responsible for what we do. If our character is determined by our early-childhood conditioning then we aren t responsible for what we do. Our character is either determined by our genetic make-up or by our early-childhood conditioning. Therefore, we aren t responsible for what we do. G = Our character is determined by our genetic make-up. R = We are responsible for what we do. C = Our character is determined by our early-childhood conditioning, G ~R, C ~R, G C - ~R 1. G ~R P 2. C ~R P 3. G C P want ~R 4. ~R 1, 2, 3 SD I have certain sense impressions. If I have sense impressions, they re either veridical or hallucinatory. If my sense impressions are veridical, then I must exist (because I m actually experiencing things external to myself). If my sense experiences are hallucinatory, then I must exist (because I m hallucinating). However, something exists, if I exist. Therefore, I can be certain that something exists. S = I have certain sense impressions. V = My sense impressions are veridical. H = My sense impressions are hallucinatory. I = I exist. E = Something exists.

44 44 S, S (V H), V I, H I, I E - E 1. S P 2. S (V H) P 3. V I P 4. H I P 5. I E P - want E 6. V H 1, 2 O 7. I 3, 4, 6 SD 8. E 5, 6 O A B, P & Q, (A & P) (Q C), B (C D) - C D 1. A B P 2. P & Q P 3. (A & P) (Q C) P 4. B (C D) P want C D, want A (C D) 5. A PP want C D 6. P 2 &O 7. A & P 5,6 O 8. Q C 3,7 O 9. Q 2 &O 10. C 8,9 O 11. C D 10 I 12. A (C D) 5-11 I 13. C D 1, 4, 12 SD Complex Dilemma 33,, P 2 2. P 3 3. P - want 4 4. PP want 2,4 5. 2, 4 O 2, I 2 7. ( ) 4 6 I 8 8. PP want 3,8 9. 3, 8 O 3, I ( ) 8-10 I

45 45 Practice 1,2, , 7, 11 SD One of my favorite quotations from Abraham Lincoln is If this is coffee, please bring me some tea; but if this is tea, please bring me some coffee. This quotation gives rise to the following argument: Either Lincoln is drinking coffee or he s drinking tea. If Lincoln is drinking coffee then he d like to have tea. If Lincoln is drinking tea then he d like to have coffee. Therefore either Lincoln would like to have tea or he d like to have coffee. C = Lincoln is drinking coffee. T = Lincoln is drinking tea. E = Lincoln would like to have tea. O = Lincoln would like to have coffee. C T, C E, T O - E O 1. C T P 2. C E P 3. T O P want E O 4. E O 1, 2, 3 CD Either things are right and wrong because God says they are or God says that things are right and wrong because they really are right and wrong. If things are right and wrong because God says they are, then ethics are arbitrary. If God says that things are right and wrong because they really are right and wrong then God is not the source of moral value. Ethics are not arbitrary. Therefore, God is not the source of moral value. S = Things are right and wrong because God says they are. R = God says that things are right and wrong because they really are right and wrong. A = Ethics are arbitrary. G = God is the source of moral value. S R, S A, R ~G, ~A - ~G 1. S R P 2. S A P 3. R ~G P 4. ~A P - want ~G 5. A ~G 1, 2, 3 CD 6. ~G 4, 5 O T only if either R or S. R is sufficient for I. E is necessary for S. Therefore, if T then I or E.