Derivations, part 2. Let s dive in to some derivations that require the use of the last four rules:

Transcription

1 Derivations, part 2 Let s dive in to some derivations that require the use of the last four rules: 1. I Derivations: Let s start with some derivations that use conditional-introduction. (a) Here s an easy one: Imagine that your friend says, If you take that job, then you ll move away. And if you move away, then I ll be sad. So, if you take that job, then I ll be sad. This is known as a hypothetical syllogism, and your textbook introduces it as the following sequent: S89: P Q, Q R Ⱶ P R Somehow, we ll need to get from (1) and (2) to (n) in the following derivation: 2 (2) Q R A 1, 2 (n) P R? We ll want to introduce the conditional on line (n), so we ll need to assume it s antecedent P. It turns out that that s all we need. Here s the derivation: 2 (2) Q R A 3 (3) P Ass. ( I) 1, 3 (4) Q 1, 3, E 1, 2, 3 (5) R 2, 4, E 1, 2 (6) P R 3, 5, I We were able to derive R from our assumption P, so we can just introduce an arrow to get P R, which discharges our assumption P. Let s do a slightly harder one: (b) Imagine that, due to another love triangle, someone says, If Paul shows up to the party, then Quinn will be really upset, but Rebecca will be thrilled. Therefore, if Paul shows up to the party, then Quinn will be really upset, and, if Paul shows up to the party, then Rebecca will be thrilled. That makes sense. Here s the sequent to be proved: S19: P (Q R) Ⱶ (P Q) (P R) Let s begin: 1

2 1 (n) (P Q) (P R)? If we assume P for the sake of later doing a I, then we can get both of the conditionals in the conclusion. Like this: 2 (2) P Ass. ( I) 1, 2 (3) Q R 1, 2, E 1, 2 (4) Q 3, E 1, 2 (5) R 3, E 1 (6) P Q 2, 4, I 1 (7) P R 2, 5, I 1 (8) (P Q) (P R) 6, 7, I Neither (6) nor (7) relies on our assumption P, since they packed P into a conditional. So, the assumption P has been discharged. Then, we simply combine the two conditionals in the last line using I to get the desired conclusion. (c) Here s another one: If Peter and Quentin both bring presents, then Rachel will have a great birthday. Therefore, if Peter brings presents, then, if Quentin brings presents, then Rachel will have a great birthday. Or: S20: (P Q) R Ⱶ P (Q R) Let s begin: 1 (1) (P Q) R A 1 (n) P (Q R)? We ll have to assume TWO sentences, P AND Q, and discharge them one at a time: 1 (1) (P Q) R A 2 (2) P Ass. ( I) #1 * 3 (3) Q Ass. ( I) #2 2, 3 (4) P Q 2, 3, I 1, 2, 3 (5) R 1, 4, E 1, 2 (6) Q R 3, 5, I (#2) ** 1 (7) P (Q R) 2, 6, I (#1) 2 * Note: To help keep track of my assumptions, I like to number them whenever there are several assumptions of the same type. ** (and I also like to note which assumption was discharged as I discharge them)

3 Here, we ve combined the two assumptions into a conjunction in order to get the antecedent of the conditional in (1). Then, once we obtained R, we discharge the assumption Q first by deriving the conditional Q R using I. Finally, we discharge the assumption P by deriving the conditional in the conclusion using another I. 2. I Derivations: Now, let s do one that involves introducing bi-conditionals. S36: P Q, Q R Ⱶ P R For instance, someone might say, Paul will go to the party if and only if Quinn goes to the party. Also, Quinn will go to the party if and only if Rebecca invites him. Therefore, Paul will go to the party if and only if Rebecca invites Quinn. We ll start: 2 (2) Q R A 1, 2 (n) P R? Let s start by breaking up the two bi-conditionals, and the breaking up the conjunctions that result (that is, let s perform E and then E ). 2 (2) Q R A 1 (3) (P Q) (Q P) 1, E 1 (4) P Q 3, E 1 (5) Q P 3, E 2 (6) (Q R) (R Q) 2, E 2 (7) Q R 6, E 2 (8) R Q 6, E 1, 2 (n) P R? Now for the tricky part. To get to the conclusion, P R, we re going to need to derive the conjunction, (P R) (R P). So, we re going to need to derive each of those two conditionals separately. We derive each one by assuming their antecedents. In other words, we ll assume P to obtain P R, and we ll assume R to get R P. Like this: 3

4 2 (2) Q R A 1 (3) (P Q) (Q P) 1, E 1 (4) P Q 3, E 1 (5) Q P 3, E 2 (6) (Q R) (R Q) 2, E 2 (7) Q R 6, E 2 (8) R Q 6, E 9 (9) P Ass. ( I) #1 1, 9 (10) Q 4, 9, E 1, 2, 9 (11) R 7, 11, E 1, 2 (12) P R 9, 11, I (#1) 13 (13) R Ass. ( I) #2 2, 13 (14) Q 8, 13, E 1, 2, 13 (15) P 5, 15, E 1, 2 (16) R P 13, 15, I (#2) 1, 2 (17) (P R) (R P) 12, 16, I 1, 2 (18) P R 17, I Whew! Take a look. We introduced P in order to get P R by using I. THEN we introduced R in order to get R P by using I again. THEN we combined the two conditionals into a conjunction of conditionals so that we could introduce the. 3. E Derivations: On to disjunction-elimination. Either Paul won t fail to show up, or Quinn won t fail to show up. Therefore, either Paul or Quinn will show up. Here s the sequent: S42: P Q Ⱶ P Q So, we have: 1 (1) P Q A 1 (n) P Q? This one will use disjunction-elimination. Remember that, in order to eliminate a disjunction, you ll need to show that, if the first disjunct WERE true, then the conclusion would follow, and also that, if the SECOND disjunct WERE true, then the conclusion would ALSO follow. So, let s assume each of the two disjuncts one at a time: 4

5 1 (1) P Q A 2 (2) P Ass. ( I) 2 (3) P 2, E 2 (4) P Q 3, I - (5) P (P Q) 2, 4, I 1 (n) P Q? Here, we ve assumed the first disjunct of (1), P, and shown by I and I that, if true, P would entail the disjunction P Q. There are no numbers to the left of (5) because we DISCHARGED the assumption that it relies on. Let s do the same for the other disjunct now: 1 (1) P Q A 2 (2) P Ass. ( I) #1 2 (3) P 2, E 2 (4) P Q 3, I - (5) P (P Q) 2, 4, I (#1) 6 (6) Q Ass. ( I) #2 6 (7) Q 6, E 6 (8) P Q 7, I - (9) Q (P Q) 6, 8, I (#2) 1 (10) P Q 1, 5, 9, E Note how we have performed exactly the same operations for the first disjunct and the second disjunct of line (1) in order to perform E. Since we know from line (5) that P entails P Q, and also from line (9) that Q ALSO entails P Q, and we ALSO know that either P or Q is true from our premise in line (1), we can infer on the final line that P Q must be true. Note that this is a little confusing because we didn t really eliminate the at the end. What the disjunction-elimination rule says is that if both disjuncts lead to SOMETHING (Δ), then we can just infer that that something (Δ) must be true. But, Δ might itself be a disjunction as is the case here in which case the symbol remains. Here s another: Peggy is bringing wine, and either Quinn or Rob is bringing beer. Therefore, either Peggy is bringing wine and Quinn is bringing beer, or Peggy is bringing wine and Rob is bringing beer. Here s the sequent to be proved: S43: P (Q R) Ⱶ (P Q) (P R) Or, in derivation form: 5

6 1 (n) (P Q) (P R)? Let s start by breaking down the conjunction in (1). 1 (2) P 1, E 1 (3) Q R 1, E 1 (n) (P Q) (P R)? Now what? Well, we have a disjunction on line (3), and what we need to do is show that, EITHER WAY, the conclusion (n) would be true. That is, whether Q is true, or R is true, it will follow that (P Q) (P R) is true. So, we re going to need to introduce some conditionals with I. Here s how: 1 (2) P 1, E 1 (3) Q R 1, E 4 (4) Q Ass. ( I) 1, 4 (5) P Q 2, 4, I 1 (n) (P Q) (P R)? So far, we ve shown that, if Q were true, then we could derive P Q. But, what we really want to get is the entire conclusion. How do we get the second disjunct ( P R )? That s easy, we can just tack it on using I. Remember, if we have ONE disjunct, we can add WHATEVER WE D LIKE as a second disjunct. (Recall that, I m either going to the bars tonight OR I m staying home to watch paint dry is still true, even if you know for sure that you re going to the bars.) Like this: 1 (2) P 1, E 1 (3) Q R 1, E 4 (4) Q Ass. ( I) 1, 4 (5) P Q 2, 4, I 1, 4 (6) (P Q) (P R) 5, I 1 (7) Q [(P Q) (P R)] 4, 6, I 6

7 1 (n) (P Q) (P R)? So, on line (7) we were able to discharge our assumption Q using I. Let s do the same for the other disjunct of line (3), R : 1 (2) P 1, E 1 (3) Q R 1, E 4 (4) Q Ass. ( I) #1 1, 4 (5) P Q 2, 4, I 1, 4 (6) (P Q) (P R) 5, I 1 (7) Q [(P Q) (P R)] 4, 6, I (#1) 8 (8) R Ass. ( I) #2 1, 8 (9) P R 2, 8, I 1, 8 (10) (P Q) (P R) 9, I 1 (11) R [(P Q) (P R)] 8, 10, I (#2) 1 (12) (P Q) (P R) 3, 7, 11, E Note how we have performed exactly the same operations for the first disjunct and the second disjunct of line (3) in order to perform E. We can infer the conclusion on line (12) because we have shown that, no matter which disjunct of the disjunction in line (3) is true, they both lead to the conclusion on line (12). 4. I Derivations: And finally, negation-introduction. Let s start with an easy one: Peggy will pass. So, it is not the case that Peggy will not pass. The sequent is: S61: P Ⱶ P Here s the derivation: 1 (1) P A 2 (2) P Ass. (Red.) 1, 2 (3) P P 1, 2, I 1 (4) P 2, 3, I In line (2), we introduced P for the sake of performing a reductio. As we can see, lines (1) and (2) combined form a syntactic contradiction in line (3). So, we can conclude that the negation of our assumption is true, using I in line (4). They get harder. Try this one: It is not the case that both Peggy and Quinn will fail. Therefore, either Peggy or Quinn will pass. The sequent is as follows: 7

8 S62: ( P Q) Ⱶ P Q In derivation form: 1 (1) ( P Q) A 1 (n) P Q? How do we begin? It doesn t look like we can do much to (1), so perhaps the best strategy is to assume the NEGATION of the conclusion and try to derive a contradiction. If we can do that, then we ll have proved the conclusion. Let s begin: 1 (1) ( P Q) A 2 (2) (P Q) Ass. (Red.)? (n-2) ( P Q) ( P Q)?? (n-1) (P Q)? 1 (n) P Q? But, how do we derive the contradiction? In other words, how do we get the first conjunct in line (n-2) that contradicts the second conjunct which is the premise from line (1)? Well, as it turns out, we re going to have to make TWO more assumptions for TWO more reductios. Let s assume P AND Q. 1 (1) ( P Q) A 2 (2) (P Q) Ass. (Red.) #1 3 (3) P Ass. (Red.) #2 3 (4) P Q 3, I 2, 3 (5) (P Q) (P Q) 2, 4, I 2 (6) P 3, 5, I 7 (7) Q Ass. (Red.) #3 7 (8) P Q 7, I 2, 7 (9) (P Q) (P Q) 2, 8, I 2 (10) Q 7, 9, I 2 (11) P Q 6, 10, I 1, 2 (12) ( P Q) ( P Q) 1, 11, I 1 (13) (P Q) 2, 12, I 1 (14) P Q 13, E The red lines denote our primary reductio proof, while the green and blue lines denote our secondary reductio proofs (to prove P and Q, respectively). 8

9 Here s the same sequent in the reverse order: S63: P Q Ⱶ ( P Q) 1 (n) ( P Q)? Since we re trying to get from a disjunction to a conclusion, it is natural to suspect that we will try to get from P (the first disjunct of the premise) to the conclusion, and also from Q (the second disjunct of the premise) to the conclusion. But, assuming P and assuming Q clearly won t be enough. We re going to ALSO have to assume the opposite of the conclusion for the purposes of a reductio TWICE. Here it is for P : 2 (2) P Ass. ( I) 3 (3) P Q Ass. (Red) 3 (4) P 3, E 2, 3 (5) P P 2, 4, I 2 (6) ( P Q) 3, 5, I - (7) P [ ( P Q)] 2, 6, I 1 (n) ( P Q)? Here, we ve assumed the first disjunct of (1) in order to show that it entails the conclusion. But, to do that, we had to assume the OPPOSITE of the conclusion in order to derive a contradiction. Follow these same steps for the second disjunct, like this: 2 (2) P Ass. ( I) 3 (3) P Q Ass. (Red) 3 (4) P 3, E 2, 3 (5) P P 2, 4, I 2 (6) ( P Q) 3, 5, I - (7) P [ ( P Q)] 2, 6, I 8 (8) Q Ass. ( I) 9 (9) P Q Ass. (Red) * 9 (10) Q 9, E 8, 9 (11) Q Q 8, 10, I 8 (12) ( P Q) 9, 11, I - (13) Q [ ( P Q)] 8, 12, I 1 (14) ( P Q) 1, 7, 13, E * Note that you do not need to assume this AGAIN. You could in fact use the assumption from line (3). If you do, this derivation will be one line shorter. (I chose to assume it again here for educational purposes, since this is slightly less confusing.) 9

10 Last one: S64: P Q Ⱶ Q P Derivation: 1 (n) Q P? Well, the obvious thing to try first is to break down (1) into two separate conditionals. And keep in mind that what we probably WANT is a conjunction of two conditionals that will give us the conclusion: 1 (2) (P Q) (Q P) 1, E 1 (3) P Q 2, E 1 (4) Q P 2, E 1 (n-1) ( Q P) ( P Q)? 1 (n) Q P? Now, if we can somehow show that Q entails P, then we ll be able to get the first conjunct of line (n-1); and vice versa for the second conjunct. To do that, we ll have to assume Q and hope that we can get it to entail P but to do THAT, we ll have to assume P for the purpose of a reductio. Like this: 1 (2) (P Q) (Q P) 1, E 1 (3) P Q 2, E 1 (4) Q P 2, E 5 (5) Q Ass. ( I) 5 (6) Q 5, E 1, 5 (7) P 4, 6, E 8 (8) P Ass. (Red) 1, 5, 8 (9) P P 7, 8, I 1, 5 (10) P 8, 9, I 1 (11) Q P 5, 10, I 1 (n-1) ( Q P) ( P Q)? 1 (n) Q P? 10

11 Now, just to the same thing for P. Like this: 1 (2) (P Q) (Q P) 1, E 1 (3) P Q 2, E 1 (4) Q P 2, E 5 (5) Q Ass. ( I) 5 (6) Q 5, E 1, 5 (7) P 4, 6, E 8 (8) P Ass. (Red) 1, 5, 8 (9) P P 7, 8, I 1, 5 (10) P 8, 9, I 1 (11) Q P 5, 10, I 12 (12) P Ass. ( I) 12 (13) P 12, E 1, 12 (14) Q 3, 13, E 15 (15) Q Ass. (Red) 1, 12, 15 (16) Q Q 14, 14, I 1, 12 (17) Q 15, 16, I 1 (18) P Q 12, 17, I 1 (19) ( Q P) ( P Q) 11, 18, I 1 (20) Q P 19, I 11