forallx Saint Louis University Solutions Booklet P.D. Magnus University at Albany, State University of New York

Transcription

1 forallx Saint Louis University Solutions Booklet P.D. Magnus University at Albany, State University of New York Tim Button University of Cambridge Kathryn Lindeman Saint Louis University

2 This booklet contains model answers to the practice exercises found in forallx slu. For several of the questions, there are multiple correct possible answers; in each case, this booklet contains at most one answer. Answers are given in blue; please contact Kathryn Lindeman at com/forallx.html if you have accessibility requirements. c by P.D. Magnus, Tim Button, and Kathryn Lindeman. Some rights reserved. This solutions booklet is based upon P.D. Magnus s forallx (version 1.29), available at fecundity.com/logic, which was released under a Creative Commons license (Attribution-ShareAlike 3.0). Tim Button s most recent version is available at You are free to copy this book, to distribute it, to display it, and to make derivative works, under the following conditions: (a) Attribution. You must give the original author credit. (b) Share Alike. If you alter, transform, or build upon this work, you may distribute the resulting work only under a license identical to this one. For any reuse or distribution, you must make clear to others the license terms of this work. Any of these conditions can be waived if you get permission from the copyright holder. Your fair use and other rights are in no way affected by the above. This is a human-readable summary of the full license, which is available on-line at In accordance with this license, Kathryn Lindeman has made minor changes to the Cambridge version Tim Button made using P.D. Magnus s original text. She offers forallx :SLU under the same Creative Commons license. This copy of forallx :SLU is current as of November 10, The most recent version and associated.tex files are available at klindeman.com/forallx.html. Typesetting was carried out entirely in L A TEX2ε. The style for typesetting proofs is based on fitch.sty (v0.4) by Peter Selinger, University of Ottawa.

3 Contents 1 Arguments 1 2 Valid arguments 2 3 Other logical notions 4 5 Connectives 5 6 Sentences of TFL Complete truth tables Semantic concepts Truth table shortcuts Partial truth tables Sentences with one quantifier Multiple generality Identity Definite descriptions Sentences of FOL Truth in FOL Using Interpretations Basic rules for TFL Additional rules for TFL Proof-theoretic concepts Derived rules Basic rules for FOL Conversion of quantifiers Rules for identity Derived rules 97 iii

4 Arguments 1 A. Highlight the phrase which expresses the conclusion of each of these arguments: 1. It is sunny. So I should take my sunglasses. 2. It must have been sunny. I did wear my sunglasses, after all. 3. No one but you has had their hands in the cookie-jar. And the scene of the crime is littered with cookie-crumbs. You re the culprit! 4. Miss Scarlett and Professor Plum were in the study at the time of the murder. And Reverend Green had the candlestick in the ballroom, and we know that there is no blood on his hands. Hence Colonel Mustard did it in the kitchen with the lead-piping. Recall, after all, that the gun had not been fired. 1

5 Valid arguments 2 B. Which of the following arguments is valid? Which is invalid? 1. Socrates is a man. 2. All men are carrots. So: Therefore, Socrates is a carrot. 1. Abe Lincoln was either born in Illinois or he was once president. 2. Abe Lincoln was never president. So: Abe Lincoln was born in Illinois. Valid Valid 1. Langston Hughes was either born in Illinois or he was a part of the Harlem Renaissance. 2. Langston Hughes was not a part of the Harlem Renaissance. So: Langston Hughes was born in Illinois. Valid 1. If I pull the trigger, Abe Lincoln will die. 2. I do not pull the trigger. So: Abe Lincoln will not die. Invalid Abe Lincoln might die for some other reason: someone else might pull the trigger; he might die of old age. 1. Maya Angelou was either from St. Louis or from Chicago. 2. Maya Angelou was not from Chicago. So: Maya Angelou was from St. Louis. Valid 1. If the world were to end today, then I would not need to get up tomorrow morning. 2. I will need to get up tomorrow morning. So: The world will not end today. Valid 1. Eliza is now 12 years old. 2. Eliza is now 3 years old. So: Oliver is now 20 years old. Valid An argument is valid if and only if it is impossible for all the premises to be true and the conclusion false. It is impossible for all the premises to be true; so it is certainly impossible that the premises are all true and the conclusion is false. 2

6 2. Valid arguments 3 C. Could there be: 1. A valid argument that has one false premise and one true premise? Yes. Example: the first argument, above. 2. A valid argument that has only false premises? Yes. Example: Socrates is a frog, all frogs are excellent pianists, therefore Socrates is an excellent pianist. 3. A valid argument with only false premises and a false conclusion? Yes. The same example will suffice. 4. A sound argument with a false conclusion? No. By definition, a sound argument has true premises. And a valid argument is one where it is impossible for the premises to be true and the conclusion false. So the conclusion of a sound argument is certainly true. 5. An invalid argument that can be made valid by the addition of a new premise? Yes. Plenty of examples, but let me offer a more general observation. We can always make an invalid argument valid, by adding a contradiction into the premises. For an argument is valid if and only if it is impossible for all the premises to be true and the conclusion false. If the premises are contradictory, then it is impossible for them all to be true (and the conclusion false). 6. A valid argument that can be made invalid by the addition of a new premise? No. An argument is valid if and only if it is impossible for all the premises to be true and the conclusion false. Adding another premise will only make it harder for the premises all to be true together. In each case: if so, give an example; if not, explain why not.

7 Other logical notions 3 A. For each of the following: Is it necessarily true, necessarily false, or contingent? 1. Willie Nelson recorded On The Road Again. Contingent 2. Someone once recorded On The Road Again. Contingent 3. No one has ever recorded On The Road Again. Contingent 4. If Willie Nelson recorded On The Road Again, then someone has. Necessarily true 5. Even though Willie Nelson recorded On The Road Again, no one has ever recorded On The Road Again. Necessarily false 6. If anyone has ever recorded On The Road Again, it was Willie Nelson. Contingent B. Look back at the sentences G1 G4 in this section (about giraffes, gorillas and martians in the wild animal park), and consider each of the following: 1. G2, G3, and G4 Jointly consistent 2. G1, G3, and G4 Jointly inconsistent 3. G1, G2, and G4 Jointly consistent 4. G1, G2, and G3 Jointly consistent Which are jointly consistent? Which are jointly inconsistent? C. Could there be: 1. A valid argument, the conclusion of which is necessarily false? Yes: = 3. So = An invalid argument, the conclusion of which is necessarily true? No. If the conclusion is necessarily true, then there is no way to make it false, and hence no way to make it false whilst making all the premises true. 3. Jointly consistent sentences, one of which is necessarily false? No. If a sentence is necessarily false, there is no way to make it true, let alone it along with all the other sentences. 4. Jointly inconsistent sentences, one of which is necessarily true? Yes = 4 and = 2. In each case: if so, give an example; if not, explain why not. 4

8 Connectives 5 A. Using the symbolization key given, symbolize each English sentence in TFL. M: Those creatures are men in suits. C: Those creatures are chimpanzees. G: Those creatures are gorillas. 1. Those creatures are not men in suits. M 2. Those creatures are men in suits, or they are not. (M M) 3. Those creatures are either gorillas or chimpanzees. (G C) 4. Those creatures are neither gorillas nor chimpanzees. (C G) 5. If those creatures are chimpanzees, then they are neither gorillas nor men in suits. (C (G M)) 6. Unless those creatures are men in suits, they are either chimpanzees or they are gorillas. (M (C G)) B. Using the symbolization key given, symbolize each English sentence in TFL. A: Mister Ace was murdered. B: The butler did it. C: The cook did it. D: The Duchess is lying. E: Mister Edge was murdered. F : The murder weapon was a frying pan. 1. Either Mister Ace or Mister Edge was murdered. (A E) 2. If Mister Ace was murdered, then the cook did it. (A C) 3. If Mister Edge was murdered, then the cook did not do it. (E C) 5

9 5. Connectives 6 4. Either the butler did it, or the Duchess is lying. (B D) 5. The cook did it only if the Duchess is lying. (C D) 6. If the murder weapon was a frying pan, then the culprit must have been the cook. (F C) 7. If the murder weapon was not a frying pan, then the culprit was either the cook or the butler. ( F (C B)) 8. Mister Ace was murdered if and only if Mister Edge was not murdered. (A E) 9. The Duchess is lying, unless it was Mister Edge who was murdered. (D E) 10. If Mister Ace was murdered, he was done in with a frying pan. (A F ) 11. Since the cook did it, the butler did not. (C B) 12. Of course the Duchess is lying! D C. Using the symbolization key given, symbolize each English sentence in TFL. T 1 : Miles is a trumpeter. T 2 : John is a trumpeter. S 1 : Miles is a saxophonist. S 2 : John is a saxophonist. C 1 : Miles has a satisfying career. C 2 : John has a satisfying career. 1. Miles and John are both trumpeters. (T 1 T 2 ) 2. If Miles is a saxophonist, then he has a satisfying career. (S 1 C 1 ) 3. Miles is a saxophonist, unless he is an trumpeter. (S 1 T 1 ) 4. John is an unsatisfied trumpeter. (T 2 C 2 ) 5. Neither Miles nor John is an trumpeter. (T 1 T 2 ) 6. Both Miles and John are trumpeters, but neither of them find it satisfying. ((T 1 T 2 ) (C 1 C 2 )) 7. John is satisfied only if he is a saxophonist. (C 2 S 2 ) 8. If Miles is not an trumpeter, then neither is John, but if Miles is, then John is too.

10 5. Connectives 7 (( T 1 T 2 ) (T 1 T 2 )) 9. Miles is satisfied with his career if and only if John is not satisfied with his. (C 1 C 2 ) 10. If John is both an trumpeter and a saxophonist, then he must be satisfied with his career. ((T 2 S 2 ) C 2 ) 11. It cannot be that John is both an trumpeter and a saxophonist. (T 2 S 2 ) 12. John and Miles are both saxophonists if and only if neither of them is an trumpeter. ((S 2 S 1 ) (T 2 T 1 )) D. Give a symbolization key and symbolize the following English sentences in TFL. E: Elizabeth is a spy. P : Philip is a spy. C: The code has been broken. R: The Russian embassy will be in an uproar. 1. Elizabeth and Phillip are both spies. (E P ) 2. If either Elizabeth or Phillip is a spy, then the code has been broken. ((E P ) C) 3. If neither Elizabeth nor Phillip is a spy, then the code remains unbroken. ( (E P ) C) 4. The Russian embassy will be in an uproar, unless someone has broken the code. (R C) 5. Either the code has been broken or it has not, but the Russian embassy will be in an uproar regardless. ((C C) R) 6. Either Elizabeth or Phillip is a spy, but not both. ((E P ) (E P )) E. Give a symbolization key and symbolize the following English sentences in TFL. W : There is work to be done by the Free French Resistance. P : Josephine Baker will perform. V : Josephine Baker is supplied with a visa. C: Charles de Gaulle will lead the Free French Resistance. G: Germany continues to occupy France.

11 5. Connectives 8 1. If there is work to be done by the Free French Resistance, then Josephine Baker will perform. (W P ) 2. Josephine Baker will not perform unless she is supplied with a visa. ( P V ) 3. Josephine Baker will either perform or she won t, but there is work to be done by the Free French Resistance regardless. ((P P ) W ) 4. Charles de Gaulle will lead the Free French Resistance if and only if Germany continues to occupy France. (C G) 5. If Charles de Gaulle will lead the Free French Resistance, Josephine Baker will be supplied with a visa. (C V ) F. For each argument, write a symbolization key and symbolize all of the sentences of the argument in TFL. 1. If Etta plays the piano in the morning, then Artis wakes up cranky. Etta plays piano in the morning unless she is distracted. So if Artis does not wake up cranky, then Etta must be distracted. P : Etta plays the Piano in the morning. C: Artis wakes up cranky. D: Etta is distracted. (P C), (P D), ( C D) 2. It will either rain or snow on Tuesday. If it rains, Dorothy will be sad. If it snows, Dorothy will be cold. Therefore, Dorothy will either be sad or cold on Tuesday. T 1 : It rains on Tuesday T 2 : It snows on Tuesday S: Dorothy is sad on Tuesday C: Dorothy is cold on Tuesday (T 1 T 2 ), (T 1 S), (T 2 C), (S C) 3. If Langston takes Zora out on the town, then things are thrilling, but not predictable. If he doesn t, then things are predictable, but not exciting. Therefore, things are either exciting or predictable; but not both. Z: Langston takes Zora out on the town. T : Things are thrilling. P : Things are predictable. (Z (T P )), ( Z (P T )), ((P T ) (P T )). G. We symbolized an exclusive or using,, and. How could you symbolize an exclusive or using only two connectives Is there any way to symbolize an exclusive or using only one connective? For two connectives, we could offer any of the following:

12 5. Connectives 9 (A B) ( A B) ( ( A B) (A B)) But if we wanted to symbolize it using only one connective, we would have to introduce a new primitive connective.

13 Sentences of TFL 6 H. For each of the following: (a) Is it a sentence of TFL, strictly speaking? (b) Is it a sentence of TFL, allowing for our relaxed bracketing conventions? 1. (A) (a) no (b) no 2. J 374 J 374 (a) no (b) yes 3. F (a) yes (b) yes 4. S (a) no (b) no 5. (G G) (a) yes (b) yes 6. (A (A F )) (D E) (a) no (b) yes 7. [(Z S) W ] [J X] (a) no (b) yes 8. (F D J) (C D) (a) no (b) no I. Are there any sentences of TFL that contain no atomic sentences? Explain your answer. No. Atomic sentences contain atomic sentences (trivially). And every more complicated sentence is built up out of less complicated sentences, that were in turn built out of less complicated sentences,..., that were ultimately built out of atomic sentences. J. What is the scope of each connective in the sentence [ (H I) (I H) ] (J K) The scope of the left-most instance of is (H I). The scope of the right-most instance of is (I H). The scope of the left-most instance of is [ (H I) (I H) ] The scope of the right-most instance of is (J K) The scope of the conjunction is the entire sentence; so conjunction is the main logical connective of the sentence. 10

14 Complete truth tables 10 A. Complete truth tables for each of the following: 1. A A 2. C C 3. (A B) (A B) A T F C T F A A T TT F T F C C T F F T F TT F 4. (A B) (B A) 5. (A B) (B A) 6. (A B) ( A B) A B (A B) (A B) T T T T T TT T F F T T F T F F T F T T T F F T F F T T F F T F T F F F T F TT F F T F A B (A B) (B A) T T T T T T T T T T F T F F T F T T F T F T T T T F F F F F T F T F T F A B (A B) (B A) T T T T T T T T T T F T F F T F T T F T F F T T T T F F F F F F T F F F 11

15 10. Complete truth tables [ (A B) (A B) ] C 8. [(A B) C] B A B (A B) ( A B) T T F T T T T F T F F T T F F T T F T F T F T F F T F F T T T T F F F T F F T F F F T T F TT F A B C [ (A B) (A B) ] C T T T T T T F F T T T F T T T F T T T F F T T T F F T F T T F F FT T F F F T T F F T F F FT T F F F F F T T F F T FT F F F F T F T F F F T FT F F T F F F F T F F F FT F F T F T F F F F F F FT F F F F F 9. [ (C A) B ] A B C [(A B) C] B T T T T T T T T T T T T F T T T F F T T T F T T F F F T T F T F F T F F F F T F F T T F F T F T T T F T F F F T F F T T F F T F F F F T T F F F F F F F F F T F A B C [ (C A) B ] T T T F T T T T T T T F F F T T T T T F T F T T T T F T F F F F T T T F F T T F T T F T T F T F F F F F T T F F T F T T F T F F F F T F F F F F B. Check all the claims made in introducing the new notational conventions in 10.3, i.e. show that: 1. ((A B) C) and (A (B C)) have the same truth table

16 10. Complete truth tables 13 A B C (A B) C A (B C) T T T T T T T T T T T T T T T F T T T F F T F T F F T F T T F F F T T F F F T T F F T F F F F T F F F F F T T F F T F T F F T T T F T F F F T F F F F T F F F F T F F F F T F F F F T F F F F F F F F F F F F F 2. ((A B) C) and (A (B C)) have the same truth table A B C (A B) C A (B C) T T T T T T T T T T T T T T T F T T T T F T T T T F T F T T T F T T T T F T T T F F T T F T F T T F F F F T T F T T T T F T T T T F T F F T T T F F T T T F F F T F F F T T F T F T T F F F F F F F F F F F F F 3. ((A B) C) and (A (B C)) do not have the same truth table A B C (A B) C A (B C) T T T T T T T T T T T T T T T F T T T F F T T T F F T F T T T F T T T T F F T T F F T T F F F T T F F F F T T F T T T T F T T T T F T F F T T F F F F T F F F F T F F F F T F F F F T F F F F F F F F F F F F F 4. ((A B) C) and (A (B C)) do not have the same truth table Also, check whether: A B C (A B) C A (B C) T T T T T T T T T T T T T T T F T T T F F T F T F F T F T T F F T T T T F T T T F F T F F T F T T F T F F T T F T T T T F T T T T F T F F T T F F F T T F F F F T F T F T T F T F T T F F F F T F F F F T F T F

17 10. Complete truth tables ((A B) C) and (A (B C)) have the same truth table Indeed they do: A B C (A B) C A (B C) T T T T T T T T T T T T T T T F T T T F F T F T F F T F T T F F F T T F F F T T F F T F F T F T T F T F F T T F F T F T F F T T T F T F F F T T F F T T F F F F T F T F T T F T F F T F F F F T F F F F F F T F

18 Semantic concepts 11 A. Revisit your answers to 10A. Determine which sentences were tautologies, which were contradictions, and which were neither tautologies nor contradictions. 1. A A Tautology 2. C C Neither 3. (A B) (A B) Tautology 4. (A B) (B A) Tautology 5. (A B) (B A) Tautology 6. (A B) ( A B) Tautology 7. [ (A B) (A B) ] C Contradiction 8. [(A B) C] B Tautology 9. [ (C A) B ] Neither B. Use truth tables to determine whether these sentences are jointly consistent, or jointly inconsistent: 1. A A, A A, A A, A A Jointly consistent (see line 1) A A A A A A A A A T T TT F T T F T T TT T TT F F T F T F TT F F F F F F F 2. A B, A C, B C Jointly consistent (see line 1) A B C A B A C B C T T T T T T T T T T T T T T F T T T T F F T F F T F T T T T T T T F T T T F F T T F T F F F T F F T T F T F F T T T T T F T F F T T F T F T F F F F T F F F F T T F T T F F F F F F F T F F T F 3. B (C A), A B, (B C) Jointly inconsistent 15

19 11. Semantic concepts 16 A B C B (C A) A B (B C) T T T T T T T T T T T F T T T T T F T T F T T T T T F T T F T F T F F T T T T F F F F T T T F F F F F T T T F F T F F F F T T T T T T F F T T F T T T F T F T F F F F F T T F T T F F F T F F T T F F T F F F T T F F F F F F F F F T F T F F F 4. A (B C), C A, A B Jointly consistent (see line 8) A B C A (B C) C A A B T T T T T T T T T F F T T F F T T T F T T T T F F T F T T F F T T F T T T F T T T F F T T TT F T F F T F F F F F T F T T TT F F T T F F T T T T TT F F T F T F T F F F T T F F TT F F T F T F F T F F F T T T TT F F TT F F F F F T F F F F TT F F TT F C. Use truth tables to determine whether each argument is valid or invalid. 1. A A.. A Invalid (see line 2) A A A A T T TT T F F T F F 2. A (A A).. A Valid A A (A A) A T T F T F F T F T F F T F FT F T F 3. A (B A).. A B Valid A B A (B A) A B T T T T T T T F T T F T T F T T F T T F T TT F F T F F T F F T F F F T F F F T F T F T F TT F 4. A B, B C, A.. B C Invalid (see line 6)

20 11. Semantic concepts 17 A B C A B B C A B C T T T T T T T T T F T T T T T T F T T T T T F F T T F F T F T T T F F T T F T F F T T F F T T F F F F F T F F F T T T F T T T T T T F T T T T T F F T T T T F T F T F F T F T F F F F T T T F F F T T F F F F F F F F T F F F F 5. (B A) C, (C A) B.. (C B) A Invalid (see line 5) A B C (B A) C (C A) B (C B) A T T T T T T T T T T T T T T T T TT T T F T T T F F F F T T T F F T TT T F T F F T T T T T T F F T F F TT T F F F F T T F F F T T F F F F TT F T T T F F T T T F F T T T T T F F F T F T F F T F F F F T T F F T T F F F T F F F T T T F F T F T F F T F F F F F F F T F F F F T F F F F T F D. Answer each of the questions below and justify your answer. 1. Suppose that A and B are tautologically equivalent. What can you say about A B? A and B have the same truth value on every line of a complete truth table, so A B is true on every line. It is a tautology. 2. Suppose that (A B) C is neither a tautology nor a contradiction. What can you say about whether A, B.. C is valid? Since the sentence (A B) C is not a tautology, there is some line on which it is false. Since it is a conditional, on that line, A and B are true and C is false. So the argument is invalid. 3. Suppose that A, B and C are jointly tautologically inconsistent. What can you say about (A B C )? Since the sentences are jointly tautologically inconsistent, there is no valuation on which they are all true. So their conjunction is false on every valuation. It is a contradiction 4. Suppose that A is a contradiction. What can you say about whether A, B C? Since A is false on every line of a complete truth table, there is no line on which A and B are true and C is false. So the entailment holds. 5. Suppose that C is a tautology. What can you say about whether A, B C? Since C is true on every line of a complete truth table, there is no line on which A and B are true and C is false. So the entailment holds.

21 11. Semantic concepts Suppose that A and B are tautologically equivalent. What can you say about (A B)? Not much. Since A and B are true on exactly the same lines of the truth table, their disjunction is true on exactly the same lines. So, their disjunction is tautologically equivalent to them. 7. Suppose that A and B are not tautologically equivalent. What can you say about (A B)? A and B have different truth values on at least one line of a complete truth table, and (A B) will be true on that line. On other lines, it might be true or false. So (A B) is either a tautology or it is contingent; it is not a contradiction. E. Consider the following principle: Suppose A and B are tautologically equivalent. Suppose an argument contains A (either as a premise, or as the conclusion). The validity of the argument would be unaffected, if we replaced A with B. Is this principle correct? Explain your answer. The principle is correct. Since A and B are tautologically equivalent, they have the same truth table. So every valuation that makes A true also makes B true, and every valuation that makes A false also makes B false. So if no valuation makes all the premises true and the conclusion false, when A was among the premises or the conclusion, then no valuation makes all the premises true and the conclusion false, when we replace A with B.

22 Truth table shortcuts 12 A. Using shortcuts, determine whether each sentence is a tautology, a contradiction, or neither. 1. B B Contradiction B B B T F F F F 2. D D Tautology D D D T T F T T 3. (A B) (B A) Neither A B (A B) (B A) T T T T T F F F F F T F F F F F F F F 4. [A (B A)] Contradiction A B [ A (B A)] T T F T T T F F T T F T F T F F F T 5. A [A (B B)] Contradiction A B A [A (B B)] T T F F FF T F F F F F T F T F F F T 19

23 12. Truth table shortcuts (A B) A Neither A B (A B) A T T F T F T F T F T F T T F F F F T F F 7. A (B C) Neither A B C A (B C) T T T T T T T F T T T F T T T T F F F F F T T T F T F T F F T T F F F T 8. (A A) (B C) Tautology A B C (A A) (B C) T T T FF T T T F FF T T F T FF T T F F FF T F T T F T F T F F T F F T F T F F F F T 9. (B D) [A (A C)] Neither

24 12. Truth table shortcuts 21 A B C D (B D) [A (A C)] T T T T T T T T T T T F F F T T T T F T T T T T T T F F F F T T T F T T F F T T T F T F F F T T T F F T F F T T T F F F F F T T F T T T T F F T F T T F F T F T F T F T T T T F F T F F F F T F F F T T F T F T F F T F F T F T F F F T F F T F F F F F F F T F

25 Partial truth tables 13 A. Use complete or partial truth tables (as appropriate) to determine whether these pairs of sentences are tautologically equivalent: 1. A, A Not tautologically equivalent A A A T T F 2. A, A A Tautologically equivalent A A A A T T T T T T 3. A A, A A Tautologically equivalent A A A A A T T T F T T 4. A B, A B Not tautologically equivalent A B A B A B T F T F 5. A A, B B Tautologically equivalent A B A A B B T T FF F F T F FF T F F T F F F F F F T F 6. (A B), A B Tautologically equivalent A B (A B) A B T T F T F F F T F T F F TT F T T F T T F F F T F T TT 22

26 13. Partial truth tables (A B), A B Not tautologically equivalent A B (A B) A B T T F T F TF 8. (A B), ( B A) Tautologically equivalent A B (A B) ( B A) T T T F T T F F T F F F T T F T F F T T TT B. Use complete or partial truth tables (as appropriate) to determine whether these sentences are jointly tautologically consistent, or jointly tautologically inconsistent: 1. A B, C B, C Jointly tautologically inconsistent A B C A B C B C T T T T F F T T T F T T F T F T F T T T T F F F T F F T T F F F T F T F F T F F F T F T T T F F F F T F 2. A B, B C, A, C Jointly tautologically inconsistent A B C A B B C A C T T T T T T F T T F T F T T T F T F T T F T F F F T T T F T T T T F F F T F T F F T F F T T T F F F F F T T F T 3. A B, B C, C A Jointly tautologically consistent A B C A B B C C A F T T T T T T 4. A, B, C, D, E, F Jointly tautologically consistent A B C D E F A B C D E F T T T F F T T T T T T T

27 13. Partial truth tables 24 C. Use complete or partial truth tables (as appropriate) to determine whether each argument is valid or invalid: 1. A [ A (A A) ].. A Invalid A A [ A (A A) ] A F T T F 2. A (B A).. A Invalid A B A (B A) A F F TF T F 3. A B, B.. A Invalid A B A B B A F T T T F 4. A B, B C, B.. A C Valid A B C A B B C B A C T T T T T T F F F T F T T T F F T F T F F T T F F F T F F F F F T F T F F F F F T F 5. A B, B C.. A C Valid A B C A B B C A C T T T T T T F T F F T F T T T F F F F F T T F F F T F T F F T T F F F F F T

28 Sentences with one quantifier 15 A. Here are the syllogistic figures identified by Aristotle and his successors, along with their medieval names. Symbolize each argument in FOL. Barbara. All G are F. All H are G. So: All H are F x(gx F x), x(hx Gx).. x(hx F x) Celarent. No G are F. All H are G. So: No H are F x(gx F x), x(hx Gx).. x(hx F x) Ferio. No G are F. Some H is G. So: Some H is not F x(gx F x), x(hx Gx).. x(hx F x) Darii. All G are H. Some H is G. So: Some H is F. x(gx F x), x(hx Gx).. x(hx F x) Camestres. All F are G. No H are G. So: No H are F. x(f x Gx), x(hx Gx).. x(hx F x) Cesare. No F are G. All H are G. So: No H are F. x(f x Gx), x(hx Gx).. x(hx F x) Baroko. All F are G. Some H is not G. So: Some H is not F. x(f x Gx), x(hx Gx).. x(hx F x) Festino. No F are G. Some H are G. So: Some H is not F. x(f x Gx), x(hx Gx).. x(hx F x) Datisi. All G are F. Some G is H. So: Some H is F. x(gx F x), x(gx Hx).. x(hx F x) Disamis. Some G is F. All G are H. So: Some H is F. x(gx F x), x(gx Hx).. x(hx F x) Ferison. No G are F. Some G is H. So: Some H is not F. x(gx F x), x(gx Hx).. x(hx F x) Bokardo. Some G is not F. All G are H. So: Some H is not F. x(gx F x), x(gx Hx).. x(hx F x) Camenes. All F are G. No G are H So: No H is F. x(f x Gx), x(gx Hx).. x(hx F x) Dimaris. Some F is G. All G are H. So: Some H is F. x(f x Gx), x(gx Hx).. x(hx F x) Fresison. No F are G. Some G is H. So: Some H is not F. x(f x Gx), x(gx Hx).. (Hx F x) 25

29 15. Sentences with one quantifier 26 B. Using the following symbolization key: domain: people Kx: x knows the combination to the safe Sx: x is a spy V x: x is a vegetarian h: Elizabeth p: Phillip symbolize the following sentences in FOL: 1. Neither Elizabeth nor Phillip is a vegetarian. V e V p 2. No spy knows the combination to the safe. x(sx Kx) 3. No one knows the combination to the safe unless Phillip does. x Kx Kp 4. Elizabeth is a spy, but no vegetarian is a spy. Se x(v x Sx) C. Using this symbolization key: domain: all animals Ax: x is an alligator. Mx: x is a monkey. Rx: x is a reptile. Zx: x lives at the zoo. a: Amos b: Bouncer c: Cleo symbolize each of the following sentences in FOL: 1. Amos, Bouncer, and Cleo all live at the zoo. Za Zb Zc 2. Bouncer is a reptile, but not an alligator. Rb Ab 3. Some reptile lives at the zoo. x(rx Zx) 4. Every alligator is a reptile. x(ax Rx) 5. Any animal that lives at the zoo is either a monkey or an alligator. x(zx (Mx Ax)) 6. There are reptiles which are not alligators. x(rx Ax) 7. If any animal is an reptile, then Amos is. xrx Ra 8. If any animal is an alligator, then it is a reptile. x(ax Rx)

30 15. Sentences with one quantifier 27 D. For each argument, write a symbolization key and symbolize the argument in FOL. 1. Dorothy is a logician. All logicians wear funny hats. So Dorothy wears a funny hat. domain: people Lx: x is a logician Hx: x wears a funny hat d: Dorothy Ld, x(lx Hx).. Hd 2. Nothing on my desk escapes my attention. There is a computer on my desk. As such, there is a computer that does not escape my attention. domain: physical things Dx: x is on my desk Ex: x escapes my attention Cx: x is a computer x(dx Ex), x(dx Cx).. x(cx Ex) 3. All my dreams are black and white. Old TV shows are in black and white. Therefore, some of my dreams are old TV shows. domain: episodes (psychological and televised) Dx: x is one of my dreams Bx: x is in black and white Ox: x is an old TV show x(dx Bx), x(ox Bx).. x(dx Ox). Comment: generic statements are tricky to deal with. Does the second sentence mean that all old TV shows are in black and white; or that most of them are; or that most of the things which are in black and white are old TV shows? I ve gone with the former, but it is not clear that FOL deals with these well. 4. Neither Langston nor Ornette has been to Australia. A person could see a kangaroo only if they had been to Australia or to a zoo. Although Langston has not seen a kangaroo, Ornette has. Therefore, Ornette has been to a zoo. domain: people Ax: x has been to Australia Kx: x has seen a kangaroo Zx: x has been to a zoo l: Langston o: Ornette Al Ao, x(kx (Ax Zx)), Ko Kl.. Zl

31 15. Sentences with one quantifier No one expects the Spanish Inquisition. No one knows the troubles I ve seen. Therefore, anyone who expects the Spanish Inquisition knows the troubles I ve seen. domain: people Sx: x expects the Spanish Inquisition T x: x knows the troubles I ve seen x Sx, x T x.. x(sx T x) 6. All babies are illogical. Nobody who is illogical can manage a crocodile. Robin is a baby. Therefore, Robin is unable to manage a crocodile. domain: people Bx: x is a baby Ix: x is illogical Cx: x can manage a crocodile b: Robin x(bx Ix), x(ix Cx), Bb.. Cb

32 Multiple generality 16 A. Using this symbolization key: domain: all animals Ax: x is an alligator Mx: x is a monkey Rx: x is a reptile Zx: x lives at the zoo Lxy: x loves y a: Amos b: Bouncer c: Cleo symbolize each of the following sentences in FOL: 1. If Cleo loves Bouncer, then Bouncer is a monkey. Lcb Mb 2. If both Bouncer and Cleo are alligators, then Amos loves them both. (Ab Ac) (Lab Lac) 3. Cleo loves a reptile. x(rx Lcx) Comment: this English expression is ambiguous; in some contexts, it can be read as a generic, along the lines of Cleo loves reptiles. (Compare I do love a good pint.) 4. Bouncer loves all the monkeys that live at the zoo. x((mx Zx) Lbx) 5. All the monkeys that Amos loves love him back. x((mx Lax) Lxa) 6. Every monkey that Cleo loves is also loved by Amos. x((mx Lcx) Lax) 7. There is a monkey that loves Bouncer, but sadly Bouncer does not reciprocate this love. x(mx Lxb Lbx) 29

33 16. Multiple generality 30 B. Using the following symbolization key: domain: all animals Dx: x is a cat. Sx: x likes samurai movies. Lxy: x is larger than y b: Oliver e: Eliza i: Ivan symbolize the following sentences in FOL: 1. Oliver is a cat who likes samurai movies. Db Sb 2. Oliver, Eliza, and Ivan are all cats. Db De Di 3. Eliza is larger than Oliver, and Ivan is larger than Eliza. Leb Lie 4. All cats like samurai movies. x(dx Sx) 5. Only cats like samurai movies. x(sx Dx) Comment: the FOL sentence just written does not require that anyone likes samurai movies. The English sentence might suggest that at least some cats do like samurai movies? 6. There is a cat that is larger than Eliza. x(dx Lxe) 7. If there is a cat larger than Ivan, then there is a cat larger than Eliza. x(dx Lxi) x(dx Lxe) 8. No animal that likes samurai movies is larger than Eliza. x(sx Lxe) 9. No cat is larger than Ivan. x(dx Lxi) 10. Any animal that dislikes samurai movies is larger than Oliver. x( Sx Lxb) Comment: this is not a great translation, though! Because dislikes does not mean the same as does not like. 11. There is an animal that is between Oliver and Eliza in size. x((lbx Lxe) (Lex Lxb)) 12. There is no cat that is between Oliver and Eliza in size. x ( Dx [ (Lbx Lxe) (Lex Lxb) ]) 13. No cat is larger than itself. x(dx Lxx)

34 16. Multiple generality Every cat is larger than some cat. x(dx y(dy Lxy)) Comment: the English sentence is potentially ambiguous here. I resolved the ambiguity by assuming it should be paraphrased by for every cat, there is a cat smaller than it. 15. There is an animal that is smaller than every cat. x y(dy Lyx) 16. If there is an animal that is larger than any cat, then that animal does not like samurai movies. x( y(dy Lxy) Sx) Comment: I ve assumed that larger than any cat here means larger than every cat. C. Using the following symbolization key: domain: people and dishes at a potluck Rx: x has run out. T x: x is on the table. F x: x is food. P x: x is a person. Lxy: x likes y. m: Miles n: Nina g: the guacamole symbolize the following English sentences in FOL: 1. All the food is on the table. x(f x T x) 2. If the guacamole has not run out, then it is on the table. Rg T g 3. Everyone likes the guacamole. x(p x Lxg) 4. If anyone likes the guacamole, then Miles does. x(p x Lxg) Lmg 5. Nina only likes the dishes that have run out. x [ (Lnx F x) Rx ] 6. Nina likes no one, and no one likes Nina. x [ P x ( Lnx Lxn) ] 7. Miles likes anyone who likes the guacamole. x((p x Lxg) Lmx) 8. Miles likes anyone who likes the people that he likes. x [( P x y[(p y Lmy) Lxy] ) Lmx ] 9. If there is a person on the table already, then all of the food must have run out. x(p x T x) x(f x Rx)

35 16. Multiple generality 32 D. Using the following symbolization key: domain: people Sx: x sings. F x: x is female. M x: x is male. Cxy: x is a child of y. Sxy: x is a sibling of y. m: Michael j: Janet e: Joe p: Paris symbolize the following arguments in FOL: 1. All of Joe s children are singers. x(cxj Sx) 2. Janet is Joe?s daughter. Cje F j 3. Joe has a daughter. x(cxj F x) 4. Joe is an only child. xsxe 5. All of Joe?s sons sing. x [ (Cxe Mx) Sx ] 6. Janet has no daughters. x(cxj F x) 7. Paris is Janet?s niece. x(sxj Cpx F p) 8. Michael is Janet?s brother. Smj Mm 9. Joe?s brothers have no children. 10. x [ (Sjx Mx) ycyx ] 10. Janet is an aunt. F j x(sxj ycyx) 11. Everyone who sings has a brother who also sings. x [ Sx y(my Syx Sy) ] 12. Every woman who sings is the child of someone who sings. x [ (F x Sx) y(cxy Sy) ]

36 Identity 17 A. Explain why: x y(ay x = y) is a good symbolization of there is exactly one apple. We might understand this in English as saying (roughly): There is something in the domain, we ll call it x, such that, regardless of what you choose from the domain, if you choose an apple, you ve chosen x, and if you choose x, you ve chosen an apple. The x in question must therefore be the one and only thing which is an apple. x y [ x = y z(az (x = z y = z) ] is a good symbolization of there are exactly two apples. Similarly to the above, we might understand this in English as saying (roughly): There are two distinct things, x and y, such that if you choose any object at all, if you chose an apple then you either chose x or y, and if you chose either x or y then you chose an apple. The x and y in question must therefore be the only things which are apples, and since they are distinct, there are two of them. 33

37 Definite descriptions 18 A. Using the following symbolization key: domain: people Kx: x knows the combination to the safe. Sx: x is a spy. V x: x is a vegetarian. T xy: x trusts y. e: Elizabeth p: Phillip symbolize the following sentences in FOL: 1. Elizabeth trusts a vegetarian. x(v x T ex) 2. Everyone who trusts Phillip trusts a vegetarian. x [ T xp y(t xy V y) ] 3. Everyone who trusts Phillip trusts someone who trusts a vegetarian. x [ T xp y ( T xy z(t yz V z) )] 4. Only Phillip knows the combination to the safe. x(kp x = p) Comment: does the English claim entail that Phillip does know the combination to the safe? If so, then we should formalize this with a. 5. Phillip trusts Elizabeth, but no one else. x(t px x = e) 6. The person who knows the combination to the safe is a vegetarian. x [ Kx y(ky x = y) V x ] 7. The person who knows the combination to the safe is not a spy. x [ Kx y(ky x = y) Sx ] Comment: the scope of negation is potentially ambiguous here; I ve read it as inner negation. 34

38 18. Definite descriptions 35 B. Using the following symbolization key: domain: cards in a standard deck Bx: x is black. Cx: x is a club. Dx: x is a deuce. Jx: x is a jack. Mx: x is a man with an axe. Ox: x is one-eyed. W x: x is wild. symbolize each sentence in FOL: 1. All clubs are black cards. x(cx Bx) 2. There are no wild cards. xw x 3. There are at least two clubs. x y( x = y Cx Cy) 4. There is more than one one-eyed jack. x y( x = y Jx Ox Jy Oy) 5. There are at most two one-eyed jacks. x y z [ (Jx Ox Jy Oy Jz Oz) (x = y x = z y = z) ] 6. There are two black jacks. x y( x = y Bx Jx By Jy) Comment: I am reading this as there are at least two.... If the suggestion was that there are exactly two, then a different FOL sentence would be required, namely: x y ( x = y Bx Jx By Jy z[(bz Jz) (x = z y = z)] ) 7. There are four deuces. w x y z( w = x w = y w = z x = y x = z y = z Dw Dx Dy Dz) Comment: I am reading this as there are at least four.... If the suggestion is that there are exactly four, then we should offer instead: w x y z ( w = x w = y w = z x = y x = z y = z Dw Dx Dy Dz v[dv (v = w v = x v = y v = z)] ) 8. The deuce of clubs is a black card. x [ Dx Cx y ( (Dy Cy) x = y ) Bx ] 9. One-eyed jacks and the man with the axe are wild. x [ (Jx Ox) W x ] x [ Mx y(my x = y) W x ]

39 18. Definite descriptions If the deuce of clubs is wild, then there is exactly one wild card. x ( Dx Cx y [ (Dy Cy) x = y ] W x ) x ( W x y(w y x = y) ) Comment: if there is not exactly one deuce of clubs, this sentence is true. Maybe that s the wrong verdict. Perhaps the sentence should be read to imply that there is exactly one deuce of clubs, and then express a conditional about wildness. If so, then we might symbolize it thus: x ( Dx Cx y [ (Dy Cy) x = y ] [ W x y(w y x = y) ]) 11. The man with the axe is not a jack. x [ Mx y(my x = y) Jx ] 12. The deuce of clubs is not the man with the axe. x y ( Dx Cx z[(dz Cz) x = z] My z(mz y = x) x = y ) C. Using the following symbolization key: domain: animals in the world Bx: x is in Farmer Brown s field. Hx: x is a horse. P x: x is a Pegasus. W x: x has wings. symbolize the following sentences in FOL: 1. There are at least three horses in the world. x y z( x = y x = z y = z Hx Hy Hz) 2. There are at least three animals in the world. x y z( x = y x = z y = z) 3. There is more than one horse in Farmer Brown s field. x y( x = y Hx Hy Bx By) 4. There are three horses in Farmer Brown s field. x y z( x = y x = z y = z Hx Hy Hz Bx By Bz) Comment: I ve read this as there are at least three.... If the suggestion was that there are exactly three, then a different FOL sentence would be required. 5. There is a single winged creature in Farmer Brown s field; any other creatures in the field must be wingless. x [ W x Bx y ( (W y By) x = y) ] 6. The Pegasus is a winged horse. x [ P x y(p y x = y) W x Hx ] 7. The animal in Farmer Brown s field is not a horse. x [ Bx y(by x = y) Hx ] Comment: the scope of negation might be ambiguous here; I ve read it as inner negation. 8. The horse in Farmer Brown s field does not have wings. x [ Hx Bx y ( (Hy By) x = y ) W x ] Comment: the scope of negation might be ambiguous here; I ve read it as inner negation.

40 18. Definite descriptions 37 D. Maya Angelou is the author of Beloved by x(ax y(ay x = y) x = m). Two equally good symbolizations would be: Am y(ay m = y) This sentence requires that Maya Angelou is an author of Beloved, and that Maya Angelou alone is an author of Beloved. Otherwise put, there is one and only one author of Beloved, namely, Maya Angelou. Otherwise put: Maya Angelou is the author of Beloved. y(ay y = m) This sentence can be understood thus: Take anything you like; now, if you chose an author of Beloved, you chose Maya Angelou, and if you chose Maya Angelou, you chose an author of Beloved. So there is one and only one author of Beloved, namely, Maya Angelou, as required. Explain why these would be equally good symbolizations.

41 Sentences of FOL 19 A. Identify which variables are bound and which are free. I shall underline the bound variables, and put free variables in blue. 1. xlxy ylyx 2. xax Bx 3. x(ax Bx) y(cx Dy) 4. x y[rxy (Jz Kx)] Ryx 5. x 1 (Mx 2 Lx 2 x 1 ) x 2 Lx 3 x 2 38

42 Truth in FOL 21 A. Consider the following interpretation: The domain comprises only Etta and Nina Ax is to be true of both Etta and Nina Bx is to be true of Nina only Nx is to be true of no one c is to refer to Etta Determine whether each of the following sentences is true or false in that interpretation: 1. Bc False 2. Ac Nc True 3. Nc (Ac Bc) True 4. xax True 5. x Bx False 6. x(ax Bx) True 7. x(ax Nx) False 8. x(nx Nx) True 9. xbx xax True B. Consider the following interpretation: The domain comprises only Ornette, Miles and John Gx is to be true of Ornette, Miles and John. Hx is to be true of and only of Miles Mx is to be true of and only of Ornette and John m is to refer to Miles j is to refer to John Determine whether each of the following sentences is true or false in that interpretation: 1. Hm True 2. Hj False 3. Mm Mj True 4. Gm Gm True 5. Mm Gm True 39

43 21. Truth in FOL xhx True 7. xhx False 8. x M x True 9. x(hx Gx) True 10. x(mx Gx) True 11. x(hx Mx) True 12. xhx xmx True 13. x(hx Mx) True 14. xgx x Gx False 15. x y(gx Hy) True C. Following the diagram conventions introduced at the end of 23, consider the following interpretation: Determine whether each of the following sentences is true or false in that interpretation: 1. xrxx True 2. xrxx False 3. x yrxy True 4. x yryx False 5. x y z((rxy Ryz) Rxz) False 6. x y z((rxy Rxz) Ryz) False 7. x y Rxy True 8. x( yrxy yryx) True 9. x y( x = y Rxy Ryx) True 10. x y(rxy x = y) True 11. x y(ryx x = y) False 12. x y( x = y Rxy z(rzx y = z)) True

44 Using Interpretations 23 A. Show that each of the following is neither a logical truth nor a contradiction: The following are examples of how to use interpretations to show each sentence is neither a logical truth nor a contradiction. Your answers will, of course, be different. 1. Da Db The following interpretation makes Da Db false: Domain comprises only Etta and Nina Dx is to be true of only Nina a is to refer to Nina b is to refer to Etta The following interpretation makes Da Db true: Domain comprises only Etta and Nina Dx is to be true of both Etta and Nina a is to refer to Nina b is to refer to Etta Because there is an interpretation in which Da Db is true and one in which it is false, it is neither a logical truth nor a contradiction. 2. xt xh The following interpretation makes xt xh false: Domain comprises only Oliver and Eliza T xy is to be true of only <Eliza, Eliza> h is to refer to Oliver e is to refer to Eliza The following interpretation makes xt xh true: Domain comprises only Oliver and Eliza T xy is to be true of only <Eliza, Oliver> h is to refer to Oliver e is to refer to Eliza Because there is an interpretation in which xt xh is true and one in which it is false, it is neither a logical truth nor a contradiction. 3. P m xp x The following interpretation makes P m xp x false: 41

45 23. Using Interpretations 42 Domain comprises only Mallory and Eliza P x is to be true of only Eliza m is to refer to Mallory e is to refer to Eliza The following interpretation makes P m xp x true: Domain comprises only Mallory and Eliza P x is to be true of only Mallory m is to refer to Mallory e is to refer to Eliza Because there is an interpretation in which P m xp x is true and one in which it is false, it is neither a logical truth nor a contradiction. 4. zjz yjy The following interpretation makes zjz yjy false: Domain comprises only Mallory and Eliza Jx is to be true of only Eliza m is to refer to Mallory e is to refer to Eliza The following interpretation makes zjz yjy true: Domain comprises only Mallory and Eliza Jx is to be true of no one. m is to refer to Mallory e is to refer to Eliza Because there is an interpretation in which zjz yjy is true and one in which it is false, it is neither a logical truth nor a contradiction. 5. x(sxmn ylxy) The following interpretation makes x(sxmn ylxy) false: Domain comprises only Noelle and Mallory Sxyz is to be true of only <Mallory, Mallory, Noelle>. Lxy is to be true of nothing. m is to refer to Mallory n is to refer to Noelle The following interpretation makes x(sxmn ylxy) true: Domain comprises only Noelle and Mallory Sxyz is to be true of <Noelle, Mallory, Noelle> and <Mallory, Mallory, Noelle>. Lxy is to be true of <Mallory, Mallory> m is to refer to Mallory n is to refer to Noelle Because there is an interpretation in which x(sxmn ylxy) is true and one in which it is false, it is neither a logical truth nor a contradiction. 6. x(gx ymy) The following interpretation makes x(gx ym y) false:

46 23. Using Interpretations 43 Domain comprises only Noelle and Mallory Gx is to be true of Noelle. Mx is to be true of only Noelle. m is to refer to Mallory n is to refer to Noelle The following interpretation makes x(gx ym y) true: Domain comprises only Noelle and Mallory Gx is to be true of no one. Mx is to be true of Noelle. m is to refer to Mallory n is to refer to Noelle Because there is an interpretation in which x(gx ymy) is true and one in which it is false, it is neither a logical truth nor a contradiction. 7. x(x = h x = i) The following interpretation makes x(x = h x = i) false: Domain comprises only Hope and Iris h is to refer to Hope i is to refer to Iris The following interpretation makes x(x = h x = i) true: Domain comprises only Hope and Iris h is to refer to Hope i is to refer to Hope Because there is an interpretation in which x(x = h x = i) is true and one in which it is false, it is neither a logical truth nor a contradiction. B. Show that the following pairs of sentences are not logically equivalent. The following are examples of how to use interpretations to show each pair of sentences are not logically equivalent. For many of the following sentences, very different interpretations might be equally correct. 1. Ja, Ka The following interpretation makes Ja true and Ka false: Domain comprises Alice and Bob. Jx is to be true of only Alice. Kx is to be true of only Bob. a is to refer to Alice. Because there is an interpretation in which Ja and Ka have different truth value assignments, they are not logically equivalent. 2. xjx, Jm The following interpretation makes xjx true and Jm false: Domain comprises Mallory and Bob. Jx is to be true of only Bob. m is to refer to Mallory.

47 23. Using Interpretations 44 Because there is an interpretation in which xjx and Jm have different truth value assignments, they are not logically equivalent. 3. xrxx, xrxx The following interpretation makes xrxx false and xrxx true: Domain comprises Mallory and Bob. Rxx is to be true of only <Bob, Bob>. Because there is an interpretation in which xrxx and xrxx have different truth value assignments, they are not logically equivalent. 4. xp x Qc, x(p x Qc) The following interpretation makes xp x Qc false and x(p x Qc) true: Domain comprises Mallory and Corey. P x is to be true of Mallory. Qx is to be true of no one. c is to refer to Corey Because there is an interpretation in which xp x Qc and x(p x Qc) have different truth value assignments, they are not logically equivalent. 5. x(p x Qx), x(p x Qx) The following interpretation makes x(p x Qx) false and x(p x Qx) true: Domain comprises Mallory and Corey. P x is to be true of Mallory. Qx is to be true of Mallory. Because there is an interpretation in which x(p x Qx) and x(p x Qx) have different truth value assignments, they are not logically equivalent. 6. x(p x Qx), x(p x Qx) The following interpretation makes x(p x Qx) false and x(p x Qx) true: Domain comprises Damien and Corey. P x is to be true of Damien. Qx is to be true of Corey. Because there is an interpretation in which x(p x Qx) and x(p x Qx) have different truth value assignments, they are not logically equivalent. 7. x(p x Qx), x(p x Qx) The following interpretation makes x(p x Qx) true and x(p x Qx) false: Domain comprises Damien and Corey. P x is to be true of no one. Qx is to be true of Corey.

48 23. Using Interpretations 45 Because there is an interpretation in which x(p x Qx) and x(p x Qx) have different truth value assignments, they are not logically equivalent. 8. x yrxy, x yrxy The following interpretation makes x yrxy true and x yrxy false: Domain comprises Damien and Peter. Rxy is to be true of <Damien, Peter>, <Peter, Damien> Because there is an interpretation in which x yrxy and x yrxy have different truth value assignments, they are not logically equivalent. 9. x yrxy, x yryx The following interpretation makes x yrxy true and x yryx false: Domain comprises Damien and Peter. Rxy is to be true of <Damien, Peter>, <Peter, Peter> Because there is an interpretation in which x yrxy and x yryx have different truth value assignments, they are not logically equivalent. C. Show that the following sentences are jointly consistent: The following are examples of how to use interpretations to show each set of sentences is jointly consistent. Your answers will, of course, be different. 1. Ma, Na, P a, Qa The following interpretation makes each of Ma, Na, P a, and Qa true: Domain comprises Anne, Peter, and Russell. Mx is to be true of Anne and Peter Nx is to be true of Peter P x is to be true of Anne and Russell Qx is to be true of no one. a is to refer to Anne. Because there is an interpretation that makes each of Ma, Na, P a, and Qa true, they are jointly consistent. 2. Lee, Leg, Lge, Lgg The following interpretation makes each of Lee, Leg, Lge, and Lgg true: Domain comprises Elias, Greg, and Russell. Lxy is to be true of <Elias, Elias>, <Elias, Greg>, and <Russell, Greg> e is to refer to Elias. g is to refer to Greg. Because there is an interpretation that makes each of Lee, Leg, Lge, and Lgg true, they are jointly consistent. 3. (Ma xax), Ma F a, x(f x Ax) The following interpretation makes each of (M a xax), M a F a, and x(f x Ax) true: Domain comprises Elias, Anne, and Russell.