Forallx - An Introduction to Formal Logic

Practice Exercises

At the end of each chapter, you will find a series of practice problems that review and explore the material covered in the chapter. There is no substitute for actually working through some problems, because logic is more about a way of thinking than it is about memorizing facts. The answers to some of the problems are provided at the end of the book in appendix B; the problems that are solved in the appendix are marked with a .

Part A Which of the following are ‘sentences’ in the logical sense?

1. England is smaller than China.
2. Greenland is south of Jerusalem.
3. Is New Jersey east of Wisconsin?
4. The atomic number of helium is 2.
5. The atomic number of helium is π.
6. I hate overcooked noodles.
7. Blech! Overcooked noodles!
8. Overcooked noodles are disgusting.
9. Take your time.
10. This is the last question.

Part B For each of the following: Is it a tautology, a contradiction, or a contingent sentence?

1. Caesar crossed the Rubicon.
2. Someone once crossed the Rubicon.
3. No one has ever crossed the Rubicon.
4. If Caesar crossed the Rubicon, then someone has.
5. Even though Caesar crossed the Rubicon, no one has ever crossed the Rubicon.
6. If anyone has ever crossed the Rubicon, it was Caesar.

Part C Look back at the sentences G1–G4 on p. 12, and consider each of the following sets of sentences. Which are consistent? Which are inconsistent?

1. G2, G3, and G4
2. G1, G3, and G4
3. G1, G2, and G4
4. G1, G2, and G3

Part D Which of the following is possible? If it is possible, give an example. If it is not possible, explain why.

1. A valid argument that has one false premise and one true premise
2. A valid argument that has a false conclusion
3. A valid argument, the conclusion of which is a contradiction
4. An invalid argument, the conclusion of which is a tautology
5. A tautology that is contingent
6. Two logically equivalent sentences, both of which are tautologies
7. Two logically equivalent sentences, one of which is a tautology and one of which is contingent
8. Two logically equivalent sentences that together are an inconsistent set
9. A consistent set of sentences that contains a contradiction
10. An inconsistent set of sentences that contains a tautology

Practice Exercises

Part A Using the symbolization key given, translate each English-language sentence into SL.

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.
2. Those creatures are men in suits, or they are not.
3. Those creatures are either gorillas or chimpanzees.
4. Those creatures are neither gorillas nor chimpanzees.
5. If those creatures are chimpanzees, then they are neither gorillas nor men in suits.
6. Unless those creatures are men in suits, they are either chimpanzees or they are gorillas.

Part B Using the symbolization key given, translate each English-language sentence into SL.

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.
2. If Mister Ace was murdered, then the cook did it.
3. If Mister Edge was murdered, then the cook did not do it.
4. Either the butler did it, or the Duchess is lying.
5. The cook did it only if the Duchess is lying.
6. If the murder weapon was a frying pan, then the culprit must have been the cook.
7. If the murder weapon was not a frying pan, then the culprit was either the cook or the butler.
8. Mister Ace was murdered if and only if Mister Edge was not murdered.
9. The Duchess is lying, unless it was Mister Edge who was murdered.
10. If Mister Ace was murdered, he was done in with a frying pan.
11. Since the cook did it, the butler did not.
12. Of course the Duchess is lying!

Part C Using the symbolization key given, translate each English-language sentence into SL.

E₁: Ava is an electrician.

E₂: Harrison is an electrician.

F₁: Ava is a firefighter.

F₂: Harrison is a firefighter.

S₁: Ava is satisfied with her career.

S₂: Harrison is satisfied with his career.

1. Ava and Harrison are both electricians.
2. If Ava is a firefighter, then she is satisfied with her career.
3. Ava is a firefighter, unless she is an electrician.
4. Harrison is an unsatisfied electrician.
5. Neither Ava nor Harrison is an electrician.
6. Both Ava and Harrison are electricians, but neither of them find it satisfying.
7. Harrison is satisfied only if he is a firefighter.
8. If Ava is not an electrician, then neither is Harrison, but if she is, then he is too.
9. Ava is satisfied with her career if and only if Harrison is not satisfied with his.
10. If Harrison is both an electrician and a firefighter, then he must be satisfied with his work.
11. It cannot be that Harrison is both an electrician and a firefighter.
12. Harrison and Ava are both firefighters if and only if neither of them is an electrician.

Part D Give a symbolization key and symbolize the following sentences in SL.

1. Alice and Bob are both spies.
2. If either Alice or Bob is a spy, then the code has been broken.
3. If neither Alice nor Bob is a spy, then the code remains unbroken.
4. The German embassy will be in an uproar, unless someone has broken the code.
5. Either the code has been broken or it has not, but the German embassy will be in an uproar regardless.
6. Either Alice or Bob is a spy, but not both.

Part E Give a symbolization key and symbolize the following sentences in SL.

1. If Gregor plays first base, then the team will lose.
2. The team will lose unless there is a miracle.
3. The team will either lose or it won’t, but Gregor will play first base regardless.
4. Gregor’s mom will bake cookies if and only if Gregor plays first base.
5. If there is a miracle, then Gregor’s mom will not bake cookies.

Part F For each argument, write a symbolization key and translate the argument as well as possible into SL.

1. If Dorothy plays the piano in the morning, then Roger wakes up cranky. Dorothy plays piano in the morning unless she is distracted. So if Roger does not wake up cranky, then Dorothy must be distracted.
2. It will either rain or snow on Tuesday. If it rains, Neville will be sad. If it snows, Neville will be cold. Therefore, Neville will either be sad or cold on Tuesday.
3. If Zoog remembered to do his chores, then things are clean but not neat. If he forgot, then things are neat but not clean. Therefore, things are either neat or clean— but not both.

Part G For each of the following: (a) Is it a wff of SL? (b) Is it a sentence of SL, allowing for notational conventions?

1. (A)
2. J₃₇₄ ∨ ¬J₃₇₄
3. ¬¬¬¬F
4. ¬ & S
5. (G & ¬G)
6. A → A
7. ( A → (A & ¬F )) ∨ (D ↔ E)
8. [(Z ↔ S) → W ] & [J ∨ X]
9. (F ↔ ¬D → J) ∨ (C & D)

Part H

1. Are there any wffs of SL that contain no sentence letters? Why or why not?
2. In the chapter, we symbolized an exclusive or using ∨, & , and ¬. How could you translate an exclusive or using only two connectives? Is there any way to translate an exclusive or using only one connective?

Practice Exercises

If you want additional practice, you can construct truth tables for any of the sentences and arguments in the exercises for the previous chapter.

Part A Determine whether each sentence is a tautology, a contradiction, or a contingent sentence. Justify your answer with a complete or partial truth table where appropriate.

1. A → A
2. ¬B & B
3. C → ¬C
4. ¬D ∨ D
5. (A ↔ B) ↔ ¬(A ↔ ¬B)
6. (A & B) ∨ (B & A)
7. (A → B) ∨ (B → A)
8. ¬[A → (B → A)]
9. (A & B) → (B ∨ A)
10. A ↔ [A → (B & ¬B)]
11. ¬(A ∨ B) ↔ (¬A & ¬B)
12. ¬(A & B) ↔ A
13. (A & B) & ¬(A & B) & C
14. A → (B ∨ C)
15. [(A & B) & C] → B
16. (A & ¬A) → (B ∨ C)
17. ¬ (C ∨ A) ∨ B
18. (B & D) ↔ [A ↔ (A ∨ C)]

Part B Determine whether each pair of sentences is logically equivalent. Justify your answer with a complete or partial truth table where appropriate.

1. A, ¬ A
2. A, A ∨ A
3. A → A, A ↔ A
4. A ∨ ¬B, A → B
5. A & ¬ A, ¬B ↔ B
6. ¬( A &  B), ¬ A ∨ ¬B
7. ¬( A → B), ¬ A → ¬B
8. ( A → B), (¬B → ¬ A)
9. [( A ∨ B) ∨ C], [ A ∨ (B ∨ C)]
10. [( A ∨ B) & C], [ A ∨ (B & C)]

Part C Determine whether each set of sentences is consistent or inconsistent. Justify your answer with a complete or partial truth table where appropriate.

1. A → A, ¬ A → ¬ A, A & A, A ∨ A
2. A & B, C → ¬B, C
3. A ∨ B, A → C, B → C
4. A → B, B → C, A, ¬C
5. B & (C ∨ A), A → B, ¬(B ∨ C)
6. A ∨ B, B ∨ C, C → ¬ A
7. A ↔ (B ∨ C), C → ¬ A, A → ¬B
8. A, B, C, ¬D, ¬E, F

Part D Determine whether each argument is valid or invalid. Justify your answer with a complete or partial truth table where appropriate.

1. A → A, .˙. A
2. A ∨ A → ( A ↔ A) , .˙. A
3. A → ( A & ¬ A), .˙. ¬ A
4. A ↔ ¬(B ↔ A), .˙. A
5. A ∨ (B → A), .˙. ¬ A → ¬B
6. A → B, B, .˙. A
7. A ∨ B, B ∨ C, ¬ A, .˙. B & C
8. A ∨ B, B ∨ C, ¬B, .˙. A & C
9. (B & A) → C, (C & A) → B, .˙. (C & B) → A
10. A ↔ B, B ↔ C, .˙. A ↔ C

Part E Answer each of the questions below and justify your answer.

1. Suppose that A and B are logically equivalent. What can you say about A↔ B?
2. Suppose that ( A &B) → C is contingent. What can you say about the argument “ A, B, .˙.C”?
3. Suppose that { A ,B ,C } is inconsistent. What can you say about ( A &B &C )?
4. Suppose that A is a contradiction. What can you say about the argument “ A, B, .˙.C”?
5. Suppose that C is a tautology. What can you say about the argument “ A, B , .˙. C ”?
6. Suppose that A and B are logically equivalent. What can you say about ( A  ∨ B)?
7. Suppose that A and B are not logically equivalent. What can you say about ( A  ∨ B)?

Part F We could leave the biconditional (↔) out of the language. If we did that, we could still write ‘ A ↔ B’ so as to make sentences easier to read, but that would be shorthand for ( A → B) & (B → A). The resulting language would be formally equivalent to SL, since A ↔ B and ( A → B) & (B → A) are logically equivalent in SL. If we valued formal simplicity over expressive richness, we could replace more of the connectives with notational conventions and still have a language equivalent to SL.

There are a number of equivalent languages with only two connectives. It would be enough to have only negation and the material conditional. Show this by writing sentences that are logically equivalent to each of the following using only parentheses, sentence letters, negation (¬), and the material conditional (→).

1. 1. A ∨ B
2. 2. A & B
3. 3. A ↔ B

We could have a language that is equivalent to SL with only negation and disjunction as connectives. Show this: Using only parentheses, sentence letters, negation (¬), and disjunction (∨), write sentences that are logically equivalent to each of the following.

1. 4. A & B
2. 5. A → B
3. 6. A ↔ B

The Sheffer stroke is a logical connective with the following characteristic truthtable:

1. 7. Write a sentence using the connectives of SL that is logically equivalent to (A|B).

Every sentence written using a connective of SL can be rewritten as a logically equivalent sentence using one or more Sheffer strokes. Using only the Sheffer stroke, write sentences that are equivalent to each of the following.

1. 8. ¬A
2. 9. (A & B)
3. 10. (A ∨ B)
4. 11. (A → B)
5. 12. (A ↔ B)

Practice Exercises

Part A Using the symbolization key given, translate each English-language sentence into QL.

1.  UD: all animals
2.   Ax: x is an alligator.
3.  Mx: x is a reptile.
4.   Rx: x lives at the zoo.
5.    Zx: y is between x and z.
6.  Lxy: x loves y.
7.       a: Amos
8.       b: Bouncer
9.       c: Cleo

1. Amos, Bouncer, and Cleo all live at the zoo.
2. Bouncer is a reptile, but not an alligator.
3. If Cleo loves Bouncer, then Bouncer is a monkey.
4. If both Bouncer and Cleo are alligators, then Amos loves them both.
5. Some reptile lives at the zoo.
6. Every alligator is a reptile.
7. Any animal that lives at the zoo is either a monkey or an alligator.
8. There are reptiles which are not alligators.
9. Cleo loves a reptile.
10. Bouncer loves all the monkeys that live at the zoo.
11. All the monkeys that Amos loves love him back.
12. If any animal is a reptile, then Amos is.
13. If any animal is an alligator, then it is a reptile.
14. Every monkey that Cleo loves is also loved by Amos.
15. There is a monkey that loves Bouncer, but sadly Bouncer does not reciprocate this love.

Part B These are syllogistic figures identified by Aristotle and his successors, along with their medieval names. Translate each argument into QL.

1. BarbaraAll Bs are Cs. All As are Bs. .˙. All As are Cs.
2. BarocoAll Cs are Bs. Some A is not B. .˙. Some A is not C.
3. Bocardo Some B is not C. All As are Bs. .˙. Some A is not C.
4. Celantes No Bs are Cs. All As are Bs. .˙. No Cs are As.
5. Celarent No Bs are Cs. All As are Bs. .˙. No As are Cs.
6. Cemestres No Cs are Bs. No As are Bs. .˙. No As are Cs.
7. Cesare No Cs are Bs. All As are Bs. .˙. No As are Cs.
8. Dabitis All Bs are Cs. Some A is B. .˙. Some C is A.
9. Darii All Bs are Cs. Some A is B. .˙. Some C is A.
10. Datisi All Bs are Cs. Some A is B. .˙. Some A is C.
11. Disamis Some B is C. All As are Bs. .˙. Some A is C.
12. Ferison No Bs are Cs. Some A is B. .˙. Some A is not C.
13. Ferio No Bs are Cs. Some A is B. .˙. Some A is not C.
14. Festino No Cs are Bs. Some A is B. .˙. Some A is not C.
15. Baralipton All Bs are Cs. All As are Bs. .˙. Some C is A.
16. FrisesomorumSome B is C. No As are Bs. .˙. Some C is not A.

Part C Using the symbolization key given, translate each English-language sentence into QL.

1.  UD: all animals
2.   Dx: x is a dog.
3.    Sx: x likes samurai movies.
4.  Lxy: x is larger than y.
5.       b: Bertie
6.       e: Emerson
7.        f: Fergis

1. Bertie is a dog who likes samurai movies.
2. Bertie, Emerson, and Fergis are all dogs.
3. Emerson is larger than Bertie, and Fergis is larger than Emerson.
4. All dogs like samurai movies.
5. Only dogs like samurai movies.
6. There is a dog that is larger than Emerson.
7. If there is a dog larger than Fergis, then there is a dog larger than Emerson.
8. No animal that likes samurai movies is larger than Emerson.
9. No dog is larger than Fergis.
10. Any animal that dislikes samurai movies is larger than Bertie.
11. There is an animal that is between Bertie and Emerson in size.
12. There is no dog that is between Bertie and Emerson in size.
13. No dog is larger than itself.
14. For every dog, there is some dog larger than it.
15. There is an animal that is smaller than every dog.
16. If there is an animal that is larger than any dog, then that animal does not like samurai movies.

Part D For each argument, write a symbolization key and translate the argument into QL.

1. 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.
2. All my dreams are black and white. Old TV shows are in black and white. Therefore, some of my dreams are old TV shows.
3. Neither Holmes nor Watson has been to Australia. A person could see a kangaroo only if they had been to Australia or to a zoo. Although Watson has not seen a kangaroo, Holmes has. Therefore, Holmes has been to a zoo.
4. 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.
5. An antelope is bigger than a bread box. I am thinking of something that is no bigger than a bread box, and it is either an antelope or a cantaloupe. As such, I am thinking of a cantaloupe.
6. All babies are illogical. Nobody who is illogical can manage a crocodile. Berthold is a baby. Therefore, Berthold is unable to manage a crocodile.

Part E Using the symbolization key given, translate each English-language sentence into QL.

1.  UD: candies
2.   Cx: x has chocolate in it.
3.  Mx: x has marzipan in it.
4.    Sx: x has sugar in it.
5.    Zx: Boris has tried x.
6.  Bxy: x is better than y.

1. Boris has never tried any candy.
2. Marzipan is always made with sugar.
3. Some candy is sugar-free.
4. The very best candy is chocolate.
5. No candy is better than itself.
6. Boris has never tried sugar-free chocolate.
7. Boris has tried marzipan and chocolate, but never together.
8. Any candy with chocolate is better than any candy without it.
9. Any candy with chocolate and marzipan is better than any candy that lacks both.

Part F Using the symbolization key given, translate each English-language sentence into QL.

1.   UD: people and dishes at a potluck
2.    Rx: x has run out.
3.    Tx: x is on the table.
4.     Fx: x is food.
5.     Px: x is a person.
6.   Lxy: x likes y.
7.       e: Eli
8.        f: Francesca
9.       g: the guacamole

1. All the food is on the table.
2. If the guacamole has not run out, then it is on the table.
3. Everyone likes the guacamole.
4. If anyone likes the guacamole, then Eli does.
5. Francesca only likes the dishes that have run out.
6. Francesca likes no one, and no one likes Francesca.
7. Eli likes anyone who likes the guacamole.
8. Eli likes anyone who likes the people that he likes.
9. If there is a person on the table already, then all of the food must have run out.

Part G Using the symbolization key given, translate each English-language sentence into QL.

1.   UD: people
2.    Dx: x dances ballet.
3.     Fx: x is female.
4.   Mx: x is male.
5.  Cxy: x is a child of y.
6.  Sxy: x is a sibling of y.
7.      e: Elmer
8.       j: Jane
9.      p: Patrick

1. All of Patrick’s children are ballet dancers.
2. Jane is Patrick’s daughter.
3. Patrick has a daughter.
4. Jane is an only child.
5. All of Patrick’s daughters dance ballet.
6. Patrick has no sons.
7. Jane is Elmer’s niece.
8. Patrick is Elmer’s brother.
9. Patrick’s brothers have no children.
10. Jane is an aunt.
11. Everyone who dances ballet has a sister who also dances ballet.
12. Every man who dances ballet is the child of someone who dances ballet.

Part H Identify which variables are bound and which are free.

1. ∃xLxy &∀yLyx
2. ∀xAx&Bx
3. ∀x(Ax&Bx)&∀y(Cx&Dy)
4. ∀x∃y[Rxy → (Jz &Kx)] ∨ Ryx
5. ∀x₁(Mx₂ ↔ Lx₂x₁)&∃x₂Lx₃x₂

Part I

1. Identify which of the following are substitution instances of ∀xRcx: Rac, Rca, Raa, Rcb, Rbc, Rcc, Rcd, Rcx
2. Identify which of the following are substitution instances of ∃x∀yLxy: ∀yLby, ∀xLbx, Lab, ∃xLxa

Part J Using the symbolization key given, translate each English-language sentence into QL with identity. The last sentence is ambiguous and can be translated two ways; you should provide both translations. (Hint: Identity is only required for the last four sentences.)

1.  UD: people
2.   Kx: x knows the combination to the safe.
3.   Sx: x is a spy.
4.   Vx: x is a vegetarian.
5.  Txy: x trusts y.
6.       h: Hofthor
7.        i: Ingmar

1. Hofthor is a spy, but no vegetarian is a spy.
2. No one knows the combination to the safe unless Ingmar does.
3. No spy knows the combination to the safe.
4. Neither Hofthor nor Ingmar is a vegetarian.
5. Hofthor trusts a vegetarian.
6. Everyone who trusts Ingmar trusts a vegetarian.
7. Everyone who trusts Ingmar trusts someone who trusts a vegetarian.
8. Only Ingmar knows the combination to the safe.
9. Ingmar trusts Hofthor, but no one else.
10. The person who knows the combination to the safe is a vegetarian.
11. The person who knows the combination to the safe is not a spy.

Part K Using the symbolization key given, translate each English-language sentence into QL with identity. The last two sentences are ambiguous and can be translated two ways; you should provide both translations for each.

1.  UD: cards in a standard deck
2.   Bx: x is black.
3.   Cx: x is a club.
4.   Dx: x is a deuce.
5.    Jx: x is a jack.
6.  Mx: x is a man with an axe.
7.   Ox: x is one-eyed.
8.  Wx: x is wild.

1. All clubs are black cards.
2. There are no wild cards.
3. There are at least two clubs.
4. There is more than one one-eyed jack.
5. There are at most two one-eyed jacks.
6. There are two black jacks.
7. There are four deuces.
8. The deuce of clubs is a black card.
9. One-eyed jacks and the man with the axe are wild.
10. If the deuce of clubs is wild, then there is exactly one wild card.
11. The man with the axe is not a jack.
12. The deuce of clubs is not the man with the axe.

Part L Using the symbolization key given, translate each English-language sentence into QL with identity. The last two sentences are ambiguous and can be translated two ways; you should provide both translations for each.

1.  UD: animals in the world
2.   Bx: x is in Farmer Brown’s field.
3.   Hx: x is a horse.
4.    Px: x is a Pegasus.
5.   Wx: x has wings.

1. There are at least three horses in the world.
2. There are at least three animals in the world.
3. There is more than one horse in Farmer Brown’s field.
4. There are three horses in Farmer Brown’s field.
5. There is a single winged creature in Farmer Brown’s field; any other creatures in the field must be wingless.
6. The Pegasus is a winged horse.
7. The animal in Farmer Brown’s field is not a horse.
8. The horse in Farmer Brown’s field does not have wings.

Practice Exercises

Part A Determine whether each sentence is true or false in the model given.

1.                  UD = {Corwin, Benedict}
2.   extension(A) = {Corwin, Benedict}
3.   extension(B) = {Benedict}
4.   extension(N) =
5.       referent(c) = Corwin

1. Bc
2. Ac ↔ ¬Nc
3. Nc → (Ac ∨ Bc)
4. ∀xAx
5. ∀x¬Bx
6. ∃x(Ax&Bx)
7. ∃x(Ax → Nx)
8. ∀x(Nx ∨ ¬Nx)
9. ∃xBx → ∀xAx

Part B Determine whether each sentence is true or false in the model given.

1.                     UD = {Waylan, Willy, Johnny}
2.      extension(H) = {Waylan, Willy, Johnny}
3.     extension(W) = {Waylan, Willy}
4.      extension(R) = {<Waylan, Willy>,<Willy, Johnny>,<Johnny, Waylan>}
5.          referent(m) = Johnny

1. ∃x(Rxm & Rmx)
2. ∀x(Rxm ∨ Rmx)
3. ∀x(Hx ↔ Wx)
4. ∀x(Rxm → Wx)
5. ∀x[Wx → (Hx & Wx)]
6. ∃xRxx
7. ∃x∃yRxy
8. ∀x∀yRxy
9. ∀x∀y(Rxy ∨ Ryx)
10. ∀x∀y∀z [(Rxy & Ryz) → Rxz]

Part C Determine whether each sentence is true or false in the model given.

1.                  UD = {Lemmy, Courtney, Eddy}
2.   extension(G) = {Lemmy, Courtney, Eddy}
3.   extension(H) = {Courtney}
4.   extension(M) = {Lemmy, Eddy}
5.       referent(c) = Courtney
6.       referent(e) = Eddy

1. Hc
2. He
3. Mc ∨ Me
4. Gc ∨ ¬Gc
5. Mc → Gc
6. ∃xHx
7. ∀xHx
8. ∃x¬Mx
9. ∃x(Hx & Gx)
10. ∃x(Mx & Gx)
11. ∀x(Hx ∨ Mx)
12. ∃xHx & ∃xMx
13. ∀x(Hx ↔ ¬Mx)
14. ∃xGx & ∃x¬Gx
15. ∀x∃y(Gx & Hy)

Part D Write out the model that corresponds to the interpretation given.

1.    UD: natural numbers from 10 to 13
2.     Ox: x is odd.
3.     Sx: x is less than 7.
4.     Tx: x is a two-digit number.
5.     Ux: x is thought to be unlucky.
6.   Nxy: x is the next number after y.

Part E Show that each of the following is contingent.

1. 1 .Da & Db
2. 2. ∃xTxh
3. 3. Pm&¬∀xPx
4.   4. ∀zJz ↔ ∃yJy
5.   5. ∀x(Wxmn ∨ ∃yLxy)
6.   6. ∃x(Gx → ∀yMy)

Part F Show that the following pairs of sentences are not logically equivalent.

1. Ja, Ka
2. ∃xJx, Jm
3. ∀xRxx, ∃xRxx
4. ∃xPx → Qc, ∃x(Px → Qc)
5. ∀x(Px → ¬Qx), ∃x(Px&¬Qx)
6. ∃x(Px&Qx), ∃x(Px → Qx)
7. ∀x(Px → Qx), ∀x(Px&Qx)
8. ∀x∃yRxy, ∃x∀yRxy 9. ∀x∃yRxy, ∀x∃yRyx

Part G Show that the following sets of sentences are consistent.

1. {Ma, ¬Na, Pa, ¬Qa}
2. {Lee, Lef, ¬Lfe, ¬Lff}
3. {¬(Ma & ∃xAx), Ma ∨ Fa, ∀x(Fx → Ax)}
4. {Ma ∨ Mb, Ma → ∀x¬Mx}
5. {∀yGy, ∀x(Gx → Hx), ∃y¬Iy}
6. {∃x(Bx ∨ Ax), ∀x¬Cx, ∀x [(Ax & Bx) → Cx] }
7. {∃xXx, ∃xY x, ∀x(Xx ↔ ¬Y x)}
8. {∀x(Px ∨ Qx), ∃x¬(Qx&Px)}
9. {∃z(Nz &Ozz), ∀x∀y(Oxy → Oyx)}
10. {¬∃x∀yRxy, ∀x∃yRxy}

Part H Construct models to show that the following arguments are invalid.