# Section 7: Practice Exercises

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

UD: 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

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.

Barbara All $$B$$s are $$C$$s. All $$A$$s are $$B$$s. .˙. All $$A$$s are $$C$$s.
Baroco All $$C$$s are $$B$$s. Some $$A$$ is not $$B$$. .˙. Some $$A$$ is not $$C$$.
Bocardo Some $$B$$ is not $$C$$. All $$A$$s are $$B$$s. .˙. Some $$A$$ is not $$C$$.
Celantes No $$B$$s are $$C$$s. All $$A$$s are $$B$$s. .˙. No $$C$$s are $$A$$s.
Celarent No $$B$$s are $$C$$s. All $$A$$s are $$B$$s. .˙. No $$A$$s are $$C$$s.
Cemestres No $$C$$s are $$B$$s. No $$A$$s are $$B$$s. .˙. No $$A$$s are $$C$$s.
Cesare No $$C$$s are $$B$$s. All $$A$$s are $$B$$s. .˙. No $$A$$s are $$C$$s.
Dabitis All $$B$$s are $$C$$s. Some $$A$$ is $$B$$. .˙. Some $$C$$ is $$A$$.
Darii All $$B$$s are $$C$$s. Some $$A$$ is $$B$$. .˙. Some $$A$$ is $$C$$.
Datisi All $$B$$s are $$C$$s. All $$A$$ is $$C$$. .˙. Some $$A$$ is $$C$$.
Disamis Some $$A$$ is $$B$$. All $$A$$s are $$C$$s. .˙. Some $$B$$ is $$C$$.
Ferison No $$B$$s are $$C$$s. Some $$A$$ is $$B$$. .˙. Some $$A$$ is not $$C$$.
Ferio No $$B$$s are $$C$$s. Some $$A$$ is $$B$$. .˙. Some $$A$$ is not $$C$$.
Festino No $$C$$s are $$B$$s. Some $$A$$ is $$B$$. .˙. Some $$A$$ is not $$C$$.
Baralipton All $$B$$s are $$C$$s. All $$A$$s are $$B$$s. .˙. Some $$C$$ is $$A$$.
Frisesomorum Some $$B$$ is $$C$$. No $$A$$s are $$B$$s. .˙. Some $$C$$ is not $$A$$.

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

UD: all animals
Dx: $$x$$ is a dog.
Sx: $$x$$ likes samurai movies.
Lxy: $$x$$ is larger than $$y$$.
b: Bertie
e: Emerson
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.

UD: candies
Cx: $$x$$ has chocolate in it.
Mx: $$x$$ has marzipan in it.
Sx: $$x$$ has sugar in it.
Tx: Boris has tried $$x$$.
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.

UD: people and dishes at a potluck
Rx: $$x$$ has run out.
Tx: $$x$$ is on the table.
Fx: $$x$$ is food.
Px: $$x$$ is a person.
Lxy: $$x$$ likes $$y$$.
e: Eli
f: Francesca
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.

UD: people
Dx: $$x$$ dances ballet.
Fx: $$x$$ is female.
Mx: $$x$$ is male.
Cxy: $$x$$ is a child of $$y$$.
Sxy: $$x$$ is a sibling of $$y$$.
e: Elmer
j: Jane
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. ∀$$x$$1($$Mx$$2 ↔ $$Lx$$2$$x$$1)&∃$$x$$2$$Lx$$3$$x$$2

Part I 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.)

UD: people
Kx: $$x$$ knows the combination to the safe.
Sx: $$x$$ is a spy.
Vx: $$x$$ is a vegetarian.
Txy: $$x$$ trusts $$y$$.
h: Hofthor
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 J 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.

UD: 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.
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 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.

UD: animals in the world
Bx: $$x$$ is in Farmer Brown’s ﬁeld.
Hx: $$x$$ is a horse.
Px: $$x$$ is a Pegasus.
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.