Solutions to Chapter 24 Sentences with one quantifier

A. Here are the syllogistic figures identified by Aristotle and his successors, along with their medieval names:

1. 1.
   ​

   Barbara. All G are F. All H are G. So: All H are F

2. $\forall x(G(x)\to F(x)),\forall x(H(x)\to G(x))\therefore \forall x(H(x)\to F(x))$

3. 2.
   ​

   Celarent. No G are F. All H are G. So: No H are F

4. $\forall x(G(x)\to \neg F(x)),\forall x(H(x)\to G(x))\therefore \forall x(H(x)\to \neg F(x))$

5. 3.
   ​

   Ferio. No G are F. Some H is G. So: Some H is not F

6. $\forall x(G(x)\to \neg F(x)),\exists x(H(x)\wedge G(x))\therefore \exists x(H(x)\wedge \neg F(x))$

7. 4.
   ​

   Darii. All G are H. Some H is G. So: Some H is F.

8. $\forall x(G(x)\to F(x)),\exists x(H(x)\wedge G(x))\therefore \exists x(H(x)\wedge F(x))$

9. 5.
   ​

   Camestres. All F are G. No H are G. So: No H are F.

10. $\forall x(F(x)\to G(x)),\forall x(H(x)\to \neg G(x))\therefore \forall x(H(x)\to \neg F(x))$

11. 6.
    ​

    Cesare. No F are G. All H are G. So: No H are F.

12. $\forall x(F(x)\to \neg G(x)),\forall x(H(x)\to G(x))\therefore \forall x(H(x)\to \neg F(x))$

13. 7.
    ​

    Baroko. All F are G. Some H is not G. So: Some H is not F.

14. $\forall x(F(x)\to G(x)),\exists x(H(x)\wedge \neg G(x))\therefore \exists x(H(x)\wedge \neg F(x))$

15. 8.
    ​

    Festino. No F are G. Some H are G. So: Some H is not F.

16. $\forall x(F(x)\to \neg G(x)),\exists x(H(x)\wedge G(x))\therefore \exists x(H(x)\wedge \neg F(x))$

17. 9.
    ​

    Datisi. All G are F. Some G is H. So: Some H is F.

18. $\forall x(G(x)\to F(x)),\exists x(G(x)\wedge H(x))\therefore \exists x(H(x)\wedge F(x))$

19. 10.
    ​

    Disamis. Some G is F. All G are H. So: Some H is F.

20. $\exists x(G(x)\wedge F(x)),\forall x(G(x)\to H(x))\therefore \exists x(H(x)\wedge F(x))$

21. 11.
    ​

    Ferison. No G are F. Some G is H. So: Some H is not F.

22. $\forall x(G(x)\to \neg F(x)),\exists x(G(x)\wedge H(x))\therefore \exists x(H(x)\wedge \neg F(x))$

23. 12.
    ​

    Bokardo. Some G is not F. All G are H. So: Some H is not F.

24. $\exists x(G(x)\wedge \neg F(x)),\forall x(G(x)\to H(x))\therefore \exists x(H(x)\wedge \neg F(x))$

25. 13.
    ​

    Camenes. All F are G. No G are H So: No H is F.

26. $\forall x(F(x)\to G(x)),\forall x(G(x)\to \neg H(x))\therefore \forall x(H(x)\to \neg F(x))$

27. 14.
    ​

    Dimaris. Some F is G. All G are H. So: Some H is F.

28. $\exists x(F(x)\wedge G(x)),\forall x(G(x)\to H(x))\therefore \exists x(H(x)\wedge F(x))$

29. 15.
    ​

    Fresison. No F are G. Some G is H. So: Some H is not F.

30. $\forall x(F(x)\to \neg G(x)),\exists x(G(x)\wedge H(x))\therefore \exists (H(x)\wedge \neg F(x))$

Symbolize each argument in FOL.

B. Using the following symbolization key:

domain:

people

$K(x)$:

blank _x knows the combination to the safe

$S(x)$:

blank _x is a spy

$V(x)$:

blank _x is a vegetarian

$h$:

Hofthor

$i$:

Ingmar

symbolize the following sentences in FOL:

1. 1.
   ​

   Neither Hofthor nor Ingmar is a vegetarian.

2. $\neg V(h)\wedge \neg V(i)$

3. 2.
   ​

   No spy knows the combination to the safe.

4. $\forall x(S(x)\to \neg K(x))$

5. 3.
   ​

   No one knows the combination to the safe unless Ingmar does.

6. $\forall x\neg K(x)\vee K(i)$

7. 4.
   ​

   Hofthor is a spy, but no vegetarian is a spy.

8. $S(h)\wedge \forall x(V(x)\to \neg S(x))$

C. Using this symbolization key:

domain:

all animals

$A(x)$:

blank _x is an alligator

$M(x)$:

blank _x is a monkey

$R(x)$:

blank _x is a reptile

$Z(x)$:

blank _x lives at the zoo

$a$:

Amos

$b$:

Bouncer

$c$:

Cleo

symbolize each of the following sentences in FOL:

1. 1.
   ​

   Amos, Bouncer, and Cleo all live at the zoo.

2. $Z(a)\wedge Z(b)\wedge Z(c)$

3. 2.
   ​

   Bouncer is a reptile, but not an alligator.

4. $R(b)\wedge \neg A(b)$

5. 3.
   ​

   Some reptile lives at the zoo.

6. $\exists x(R(x)\wedge Z(x))$

7. 4.
   ​

   Every alligator is a reptile.

8. $\forall x(A(x)\to R(x))$

9. 5.
   ​

   Any animal that lives at the zoo is either a monkey or an alligator.

10. $\forall x(Z(x)\to (M(x)\vee A(x)))$

11. 6.
    ​

    There are reptiles which are not alligators.

12. $\exists x(R(x)\wedge \neg A(x))$

13. 7.
    ​

    If any animal is an reptile, then Amos is.

14. $\exists xR(x)\to R(a)$

15. 8.
    ​

    If any animal is an alligator, then it is a reptile.

16. $\forall x(A(x)\to R(x))$

D. For each argument, write a symbolization key and symbolize the argument in FOL.

1. 1.
   ​

   Willard is a logician. All logicians wear funny hats. So Willard wears a funny hat

   domain:

   people

   $L(x)$:

   blank _x is a logician

   $H(x)$:

   blank _x wears a funny hat

   $i$:

   Willard

   $L(i),\forall x(L(x)\to H(x))\therefore H(i)$

2. 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

   $D(x)$:

   blank _x is on my desk

   $E(x)$:

   blank _x escapes my attention

   $C(x)$:

   blank _x is a computer

   $\forall x(D(x)\to \neg E(x)),\exists x(D(x)\wedge C(x))\therefore \exists x(C(x)\wedge \neg E(x))$

3. 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)

   $D(x)$:

   blank _x is one of my dreams

   $B(x)$:

   blank _x is in black and white

   $O(x)$:

   blank _x is an old TV show

   $\forall x(D(x)\to B(x)),\forall x(O(x)\to B(x))\therefore \exists x(D(x)\wedge O(x))$.
   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 have gone with the former, but it is not clear that FOL deals with these well.

4. 4.
   ​

   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.

   domain:

   people

   $A(x)$:

   blank _x has been to Australia

   $K(x)$:

   blank _x has seen a kangaroo

   $Z(x)$:

   blank _x has been to a zoo

   $h$:

   Holmes

   $a$:

   Watson

   $\neg A(h)\wedge \neg A(a),\forall x(K(x)\to (A(x)\vee Z(x))),\neg K(a)\wedge K(h)\therefore Z(h)$

5. 5.
   ​

   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

   $S(x)$:

   blank _x expects the Spanish Inquisition

   $T(x)$:

   blank _x knows the troubles I’ve seen

   $\forall x\neg S(x),\forall x\neg T(x)\therefore \forall x(S(x)\to T(x))$

6. 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.

   domain:

   people

   $B(x)$:

   blank _x is a baby

   $I(x)$:

   blank _x is illogical

   $C(x)$:

   blank _x can manage a crocodile

   $b$:

   Berthold

   $\forall x(B(x)\to I(x)),\forall x(I(x)\to \neg C(x)),B(b)\therefore \neg C(b)$