Solutions to Chapter 25 Multiple generality

A. 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

$L(x,y)$:

blank _x loves blank _y

$a$:

Amos

$b$:

Bouncer

$c$:

Cleo

symbolize each of the following sentences in FOL:

1. 1.
   ​

   If Cleo loves Bouncer, then Bouncer is a monkey.

2. $L(c,b)\to M(b)$

3. 2.
   ​

   If both Bouncer and Cleo are alligators, then Amos loves them both.

4. $(A(b)\wedge A(c))\to (L(a,b)\wedge L(a,c))$

5. 3.
   ​

   Cleo loves a reptile.

6. $\exists x(R(x)\wedge L(c,x))$
   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’.)

7. 4.
   ​

   Bouncer loves all the monkeys that live at the zoo.

8. $\forall x((M(x)\wedge Z(x))\to L(b,x))$

9. 5.
   ​

   All the monkeys that Amos loves love him back.

10. $\forall x((M(x)\wedge L(a,x))\to L(x,a))$

11. 6.
    ​

    Every monkey that Cleo loves is also loved by Amos.

12. $\forall x((M(x)\wedge L(c,x))\to L(a,x))$

13. 7.
    ​

    There is a monkey that loves Bouncer, but sadly Bouncer does not reciprocate this love.

14. $\exists x(M(x)\wedge L(x,b)\wedge \neg L(b,x))$

B. Using the following symbolization key:

domain:

all animals

$D(x)$:

blank _x is a dog

$S(x)$:

blank _x likes samurai movies

$L(x,y)$:

blank _x is larger than blank _y

$r$:

Rave

$h$:

Shane

$d$:

Daisy

symbolize the following sentences in FOL:

1. 1.
   ​

   Rave is a dog who likes samurai movies.

2. $D(r)\wedge S(r)$

3. 2.
   ​

   Rave, Shane, and Daisy are all dogs.

4. $D(r)\wedge D(h)\wedge D(d)$

5. 3.
   ​

   Shane is larger than Rave, and Daisy is larger than Shane.

6. $L(h,r)\wedge L(d,h)$

7. 4.
   ​

   All dogs like samurai movies.

8. $\forall x(D(x)\to S(x))$

9. 5.
   ​

   Only dogs like samurai movies.

10. $\forall x(S(x)\to D(x))$
    Comment: the FOL sentence just written does not require that anyone likes samurai movies. The English sentence might suggest that at least some dogs do like samurai movies?

11. 6.
    ​

    There is a dog that is larger than Shane.

12. $\exists x(D(x)\wedge L(x,h))$

13. 7.
    ​

    If there is a dog larger than Daisy, then there is a dog larger than Shane.

14. $\exists x(D(x)\wedge L(x,d))\to \exists x(D(x)\wedge L(x,h))$

15. 8.
    ​

    No animal that likes samurai movies is larger than Shane.

16. $\forall x(S(x)\to \neg L(x,h))$

17. 9.
    ​

    No dog is larger than Daisy.

18. $\forall x(D(x)\to \neg L(x,d))$

19. 10.
    ​

    Any animal that dislikes samurai movies is larger than Rave.

20. $\forall x(\neg S(x)\to L(x,r))$
    Comment: this is very poor, though! For ‘dislikes’ does not mean the same as ‘does not like’.

21. 11.
    ​

    There is an animal that is between Rave and Shane in size.

22. $\exists x((L(r,x)\wedge L(x,h))\vee (L(h,x)\wedge L(x,r)))$

23. 12.
    ​

    There is no dog that is between Rave and Shane in size.

24. $\forall x\left(D(x)\to \neg \left[(L(b,x)\wedge L(x,h))\vee (L(h,x)\wedge L(x,r))\right]\right)$

25. 13.
    ​

    No dog is larger than itself.

26. $\forall x(D(x)\to \neg L(x,x))$

27. 14.
    ​

    Every dog is larger than some dog.

28. $\forall x(D(x)\to \exists y(D(y)\wedge L(x,y)))$
    Comment: the English sentence is potentially ambiguous here. I have resolved the ambiguity by assuming it should be paraphrased by ‘for every dog, there is a dog smaller than it’.

29. 15.
    ​

    There is an animal that is smaller than every dog.

30. $\exists x\forall y(D(y)\to L(y,x))$

31. 16.
    ​

    If there is an animal that is larger than any dog, then that animal does not like samurai movies.

32. $\forall x(\forall y(D(y)\to L(x,y))\to \neg S(x))$
    Comment: I have assumed that ‘larger than any dog’ here means ‘larger than every dog’.

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

domain:

candies

$C(x)$:

blank _x has chocolate in it

$M(x)$:

blank _x has marzipan in it

$S(x)$:

blank _x has sugar in it

$T(x)$:

Boris has tried blank _x

$B(x,y)$:

blank _x is better than blank _y

1. 1.
   ​

   Boris has never tried any candy.

2. 2.
   ​

   Marzipan is always made with sugar.

3. 3.
   ​

   Some candy is sugar-free.

4. 4.
   ​

   The very best candy is chocolate.

5. 5.
   ​

   No candy is better than itself.

6. 6.
   ​

   Boris has never tried sugar-free chocolate.

7. 7.
   ​

   Boris has tried marzipan and chocolate, but never together.

8. 8.
   ​

   Any candy with chocolate is better than any candy without it.

9. 9.
   ​

   Any candy with chocolate and marzipan is better than any candy that lacks both.

D. Using the following symbolization key:

domain:

people and dishes at a potluck

$R(x)$:

blank _x has run out

$T(x)$:

blank _x is on the table

$F(x)$:

blank _x is food

$P(x)$:

blank _x is a person

$L(x,y)$:

blank _x likes blank _y

$e$:

Eli

$f$:

Francesca

$g$:

the guacamole

symbolize the following English sentences in FOL:

1. 1.
   ​

   All the food is on the table.

2. $\forall x(F(x)\to T(x))$

3. 2.
   ​

   If the guacamole has not run out, then it is on the table.

4. $\neg R(g)\to T(g)$

5. 3.
   ​

   Everyone likes the guacamole.

6. $\forall x(P(x)\to L(x,g))$

7. 4.
   ​

   If anyone likes the guacamole, then Eli does.

8. $\exists x(P(x)\wedge L(x,g))\to L(e,g)$

9. 5.
   ​

   Francesca only likes the dishes that have run out.

10. $\forall x\left[(L(f,x)\wedge F(x))\to R(x)\right]$

11. 6.
    ​

    Francesca likes no one, and no one likes Francesca.

12. $\forall x\left[P(x)\to (\neg L(f,x)\wedge \neg L(x,f))\right]$

13. 7.
    ​

    Eli likes anyone who likes the guacamole.

14. $\forall x((P(x)\wedge L(x,g))\to L(e,x))$

15. 8.
    ​

    Eli likes anyone who likes the people that he likes.

16. $\forall x\left[\left(P(x)\wedge \forall y[(P(y)\wedge L(e,y))\to L(x,y)]\right)\to L(e,x)\right]$

17. 9.
    ​

    If there is a person on the table already, then all of the food must have run out.

18. $\exists x(P(x)\wedge T(x))\to \forall x(F(x)\to R(x))$

E. Using the following symbolization key:

domain:

people

$D(x)$:

blank _x dances ballet

$F(x)$:

blank _x is female

$M(x)$:

blank _x is male

$C(x,y)$:

blank _x is a child of blank _y

$S(x,y)$:

blank _x is a sibling of blank _y

$e$:

Elmer

$j$:

Jane

$p$:

Patrick

symbolize the following sentences in FOL:

1. 1.
   ​

   All of Patrick’s children are ballet dancers.

2. $\forall x(C(x,p)\to D(x))$

3. 2.
   ​

   Jane is Patrick’s daughter.

4. $C(j,p)\wedge F(j)$

5. 3.
   ​

   Patrick has a daughter.

6. $\exists x(C(x,p)\wedge F(x))$

7. 4.
   ​

   Jane is an only child.

8. $\neg \exists xS(x,j)$

9. 5.
   ​

   All of Patrick’s sons dance ballet.

10. $\forall x\left[(C(x,p)\wedge M(x))\to D(x)\right]$

11. 6.
    ​

    Patrick has no sons.

12. $\neg \exists x(C(x,p)\wedge M(x))$

13. 7.
    ​

    Jane is Elmer’s niece.

14. $\exists x(S(x,e)\wedge C(j,x)\wedge F(j))$

15. 8.
    ​

    Patrick is Elmer’s brother.

16. $S(p,e)\wedge M(p)$

17. 9.
    ​

    Patrick’s brothers have no children.

18. $\forall x\left[(S(p,x)\wedge M(x))\to \neg \exists yC(y,x)\right]$

19. 10.
    ​

    Jane is an aunt.

20. $F(j)\wedge \exists x(S(x,j)\wedge \exists yC(y,x))$

21. 11.
    ​

    Everyone who dances ballet has a brother who also dances ballet.

22. $\forall x\left[D(x)\to \exists y(M(y)\wedge S(y,x)\wedge D(y))\right]$

23. 12.
    ​

    Every woman who dances ballet is the child of someone who dances ballet.

24. $\forall x\left[(F(x)\wedge D(x))\to \exists y(C(x,y)\wedge D(y))\right]$