Solutions to Chapter 28 Definite descriptions

A. 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

$T(x,y)$:

blank _x trusts blank _y

$h$:

Hofthor

$i$:

Ingmar

symbolize the following sentences in FOL:

1. 1.
   ​

   Hofthor trusts a vegetarian.

2. $\exists x(V(x)\wedge T(h,x))$

3. 2.
   ​

   Everyone who trusts Ingmar trusts a vegetarian.

4. $\forall x\left[T(x,i)\to \exists y(T(x,y)\wedge V(y))\right]$

5. 3.
   ​

   Everyone who trusts Ingmar trusts someone who trusts a vegetarian.

6. $\forall x[T(x,i)\to \exists y(T(x,y)\wedge \exists z(T(y,z)\wedge V(z)))]$

7. 4.
   ​

   Only Ingmar knows the combination to the safe.

8. $\forall x(K(x)\to x=i)$
   Comment: does the English claim entail that Ingmar does know the combination to the safe? If so, then we should formalize this with a ‘$\leftrightarrow$’.

9. 5.
   ​

   Ingmar trusts Hofthor, but no one else.

10. $\forall x(T(i,x)\leftrightarrow x=h)$

11. 6.
    ​

    The person who knows the combination to the safe is a vegetarian.

12. $\exists x\left[K(x)\wedge \forall y(K(y)\to x=y)\wedge V(x)\right]$

13. 7.
    ​

    The person who knows the combination to the safe is not a spy.

14. $\exists x\left[K(x)\wedge \forall y(K(y)\to x=y)\wedge \neg S(x)\right]$
    Comment: the scope of negation is potentially ambiguous here; I have read it as inner negation.

B. Using the following symbolization key:

domain:

cards in a standard deck

$B(x)$:

blank _x is black

$C(x)$:

blank _x is a club

$D(x)$:

blank _x is a deuce

$J(x)$:

blank _x is a jack

$M(x)$:

blank _x is a man with an axe

$O(x)$:

blank _x is one-eyed

$W(x)$:

blank _x is wild

symbolize each sentence in FOL:

1. 1.
   ​

   All clubs are black cards.

2. $\forall x(C(x)\to B(x))$

3. 2.
   ​

   There are no wild cards.

4. $\neg \exists xW(x)$

5. 3.
   ​

   There are at least two clubs.

6. $\exists x\exists y(\neg x=y\wedge C(x)\wedge C(y))$

7. 4.
   ​

   There is more than one one-eyed jack.

8. $\exists x\exists y(\neg x=y\wedge J(x)\wedge O(x)\wedge J(y)\wedge O(y))$

9. 5.
   ​

   There are at most two one-eyed jacks.

10. $\forall x\forall y\forall z\left[(J(x)\wedge O(x)\wedge J(y)\wedge O(y)\wedge J(z)\wedge O(z))\to (x=y\vee x=z\vee y=z)\right]$

11. 6.
    ​

    There are two black jacks.

12. $\exists x\exists y(\neg x=y\wedge B(x)\wedge J(x)\wedge B(y)\wedge J(y))$
    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:
    $\exists x\exists y\left(\neg x=y\wedge B(x)\wedge J(x)\wedge B(y)\wedge J(y)\wedge \forall z[(B(z)\wedge J(z))\to (x=z\vee y=z)]\right)$

13. 7.
    ​

    There are four deuces.

14. $\exists w\exists x\exists y\exists z(\neg w=x\wedge \neg w=y\wedge \neg w=z\wedge \neg x=y\wedge \neg x=z\wedge \neg y=z\wedge D(w)\wedge D(x)\wedge D(y)\wedge D(z))$
    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:
    $\exists w\exists x\exists y\exists z\left(\neg w=x\wedge \neg w=y\wedge \neg w=z\wedge \neg x=y\wedge \neg x=z\wedge \neg y=z\wedge D(w)\wedge D(x)\wedge D(y)\wedge D(z)\wedge \forall v[D(v)\to (v=w\vee v=x\vee v=y\vee v=z)]\right)$

15. 8.
    ​

    The deuce of clubs is a black card.

16. $\exists x\left[D(x)\wedge C(x)\wedge \forall y\left((D(y)\wedge C(y))\to x=y\right)\wedge B(x)\right]$

17. 9.
    ​

    One-eyed jacks and the man with the axe are wild.

18. $\forall x\left[(J(x)\wedge O(x))\to W(x)\right]\wedge \exists x\left[M(x)\wedge \forall y(M(y)\to x=y)\wedge W(x)\right]$

19. 10.
    ​

    If the deuce of clubs is wild, then there is exactly one wild card.

20. $\exists x\left(D(x)\wedge C(x)\wedge \forall y\left[(D(y)\wedge C(y))\to x=y\right]\wedge W(x)\right)\to \exists x\left(W(x)\wedge \forall y(W(y)\to x=y)\right)$
    Comment: if there is not exactly one deuce of clubs, then the above sentence is true. Maybe that’s the wrong verdict. Perhaps the sentence should definitely be taken to imply that there is one and only one deuce of clubs, and then express a conditional about wildness. If so, then we might symbolize it thus:
    $\exists x(D(x)\wedge C(x)\wedge \forall y[(D(y)\wedge C(y))\to x=y]\wedge [W(x)\to \forall y(W(y)\to x=y)])$

21. 11.
    ​

    The man with the axe is not a jack.

22. $\exists x\left[M(x)\wedge \forall y(M(y)\to x=y)\wedge \neg J(x)\right]$

23. 12.
    ​

    The deuce of clubs is not the man with the axe.

24. $\exists x\exists y\left(D(x)\wedge C(x)\wedge \forall z[(D(z)\wedge C(z))\to x=z]\wedge M(y)\wedge \forall z(M(z)\to y=z)\wedge \neg x=y\right)$

C. Using the following symbolization key:

domain:

animals in the world

$B(x)$:

blank _x is in Farmer Brown’s field

$H(x)$:

blank _x is a horse

$P(x)$:

blank _x is a Pegasus

$W(x)$:

blank _x has wings

symbolize the following sentences in FOL:

1. 1.
   ​

   There are at least three horses in the world.

2. $\exists x\exists y\exists z(\neg x=y\wedge \neg x=z\wedge \neg y=z\wedge H(x)\wedge H(y)\wedge H(z))$

3. 2.
   ​

   There are at least three animals in the world.

4. $\exists x\exists y\exists z(\neg x=y\wedge \neg x=z\wedge \neg y=z)$

5. 3.
   ​

   There is more than one horse in Farmer Brown’s field.

6. $\exists x\exists y(\neg x=y\wedge H(x)\wedge H(y)\wedge B(x)\wedge B(y))$

7. 4.
   ​

   There are three horses in Farmer Brown’s field.

8. $\exists x\exists y\exists z(\neg x=y\wedge \neg x=z\wedge \neg y=z\wedge H(x)\wedge H(y)\wedge H(z)\wedge B(x)\wedge B(y)\wedge B(z))$
   Comment: I have 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.

9. 5.
   ​

   There is a single winged creature in Farmer Brown’s field; any other creatures in the field must be wingless.

10. $\exists x\left[W(x)\wedge B(x)\wedge \forall y((W(y)\wedge B(y))\to x=y)\right]$

11. 6.
    ​

    The Pegasus is a winged horse.

12. $\exists x\left[P(x)\wedge \forall y(P(y)\to x=y)\wedge W(x)\wedge H(x)\right]$

13. 7.
    ​

    The animal in Farmer Brown’s field is not a horse.

14. $\exists x\left[Bx\wedge \forall y(B(y)\to x=y)\wedge \neg H(x)\right]$
    Comment: the scope of negation might be ambiguous here; I have read it as inner negation.

15. 8.
    ​

    The horse in Farmer Brown’s field does not have wings.

16. $\exists x\left[H(x)\wedge B(x)\wedge \forall y\left((H(y)\wedge B(y))\to x=y\right)\wedge \neg W(x)\right]$
    Comment: the scope of negation might be ambiguous here; I have read it as inner negation.

D. In this chapter, we symbolized ‘Nick is the traitor’ by ‘$\exists x(T(x)\wedge \forall y(T(y)\to x=y)\wedge x=n)$’. Explain why these would be equally good symbolisations:

• ‣
  ​

  $T(n)\wedge \forall y(T(y)\to n=y)$

• This sentence requires that Nick is a traitor, and that Nick alone is a traitor. Otherwise put, there is one and only one traitor, namely, Nick. Otherwise put: Nick is the traitor.

• ‣
  ​

  $\forall y(T(y)\leftrightarrow y=n)$

• This sentence can be understood thus: Take anything you like; now, if you chose a traitor, you chose Nick, and if you chose Nick, you chose a traitor. So there is one and only one traitor, namely, Nick, as required.