# Logic Text Chapter 4 Solutions

## Chapter 12: Definite Descriptions

### Basic:

Question {12.1}

1. (Ix)(Vx,Fx)

2. (Ix)(Lx,Hx)

3. (Ix)(Fx & Hx,Sxb)

4. La

5. (Ix)(Lx,x=a)

6. (Ix)(Hx & ~ Fx, Scx)

7. (Ix)(Fx & Hx, (Iy)(Hy & ~ Fy,Sxy))

8. (∃ y)(Hy & ~ Fy & (Ix)(Fx & Hx, Sxy))

9. (∀ x)((Iy)(Fy & Hy, xy) ⊃ (Iy)(Fy & Hy, Sxy))

10. (Ix)(Vx, Sax) & (Ix)(Fx & Hx, Sbx)

Question {12.2}

1. (∃ x)(Vx & (∀ y)(Vyx=y) & Fx)

2. (∃ x)(Lx & (∀ y)(Lyx=y) & Hx)

3. (∃ x)(Fx & Hx & (∀ y)((Fy & Hy) ⊃ x=y) & Sxb)

4. La

5. (∃ x)(Lx & (∀ y)(Lyx=y) & x=a)

6. (∃ x)(Hx & ~ Fx & (∀ y)((Hy & ~ Fy) ⊃ x=y) & Scx)

7. (∃ x)(Fx & Hx & (∀ y)((Fy & Hy) ⊃ x=y) & ((∃ y)(Hy & ~ Fy & (∀ z)((Hz & ~ Fz) ⊃ y=z) & Sxy)))

8. (∃ y)(Hy & ~ Fy & (∃ x)(Fx & Hx & (∀ z)((Fz & Hz) ⊃ x=z) & Sxy))

9. (∀ x)((∃ y)(Fy & Hy & (∀ z)((Fz & Hz) ⊃ y=z) & xy) ⊃ (∃ y)(Fy & Hy & (∀ z)((Fz & Hz) ⊃ y=z) & Sxy))

10. (∃ x)(Vx & (∀ y)(Vyx=y) & Sax) & (∃ x)(Fx & Hx & (∀ y)((Fy & Hy) ⊃ x=y) & Sbx)

Question {12.3}

1. (Ix)(Fx, Gx) & (Ix)(Fx, Hx), therefore (Ix)(Fx, Gx & Hx) is valid.

 (Ix)(Fx, Gx) & (Ix)(Fx, Hx) ~(Ix)(Fx, Gx & Hx) / b | (Ix)(Fx, Gx) / a (Ix)(Fx, Hx) / b | Fa (∀ x)(Fx ⊃ x=a) \b Ga | Fb (∀ x)(Fx ⊃ x=b) \c Hb | Fb ⊃ b=a / \ ~ Fb b=a X / | \ ~ Fb ~(∀ x)(Fx ⊃ b=x) / c ~(Gb & Hb) X | / \ ~(Fc ⊃ b=c) ~ Gb ~ Hb | | X Fc ~ Ga b ≠ c X | Fc ⊃ c=b / \ ~ Fc c=b X | b ≠ b X

2.

(Ix)(Fx & Gx, Hx), therefore (Ix)(Fx, Hx). This is invalid.

 (Ix)(Fx & Gx, Hx) / a ~(Ix)(Fx, Hx) \a | Fa & Ga (∀ x)((Fx & Gx) ⊃ x=a) \b Ha | Fa Ga / | \ ~ Fa ~(∀ x)(Fx ⊃ x=a) / b ~ Ha X | X ~(Fb ⊃ b=a) | Fb b ≠ a | (Fb & Gb) ⊃ b=a / \ ~(Fb & Gb) b=a / \ X ~ Fb ~ Gb X

The remaining branch is open. We could substitute a in the universal quantifier, but that would just give us (Fa & Ga) ⊃ a=a, which is a tautology anyway.

The open branch gives us a model D = {a,b}

 I(F) I(G) I(H) a 1 1 1 b 1 0

There’s nothing to tell us whether b has property H or not. So let’s just say Hb is false.

 I(F) I(G) I(H) a 1 1 1 b 1 0 0

3. (Ix)(Fx, Gx), therefore (∀ x)(~ Gx ⊃ ~ Fx) is valid.

 (Ix)(Fx, Gx) / a ~(∀ x)(~ Gx ⊃ ~ Fx) / b | Fa (∀ x)(Fx ⊃ x=a) \b Ga | ~(~ Gb ⊃ ~ Fb) | ~ Gb ~~ Fb | Fb ⊃ b=a / \ ~ Fb b=a X | ~ Ga X

4. (Ix)(Fx, ~ Gx), therefore ~(Ix)(Fx, Gx), is valid.

 (Ix)(Fx, ~ Gx) / a ~~(Ix)(Fx, Gx) | (Ix)(Fx, Gx) / b | Fa (∀ x)(Fx ⊃ x=a) \b ~ Ga | Fb (∀ x)(Fx ⊃ x=b) Gb | Fb ⊃ b=a / \ ~ Fb b=a X | Ga X

5. Fa, (Ix)(Fx, Gx), therefore (Ix)(Fx, x=a), is valid.

 Fa (Ix)(Fx, Gx) / b ~(Ix)(Fx, x=a) / b | Fb (∀ x)(Fx ⊃ x=b) \a Gb | Fa ⊃ a=b / \ ~ Fa a=b X / | \ ~ Fb ~(∀ x)(Fx ⊃ x=b) b ≠ a X X | b ≠ b X

Question {12.4}

(Ix)(Fx, Fx) is “the F is an F”, which means, there’s exactly one thing with property F.