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)(Fxx=a) \b
Ga
|
Fb
(∀ x)(Fxx=b) \c
Hb
|
Fbb=a
/\
~ Fbb=a
X /|\
~ Fb~(∀ x)(Fxb=x) / c~(Gb & Hb)
X|/\
~(Fcb=c)~ Gb~ Hb
||X
Fc~ Ga
bcX
|
Fcc=b
/\
~ Fcc=b
X |
bb
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)(Fxx=a) / b~ Ha
X |X
~(Fbb=a)
|
Fb
ba
|
(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 111
b 10

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 111
b 100

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

(Ix)(Fx, Gx) / a
~(∀ x)(~ Gx ⊃ ~ Fx) / b
|
Fa
(∀ x)(Fxx=a) \b
Ga
|
~(~ Gb ⊃ ~ Fb)
|
~ Gb
~~ Fb
|
Fbb=a
/\
~ Fbb=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)(Fxx=a) \b
~ Ga
|
Fb
(∀ x)(Fxx=b)
Gb
|
Fbb=a
/\
~ Fbb=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)(Fxx=b) \a
Gb
|
Faa=b
/\
~ Faa=b
X /|\
~ Fb~(∀ x)(Fxx=b)ba
XX |
bb
X

Question {12.4}

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