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, x ≠ y) ⊃ (Iy)(Fy & Hy, Sxy))
10. (Ix)(Vx, Sax) & (Ix)(Fx & Hx, Sbx)
Question {12.2}
1. (∃ x)(Vx & (∀ y)(Vy ⊃ x=y) & Fx)
2. (∃ x)(Lx & (∀ y)(Ly ⊃ x=y) & Hx)
3. (∃ x)(Fx & Hx & (∀ y)((Fy & Hy) ⊃ x=y) & Sxb)
4. La
5. (∃ x)(Lx & (∀ y)(Ly ⊃ x=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) & x ≠ y) ⊃ (∃ y)(Fy & Hy & (∀ z)((Fz & Hz) ⊃ y=z) & Sxy))
10. (∃ x)(Vx & (∀ y)(Vy ⊃ x=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.