Question {11.1}
1. Alphonzo is Bernadette.
2. Alphonzo is Bernadette and Alphonzo isn’t Candide.
3. There’s someone (other than Candide) who is rich.
4. Everyone (other than Candide) is rich.
5. Everyone (other than Bernadette) is smaller than Bernadette.
6. No-one (other than Bernadette) is smaller than Candide.
7. Everyone works for someone.
8. Everyone works for someone other than themselves.
9. Someone works for someone.
10. Someone works for someone other than themselves.
Question {11.2}
1. Ra & ~ Rc
2. (∃ x)(Px & Rx & Sxc)
3. (∃ x)(Px & x ≠ c & Sxa)
4. (∀ x)(Px ⊃ (∃ y)(Py & Sxy))
5. (∀ x)(Px ⊃ (∃ y)(Py & x ≠ y & Sxy))
6. (∀ x)((Px & x ≠ a) ⊃ (∃ y)(Py & Wxy))
7. ~(∃ x)((Px & x ≠ b) & Wxa)
8. (∃ x)(∃ y)(x ≠ y & Px & Py)
9. (∀ x)(∀ y)(∀ z)(∀ w)((Px & Py & Pz & Pw) ⊃ (x=y ∨ x=z ∨ x=w ∨ y=z ∨ y=w ∨ z=w))
10. (∃ x)((Px & Wxa) & (∀ y)((Py & Way) ⊃ x=y))
Question {11.3}
The first four are tautologies, and the last is not.
1. Tautology.
| ~(∀ x)(x=x) / a |
| | |
| a ≠ a |
| X |
2. Tautology
| ~(∀ x)(∀ y)(x=y ⊃ y=x) / a |
| | |
| ~(∀ x)(a=y ⊃ y=a) / b |
| | |
| ~(a=b ⊃ b=a) |
| | |
| a=b |
| b ≠ a |
| | |
| b ≠ b |
| X |
3. Tautology
| ~(∀ x)(∃ y)x=y / a |
| | |
| ~(∃ y)a=y \a |
| | |
| a ≠ a |
| X |
4. Tautology
| ~(∀ x)(∀ y)(∀ z)((x=y & y=z) ⊃ x=z) / a |
| | |
| ~(∀ y)(∀ z)((a=y & y=z) ⊃ a=z) / b |
| | |
| ~(∀ z)((a=b & b=z) ⊃ a=z) / c |
| | |
| ~((a=b & b=c) ⊃ a=c) |
| | |
| a=b & b=c |
| a ≠ c |
| | |
| a=b |
| b=c |
| | |
| b ≠ c |
| | |
| c ≠ c |
| X |
5. Not a tautology
| ~(∀ x)(∀ y)(∀ z)(∀ w)((Rxy & Rzw) ⊃ (y ≠ z ∨ Rxw)) / a |
| | |
| ~(∀ y)(∀ z)(∀ w)((Ray & Rzw) ⊃ (y ≠ z ∨ Raw)) / b |
| | |
| ~(∀ z)(∀ w)((Rab & Rzw) ⊃ (b ≠ z ∨ Raw)) / c |
| | |
| ~(∀ w)((Rab & Rcw) ⊃ (b ≠ c ∨ Raw)) / d |
| | |
| ~((Rab & Rcd) ⊃ (b ≠ c ∨ Rad)) |
| | |
| Rab & Rcd |
| ~(b ≠ c ∨ Rad) |
| | |
| Rab |
| Rcd |
| | |
| ~(b ≠ c) |
| ~ Rad |
| | |
| b=c |
| | |
| Rac |
This branch remains open. The branch contains the names a, b, c, and d, and we know that b=c. This means we have a domain D = {a,c,d}, and R works at least like this:
| I(R) | a | c | d |
| a | 1 | 0 | |
| c | 1 | ||
| d |
Any interpretation which has this form will make our original formula false. Here is an instance of the quantified formula:
(Rac & Rcd) ⊃ (c ≠ c ∨ Rad)
which is false. So the original formula is false. So, here is one interpretation in which the formula is false.
| I(R) | a | c | d |
| a | 1 | 1 | 0 |
| c | 1 | 1 | 1 |
| d | 1 | 1 | 1 |
Question {11.4}
Argument 1 is invalid, and argument 2 is valid.
1.
| (∀ x)(Fx ⊃ Gx) \a | ||||||
| (∀ x)(Fx ⊃ x=a) \a | ||||||
| ~ Ga | ||||||
| | | ||||||
| Fa ⊃ Ga | ||||||
| Fa ⊃ a=a | ||||||
| / | \ | |||||
| ~ Fa | Ga | |||||
| / | \ | X | ||||
| ~ Fa | a=a |
Both branches on the left here are open. They give rise to this interpretation
| I(F) | I(G) | |
| a | 0 | 0 |
The premises are true, but the conclusion is false.
2.
| (∀ x)(∀ y)x=y \a |
| ~((∀ x)Gx ∨ (∀ x)~ Gx) |
| | |
| ~(∀ x)Gx / a |
| ~(∀ x)~ Gx / b |
| | |
| ~ Ga |
| | |
| ~~ Gb |
| | |
| (∀ y)a=y \b |
| | |
| a=b |
| | |
| ~ Gb |
| X |
The tree closes, the argument is valid.
Question {11.5}
1. At most two frogs are blue, therefore at most three frogs are blue. (∀ x)(∀ y)(∀ z)(((Fx & Bx) & (Fy & By) & (Fz & Bz)) ⊃ (x=y ∨ y=z ∨ x=z)) therefore,
(∀ x)(∀ y)(∀ z)(∀ w)(((Fx & Bx) & (Fy & By) & (Fz & Bz) & (Fw & Bw)) ⊃ (x=y ∨ x=z ∨ x=w ∨ y=z ∨ y=w ∨ z=w))
This is a valid argument. I will write the tree in shorthand, skipping obvious steps or running things together (such as the long disjunctions or conjunctions).
| (∀ x)(∀ y)(∀ z)(((Fx & Bx) & (Fy & By) & (Fz & Bz)) ⊃ (x=y ∨ y=z ∨ x=z)) | ||||||
| ~(∀ x)(∀ y)(∀ z)(∀ w)(((Fx & Bx) & (Fy & By) & (Fz & Bz) & (Fw & Bw)) ⊃ (x=y ∨ x=z ∨ x=w ∨ y=z ∨ y=w ∨ z=w)) | ||||||
| | | ||||||
| ~(((Fa & Ba) & (Fb & Bb) & (Fc & Bc) & (Fd & Bd)) ⊃ (a=b ∨ a=c ∨ a=d ∨ b=c ∨ b=d ∨ c=d)) | ||||||
| | | ||||||
| Fa & Ba | ||||||
| | | ||||||
| Fb & Bb | ||||||
| | | ||||||
| Fc & Bc | ||||||
| | | ||||||
| Fd & Bd | ||||||
| | | ||||||
| ~(a=b ∨ a=c ∨ a=d ∨ b=c ∨ b=d ∨ c=d) | ||||||
| | | ||||||
| a ≠ b | ||||||
| a ≠ c | ||||||
| a ≠ d | ||||||
| b ≠ c | ||||||
| b ≠ d | ||||||
| c ≠ d | ||||||
| | | ||||||
| ((Fa & Ba) & (Fb & Bb) & (Fc & Bc)) ⊃ (a=b ∨ a=c ∨ b=c) | ||||||
| / | \ | |||||
| ~((Fa & Ba) & (Fb & Bb) & (Fc & Bc)) | a=b ∨ a=c ∨ b=c | |||||
| / | | | \ | / | | | \ | |
| ~(Fa & Ba) | ~(Fb & Bb) | ~(Fc & Bc) | a=b | a=c | b=c | |
| X | X | X | X | X | X |
2. At least two frogs are blue, therefore at least three frogs are blue. (∃ x)(∃ y)((Bx & Fx) & (By & Fy) & x ≠ y) therefore (∃ x)(∃ y)(∃ z)((Bx & Fx) & (By & Fy) & (Bz & Fz) & x ≠ y & x ≠ z & y ≠ z)
| (∃ x)(∃ y)((Bx & Fx) & (By & Fy) & x ≠ y) / a,b |
| ~(∃ x)(∃ y)(∃ z)((Bx & Fx) & (By & Fy) & (Bz & Fz) & x ≠ y & x ≠ z & y ≠ z) / a,b |
| | |
| Ba & Fa |
| | |
| Bb & Fb |
| | |
| a ≠ b |
| | |
| ~(∃ z)((Ba & Fa) & (Bb & Fb) & (Bz & Fz) & a ≠ b & a ≠ z & b ≠ z) |
| | |
| : |
| : |
This tree remains open. There’s nothing that we can do to close it. Here’s an interpretation read off the tree which makes the premise true and the conclusion false. D = {a,b}
| I(B) | I(F) | |
| a | 1 | 1 |
| b | 1 | 1 |
Question {11.6}
(You don’t need to know how to do this one.)
Part 1:
| a = b |
| f(a) ≠ f(b) |
| | |
| f(b) ≠ f(b) |
| X |
Part 2:
f(a) = f(b) ⊬ a=b
Consider the counterexample:
D={a,b}
| I(f) | |
| a | a |
| b | a |