8.4.xa. Exercise answers
| 1. | a. |
using Russell’s analysis:
Sam guessed the winning number the winning number is such that (Sam guessed it) (∃x: x is a winning number ∧ only x is a winning number) Sam guessed x (∃x: Wx ∧ (∀y: ¬ y = x) ¬ y is a winning number) Gsx
(∃x: Wx ∧ (∀y: ¬ y = x) ¬ Wy) Gsx
∃x (Wx ∧ ∀y (¬ y = x → ¬ Wy) ∧ Gsx) or: (∃x: Wx ∧ (∀y: Wy) x = y) Gsx ∃x (Wx ∧ ∀y (Wy → x = y) ∧ Gsx)
G: [ _ guessed _ ]; W: [ _ is a winning number]; s: Sam
with the description operator:
Sam guessed the winning number G Sam the winning number Gs(Ix x is a winning number)
Gs(Ix Wx)
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| b. |
using Russell’s analysis:
The winner who spoke to Tom was well-known The winner who spoke to Tom is such that (he or she was well-known) (∃x: x is a winner who spoke to Tom ∧ only x is a winner who spoke to Tom) x was well-known (∃x: (x is a winner ∧ x spoke to Tom) ∧ (∀y: ¬ y = x) ¬ (y is a winner ∧ y spoke to Tom)) Kx
(∃x: (Wx ∧ Sxt) ∧ (∀y: ¬ y = x) ¬ (Wy ∧ Syt)) Kx
∃x ((Wx ∧ Sxt) ∧ ∀y (¬ y = x → ¬ (Wy ∧ Syt)) ∧ Kx) or: (∃x: (Wx ∧ Sxt) ∧ (∀y: Wy ∧ Syt) x = y) Kx ∃x ((Wx ∧ Sxt) ∧ ∀y ((Wy ∧ Syt) → x = y) ∧ Kx) K: [ _ was well-known]; S: [ _ spoke to _ ]; W: [ _ is a winner]; t: Tom] |
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with the description operator:
The winner who spoke to Tom was well-known The winner who spoke to Tom was well-known K the winner who spoke to Tom K(Ix (x is a winner who spoke to Tom)) K(Ix (x is a winner ∧ x spoke to Tom))
K(Ix (Wx ∧ Sxt))
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| c. |
using Russell’s analysis:
The winner, who spoke to Tom, was well-known. The winner is such that (he or she, who spoke to Tom, was well-known). (∃x: x is a winner ∧ only x is a winner) x, who spoke to Tom, was well-known) (∃x: x is a winner ∧ (∀y: ¬ y = x) ¬ y is a winner) (x spoke to Tom ∧ x was well-known)
(∃x: Wx ∧ (∀y: ¬ y = x) ¬ Wy) (Sxt ∧ Kx)
∃x (Wx ∧ ∀y (¬ y = x → ¬ Wy) ∧ (Sxt ∧ Kx)) or: (∃x: Wx ∧ (∀y: Wy) x = y) (Sxt ∧ Kx) ∃x (Wx ∧ ∀y (Wy → x = y) ∧ (Sxt ∧ Kx)) K: [ _ was well-known]; S: [ _ spoke to _ ]; W: [ _ is a winner]; t: Tom
with the description operator:
The winner, who spoke to Tom, was well-known. The winner spoke to Tom ∧ the winner was well-known S the winner Tom ∧ K the winner S(Ix x is a winner)t ∧ K(Ix x is a winner)
S(Ix Wx)t ∧ K(Ix Wx)
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| d. |
using Russell’s analysis:
Every number greater than one is greater than its positive square root (∀x: x is a number greater than one) x is greater than its positive square root (∀x: x is a number ∧ x is greater than one) x is greater than the positive square root of x (∀x: Nx ∧ Gxo) the positive square root of x is such that (x is greater than it) (∀x: Nx ∧ Gxo) (∃y: y is a positive square root of x ∧ only y is a positive square root of x) x is greater than y (∀x: Nx ∧ Gxo) (∃y: (y is positive ∧ y is a square root of x) ∧ (∀z: ¬ z = y) ¬ (z is positive ∧ z is a square root of x)) Gxy
(∀x: Nx ∧ Gxo) (∃y: (Py ∧ Syx) ∧ (∀z: ¬ z = y) ¬ (Pz ∧ Szx)) Gxy
or:
(∀x: Nx ∧ Gxo) (∃y: (Py ∧ Syx) ∧ (∀z: Pz ∧ Szx) y = z) Gxy
G: [ _ is greater than y]; N: [ _ is a number]; P: [ _ is positive]; S: [ _ is a square root of _ ] |
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with the description operator:
Every number greater than one is greater than its positive square root (∀x: x is a number ∧ x is greater than one) x is greater than the positive square root of x (∀x: Nx ∧ Gxo) G x the positive square root of x (∀x: Nx ∧ Gxo) Gx(Iy y is a positive square root of x) (∀x: Nx ∧ Gxo) Gx(Iy (y is a positive ∧ y is a square root of x))
(∀x: Nx ∧ Gxo) Gx(Iy (Py ∧ Syx))
∀x ( (Nx ∧ Gxo) → Gx(Iy (Py ∧ Syx)) ) |
| 2. | a. |
(∃x: x owns Spot ∧ (∀y: ¬ y = x) ¬ y owns Spot) x called (∃x: x owns Spot ∧ only x owns Spot) x called The owner of Spot is such that (it called) Spot’s owner called |
| b. |
John found (Ix (x is a photographer ∧ x enlarged (Iy y is a picture of John))) John found (Ix (x is a photographer ∧ x enlarged the picture of John)) John found (Ix (x is a photographer who enlarged the picture of John)) John found the photographer who enlarged the picture of him |