Phi 270 F98 test 5

F98 test 5 questions

(These questions are from the last of the 6 quizzes given in F98.)

Analyze the following sentences in as much detail as possible, providing a key to the non-logical vocabulary (upper and lower case letters) appearing in your answer.
1. George traveled to LA by way of some town in Wyoming. [Give this analysis also using an unrestricted quantifier.]
answer
2. Everyone is afraid of something. [This sentence is ambiguous. Analyze it in two different ways, and describe a situation in which the sentence is true on one of your interpretations and false on the other.]
answer
3. Spot knew exactly one trick.
answer
Analyze the sentence below using each of the two ways of analyzing definite descriptions. That is, analyze it using Russell’s analysis of definite descriptions as quantifier phrases and then analyze it again using the description operator.
4. Tom opened the letter from Bulgaria
answer
Use derivations to show that the following argument is valid. You may use any rules.
5.
∃x (Fx ∧ ∃y ¬ x = y)
∃x ∃y (¬ y = x ∧ Fy)

That is: Some finding is different from something ⊨ Something is such that something different from it is a finding [but don’t hesitate to ignore the English if it doesn’t help].

answer
Use a derivation to show that the following argument is not valid and describe a structure dividing an open gap.
6.
∃x ∃y Rxy
∃x Rxx
answer
Complete the following to give a definition of equivalence in terms of truth values and possible worlds:
7. A sentence φ is equivalent to a sentence ψ if and only if …
answer
Describe a structure (i.e., an assignment of extensions to the non-logical vocabulary) which makes the 8 sentences below all true.
8.
fab = fba, ga = fab, fba = c, Fb, F(ga), Rab, ¬ Rba, R(ga)c
answer
9. [This question was on a topic not covered this year]

F98 test 5 answers

1.

George traveled to LA by way of some town in Wyoming

some town in Wyoming is such that (George traveled to LA by way of it)

(∃x: x is a town in Wyoming) George traveled to LA by way of x

(∃x: x is a town ∧ x is in Wyoming) George traveled to LA by way of x

(∃x: Tx ∧ Nxm) Rglx
∃x ((Tx ∧ Nxm) ∧ Rglx)

N: [ _ is in _ ]; R: [ _ traveled to _ by way of _ ]; T: [ _ is a town]; g: George; l: LA; m: Wyoming

2.

first analysis:

Everyone is afraid of something

everyone is such that (he or she is afraid of something)

(∀x: x is a person) x is afraid of something

(∀x: Px) something is such that (x is afraid of it)

(∀x: Px) ∃y x is afraid of y

(∀x: Px) ∃y Axy

second analysis:

Everyone is afraid of something

something is such that (everyone is afraid of it)

∃x everyone is afraid of x

∃x everyone is such that (he or she is afraid of x)

∃x (∀y: y is a person) y is afraid of x

∃x (∀y: Py) Ayx

A: [ _ is afraid of _ ]; P: [ _ is a person]

The first is true and the second false if all people are fearful but not all fearful of the same thing

3.

Spot knew exactly one trick

Spot knew a trick ∧ ¬ Spot knew at least two tricks

(∃ x: x is a trick) Spot knew x ∧ ¬ (∃ x: x is a trick) (∃ y: y is a trick ∧ ¬ y = x) (Spot knew x ∧ Spot knew y)

(∃ x: Tx) Ksx ∧ ¬ (∃ x: Tx) (∃ y: Ty ∧ ¬ y = x) (Ksx ∧ Ksy)
or
(∃ x: Tx) (Ksx ∧ (∀ y: Ty ∧ ¬ y = x) ¬ Ksy)
or
(∃ x: Tx) (Ksx ∧ (∀ y: Ty ∧ Ksy) x = y)

K: [ _ knew _ ]; T: [ _ is a trick]; s: Spot

4.

using Russell’s analysis:

Tom opened the letter from Bulgaria

the letter from Bulgaria is such that (Tom opened it)

(∃x: x and only x is a letter from Bulgaria) Tom opened x

(∃x: x is a letter from Bulgaria ∧ (∀y: ¬ y = x) ¬ y is a letter from Bulgaria) Otx

(∃x: x is a letter ∧ x is from Bulgaria ∧ (∀y: ¬ y = x) ¬ y is a letter ∧ y is from Bulgaria) Otx

(∃x: (Lx ∧ Fxb) ∧ (∀y: ¬ y = x) ¬ (Ly ∧ Fyb)) Otx
or: (∃x: (Lx ∧ Fxb) ∧ (∀y: Ly ∧ Fyb) x = y) Otx

using the description operator:

Tom opened the letter from Bulgaria

Ot(the letter from Bulgaria)

Ot(Ix x is a letter from Bulgaria)

Ot(Ix (x is a letter ∧ x is from Bulgaria))

Ot(Ix (Lx ∧ Fxb))

F: [ _ is from _ ]; L: [ _ is a letter]; O: [ _ opened _ ]; b: Bulgaria; t: Tom

5.
│∃x (Fx ∧ ∃y ¬ x = y) 1
├─
│ⓐ
││Fa ∧ ∃y ¬ a = y 2
│├─
2 Ext ││Fa (4)
2 Ext ││∃y ¬ a = y 3
││
││ⓑ
│││¬ a = b (4)
││├─
4 Adj │││¬ a = b ∧ Fa X, (5)
5 EG │││∃y (¬ y = b ∧ Fy) X, (6)
6 EG │││∃x ∃y (¬ y = x ∧ Fy) X, (7)
│││●
││├─
7 QED │││∃x ∃y (¬ y = x ∧ Fy) 3
│├─
3 PCh ││∃x ∃y (¬ y = x ∧ Fy) 1
├─
1 PCh │∃x ∃y (¬ y = x ∧ Fy)
6.
│∃x ∃y Rxy 1
├─
│ⓐ
││∃y Ray 2
│├─
││ⓑ
│││Rab
││├─
││││∀x ¬ Rxx a:4,b:5
│││├─
4 UI ││││¬ Raa
5 UI ││││¬ Rbb
││││○ Rab, ¬Raa, ¬Rbb ⊭ ⊥
│││├─
││││⊥ 3
││├─
3 NcP │││∃x Rxx 2
│├─
2 PCh ││∃x Rxx 1
├─
1 PCh │∃x Rxx
7. A sentence φ is equivalent to a sentence ψ if and only if there is no possible world in which φ and ψ have different truth values
8. range: 1, 2, 3
abc
123
f123
1131
2311
3111
τgτ
13
21
31
τFτ
1F
2T
3T
R123
1FTF
2FFF
3FFT
 

Only non-arbitrary values are shown for f and g

  The diagram provides a complete answer, as do the tables above it. The tables below are a way of finding this structure.
 
alias sets  IDs  values
a 1 a: 1
b 2 b: 2
c
fab
fba
ga		 3 c: 3
f12: 3
f21: 3
g1: 3
resources values
Fb
F(ga)
Rab
¬ Rba
R(ga)c 		F2: T
F3: T
R12: T
R21: F
R33: T
9. [This question was on a topic not covered this year]