Phi 270
Fall 2013
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Phi 270 F99 test 5

F99 test 5 questions

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. Notice the additional instructions given for each of the first two.
1. Sam mentioned someone Tina didn’t know. [Give this analysis also using an unrestricted quantifier.]
answer
2. Every shoe fit someone. [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. Sam found at least two pieces.
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. The elephant standing on Sam sighed.
answer
5. [This question was on a topic not covered this year]
Use derivations to show that the following argument is valid. You may use attachment rules (but not replacement by equivalence).
6.
∀x ∀y (Rxy → (Ryx → Rxx))
∃x ∃y (Rxy ∧ Ryx)
∃x Fxx
answer
Use a derivation to show that the following argument is not valid and describe a counterexample lurking in an open gap.
7.
∃x Fx
∃x (Gx ∧ Hx)
∃x (Fx ∧ Hx)
answer
Complete the following to give a definition of entailment by a single sentence (i.e., implication) in terms of truth values and possible worlds:
8. A sentence φ entails a sentence ψ if and only if …
answer
Complete the following truth table by calculating the truth value of the sentence on the given assignment. Show the value of each component by writing it under the main connective of that component.
9.
ABCD¬(A∧B)→(C∨¬D)
TFFT
answer
Describe a structure (i.e., an assignment of extensions to the non-logical vocabulary) which makes the sentences below all true. (You may use either tables or a diagram.)
10.
a = fb, fb = fc, fa = c, Pa, Pb, ¬ Pc, Rab, Rbc, Rc(fb)
answer

F99 test 5 answers

1.

Sam mentioned someone Tina didn’t know

someone Tina didn’t know is such that (Sam mentioned him or her)

(∃x: x is a person Tina didn’t know) Sam mentioned x

(∃x: x is a person ∧ ¬ Tina knew x) Sam mentioned x

(∃x: Px ∧ ¬ Ktx) Msx
∃x ((Px ∧ ¬ Ktx) ∧ Msx)

K: [ _ knew _ ]; M: [ _ mentioned _ ]; P: [ _ is a person]; s: Sam; t: Tina

2.

first analysis:

Every shoe fit someone

every shoe is such that (it fit someone)

(∀x: x is a shoe) x fit someone

(∀x: Sx) someone is such that (x fit him or her)

(∀x: Sx) (∃y: y is a person) x fit y

(∀x: Sx) (∃y: Py) Fxy

 

second analysis:

Every shoe fit someone

someone is such that (every shoe fit him or her)

(∃x: x is a person) every shoe fit x

(∃x: Px) every shoe is such that (it fit x)

(∃x: Px) (∀y: y is a shoe) y fit x

(∃x: Px) (∀y: Sy) Fyx

F: [ _ fit _ ]; P: [ _ is a person]; S: [ _ is a shoe]

The sentence is true on the first analysis and false on the second if every shoe could be worn but not all by the same person

3.

Sam found at least two pieces

at least two pieces are such that (Sam found them)

(∃x: x is a piece) (∃y: y is a piece ∧ ¬ y = x) (Sam found x ∧ Sam found y)

(∃ x: Px) (∃ y: Py ∧ ¬ y = x) (Fsx ∧ Fsy)

F: [ _ found _ ]; P: [ _ is a piece]; s: Sam

4.

using Russell’s analysis:

The elephant standing on Sam sighed

The elephant standing on Sam is such that (it sighed)

(∃x: x and only x is an elephant standing on Sam) x sighed

(∃x: x is an elephant standing on Sam ∧ (∀y: ¬ y = x) ¬ y is an elephant standing on Sam) Sx

(∃x: (x is an elephant ∧ x is standing on Sam) ∧ (∀y: ¬ y = x) ¬ (y is an elephant ∧ y is standing on Sam)) Sx

(∃x: (Ex ∧ Txs) ∧ (∀y: ¬ y = x) ¬ (Ey ∧ Tys)) Sx
or:
(∃x: (Ex ∧ Txs) ∧ (∀y: Ey ∧ Tys) x = y) Sx

 

using the description operator:

The elephant standing on Sam sighed

S (the elephant standing on Sam)

S (Ix x is an elephant standing on Sam)

S (Ix (x is an elephant ∧ x is standing on Sam))

S(Ix (Ex ∧ Txs))

E: [ _ is an elephant]; S: [ _ sighed]; T: [ _ is standing on _ ]; s: Sam

5. [This question was on a topic not covered this year]
6.
│∀x ∀y (Rxy → (Ryx → Rxx)) a:4
│∃x ∃y (Rxy ∧ Ryx) 1
├─
│ⓐ
││∃y (Ray ∧ Rya) 2
│├─
││ⓑ
│││Rab ∧ Rba 3
││├─
3 Ext │││Rab (6)
3 Ext │││Rba (7)
4 UI │││∀y (Ray → (Rya → Raa)) b:5
5 UI │││Rab → (Rba → Raa) 6
6 MPP │││Rba → Raa 7
7 MPP │││Raa (8)
8 EG │││∃x Rxx X, (9)
│││●
││├─
9 QED │││∃x Rxx 2
│├─
2 PCh ││∃x Rxx 1
├─
1 PCh │∃x Rxx
7.
│∃x Fx 1
│∃x (Gx ∧ Hx) 2
├─
│ⓐ
││Fa (7)
│├─
││ⓑ
│││Gb ∧ Hb 3
││├─
3 Ext │││Gb
3 Ext │││Hb (8)
│││
││││∀x ¬ (Fx ∧ Hx) a:5, b:6
│││├─
5 UI ││││¬ (Fa ∧ Ha) 7
6 UI ││││¬ (Fb ∧ Hb) 8
7 MPT ││││¬ Ha
8 MPT ││││¬ Fb
││││○ Fa,Gb,Hb,¬Ha,¬Fb ⊭ ⊥
│││├─
││││⊥ 4
││├─
4 NcP │││∃x (Fx ∧ Hx) 2
│├─
2 PCh ││∃x (Fx ∧ Hx) 1
├─
1 PCh │∃x (Fx ∧ Hx)
 
8. A sentence φ entails a sentence ψ if and only if there is no possible world in which φ is true but ψ is false (or: if and only if ψ is true in every possible world in which φ is true)
9.
ABCD¬(A∧B)→(C∨¬D)
TFFTTFⒻFF
10.
range: 1, 2, 3
abc
123
τfτ
13
21
31
τPτ
1T
2T
3F
R123
1FTF
2FFT
3TFT
 

The diagram above provides a complete answer, as do the tables to its left. The tables below illustrate a way of finding this structure.

 
alias sets IDs values
a
fb
fc		 1 a: 1
f2: 1
f3: 1
b 2 b: 2
c
fa		 3 c: 3
f1: 3
resources values
Pa
Pb
¬ Pc
Rab
Rbc
Rc(fb) 		P1: T
P2: T
P3: F
R12: T
R23: T
R31: T