Phi 270
Fall 2013
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Phi 270 F00 test 4

F00 test 4 questions

Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. Notice the special instructions for 2.

1. Only necessary projects were funded. [Different interpretations of the scope of only are possible here; any of them will do.]
answer
2. Tom can solve the puzzle if anyone can. [In this case, restate your answer using an unrestricted quantifier.]
answer
3. No one received every vote
answer

Use derivations to establish the validity of the following arguments. You may use attachment rules. English interpretations are suggested but remember that they play no role in derivations, and don’t hesitate to ignore them if they don’t help you think about the derivations.

4.
∀x (Dx → Mx)
∀x (¬ Ax → ¬ Mx)
∀x (Dx → Ax)
answer
A: [ _ is an animal]; D: [ _ is dog]; M: [ _ is a mammal]
5.
∀x ∀y ((Py ∧ Byx) → Dyx)
∀x (Px → ∀y (Bxy → Dxy))
answer
Everyone who has built anything is proud of it / Everyone is proud of everything he or she has built

Use a derivation to show that the following argument is not valid and describe a structure (by using either a diagram or tables) that is a counterexample lurking in one of the derivation’s open gaps.

6.
∀x (Rxx → ¬ Fx)
∀x Rxc
∀x ∀y (Fy → ¬ Rxy)
answer

F00 test 4 answers

1.

Only necessary projects were funded

(∀x: ¬ x was a necessary project) ¬ x was funded

(∀x: ¬ (x was a project ∧ x was necessary)) ¬ x was funded

(∀x: ¬ (Px ∧ Nx)) ¬ Fx

F: [ _ was funded]; N: [ _ was necessary]; P: [ _ was a project]

(∀x: Px ∧ ¬ Nx) ¬ Fx—i.e., No unnecessary projects were funded—and (∀x: Nx ∧ ¬ Px) ¬ Fx—i.e., Among the necessities only projects were funded—are not equivalent but are possible interpretations that would be marked by emphasis on necessary and projects, respectively.

2.

Tom can solve the puzzle if anyone can

(∀x: x is a person) Tom can solve the puzzle if x can

(∀x: Px) (Tom can solve the puzzle ← x can solve the puzzle)

(∀x: Px) (S Tom the puzzle ← S x the puzzle)

(∀x: Px) (Stp ← Sxp) [or: (∀x: Px) (Sxp → Stp)]
∀x (Px → (Stp ← Sxp)) [or: ∀x (Px → (Sxp → Stp))]

P: [ _ is a person]; S: [ _ can solve _ ]; p: the puzzle; t: Tom

3.

No one received every vote

(∀x: x is a person) ¬ x received every vote

(∀x: Px) ¬ x received every vote

(∀x: Px) ¬ (∀y: y is a vote) x received y

(∀x: Px) ¬ (∀y: Vy) Rxy

P: [ _ is a person]; R: [ _ received _ ]; V: [ _ is a vote]

Incorrect answers:

(∀x: Px) (∀y: Vy) ¬ Rxy says No one received any vote

¬ (∀x: Px) (∀y: Vy) Rxy says Not everyone received every vote

(∀y: Vy) ¬ (∀x: Px) Rxy says No vote is such that everyone received it

4.
│∀x (Dx → Mx) a:3
│∀x (¬ Ax → ¬ Mx) a:5
├─
│ⓐ
│││Da (4)
││├─
3 UI │││Da → Ma 4
4 MPP │││Ma (6)
5 UI │││¬ Aa → ¬ Ma 6
6 MTT │││Aa (7)
│││●
││├─
7 QED │││Aa 2
│├─
2 CP ││Da → Aa 1
├─
1 UG │∀x (Dx → Ax)
5.
│∀x ∀y ((Py ∧ Byx) → Dyx) b:5
├─
│ⓐ
│││Pa (9)
││├─
│││ⓑ
│││││Bab (10)
││││├─
5 UI │││││∀y ((Py ∧ Byb) → Dyb) a:6
6 UI │││││(Pa ∧ Bab) → Dab 8
│││││
││││││¬ Dab (8)
│││││├─
8 MTT ││││││¬ (Pa ∧ Bab) 9
9 MPT ││││││¬ Bab (10)
│││││├─
10 Nc ││││││⊥ 7
││││├─
7 IP │││││Dab 4
│││├─
4 CP ││││Bab → Dab 3
││├─
3 UG │││∀y (Bay → Day) 2
│├─
2 CP ││Pa → ∀y (Bay → Day) 1
├─
1 UG │∀x (Px → ∀y (Bxy → Dxy))
[This can be done without the reductio argument begun at stage 7 by using Adj to derive Pa ∧ Bab in order to exploit (Pa ∧ Bab) → Dab for a]

 

6.
│∀x (Rxx → ¬ Fx) b:4, c:9, a:11
│∀x Rxc a:6, b:7, c:8
├─
│ⓐ
││ⓑ
││││Fb (5)
│││├─
4 UI ││││Rbb → ¬ Fb 5
5 MTT ││││¬ Rbb
6 UI ││││Rac
7 UI ││││Rbc
8 UI ││││Rcc (10)
9 UI ││││Rcc → ¬ Fc 10
10 MPP ││││¬ Fc
11 UI ││││Raa → ¬ Fa 13
││││
│││││Rab
││││├─
│││││││¬ Raa
││││││├─
│││││││○ Fb,¬Fc,¬Raa,Rab,Rac,
│││││││        ¬Rbb,Rbc,Rcc ⊭ ⊥
││││││├─
│││││││⊥ 14
│││││├─
14 IP ││││││Raa 13
│││││
││││││¬ Fa
│││││├─
││││││○ ¬Fa,Fb,¬Fc,Rab,Rac,
││││││        ¬Rbb,Rbc,Rcc ⊭ ⊥
││││││
│││││├─
││││││⊥ 13
││││├─
13 RC │││││⊥ 12
│││├─
12 RAA ││││¬ Rab 3
││├─
3 CP │││Fb → ¬ Rab 2
│├─
2 UG ││∀y (Fy → ¬ Ray) 1
├─
1 UG │∀x ∀y (Fy → ¬ Rxy)
 

lurks in both open gaps