Phi 270
Fall 2013
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Phi 270 F03 test 4

F03 test 4 topics

The following are the topics to be covered. The proportion of the test covering each will approximate the proportion of the classes so far that have been devoted to that topic. Your homework and the collection of old tests will provide specific examples of the kinds of questions I might ask.

Analysis. Be ready to handle any of the key issues discussed in class--for example, the proper analysis of every, no, and only (§7.2), how to incorporate bounds and exceptions (§7.2), ways of handling compound quantifier phrases (such as only cats and dogs, §7.3), the distinction between every and any (§§7.3 and 7.4), how to represent multiple quantifier phrases with overlapping scope (§7.4). Be able restate you analysis using unrestricted quantifiers, but you will not need to present it in English notation.

Synthesis. You may be given a symbolic form and an interpretation of its non-logical vocabulary and asked to express the sentence in English. (This sort of question is less likely to appear than a question about analysis and there would certainly be substantially fewer such questions.)

Derivations. Be able to construct derivations to show that entailments hold and to show that they fail (derivations that hold are more likely). I may tell you in advance whether an entailment holds or leave it to you to check that using derivations. If a derivation fails, you may be asked to present a counterexample, which will involve describing a structure. You will not be responsible for the rules introduced in §7.8.


F03 test 4 questions

Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. Restate 2 using an unrestricted quantifier.
1. No one called the new number
answer
2. Sam asked everyone he could think of [Remember to restate this one using an unrestricted quantifier.]
answer
3. If any door was opened, the alarm sounded
answer
4. Only people who’d read everything the author had written were asked to review the book
answer
Use derivations to show that the following arguments are valid. You may use any rules.
5.
∀x (Fx ∧ Gx)
∀x Gx
answer
6.
∀x (Fx → Gx)
∀x ∀y (Gy → Rxy)
∀x ∀y (Fy → Rxy)
answer
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 an open gap.
7.
∀x (Fx → Rxa)
Fa → ∀x Rxx
answer

F03 test 4 answers

1.

No one called the new number

No one is such that (he or she called the new number)

(∀x: x is a person) ¬ x called the new number)

(∀x: Px) ¬ Cxn

C: [ _ called _ ]; P: [ _ is a person]; n: the new number

2.

Sam asked everyone he could think of

everyone Sam could think of is such that (Sam asked him or her)

(∀x: x is a person Sam could think of) Sam asked x

(∀x: x is a person ∧ Sam could think of x) Asx

( ∀x: Px ∧ Tsx) Asx
∀x ((Px ∧ Tsx) → Asx)

A: [ _ asked _ ]; P: [ _ is a person]; T: [ _ could think of _ ]; s: Sam

3.

If any door was opened, the alarm sounded

every door is such that (if it was opened, the alarm sounded)

(∀x: x is a door) if x was opened, the alarm sounded

(∀x: Dx) (x was opened → the alarm sounded)

( ∀x: Dx) (Ox → Sa)

D: [ _ is a door]; O: [ _ was opened]; S: [ _ sounded]; a: the alarm

4.

Only people who’d read everything the author had written were asked to review the book

Only people who’d read everything the author had written are such that (they were asked to review the book)

(∀x: ¬ x is a person who’d read everything the author had written) ¬ x was asked to review the book

(∀x: ¬ (x is a person ∧ x had read everything the author had written)) ¬ Axb

(∀x: ¬ (x is a person ∧ everything the author had written is such that (x had read it))) ¬ Axb

(∀x: ¬ (Px ∧ (∀y: y is a thing the author had written) x had read y)) ¬ Axb

(∀x: ¬ (Px ∧ (∀y: the author had written y) Rxy)) ¬ Axb

(∀x: ¬ (Px ∧ (∀y: Way) Rxy)) ¬ Axb

A: [ _ was asked to review _ ]; P: [ _ is a person]; R: [ _ had read _ ]; R: [ _ had written _ ]; a: the author; b: the book

5.
│∀x (Fx ∧ Gx) a: 2
├─
│ⓐ
2 UI ││Fa ∧ Ga 3
3 Ext ││Fa
3 Ext ││Ga (4)
││ ●
│├─
4 QED ││Ga 1
├─
1 UG │∀x Gx
6.
│∀x (Fx → Gx) b:4
│ ∀x ∀y (Gy → Rxy) a:6
├─
│ⓐ
││ⓑ
││││Fb (5)
│││├─
4 UI ││││Fb → Gb 5
5 MPP ││││Gb (8)
6 UI ││││∀y (Gy → Ray) b:7
7 UI ││││Gb → Rab 8
8 MPP ││││Rab (9)
││││●
│││├─
9 QED ││││Rab 3
││├─
3 CP │││Fb → Rab 2
│├─
2 UG ││∀y (Fy → Ray) 1
├─
1 UG │∀x ∀y (Fy → Rxy)
7.
│∀x (Fx → Rxa) a:2, b:5
├─
││Fa (3)
│├─
2 UI ││Fa → Raa 3
3 MPP ││Raa
││ⓑ
5 UI │││Fb → Rba 7
│││
││││¬ Rbb
│││├─
││││││¬ Fb
│││││├─
││││││○ Fa,Raa,¬Rbb,¬Fb ⊭ ⊥
│││││├─
││││││⊥ 8
││││├─
8 IP │││││Fb 7
││││
│││││Rba
││││├─
│││││○ Fa,Raa,¬Rbb,Rba ⊭ ⊥
││││├─
│││││⊥ 7
│││├─
7 RC ││││⊥ 6
││├─
6 IP │││Rbb 4
│├─
4 UG ││∀x Rxx 1
├─
1 CP │Fa → ∀x Rxx

Counterexample presented by tables

range: 1, 2
ab
12
τ
1T
2F
R12
1TF
2TF

(This counterexample lurks in both gaps; the value of F2 is needed only for the 1st and the value of R21 only for the 2nd.)

Counterexample presented by a diagram