Phi 270
Fall 2013
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Phi 270 F08 test 4

F08 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 (see §7.2.2), how to incorporate bounds on complementary generalizations (see §7.2.3), ways of handling compound quantifier phrases (such as only cats and dogs, see §7.3.2), the distinction between every and any (see §§7.3.3 and 7.4.2), how to represent multiple quantifier phrases with overlapping scope (see §7.4.1). You should be able restate your analysis using unrestricted quantifiers (see §7.2.1), 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. Remember that the distinction between every and any can be important here, too.

Derivations. Be able to construct derivations to show that entailments hold and to show that they fail. 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 §7.8.1.


F08 test 4 questions

Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. State your analysis also in a form that expresses any generalizations using unrestricted quantifiers.

1.

No cover fit the container.

answer
2.

Everyone who Sam spoke to had seen the movie.

answer
3.

Only dogs chewed every bone.

answer
4.

No one who everyone knew bought anything.

answer

Use derivations to show that the following arguments are valid. You may use any rules.

5.
∀x (Fx → Hx)
∀x ((Fx ∧ Gx) → Hx)
answer
6.
∀x (Px → ∀y (Rxy → Txy))
∀x ∀y ((Px → Rxy) → (Px → Txy))
answer

Use a derivation to show that the following argument is not valid and present a counterexample by using a diagram to describe a structure that is a counterexample lurking an open gap.

7.
∀x Rax
∀x (Rxx → Rxa)
answer

F08 test 4 answers

1.

no cover fit the container

no cover is such that (it fit the container)

(∀x: x is a cover) ¬ x fit the container

(∀x: Cx) ¬ Fxc
∀x (Cx→ ¬ Fxc)

C: [ _ is a cover]; F: [ _ fit _ ]; c: the container

2.

everyone who Sam spoke to had seen the movie

everyone who Sam spoke to is such that (he or she had seen the movie)

(∀x: x is a person who Sam spoke to) x had seen the movie

(∀x: x is a personSam spoke to x)) Sxm

(∀x: Px ∧ Ksx) Sxm
∀x ((Px ∧ Ksx) → Sxm)

K: [ _ spoke to _ ]; P: [ _ is a person]; S: [ _ had seen _ ]; m: the movie; s: Sam

3.

only dogs chewed every bone

only dogs are such that (they chewed every bone)

(∀x: ¬ x is a dog) ¬ x chewed every bone

(∀x: ¬ Dx) ¬ every bone is such that (x chewed it)

(∀x: ¬ Dx) ¬ (∀y: y is a bone) x chewed y

(∀x: ¬ Dx) ¬ (∀y: By) Cxy
∀x (¬ Dx → ¬ ∀y (By → Cxy))

B: [ _ is a bone]; C: [ _ chewed _ ]; D: [ _ is a dog]

4.

No one who everyone knew bought anything

everything is such that (no one who everyone knew bought it)

∀x no one who everyone knew bought x

∀x no one who everyone knew is such that (he or she bought x)

∀x (∀y: y is a person who everyone knew) ¬ y bought x

∀x (∀y: y is a personeveryone knew y) ¬ Byx

∀x (∀y: Py ∧ everyone is such that (he or she knew y)) ¬ Byx

∀x (∀y: Py ∧ (∀z: z is a person) z knew y) ¬ Byx

∀x (∀y: Py ∧ (∀z: Pz) Kzy) ¬ Byx
∀x ∀y ((Py ∧ ∀z (Pz → Kzy)) → ¬ Byx)

B: [ _ bought _ ]; K: [ _ knew _ ]; P: [ _ is person]

5.
│∀x (Fx → Hx) a:4
├─
│ⓐ
│││Fa ∧ Ga 3
││├─
3 Ext │││Fa (5)
3 Ext │││Ga
4 UI │││Fa → Ha 5
5 MPP │││Ha (6)
│││●
││├─
6 QED │││Ha 2
│├─
2 CP ││(Fa ∧ Ga) → Ha 1
├─
1 UG │∀x ((Fx ∧ Gx) → Hx)
6.
│∀x (Px → ∀y (Rxy → Txy)) a:6
├─
│ⓐ
││ⓑ
││││Pa → Rab 5
│││├─
│││││Pa (5), (7)
││││├─
5 MPP │││││Rab (9)
6 UI │││││Pa → ∀y (Ray → Tay) 7
7 MPP │││││∀y (Ray → Tay) b:8
8 UI │││││Rab → Tab 9
9 MPP │││││Tab (10)
│││││●
││││├─
10 QED │││││Tab 4
│││├─
4 CP ││││Pa → Tab 3
││├─
3 CP │││(Pa → Rab) → (Pa → Tab) 2
│├─
2 UG ││∀y ((Pa → Ray) → (Pa → Tay)) 1
├─
1 UG │∀x ∀y ((Px → Rxy) → (Px → Txy))
7.
│∀x Rax a:3, b:4
├─
│ⓑ
│││Rbb
││├─
3 UI │││Raa
4 UI │││Rab
│││
││││¬ Rba
│││├─
││││○ ¬ Rba, Rab, Raa, Rbb ⊭ ⊥
│││├─
││││⊥ 5
││├─
5 IP │││Rba 2
│├─
2 CP ││Rbb → Rba 1
├─
1 UG │∀x (Rxx → Rxa)

Counterexample presented by a diagram