Phi 270 F11 test 4

F11 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 analyze 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.


F11 test 4 questions

Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. Also restate your analyses using unrestricted quantifiers.

1.

Everyone who called Bill praised Al.

answer
2.

The show didn’t please any critic.

answer
3.

No bid met every specification.

answer

Synthesize an English sentence that has the following logical form; that is, devise a sentence that would have the following analysis:

4.

(∀x: (Tx ∧ Nxs) ∧ Pxr) ¬ Ftx

F: [ _ found _ ]; N: [ _ was in _ ]; P: [ _ predated _ ]; T: [ _ was a tree]; r: the storm s: the stand t: Tom

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 (Rxa → ∀y Rxy)
∀x Rxx
∀x Rax
answer

Use a derivation to show that the following argument is not valid and present a counterexample that divides an open gap. (You may present the counterexample either by a diagram or by tables.)

7.
∀x Rxa
∀x ∀y Rxy
answer

F11 test 4 answers

1.

Everyone who called Bill praised Al

Everyone who called Bill is such that (he or she praised Al)

(∀x: x is a person who called Bill) x praised Al

(∀x: x is a person ∧ x called Bill) x praised Al

(∀x: Px ∧ Cxb) Rxa
∀x ((Px ∧ Cxb) → Rxa)

C: [ _ is a person _ ]; P: [ _ is a person]; R: [ _ praised _ ]; a: Al; b: Bill

2.

The show didn’t please any critic

every critic is such that (the show didn’t please him or her)

(∀x: x is a critic) the show didn’t please x

(∀x: x is a critic) ¬ the show pleased x

(∀x: Cx) ¬ Psx
∀x (Cx → ¬ Psx)

C: [ _ is a critic]; P: [ _ pleased _ ]; s: the show

3.

No bid met every specification

No bid is such that (it met every specification)

(∀x: x is a bid) ¬ x met every specification

(∀x: Bx) ¬ every specification is such that (x met it)

(∀x: Bx) ¬ (∀y: y is a specification) x met y

(∀x: Bx) ¬ (∀y: Sy) Mxy
∀x (Bx → ¬ ∀y (Sy → Mxy))

B: [ _ is a bid]; M: [ _ met _ ]; S: [ _ is a specification]

4.

(∀x: (x was a tree ∧ x was in the stand) ∧ x predated the storm) ¬ Tom found x

(∀x: x was a tree in the stand ∧ x predated the storm) ¬ Tom found x

(∀x: x was a tree in the stand that predated the storm) ¬ Tom found x

no tree in the stand that predated the storm it such that (Tom found it)
every tree in the stand that predated the storm it such that (Tom didn’t find it)

Tom found no tree in the stand that predated the storm
or: Tom didn’t find any tree in the stand that predated the storm
incorrect: Tom didn’t find every tree in the stand that predated the storm

5.
│∀x (Fx ∧ Gx)a:2
├─
│ⓐ
2 UI││Fa ∧ Ga3
3 Ext││Fa
3 Ext││Ga(4)
││●
│├─
4 QED││Ga1
├─
1 UG│∀x Gx
6.
│∀x (Rxa → ∀y Rxy)a:2
│∀x Rxxa:3
├─
│ⓑ
2 UI││Raa → ∀y Ray4
3 UI││Raa(4)
4 MPP││∀y Rayb:5
5 UI││Rab(6)
││●
│├─
6 QED││Rab1
├─
1 UG│∀x Rax
7.
│∀x Rxaa:3, b:4, c:5
├─
│ⓑ
││ⓒ
3 UI│││Raa
4 UI│││Rba
5 UI│││Rca
││││¬ Rbc
│││├─
││││○¬ Rbc, Rca, Rba, Raa ⊭ ⊥
│││├─
││││⊥6
││├─
6 IP│││Rbc2
│├─
2 UG││∀y Rby1
├─
1 UG│∀x ∀y Rxy

Counterexample presented by a diagram

Counterexample presented by tables

abc
123
R123
1TFF
2TFF
3TFF