phi270F12tst4

Phi 270 F12 test 4

F12 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.

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Analysis. Be ready to handle any of the key issues discussed in class—for example,

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the proper analysis of every, no, and only (see §7.2.2),

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how to incorporate bounds on complementary generalizations (see §7.2.3),

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ways of handling compound quantifier phrases (such as only cats and dogs, see §7.3.2),

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the distinction between every and any (see §§7.3.3 and 7.4.2),

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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.

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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.

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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.

F12 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.	`Every book that Al read was older than him.` answer
2.	`Al didn’t read every chapter of the book.` answer
3.	`No book that Al liked won any award.` answer

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

(∀x: Cx ∧ ¬ Bx(ga)) ¬ (Rx ∧ Hx)

B: [ _ is beyond _ ]; C: [ _ is a car]; H: [ _ is cheap to operate]; R: [ _ is reliable]; a: Al; g: [ _’s budget]

answer

Use derivations to show that the following entailments hold. You may use any rules.
5.	∀x (Fa ∧ Rax) ⊨ ∀x Rax answer
6.	∀x ∀y ∀z Ryzx ⊨ ∀x ∀y Rxyy answer

Use a derivation to show that the following claim of entailment fails, and present a counterexample that lurks in an open gap. You may present the counterexample either by a diagram or by tables.
7.	Rba ∧ ¬ Fa, ∀x (Rxb → Fx), Rbb ⊨ ∀x (Rax → Fx) answer

F12 test 4 answers

Every book that Al read was older than him

Every book that Al read is s.t. (it was older than him)

(∀x: x is book that Al read) x was older than Al

(∀x: x is book ∧ Al read x) x was older than Al

(∀x: Bx ∧ Rax) Oxa
∀x ((Bx ∧ Rax) → Oxa)

B: [ _ is a book]; O: [ _ read _ ]; R: [ _ was older than _ ]; a: Al

Al didn’t read every chapter of the book

¬ Al read every chapter of the book

¬ every chapter of the book is s.t. (Al read it)

¬ (∀x: x is a chapter of the book) Al read x

¬ (∀x: Cxb) Rax
¬ ∀x (Cxb → Rax)

C: [ _ is a chapter of _ ]; R: [ _ read _ ]; a: Al; b: the book

(∀x: Cxb) ¬ Rax is wrong; it is the analysis of Al didn’t read any chapter of the book

No book that Al liked won any award

every award is s.t. (no book that Al liked won it)

(∀x: x is an award) no book that Al liked won x

(∀x: Ax) no book that Al liked is s.t. (it won x)

(∀x: Ax) (∀y: y is a book that Al liked) ¬ y won x

(∀x: Ax) (∀y: y is a book ∧ Al liked y) ¬ Wyx

(∀x: Ax) (∀y: By ∧ Lay) ¬ Wyx
∀x (Ax → ∀y ((By ∧ Lay) → ¬ Wyx))

A: [ _ is an award]; B: [ _ is a book]; L: [ _ liked _ ]; W: [ _ won _ ]; a: Al

(∀y: By ∧ Lay) ¬ (∀x: Ax) Wyx is wrong; it is the analysis of No book that Al liked won every award. And ¬ (∀y: By ∧ Lay) (∀x: Ax) Wyx is the analysis of Not every book that Al liked won every award. On the other hand, merely switching the order of universals (with no negation between them) does not change the meaning, so (∀y: By ∧ Lay) (∀x: Ax) ¬ Wyx is not wrong (though it is less close to the form of the English).

(∀x: Cx ∧ ¬ Bx(ga)) ¬ (Rx ∧ Hx) with B: [ _ is beyond _ ]; C: [ _ is a car]; H: [ _ is cheap to operate]; R: [ _ is reliable]; a: Al; g: [ _’s budget]

(∀x: x is a car ∧ ¬ x is beyond Al’s budget) ¬ (x is reliable ∧ x is cheap to operate)

(∀x: x is a car ∧ x isn’t beyond Al’s budget) ¬ x is both reliable and cheap to operate

(∀x: x is a car that isn’t beyond Al’s budget) ¬ x is both reliable and cheap to operate

No car that isn’t beyond Al’s budget is s.t. (it is both reliable and cheap to operate)

No car that isn’t beyond Al’s budget is both reliable and cheap to operate

Or: Among cars, only ones that are beyond Al’s budget are both reliable and cheap to operate

	│∀x (Fa ∧ Rax)	b:2
	├─
	│ⓑ
2 UI	││Fa ∧ Rab	3
3 Ext	││Fa
3 Ext	││Rab	(4)
	││●
	│├─
4 QED	││Rab	1
	├─
1 UG	│∀x Rax

	│∀x ∀y ∀z Ryzx	b:3
	├─
	│ⓐ
	││ⓑ
3 UI	│││∀y ∀z Ryzb	a:4
4 UI	│││∀z Razb	b:5
5 UI	│││Rabb	(6)
	│││●
	││├─
6 QED	│││Rabb	2
	│├─
2 UG	││∀y Rayy	1
	├─
1 UG	│∀x ∀y Rxyy

	│Rba ∧ ¬ Fa	3
	│∀x (Rxb → Fx)	a:4, b:6, c:8
	│Rbb	(7)
	├─
	│ⓒ
	│││Rac
	││├─
3 Ext	│││Rba
3 Ext	│││¬ Fa	(5)
4 UI	│││Rab → Fa	5
5 MTT	│││¬ Rab
6 UI	│││Rbb → Fb	7
7 MPP	│││Fb
8 UI	│││Rcb → Fc	10
	│││
	││││¬ Fc	(10)
	│││├─
10 MTT	││││¬ Rcb
	││││	¬ Fa, Fb, ¬ Fc, ¬ Rab,
	││││○	¬ Fa, Rac, Rba, Rbb,
	││││	¬ Fa, ¬ Rcb ⊭ ⊥
	│││├─
	││││⊥	9
	││├─
9 IP	│││Fc	2
	│├─
2 CP	││Rac → Fc	1
	├─
1 UG	│∀x (Rax → Fx)

Only one of the following styles of presentation was needed:

Counterexample presented by a diagram

Counterexample presented by tables

a	b	c
1	2	3

τ	Fτ
1	F
2	T
3	F

R	1	2	3
1	F	F	T
2	T	T	F
3	F	F	F

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