Phi 270
Fall 2013
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Phi 270 F97 test 3

F97 test 3 questions

(Questions 1-6 are from quiz 3 and 7-9 are from quiz 4 out of 6 quizzes—these two quizzes addressed the part of the course your test is designed to cover.)

Analyze the sentences below in as much detail as possible without going below the level of sentences (i.e., without recognizing individual terms and predicates). Be sure that the unanalyzed components of your answer are complete and independent sentences and that you respect any grouping in the English.
  1. The creek will be high enough only if it rains.
answer
  2. Unless you object, Al will show the letter to Barb if she asks to see it.
answer
Check each of the following for validity using the basic system of derivations (i.e., do not use attachment rules but you may use detachment rules). If a derivation fails, confirm a counterexample that lurks in an open gap.
  3.
A → (B ∨ C)
¬ C → (A → B)
answer
  4.
A → (B → C)
(C ∧ A) → B
answer
5. Analyze the sentence below in as much detail as possible, continuing the analysis when there are no more connectives by identifying predicates, functors, and individual terms. Be sure that the unanalyzed expressions in your answer are independent and that you respect any grouping in the English.
If Dan’s wife received the message, she is the person who called.
answer
6. a. Give two different expansions (using predicate abstracts) of the sentence: Raba.
answer
  b. Put the following into reduced form: [Pxa ∧ Qbx]xa.
answer
7. Describe a structure (i.e., an assignment of extensions to the non-logical vocabulary) which makes the following sentences all true. (You may present the structure either using tables or, were possible, using diagrams.)
a = fb, fa = fb, b = c, Fa, ¬ F(gc), Rb(fa), ¬ Ra(fb), R(gc)c
answer
Use derivations to check each of the claims of entailment below. You need not present counterexamples to dead-end gaps.
  8. Fa ∧ ¬ Fb ⊨ b = c → ¬ a = c
answer
  9. fa = c, fb = c, Rc(fa) → Ra(fa) ⊨ R(fa)(fb) → Rb(fb)
answer

F97 test 3 answers

1.

the creek will be high enough only if it rains

¬ the creek will be high enough ← ¬ it will rain

¬ H ← ¬ R or  ¬ R → ¬ H
if not R then not H

H: the creek will be high enough; R: it will rain

2.

¬ you will object → Al will show the letter to Barb if she asks to see it

¬ you will object → (Al will show the letter to Barb ← Barb will ask to see the letter)

¬ O → (S ← A) or ¬ O → (A → S)
if not O then if A then S

A: Barb will ask to see the letter; O: you will object; S: Al will show the letter to Barb

3.
│A → (B ∨ C) 3
├─
││¬ C (4)
│├─
│││A (3)
││├─
3 MPP │││B ∨ C 4
4 MTP │││B (5)
│││●
││├─
5 QED │││B 2
│├─
2 CP ││A → B 1
├─
1 CP │¬ C → (A → B)
4.
│A → (B → C) 3
├─
││C ∧ A 2
│├─
2 Ext ││C
2 Ext ││A (3)
3 MPP ││B → C 5
││
│││¬ B
││├─
│││││¬ B
││││├─
│││││○ A,¬ B,C ⊭ ⊥
││││├─
│││││⊥ 6
│││├─
6 IP ││││B 5
│││
││││C
│││├─
││││○ A,¬ B,C ⊭ ⊥
│││├─
││││⊥ 5
││├─
5 RC │││⊥ 4
│├─
4 IP ││B 1
├─
1 CP │(C ∧ A) → B
ABCA(BC)/(CA)B
TFTTT
5.

Dan’s wife received the message → Dan’s wife is the person who called

[ _ received _ ] Dan’s wife the message → Dan’s wife = the person who called

R(Dan’s wife)m → [ _’s wife] Dan = p

R(fd)m → fd = p

R: [ _ received _ ]; d: Dan; f: [ _’s wife]; m: the message; p: the person who called

6. a. The following are the possibilities; in the last, τ may be any term:
[Rxbx]xa, [Rxba]xa, [Rabx]xa, [Raxa]xb, [Raba]xτ
  b. Paa ∧ Qba
7.
range: 1, 2, 3
abc
122
τ
11
21
33
τ
13
23
33
τ
1T
2F
3F
R123
1FFF
2TFF
3TTF
 

(The diagram at the left provides a complete answer, and so do the tables above. The tables below show a way of arriving at these answers.)

 
alias sets IDs values
a
fa
fb
1 a: 1
f1: 1
f2: 1
b
c
2 b: 2
c: 2
gc 3 g2: 3
resources values
Fa
¬ F(gc)
Rb(fa)
¬ Ra(fb)
R(gc)c
F1: T
F3: F
R21: T
R11: F
R32: T
8.
│Fa ∧ ¬ Fb 1
├─
1 Ext │Fa (4)
1 Ext │¬ Fb (4)
││b = c a,b-c
│├─
│││a = c a-b-c
││├─
│││●
││├─
4 Nc= │││⊥ 3
│├─
3 RAA ││¬ a = c 2
├─
2 CP │b = c → ¬ a = c
9.
│fa = c
│fb = c a,b,c-fa-fb
│Rc(fa) → Ra(fa) 3
├─
││R(fa)(fb) (4)
│├─
│││¬ Rb(fb)
││├─
││││●
│││├─
4 QED= ││││Rc(fa) 3
│││
││││Ra(fa)
│││├─
││││○ fa=c,fb=c,R(fa)(fb),
││││ fa=c,¬Rb(fb),Ra(fa) ⊭ ⊥
│││├─
││││⊥ 3
││├─
3 RC │││⊥ 2
│├─
2 IP ││Rb(fb) 1
├─
1 CP │R(fa)(fb) → Rb(fb)