Phi 270 F06 test 2

F06 test 2 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 able to analyze the logical form of a sentence as fully as possible using conjunction and present the form in both symbolic and English notation.

Synthesis. Be able to synthesize an English sentence that has a given logical form.

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. There will be at least one derivation where detachment and attachment rules may be used and where they will shorten the proof. But there will be other derivations where you must rely on other rules, either because detachment and attachment rules do not apply or because I tell you not to use them.

I may also ask you to explain why a derivation rule works or does not work. This may be one of the rules you have available for use or another conceivable rule that is not part of the system we use. (For an example of a question of this sort, see question 8 of test 1 for 2000.)


F06 test 2 questions

Analyze each sentence below in as much detail as possible, presenting the result in both symbols and English notation (using both … and …, etc., as well as ∧, etc.). Be sure that the unanalyzed components of your answer are complete and independent sentences; also try to respect any grouping in the English.

1.

Sam was cool, but he was not both calm and collected.

answer
2.

Tom spoke to either Al or Barb but to neither Carol nor Dave.

answer

Synthesize an English sentence (the more idiomatic the better) that has the following analysis:

3.

¬ E ∨ F (E: Ed worked last weekend; F: Fred worked last weekend)

answer

Use derivations to check whether each of the entailments below holds. If one fails, present a counterexample by providing a table in which you calculate the truth values of the premises and conclusion on an extensional interpretation (i.e., an assignment of truth values) that divides an open gap.

Do not use attachment or detachment rules in 4-6. That is, do not use Adj or the rules MTP, MPT, and Wk of §4.3; instead use only the basic rules for exploiting resources, planning for goals, and closing gaps.

4. A ∧ B ⊨ ¬ (C ∧ ¬ B)
answer
5. ¬ (A ∧ B), ¬ A ⊨ B
answer
6. C, A ∨ B ⊨ A ∨ (B ∧ C)
answer

In 7 you may use attachment and detachment rules (and their use can simplify the derivation).

7. ¬ (A ∧ C), A ∨ B ⊨ B ∨ ¬ C
answer

F06 test 2 answers

1.

Sam was cool, but he was not both calm and collected

Sam was coolSam was not both calm and collected

Sam was cool ∧ ¬ Sam was both calm and collected

Sam was cool ∧ ¬ (Sam was calmSam was collected)

L ∧ ¬ (M ∧ T)
both L and not both M and T

L: Sam was cool; M: Sam was calm; T: Sam was collected

2.

Tom spoke to either Al or Barb but to neither Carol nor Dave

Tom spoke to either Al or BarbTom spoke to neither Carol nor Dave

(Tom spoke to AlTom spoke to Barb) ∧ ¬ Tom spoke to either Carol or Dave

(Tom spoke to AlTom spoke to Barb) ∧ ¬ (Tom spoke to CarolTom spoke to Dave)

(A ∨ B) ∧ ¬ (C ∨ D)
both either A or B and not either C or D

A: Tom spoke to Al; B: Tom spoke to Barb; C: Tom spoke to Carol; D: Tom spoke to Dave

3.

¬ E ∨ F (E: Ed worked last weekend; F: Fred worked last weekend)

¬ Ed worked last weekendFred worked last weekend

Ed didn’t work last weekendFred worked last weekend

Either Ed didn’t work last weekend or Fred did

4.
│A ∧ B 1
├─
1 Ext │A
1 Ext │B (4)
││C ∧ ¬ B 3
│├─
3 Ext ││C
3 Ext ││¬ B (4)
││●
│├─
4 Nc ││⊥ 2
├─
2 RAA │¬ (C ∧ ¬ B)
5.
│¬ (A ∧ B) 2
│¬ A 1
├─
││¬ B 3
│├─
│││││¬ A
││││├─
│││││○ ¬ A, ¬ B ⊭ ⊥
││││├─
│││││⊥ 4
│││├─
4 IP ││││A 3
│││
│││││¬ B
││││├─
│││││○ ¬ A, ¬ B ⊭ ⊥
││││├─
│││││⊥
│││├─
5 IP ││││B 3
││├─
3 Cnj │││A ∧ B 2
│├─
2 CR ││⊥ 1
├─
1 IP │B
  A  B    ¬ (A ∧ B) ¬ A  /  B 
  F  F     F          
6.
│ C (3)
│ A ∨ B 4
├─
││¬ A (7)
│├─
││││A (7)
│││├─
│││││¬ B
││││├─
│││││●
││││├─
7 Nc │││││⊥ 6
│││├─
6 IP ││││B 4
│││
││││B (5)
│││├─
││││●
│││├─
5 QED ││││B 4
││├─
4 PC │││B 2
││
│││●
││├─
3 QED │││C 2
│├─
2 Cnj ││B ∧ C 1
├─
1 PE │A ∨ (B ∧ C)
7.
│ ¬ (A ∧ C) 2
│ A ∨ B 3
├─
││C (2)
│├─
2 MPT ││¬ A (3)
3 MTP ││B (4)
││●
│├─
4 QED ││B 1
├─
1 PE │B ∨ ¬ C