Phi 270 F06 test 1
1.

Define tautologousness by completing the following with a definition in terms of truth values and possible worlds:

φ is a tautology if and only if …
answer
2.

Explain what truth values are possible for sentences φ and ψ that are both mutually exclusive (i.e., φ, ψ ⇒) and jointly exhaustive (i.e., ⇒ φ, ψ).

answer

The nursery rhyme “Jack and Jill” contains the line

Jack fell down and broke his crown

Even when this is taken out of context, it is natural to suppose that Jack broke his crown as a result of falling down rather than that falling down and the injury were simply two things that happened to him. I would claim that this tie between the two events is an implicature rather than an implication of the sentence.

  3.

Explain what I mean when I make that claim in a way that shows you understand the definition of implicature. (You need not support or reject my claim; I’m asking you only to explain what it means.)

answer
  4.

If the line did imply (rather than merely implicate) that Jack’s broken crown was the result of the fall, the sentence would not be a conjunction of Jack fell down and Jack broke his crown. Explain why this is so in a way that shows you understanding the meaning of implication and the conditions under which conjunctions are true.

answer

Analyze the sentence below in as much detail as possible, presenting the result using symbolic notation and also English notation (i.e., using bothand). Be sure that the unanalyzed components of your answer are complete and independent sentences (and give a key to the abbreviations you use for them); also try to respect any grouping in the English.

  5. The building was completed on time and with no cost overruns, but not everyone was satisfied with it.
answer

Use derivations to check whether each of the claims of entailment below holds. If an entailment fails, present a counterexample by providing a table in which you calculate the truth values of the premises and conclusion on an assignment of truth values that divides an open gap. Your table should show the value of any component of any component of the premises and conclusion that is also compound by writing this value under the main connective of the component. Do not use the rule Adj in the first derivation, but you may use it in the second.

  6. A ∧ C, B ∧ D ⇒ B ∧ C
answer
  7. A ∧ (D ∧ E), B ∧ F ⇒ (A ∧ B) ∧ C
answer

Phi 270 F06 test 1 answers
1.

φ is a tautology if and only if there is no possible world in which φ is false (or: if and only if φ is true in every possible world)

2.

Sentences that are both mutually exclusive and jointly exhaustive can only have different truth values because mutually exclusive sentences cannot be both true and jointly exhaustive sentences cannot be both false.

3.

To say that Jack fell down and broke his crown implicates but does not imply that breaking his crown is the result of falling down is to say that the sentence would be inappropriate if the two events were not connected in this way but the sentence doesn’t actually say that they were so connected.

4.

The conjunction Jack fell downJack broke his crown is true whenever both components are true whether or not one led to the other, so Jack fell downJack broke his crown does not imply that one led to the other because it can be true even if the sentence Jack’s broken crown is the result of his fall is false. So, if Jack fell down and broke his crown did imply this connection, it would not say the same thing as the conjunction.

5.

The building was completed on time and with no cost overruns, but not everyone was satisfied with it

The building was completed on time and with no cost overrunsnot everyone was satisfied with the building

(the building was completed on timethe building was completed with no cost overruns) ∧ not everyone was satisfied with the building

(T ∧ C) ∧ S
both both T and C and S
C: the building was completed with no cost overruns; S: not everyone was satisfied with the building; T: the building was completed on time
6.
│A ∧ C 1
│B ∧ D 2
├─
1 Ext │A
1 Ext │C (5)
2 Ext │B (4)
2 Ext │D
││●
│├─
4 QED ││B 3
││●
│├─
5 QED ││C 3
├─
3 Cnj │B ∧ C
   
7.
│ A ∧ (D ∧ E) 1
│ B ∧ F 3
├─
1 Ext │A (6)
1 Ext │D ∧ E 2
2 Ext │D
2 Ext │E
3 Ext │B (7)
3 Ext │F
│││●
││├─
6 QED │││A 5
││
│││●
││├─
7 QED │││B 5
│├─
5 Cnj ││A ∧ B 4
││○ A, B, D, E, F ⇏ C
│├─
││C 4
├─
4 Cnj │(A ∧ B) ∧ C
  A   B   C   D   E   F     A ∧  (DE)    B ∧  F   /   (A  ∧ B) ∧  C   
  T   T   F   T   T   T       T             T