| 1. |
Define entailment by completing the following: Γ ⇒ φ if and only if ... . (Your answer need not replicate the wording of the text's definitions, but it should define entailment in terms of truth values and possible worlds. Remember that Γ is a set, not a sentence, so it does not itself have a truth-value.)
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| 2. |
Suppose you know that ⇒ φ (i.e., φ is a tautology) and that φ ⇒ ψ. Can you conclude that φ ⇔ ψ ? Explain why or why not by considering possibilities of truth and falsity.
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| 3. |
Explain how a sentence can implicate something it doesn't imply. (While you don't need to state the definitions of implication and implicature, you will need to employ the ideas used in them.)
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| 4. |
Analyze the sentence below in as much detail as possible, presenting the result in both symbolic and English notation. Be sure that the unanalyzed components of your answer are complete and independent sentences; also try to respect any grouping in the English.
Although Carol called Dave, she didn't reach him; but Ed stopped by and helped her finish the job.
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| 5. |
Use the basic system of derivations (i.e., no replacement rules) to check whether the entailment below holds. If it fails, provide 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) which makes the premises true and conclusion false.
A ∧ (B ∧ C), B ∧ D ⇒ (C ∧ D) ∧ E
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| 6. |
[This question was on a topic not covered in F05]
Use replacement principles to put the following sentence into list normal form (in which no conjunction is the left component of a conjunction and letters appear in alphabetical order without repetition):
(B ∧ C) ∧ (A ∧ B)
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| 1. | Γ ⇒ φ if and only if there is no possible world in which φ is false while every member of Γ is true. | ||
| 2. | Yes. In every possible world, φ is true (because it is a tautology) so ψ is true too (since φ ⇒ ψ), so there is no possible world in which φ and ψ have different truth values. | ||
| 6. |
[This question was on a topic not covered in F05]
(B ∧ C) ∧ (A ∧ B)
⇔ (A ∧ B) ∧ (B ∧ C) ⇔ A ∧ (B ∧ (B ∧ C)) ⇔ A ∧ ((B ∧ B) ∧ C) ⇔ A ∧ (B ∧ C) |