1.2.xa. Exercise answers

1. a. The package will arrive next Tueday entails The package will arrive next week because the package arriving next Tuesday is one of ways for it to be true that it arrives next week
  b. The package will arrive next week does not entail The package will arrive next Tuesday because the premise would still be true if it arrived another day next week
  c. The package will arrive next Tuesday and The package will arrive next week are not mutually exclusive because both will be true if it does arrive next Tuesday
  d. The package will arrive next Tuesday and The package will arrive next Wednesday are mutually exclusive since the package cannot arrive both days
  e. The package will arrive before next Tueday and The package will arrive after next Tuesday are not jointly exhaustive since both will be false if it arrives on next Tuesday
  f. The package will arrive next Tuesday or before and The package will not arrive before next Wednesday are jointly exhaustive because, if the second is false—i.e., if it does arrive before next Wednesday—then the first must be true
  g. The package will arrive after next Tuesday is equivalent to The package will arrive next Wednesday or later because arriving next Wednesday or later than that are the two ways in which a package could arrive after next Tuesday
  h. The bridge will open at the end of May is not equivalent to The bridge will open before June since it is not now the end of May so the bridge could open before June by opening even earlier than the end of May
  i. The package will arrive before next Wednesday and The package will arrive after next Wednesday are not contradictory because both will be false if it arrives on next Wednesday
  j. The bridge will open before June and The bridge will open in June or later or never at all are contradictory because opening before June, opening in June, opening later than June, and not opening at all exhaust all possibilities and are mutually incompatible
2. a. φ ⇒ φ holds always because φ cannot fail to be true if it is true
  b. φ ⇒ ⊤ holds always because ⊤ cannot fail to be true no matter what φ is like
  c. φ ⇒ ⊥ holds only when φ is absurd because, if there is any possibility of φ being true, there is a possibility of ⊥ being false when φ is true
  d. φ ⇐ φ holds always because the truth of φ is guaranteed by its own truth
  e. φ ⇐ ⊤ holds only when φ is a tautology because if there is any possibility of φ being false, there is a possibility of it being false when ⊤ is true
  f. φ ⇐ ⊥ holds always because there is no possibility of ⊥ being true so no possibility of φ being false when ⊥ is true
  g. φ ◯ φ holds only when φ is a tautology because if there is any possibility of φ being false, it does not, together with itself exhaust all possibilities
  h. φ ◯ ⊤ holds always becuase ⊤ covers all possibilities by itself, so it certainly exhausts them when taken together with φ
  i. φ ◯ ⊥ holds only when φ is a tautology becuase, since ⊥ leaves open no possibilities, it contributes nothing to exhausting them all and φ must do that all by itself
  j. φ × φ holds only when φ is absurd because, unless φ rules out all possibilities, there will be a possibility of it being true along with itself
  k. φ × ⊤ holds only when φ is absurd because, since ⊤ is bound to be true, any possibility of φ being true will be a possibility of both being true
  l. φ × ⊥ holds always because, since ⊥ cannot be true, it cannot be true together with any sentence (even itself)
  m. φ ⇔ φ holds always since a sentence must have the same truth value as itself
  n. φ ⇔ ⊤ holds only when φ is a tautology because, if φ is bound to have the same truth value as a tautology, it must be one
  o. φ ⇔ ⊥ holds only when φ is absurd because, if φ is bound to have the same truth value as an absurd sentence, it must be one
  p. φ ⊗ φ never holds because no sentence can be both true and false at the same time
  q. φ ⊗ ⊤ holds only when φ is absurd because φ is bound to be false if its value is opposite that of a sentence that is bound to be true
  r. φ ⊗ ⊥ holds only when φ is a tautology because φ is bound to be true if its value is opposite that of a sentence that is bound to be false
3.

The appearance of — in a cell in the table below indicates that nothing can be concluded in general about the relation between φ and χ.

 
  ψ ⇒ χ ψ ⇐ χ ψ ⇔ χ ψ × χ ψ ◯ χ ψ ⊗ χ
φ ⇒ ψ φ ⇒ χ —† φ ⇒ χ φ × χ —† φ × χ
φ ⇐ ψ —* φ ⇐ χ φ ⇐ χ —* φ ◯ χ φ ◯ χ
φ ⇔ ψ φ ⇒ χ φ ⇐ χ φ ⇔ χ φ × χ φ ◯ χ φ ⊗ χ
φ × ψ —* φ × χ φ × χ —* φ ⇒ χ φ ⇒ χ
φ ◯ ψ φ ◯ χ —† φ ◯ χ φ ⇐ χ —† φ ⇐ χ
φ ⊗ ψ φ ◯ χ φ × χ φ ⊗ χ φ ⇐ χ φ ⇒ χ φ ⇔ χ

In cells marked with †, the fact that no relations hold in general can be seen by noting that if ψ is a tautology, the relations between it and φ and χ will hold no matter what sentences they are. And, similarly, in the cells marked with *, the relations between ψ and each of the other sentences will hold no matter what they are if ψ is absurd. There are various considerations which can be used to show that nothing more can be said in other cases, but it is probably easiest just to confirm for yourself that no further possibilities for the truth values of φ and χ are ruled out by the given information.

Glen Helman 28 Aug 2008