5.1 A; Some

• Proposal to be amended
  1. John loves someone.
  2. $\mathrm{\:\bf Love\:}(\mathrm{\:j\:},\: x)$

• Next step in the proposal to be amended: Treat nouns and adjectives as above. Treat a like some:
  1. A dog is happy.
  2. ${Dog}(x)\wedge{Happy}(x)$

  Ditto for other predicate words

  1. A dog barked.
  2. $\mathrm{\:\bf Dog\:}(x)\wedge\mathrm{\:\bf Bark\:}(x)$
  1. John drives a Buick.
  2. $\mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: x)\wedge\mathrm{\:\bf Buick\:}(x)$
  • We're using $\wedge$ even though the word and hasn't occurred in the sentence.
  • $\wedge$ is going to turn out to have a lot more uses in our logical translations than just as a translation of and
  • Other sentential logical connectives will also turn up in surprising places

• The problem
  1. John doesn't drive a Buick.
  2. $\neg \lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: x)\wedge\mathrm{\:\bf Buick\:}(x) \rbrack$

  Does this mean the right thing?

  

  Capturing truth conditions: (5.1.2) gives the wrong truth-conditions for (5.1.1)

  1. Suppose B11 and B12 are both Buicks. John drives B11 and John doesn't drive B12.
  2. Then there is an x such that it's not the case both that x is a Buick and John drives x. Namely B12. While B12 is a Buick John doesnt drive it.
  3. So the logical formula (5.1.2) comes out true in these circumstances.
  4. But the English sentence (5.1.1) is not true in these circumstances. John shouldn't be driving ANY Buicks, yet he's driving B11.
  5. The logical formula (5.1.2) misdescribes the truth conditions of (5.1.1) .
  6. This is the semantic analogue of the grammar mis-describing the grammaticality of a sentence.

• Solution
  1. John drives a Buick.
  2. $\exists x \lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: x)\wedge\mathrm{\:\bf Buick\:}(x)\rbrack$
  3. $\exists x \lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: x)\wedge\mathrm{\:\bf Buick\:}(x)\rbrack$ is true iff there is some entity b such that $\lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: \mathrm{\:\bf b\:})\wedge\mathrm{\:\bf Buick\:}(\mathrm{\:\bf b\:})\rbrack$ is true.
  4. True whenever John drives any entity that is a Buick
  5. False only if there is NO entity that is a Buick that John drives
  6. $\lbrack\!\lbrack \exists x \phi(x) \rbrack\!\rbrack = \mathrm{\:true\:}$ iff there is some entity b such that $\lbrack\!\lbrack \phi(x) \rbrack\!\rbrack ^{\mathrm{\:\footnotesize\bf b\:}/x} = \mathrm{\:true\:}$
  1. John doesn't drive a Buick.
  2. $\neg \exists x \lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: x)\wedge\mathrm{\:\bf Buick\:}(x) \rbrack$
  3. $\neg \exists x \lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: x)\wedge\mathrm{\:\bf Buick\:}(x) \rbrack$ is true iff it is not the case that there exists some entity b such that $\lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: \mathrm{\:\bf b\:})\wedge\mathrm{\:\bf Buick\:}(\mathrm{\:\bf b\:})\rbrack$ is true.
  4. Previously: $\neg \lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: x)\wedge\mathrm{\:\bf Buick\:}(x) \rbrack$ is true iff there exists some entity b such that it is not the case that $\lbrack \mathrm{\:\bf Drive\:}(\mathrm{\:j\:},\: \mathrm{\:\bf b\:})\wedge\mathrm{\:\bf Buick\:}(\mathrm{\:\bf b\:})\rbrack$ is true.

  Other Fixes

  1. A dog is happy.
  2. $\exists x\lbrack \mathrm{\:\bf Dog\:}(x)\wedge\mathrm{\:\bf Happy\:}(x)\rbrack$
  3. A dog barked.
  4. $\exists x\lbrack \mathrm{\:\bf Dog\:}(x)\wedge\mathrm{\:\bf Bark\:}(x)\rbrack$
  5. Fido is a dog.
     • $\exists x\lbrack \mathrm{\:\bf Dog\:}(x)\wedge x = \mathrm{\:fido\:}\rbrack$
     • Equivalent to: $\mathrm{\:dog\:}(\mathrm{\:fido\:})$
     • Uniform treatment of a dog