Lecture Notes, 22 Sep 2010

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§1.3: Predicates & Quantifiers (cont'd)

Knowledge Representation & Translation:

1. Some useful terminology:
   In "∀x[A]" and ∃x[A]":
   1. The "∀" and "∃" are called the quantifiers;
   2. The "x" that follows the quantifier is called the quantified variable (or the bound variable);
   3. And the bracketed part is called the scope of the quantifier.

      E.g., the scope of ∀x[P(x) ∧ Q(x)] is [P(x) ∧ Q(x)]

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   • There is no known algorithm for translating directly
     from the surface structure of an English sentence
     to an equivalent FOL proposition.
     • There are only "heuristics" (suggestions) for doing so,
       based on grammatical structure & meaning

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   • Lots of examples:
     • "Dictionary" (syntax & semantics) of predicates and terms (for use in the examples):

       Let Dog(x) = (x) is a dog
       Let Fish(x) = (x) is a fish
       Let Glitters(x) = (x) glitters
       Let Gold(x) = (x) is gold
       Let Mammal(x) = (x) is a mammal
       Let Man(x) = (x) is a man
       Let Pet(x) = (x) is a pet
       Let Wise(x) = (x) is wise
       Let mark-twain = Mark Twain
       Let sam-clemens = Samuel Langhorne Clemens (as in Clemens Hall)
       

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     1. Mark Twain is wise = "Wise(mark-twain)"
     2. Mark Twain was a wise man = "Wise(mark-twain) ∧ Man(mark-twain)"
        • Note: The tense of "was" is ignored in FOL!

     3. Mark Twain is Samuel Langhorne Clemens = "mark-twain = sam-clemens"
     4. All dogs are pets = "∀x[Dog(x) → Pet(x)]"

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     5. Some dogs are pets = "∃x[Dog(x) ∧ Pet(x)]"
        Notes:
        1. NOT:  "∃x[Dog(x) → Pet(x)]" !!!
           which is true about my cat!

           i.e., even if there were no dogs, but only cats, "∃x[Dog(x) → Pet(x)]" would be T,
           but the English would be false.

           See "On the Translation of ‘Some Dogs Are Pets’" for further explanation.

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        2. Here's yet another explanation:
           1. Semantics for ∀x[Dog(x) → Pet(x)]:
              • Consider a giant grab-bag of everything in the universe (i.e., everything in the domain)
              • Pick any thing at random from the bag.
              • Is it a dog?
              • If not, then (Dog(x) → Pet(x)) has not been refuted;
                i.e., it has not (yet) been determined to have tval=F
                ∴ assume its tval=T, & continue for-loop through bag.

              • If it is a dog, then is it a pet?
              • If so, then tval=T else tval=F

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           2. Semantics for ∃x[Dog(x) ∧ Pet(x)]:
              • Consider grab-bag of everything in univ.
              • Search in it for a dog.
              • If you find a dog, then tval=T
                i.e., you found an x in the domain that is both a dog and a pet
                
                else continue searching for a dog

              • If you don't find any dogs, then tval=F

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           3. Semantics for ∃[Dog(x) → Pet(x)]:
              • Consider grab-bag of everything in univ.
              • Search in it for a dog.
              • If you find a dog, then is it a pet?
              • If so, then tval=T else continue searching for a dog.
              • If you don't find any dogs, then tval=T (!)
                (because everything you found is not a dog,
                ∴ the antecedent's tval=F
                ∴ the conditional's tval=T (!)

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        3. I didn't get a chance to mention this in lecture, so I'll mention it here, for the sake of completeness:

           Some other versions of FOL use "restricted quantifiers":

           "Some dogs are pets" becomes:

           (∃x : Dog(x))[Pet(x)]

           where the "restriction" on x that Dog(x) is like a type declaration.

           "All dogs are pets" becomes:

           (∀x : Dog(x))[Pet(x)]

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     6. No dogs are fish = ¬∃x[Dog(x) ∧ Fish(x)]
        • But this could also be represented as: ∀x[Dog(x) → ¬Fish(x)]

        • So which is "correct"? Both! They are logically equivalent (you will prove this for HW).

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     7. To be continued…

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