Lecture Notes, 24 Sep 2010

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1. §1.3: Predicates & Quantifiers (cont'd):
   Knowledge Representation & Translation (cont'd):
   1. Lots of examples:

      In the interest of completeness, here I repeat all of the examples from last time,
      and add some new ones.

      • "Dictionary" 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)
        

      ---

      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)]"

         ---

      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.

            ---

         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

               ---

            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

               ---

            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 (!)

            ---

         3. 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)]

         ---

      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.

         ---

      7. All dogs are not fish = ∀x[Dog(x) → ¬Fish(x)]
         • But: Shakespeare's "All that glitters is not gold" = ¬∀x[Glitters(x) → Gold(x)] !

      8. Some dogs are not pets = ∃x[Dog(x) ∧ ¬Pet(x)]
      9. No dogs are not mammals = ¬∃x[Dog(x) ∧ ¬Mammal(x)]
         • This is ≡ ∀x[Dog(x) → Mammal(x)]

      10. Not all dogs are pet = ¬∀x[Dog(x) → Pet(x)]
          • This is ≡ ∃x[Dog(x) ∧ ¬Pet(x)]

      ---

   2. For more examples (in slightly different notation), see "Representing English Sentences in FOL"

      For more suggestions on how to translate, see:

      • Suber, Peter (2002), "Translation Tips" (converted to Rosen's notation)
      • "Translating from English to a First-Order Language"

   ---

2. §1.4: Nested Quantifiers: Propositional functions with ≥2 variables can have ≥2 quantifiers:
   1. Consider the propositional function "x > y":
      1. By itself, this is a propositional function (an "open sentence"), not a proposition
         (it has no tval)

      2. It can become a proposition (it can gain a tval) by:
         1. assigning constant values to both variables: "3 > 2"
         2. or by quantifying ("binding") both variables:
            • ∀x∀y[x > y]
              • "Every number is > every number"
              • For all numerical domains (W,N,Z,R,C), this is false.

      ---

   2. ∀x∀y[x > y] ≡ ∀y∀x[x > y]
      • and it is ≡ ∀y∀x[y > x]

      • and it is ≡ ∀x∀y[y > x]
        • I.e., the order of the same quantifiers doesn't matter.

      ---

   3. Ditto for: ∃x∃y[x > y]
      • "Some number is > some number"
      • For all numerical domains, this is true

      ---

   4. But consider: ∀x∃y[x > y]
      • "For every number, there is a number that it's > than"
      • This is true for Z
      • But false for N
        • because x=0 → ¬∃y[x > y]

        ---

      • When ∀ and ∃ quantifiers are mixed, their order does matter:

        1. Consider: ∃y∀x[x > y]
           • "There is a strictly smallest number"
           • For all numerical domains, this is false

        2. Consider: ∀y∃x[x > y]
           • "For every number, there's a number > it"
           • For all numerical domains, this is true

        3. Consider: ∃x∀y[x > y]
           • "There is a biggest number"
           • For all infinite numerical domains, this is false

      ---

   5. Uniqueness:

      "There is exactly one dog" can be represented as:

      ∃x[Dog(x) ∧ ∀y[Dog(y) → y = x]]
      • I.e., There is (at least) one thing in the domain that is a dog,
        and, any (other) dog in the domain is identical to that one.

      • Sometimes, logicians abbreviate this (with the "E-shriek!" quantifier) as:
        ∃!xDog(x)

        (read "There is a unique x in the domain such that x is a dog".)

      1. The scope of "∃x" is [Dog(x) ∧ ∀y[Dog(y) → y = x]]

         … and I will tell you more about this on Monday

Next lecture…