Lecture Notes, 6 Oct 2010

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§1.5: Rules of Inference (cont'd)

1. Rules of Inference for Quantifiers:
   1. Notation:
      1. Let A(t) represent any proposition or propositional function
         (even one with a multiple-place predicate)
         that contains at least one occurrence of the term t.

      2. Let A(t₁ := t₂) represent any proposition or propositional function in which all free occurrences of term t₁
         are replaced by (i.e., are assigned the value) term t₂.
      • E.g., if A(x) is: (P(x) → Q(x,y)),
                then A(x := c) is: (P(c) → Q(c,y))

   2. The rules:
      • Universal Instantiation (UI)

        ∀xA(x)
        A(x := c)    i.e., A(c)

        where c could name any object in the domain

      • Universal Generalization (UG)

        A(c)          
        ∀xA(c := x)    i.e., ∀xA(x)

        where c must name an arbitrary object in the domain
        not any particular or special one
        (cf. proofs in geometry)

      • Existential Instantiation (EI)

        ∃xA(x)   
        A(x := c)    i.e., A(c)

        where c must name an arbitrary object in the domain,
        not any particular one

        i.e., c must be a brand-new name.

      • Existential Generalization (EG)

        A(c)         
        ∃xA(c := x)    i.e., ∃xA(x)

        where c names some actual object in the domain that is known to satisfy A.

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2. Sample Proof of Validity:
   1. Argument:
      P1: Every student in 191 is a freshman.
      P2: Every freshman will pass the midterm.
      P3: Fred is a student in 191.
      C : ∴ Some student in 191 will pass the midterm.

   2. Syntax & Semantics of Representation:
      191(x) = x is a student
      Frosh(x) = x is a freshman
      Pass(x) = x will pass the midterm
      fred = Fred

   3. Translation:
      P1: ∀x[191(x) → Frosh(x)]
      P2: ∀x[Frosh(x) → Pass(x)]
      P3: 191(fred)
      C : ∴ ∃x[191(x) ∧ Pass(x)]

   4. Strategy:
      Want to show C (i.e., ∃x[191(x) ∧ Pass(x)]
      • Can show C if can find (remember: ∃ is a search) a 191 student who passes
      • We know about one individual: Fred
      • ∴ try to show (191(fred) ∧ Pass(fred)):
        • to do that, need to show 191(fred) and need to show Pass(fred)
        1. To show 191(fred) is easy: It's P3
        2. To show Pass(fred):
           use UI (x := fred) on P2, then MP if can show Frosh(fred)
           Can show Frosh(fred) from P1 by UI (x := fred)
           then MP if can show 191(fred)
           But 191(fred) is easy: It's P3 (again)

   5. Proof of validity:

      1	∀x[191(x) → Frosh(x)]	:	P1
      2	(191(fred) → Frosh(fred))	:	1; UI (x := fred)
      3	191(fred)	:	P3
      4	Frosh(fred)	:	2,3: MP
      5	∀x[Frosh(x) → Pass(x)]	:	P2
      6	(Frosh(fred) → Pass(fred))	:	5; UI (x := fred)
      7	Pass(fred)	:	4,6; MP
      8	(191(fred) ∧ Pass(fred))	:	3,7; Conj
      9	∃x[191(x) ∧ Pass(x)]	:	8, EG (fred := x)

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3. Understanding & Creating Proofs
   • (click on link & read the .ppt, .pdf, or .html file)
   • (you might also find the article by Bowling (1977) of interest)
   • (we'll look at "Proof Strategies" next time)

Next lecture…