Discrete Structures

Lecture Notes, 6 Oct 2010

Last Update: 6 October 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)

        A(x := c)    i.e., A(c)

        where c could name any object in the domain

      • Universal Generalization (UG)

        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)

        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)

        xA(c := x)    i.e., ∃xA(x)

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

  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)

  3. Understanding & Creating Proofs

Next lecture…

Text copyright © 2010 by William J. Rapaport (rapaport@buffalo.edu)
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