Discrete Structures

Lecture Notes, 13 Oct 2010

Last Update: 13 October 2010

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§§1.6: Proof Strategies (cont'd)

  1. The Example from Last Time, concluded:

    1. Proof by Contradiction:

      To show any proposition A:
      supp. ¬A & try to derive a contradiction;
      i.e., try to find B such that (B ∧ ¬B)

    2. E.g.: Show √2 is irrational:

      • Let Q(x) = "x is rational"

        • Remember: Q is the mathematical symbol for the set of rational numbers.

        So we need to show: ¬Q(√2)

        • Suppose, by way of contradiction, Q(√2)
          & try to find B such that (B ∧ ¬B):

          • Because we are assuming that Q(√2),
            it follows that (∃ int a,b)[(√2 = a/b) ∧ (a/b is in lowest terms)],

            • i.e., a,b have no common factors

              • Call this B

            ∴ 2 = a2/b2

            ∴ 2b2 = a2

            ∴ Even(a2)

            ∴ Even(a), by lemma (Rosen, p.85: 16)

            • (Hint: If Even(a2), could Odd(a)?)

            ∴ ∃j[a=2j]; by EI, call it j1

            ∴ 2b2 = (2j1)2 = 4j12

            b2 = 2j12

            ∴ Even(b2)

            ∴ Even(b)

            But now we have Even(a) ∧ Even(b).

            a,b do have a common factor, namely: 2

            • i.e., ¬B!

            So, we have our contradiction, (B ∧ ¬B)
            which means that our assumption for the proof by contradiction, Q(√2), was wrong.

            ∴ ¬Q(√2). QED

  2. More Strategies:

    1. (not covered in lecture, but included here for reference)

      Another strategy to prove (AB)
      (another form of proof by contradiction):

      • Suppose A
        Suppose ¬B
        & try to find C such that (C ∧ ¬C)

      Works because:

      • previous strategy for (AB) was to suppose A & show B, and
      • previous strategy to show B was to suppose ¬B & try to derive a contra.
      • this strategy combines them.

    2. To prove (AB):

      • Show (AB), and
        show (BA)

    3. To prove (A1A2A3 ↔ … ↔ An):

      • could try to show that all possible pairs are equivalent, but it's easier to…

        show (A1A2), and
        show (A2A3), and
        …, and
        show (An–1An), and
        show (AnA1)

      • E.g.: The Church-Turing Thesis says that all models of computation are equivalent to each other:

          Turing machines ↔ lambda-calculus ↔ Post production systems ↔ Markov algorithms ↔ register machines, etc.

          To prove this, compile each into the next, and the last into the first.

    4. To prove ¬∀xP(x):

      • Find (i.e., search for; i.e., ∃) a counterexample
        i.e., show ∃x¬P(x)

  3. There are lots of ways to have invalid proofs!

    1. See Rosen, pp. 83–84.

    2. More importantly:

      1. It's possible for there to be invalid proofs of true conclusions!

        • That doesn't make the proof valid!
        • And it doesn't prove the conclusion!

        • E.g.:

            All birds fly.
            Tweety the canary flies.
            ∴ Tweety is a bird.

            • The structure of this argument is:

                x[P(x) → Q(x)]
                ∴ P(c)

            • This is an invalid argument with all true premises
              and a true conclusion
                (at least, according to Warner Bros. cartoons!)

            • To show it's invalid, consider this argument with:

              • the same structure,
              • all true premises,
              • but a false conclusion:

              All dogs are mammals.
              Garfield the cat is a mammal.
              ∴ Garfield is a dog.

      2. And it's possible for there to be valid proofs of false conclusions!
        (But they're not sound!)

        • E.g.:
            (2+2=5) → Pope(bertrand-russell)
            ∴ Pope(bertrand-russell)

          proof #1:

            This is argument is an instance of the valid argument form modus ponens!

          proof #2: Show (2+2=5) → Pope(bertrand-russell) by Direct Proof:

          Suppose 2+2=5 & show Pope(b-r):

          1.2+2=5: assumption
          2.(2+2=5) ∨ Pope(b-r): 1; Addition
          3.¬(2+2=5): theorem of arithmetic
          4.Pope(b-r): 2,3; DS

        • This argument works because of the truth-table for the material conditional,
          which, as I said when I introduced it, doesn't exactly match the conditional of ordinary English.

        • For more on this, see "Paradoxes of Material Conditional"

        • The branch of logic that disallows this is called "relevance logic".

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