Discrete Structures

Lecture Notes, 11 Oct 2010

Last Update: 11 October 2010

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§§1.6: Proof Strategies (cont'd)

  1. The Example from Last Time, concluded:

    1. Prove: If n is odd, then n+1 is even.
      Hidden assumption: n is an integer
      Logical form (FOL transation):
        (*)   (∀ integer n)[Odd(n) → Even(n+1)]

      Strategy & Proof:

        Let c be an arbitrary integer.
        Show Odd(c) → Even(c+1) (and then use UG to show (*))
          Suppose Odd(c) & show Even(c+1):
            i.e., show ∃j[c+1=2j] (replace "Even" by its definition)
            But Odd(c)
            ∴ ∃k[c=2k+1] (by def of Odd)
            Call the k that works "k1" (by EI)
            Now show (2k1+1)+1 is even:
              Use algebra: (2k1+1)+1 = 2k1+2 = 2(k1+1)
            So, take j = k1+1
            i.e., ∃j[c+1=2j], by EG, namely: j = k1+1
            c+1 is even,
            i.e., Even(c+1).
          Now that we know Even(c+1) on the assumption that Odd(c),
          we also know Odd(c) → Even(c+1)
        ∴ we can infer (∀ integer n)[Odd(n) → Even(n+1)], by UG

    2. Here is a more formal version:

        Show: (∀ integer n)[Odd(n) → Even(n+1)]
        Let c be an arbitrary integer
        Show Odd(c) → Even(c+1):

        1. Odd(c) :  temporary assumption to show Even(c+1) by Direct Proof
        2. k[c=2k+1] :  1; def of Odd
        3. c=2k1+1 :  2; EI
        4. c+1=2k1+1+1 :  3; algebra
        5. c+1=2k1+2 :  4; algebra
        6. c+1=2(k1+1) :  5; algebra
        7. j[c+1=2j] :  6; EG
        8. Even(c+1) :  7; def of Even
        9. Odd(c) → Even(c+1) :  1–8; Direct Proof
        10. x[Odd(x) → Even(x+1)] :  9; UG

      • Only logicians and computer scientists write proofs like this,
        paying attention to every detail

      • mathematicians are sloppier :-)

    3. Simpler ("mathematical") version:

        Let c be an arbitrary integer.
        Suppose Odd(c) & show Even(c+1).
        Because Odd(c), we know c=2k+1, for some k.
        ∴ show (2k+1)+1 is even.
        But (2k+1)+1 = 2(k+1)
        ∴ Even(c+1). QED

  2. Proof by Contraposition (or: Indirect Proof):

    1. Another way to prove (AB):

      • try to show the logically equivalent (¬B → ¬A)

    2. How? By Direct Proof:

      • Suppose ¬B
        & try to show ¬A

    3. This is especially useful if B is "simpler" than A

    4. E.g., show: If 7n–5 is odd, then n is even.
      Let domain = Z

      Strategy & proof:

        Show (∀ int n)[Odd(7n–5) → Even(n)]

          Let c be an arb. int.
          & show Odd(7c–5) → Even(c):

            We could supp. Odd(7c–5) & show Even(c)

            • (Try it! Click on link to see what happens.)

            But might be easier to show the contrapositive: (¬Even(c) → ¬Odd(7c–5)):

            ∴ supp. ¬Even(c), & show ¬Odd(7c–5):

            But now we can use another strategy that I didn't mention:

            • Get rid of negations!

            i.e., instead of supposing ¬Even(c), let's supp. Odd(c),
            which is mathematically equivalent (you should know that!)

            and instead of trying to show ¬Odd(7c–5),
            let's try to show Even(7c–5)
            which is mathematically equivalent.

            • Note that we are not proving anything different!
              We're just restating—in a simpler form—what we're proving

            So: supp. ∃k[c=2k+1]
            & show ∃j[7c–5 = 2j]:

            i.e., find j such that 7c–5 = 2j:

              We know c=2k+1 (for some k)

              ∴ 7c–5 = 7(2k+1)–5
                           = 14k+7–5
                           = 14k+2
                           = 2(7k+1)

              ∴ take j = 7k+1.

  3. When should you use Direct Proof
    and when should you use Indirect Proof (i.e., Proof by Contrapositive)?

    1. In (AB), if B is simpler than A, then use Indirect Proof.

    2. Else, try one strategy, and, if it goes nowhere, then try the other!

  4. Proof by Contradiction:

    1. Another kind of indirect proof

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

    3. Why should this work? Consider this truth-table analysis:

      • (let "FALSE" name any contradiction, i.e., any proposition that is always false)

        Supp. that the argument from ¬A to (B ∧ ¬ B) is valid;
        ∴ (¬A → (B ∧ ¬ B)) is a tautology.
        but it ≡ ¬A → FALSE
        ∴ tval(¬A) = F (else it wouldn't be a tautology)
        ∴ tval(A) = T

    4. 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)?)

    …to be continued…

Next lecture…

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