Lecture Notes, 11 Oct 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
      QED

   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 (A → B):
      • 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.
      QED

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3. When should you use Direct Proof
   and when should you use Indirect Proof (i.e., Proof by Contrapositive)?
   1. In (A → B), 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
      • The "look under the street lamp" strategy
      • i.e., instead of trying to prove what you're trying to prove,
        try to prove something else!

   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…

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