Lecture Notes, 8 Oct 2010

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§§1.6: Proof Strategies

1. Top-Level Proof Strategy:
   1. Look at the logical structure
      (i.e., at the principal connective or quantifier)
      of the proposition to be proved:
      • Is it ∀, ∃, →, ¬, etc.?

   2. Then use a lower-level strategy for proving propositions of that form.

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2. A General Strategy Applicable at Any Time:
   • Replace predicates by their definitions whenever that's useful
     • i.e., to prove P(x), where P(x) =def D(x),
       try to show D(x) instead.

     • e.g., to show Even(n), where Even(x) =def ∃j[x=2j],
       try to show ∃j[n=2j]
       • That's a search statement, so:
       • try to find j such that n=2j

     • e.g., to show Odd(n), where Odd(x) =def ∃j[x=2j+1],
       try to show ∃j[n=2j]
       • That's a search statement, so:
       • try to find j such that n=2j+1

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3. Direct Proof:
   • To show (A → B):
     Assume/suppose/make believe that A is the case
     and try to show B

   • Can combine strategies:
     • To show (P(n) → B),
       where P(x) =def D(x),
       suppose D(n) & try to show B

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4. To show ∀x[P(x) → Q(x)]:
   • Choose an arbitrary object c in the domain,
   • try to show (P(c) → Q(c))
     • How? By using the Direct Proof strategy above!

   • and then apply UG.

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5. An Example:
   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:

      • Proofs are stories;
        you don't just want to say what happens,
        but also how & why
      Let c be an arbitrary integer.
      • i.e., instead of trying to show that every odd integer is such that its successor is even,
        just show it for one "typical" or "arbitrary" integer.

      Show Odd(c) → Even(c+1) (and then use UG to show (*))

      Suppose Odd(c) & show Even(c+1):
      • i.e., instead of showing that the successor of odd c is even,
        suppose an extra fact, namely, that c is odd,
        & just show that c+1 is even
      i.e., show ∃j[c+1=2j] (replace "Even" by its definition)
      But Odd(c)
      ∴ ∃k[c=2k+1] (by def of Odd)
      • (using "k" to avoid confusion with "j")

      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 ("cleaned up") 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 for 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

      …(cue the closing music and a deep voice: "to be continued next time")

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