Lecture Notes, 13 Oct 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

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2. More Strategies:
   1. (not covered in lecture, but included here for reference)

      Another strategy to prove (A → B)
      (another form of proof by contradiction):

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

      Works because:

      • previous strategy for (A → B) 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 (A ↔ B):
      • Show (A → B), and
        show (B → A)

   3. To prove (A₁ ↔ A₂ ↔ A₃ ↔ … ↔ A_n):
      • could try to show that all possible pairs are equivalent, but it's easier to…

        show (A₁ → A₂), and
        show (A₂ → A₃), and
        …, and
        show (A_n–1 → A_n), and
        show (A_n → A₁)

      • 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)]
             Q(c)
             ∴ 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
           (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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