Proving an Implication

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Proving an Implication
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9 Common Methods

• Proofs generally entail proving an implication.
• Here are 9 common methods
• Trivial proof of p ? q
• If you can prove q without assuming p, then p ? q
  is called trivial.
• This situation rarely arises.
• Vacuous proof of p ? q
• If you show that p is false, then you are done.
• This situation rarely arises.

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Direct proof of p ? q

• Assume that p is true.
• Using that, show that q is true (perhaps using
  other information) by valid inference.
• Example Prove (a ? Z, 3 a - 2) ? (3 a2-1).
• 3 a 2 (Assumption)
• a - 2 ? 0 modulo 3.
• a ? 2 modulo 3.
• 4. a2 ? 1 mod 3.
• a2 - 1 ? 0 mod 3.
• 3 a2 - 1.

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9 Common Methods ...

• Indirect Proof of p ? q
• Direct proof of the contrapositive
• Assume q.
• Using that, show p by valid inference.
• Proof of p ? q by contradiction
• (p ? q) (p ? q).
• Show that (p ? q) p ? q is false.

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Proof by Contrapositive

• n ab, for a, b ? ? ? a ? n1/2 ? b ? n1/2 .
• 1. Assume a gt n1/2 ? b gt n1/2 .
• 2. Then, ab gt n.

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Proof by Contradiction Example

• 1, 2, , 10 are placed randomly in a circle ? the
  sum of some 3 adjacent numbers ? 17.
• 1. Assume 1, 2, , 10 are placed randomly in a
  circle and the sum of no 3 adjacent numbers ? 17.
• 2. x1 x2 x3 ? 16
• x2 x3 x4 ? 16
• x10 x1 x2 ? 16

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• 3. 3(x1 x2 . . .x10) ? 10 ? 16.
• 4. 3(x1 x2 . . .x10) 3(1 2 10) 3
  ? 55, a contradiction.
• Which 3 adjacent numbers have this big sum?
• This is called a non-constructive proof.

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9 Common Methods ...

• Proof of p ? q by cases
• Let p p1 ? p2 ? . . . ? pn.
• (p1 ? p2 ? . . . ? pn )? q (p1 ? q) ? (p2 ? q)
  ? . . . ? (pn ? q).
• Thus, you can prove the compound implication by
  proving each of the single implications.
• Why do you have to prove all of them?
• This form is rare.

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9 Common Methods ...

• Proof by elimination of cases
• (p ? (q1? q2? . . . ?qn) ? q1? q2? . . . ?
  qn-1) ? (p ? qn) .
• For example, if p implies q1 ? q2, and q1 is
  false, then p implies q2.

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Proof by elimination of cases

• Prove p is an odd prime ? p has the form 6n 1
  or 6n 5 or p 3.
• 1. Assume p is an odd prime.
• 2. p / 6 has only 6 possible remainders.
• 3. p thus is of the form 6n, 6n 1, , or 6n
  5.
• 4. Since p is odd, eliminate 6n, 6n 2, and 6n
  4.
• 5. Since p is prime, eliminate 6n 3 and p ? 3.
  (Since in that case 3 p)
• 6. Only 6n 1, 6n 5, and p 3 remain as
  possibilities.

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9 Common Methods ...

• Conditional proof
• p ? (q ? r) (p ? q) ? r.
• Java style tip
• if (p)
• if (q)
• r
• can be rewritten
• if (p q)
• r

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9 Common Methods ...

• Proof of equivalence (if and only if)
• p ?q (p ? q) ? (q ? p).
• In other words, prove 2 cases
• p if q (i.e., q ? p)
• p only if q (i.e., p ? q)

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Proof of Equivalence

• 2 integers a b have the same remainder when
  divided by positive integer n iff n a - b.
• The only if part
• 1. Assume 2 integers a b have the same
  remainder when divided by positive integer n.
• 2. Show n a - b.
• 3. Let a nj r, for 0 ? r lt n.
• b nk r, for 0 ? r lt n.
• 4. Then, a - b n(j - k), a multiple of n.

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The if part

• 1. Assume n a - b.
• 2. Show 2 integers a b have the same remainder
  when divided by positive integer n.
• 3. a - b ni, for some integer i.
• 4. Let a nj r1, for 0 ? r1 lt n.
• b nk r2, for 0 ? r2 lt n.
• 5. a - b ni. (By assumption)
• 6. a ni b ni (nk r2) n(i k) r2 .
• 7. But, r1 is the unique remainder of a / n.
• 8. ? r1 r2.

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