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Tutorial 2: First O rder Logic and Methods of Proofs

Tutorial 2: First O rder Logic and Methods of Proofs. Peter Poon. Agenda. First Order Logic Order of quantifier Formulation Negation Methods of Proofs Direct Proof Contrapositive Contradiction. First Order Logic. Order of quantifier. Which one are equivalent?. Order of quantifier.

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Tutorial 2: First O rder Logic and Methods of Proofs

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  1. Tutorial 2: First Order Logic and Methods of Proofs Peter Poon

  2. Agenda • First Order Logic • Order of quantifier • Formulation • Negation • Methods of Proofs • Direct Proof • Contrapositive • Contradiction

  3. First Order Logic

  4. Order of quantifier • Which one are equivalent?

  5. Order of quantifier • Which one are equivalent?

  6. Formulation • Express the following using first order logic • Let be the set of all positive integers be the set of all real numbers be “x is prime”

  7. Formulation • Express the following using first order logic

  8. Negation • You know that • Write down the negation of the following statements

  9. Negation • Write down the negation of the following statements

  10. Method of Proof

  11. Direct Proof • For every positive integer n, is even

  12. Direct Proof • For every positive integer n, is even

  13. Contrapositive • If n2 is divisible by 3, then n is divisible by 3

  14. Contrapositive • If n2 is divisible by 3, then n is divisible by 3 • Contrapositive form • If n is not divisible by 3, then n2 is not divisible by 3 • Case 1: n = 3k + 1 • n2= (3k + 1)2= 9k2+ 6k + 1 = 3(3k2 + 2k) + 1 • Case 2: n = 3k + 2 • n2 = (3k + 2)2= 9k2+ 12k + 4 = 3(3k2+ 4k + 1) + 1 • Both are not divisible by 3

  15. Contradiction • Show that is not rational. • Given If n2 is divisible by 3, then n is divisible by 3

  16. Contradiction • Show that is not rational. • Given If n2 is divisible by 3, then n is divisible by 3 • If is rational • Since , which is divisible by 3 • So p = 3k, k is positive integer • Also p2 = 3q2 • so 9k2 = 3q2 • q2 = 3k2 (p and q have the common factor 3 contradiction!!!)

  17. Contradiction • If there 40 pigeons sharing 7 pigeonholes, then at least 1 pigeonhole have more then 5 pigeons.

  18. Contradiction • If there 40 pigeons sharing 7 pigeonholes, then at least 1 pigeonhole have more then 5 pigeons. • Assume it is false • Then every pigeonhole have at most 5 pigeons • Total number of pigeons <= 5 * 7 = 35 • Contradiction!!! • Pigeonhole principle • http://en.wikipedia.org/wiki/Pigeonhole_principle

  19. Conclusion • Contrapositive • Find the contrapositive form • Prove it • Contradiction • Assume it is false • Show it is impossible by finding contradiction

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