1. Proofs1 Advanced Discrete Mathematics Jim Skon

2. Proofs2  Definition: A theorem is a valid logical assertion which can be proved using other theorems axioms (statements which are given to be true) and rules of inference (logical rules which allow the deduction of conclusions from premises).

3. Proofs3  A lemma (not a “lemon”) is a 'pre- theorem' or a result which is needed to prove a theorem.  A corollary is a 'post-theorem' or a result which follows directly from a theorem.

4. Proofs4 Valid reasoning in proofs §A mathematical proof is a sequence of statements, such that each statement: 1. is an assumption, or 2. is a proposition already proved, or 3. Follow logically from one or more previous statements in proof.

5. Proofs5 Valid Inference  Consider: H 1  H 2 .....  H n  C where  H i are called the hypotheses and  C is the conclusion.

6. Proofs6 Valid Inference : Argument §Argument - consists of a collection of statements, called premises of the argument, followed by a conclusion statement. A 1 A 2 : A n  A where  means 'therefore' or 'it follows that.' } Premises Conclusion

7. Proofs7 Valid reasoning in proofs §Example: modus ponens P  (P  Q)  Q modus ponens P: The car is running Q: The car has gas. If we know that the car is running (P), we can prove that (Q) it has gas.

8. Proofs8 Rules of Inference §Rules of Inference - used in proofs, or arguments, to move from what is known to what we want to prove. §modus ponens is a valid rule of inference.

9. Proofs9 Valid Argument §An argument is said to be valid if whenever all the premises are true, the conclusion is also true. § If the premises are true, but the conclusion false, the argument is said to be invalid.

10. Proofs10 Rules of Inference  Other famous rules of inference: P  P  Q Addition ________________________________ ____ P  Q  P Simplification

11. Proofs11 Rules of Inference   Q P  Q  P Modus Tollens ________________________________ ____ P  Q Q  R  P  R Hypothetical syllogism

12. Proofs12 Rules of Inference  P  Q  P  Q Disjunctive syllogism ________________________________ ____ P Q  P  Q Conjunction

13. Proofs13 Rules of Inference (P  Q)  (R  S) P  R  Q  S Constructive dilemma

14. Proofs14 Proofs §Three Techniques l Show true using logical inference Assume the hypotheses are true Use the rules of inference and logical equivalences to determine that the conclusion is true. l Show true by showing that no way exists to make all premises true but conclusion false l Show false by finding a way to make premises true but conclusion false.

15. Proofs15 Formal Proofs  To prove an argument is valid or the conclusion follows logically from the hypotheses: Assume the hypotheses are true Use the rules of inference and logical equivalences to determine that the conclusion is true.

16. Proofs16 Proof Example  Consider the following logical argument:  If horses fly or cows eat artichokes, then the mosquito is the national bird. If the mosquito is the national bird then peanut butter takes good on hot dogs. But peanut butter tastes terrible on hot dogs. Therefore, cows don't eat artichokes.

17. Proofs17 Proof Example  F Horses fly  A Cows eat artichokes  M The mosquito is the national bird  P Peanut butter tastes good on hot dogs

18. Proofs18 Proof Example  Represent the formal argument using the variables 1.(F  A)  M 2.M  P 3.  P  A

19. Proofs19 Proof Example  Use the hypotheses 1., 2., and 3. and the above rules of inference and any logical equivalencies to construct the proof.

20. Proofs20 Proof Example Assertion Reasons 1.(F  A)  M Hypothesis 1. 2.M  P Hypothesis 2. 3.(F  A)  P` steps 1 and 2 and hypothetical syll. 4.  P Hypothesis 3. 5.  (F  A) steps 3 and 4 and modus tollens 6.  F  A step 5 and DeMorgan 7.  A  F step 6 and commutativity of 'and' 8.  A step 7 and simplification Q. E. D.

21. Proofs21 Proof Example §Consider: p  (q  r) q  p  r

22. Proofs22 Proof by contradiction §Consider: r  s p  s r  q  p  q

23. Proofs23 Rules of Inference for Quantifiers   xP(x)   P(c) Universal Instantiation (UI)  _______________________________  P(x)   xP(x) Universal Generalization (UG)

24. Proofs24 Rules of Inference for Quantifiers  P(c)   xP(x) Existential Generalization (EG)  _______________________________   xP( x)   P(c) Existential Instantiation (EI)

25. Proofs25 Example Every man has two legs. John Smith is a man. Therefore, John Smith has two legs.

26. Proofs26 Example Define the predicates: M(x): x is a man L(x): x has two legs J: John Smith, a member of the universe The argument becomes 1.  x[M(x)  L(x)] 2.M( J )  L( J)

27. Proofs27 Example The proof is 1.  x[M(x)  L(x)] Hypothesis 1 2.M( J )  L(J ) step 1 and UI 3.M(J) Hypothesis 2 4. L( J) steps 2 and 3 and modus ponens Q. E. D.

28. Proofs28 Fallacies  Fallacies are incorrect inferences.  Some common fallacies: The Fallacy of Affirming the Consequent The Fallacy of Denying the Antecedent Begging the question or circular reasoning

29. Proofs29 The Fallacy of Affirming the Consequent If the butler did it he has blood on his hands. The butler had blood on his hands. Therefore, the butler did it.  This argument has the form: P  Q Q or [(P  Q)  Q]  P  P  which is not a tautology and therefore not a rule of inference!

30. Proofs30 The Fallacy of Denying the Antecedent (or the hypothesis)  If the butler is nervous, he did it. The butler is really mellow. Therefore, the butler didn't do it.  This argument has the form P  Q  P or [(P  Q)  P]  Q  Q which is also not a tautology and hence not a rule of inference.

31. Proofs31 Begging the question or circular reasoning  This occurs when we use the truth of statement being proved (or something equivalent) in the proof itself.  Example: Conjecture: if x 2 is even then x is even. Proof: If x 2 is even then x 2 = 2k for some k. Then x = 2l for some l. Hence, x must be even.

32. Proofs32 Proof Example §If the law is sufficient, then Christ died in vain §The law is sufficient §Therefore Christ died in vain.

33. Proofs33 Example §Babies are illogical §Nobody is despised who can manage a crocodile §Illogical persons are despised §Therefore, babies cannot manage crocodiles

34. Proofs34 Methods of Proof  We wish to establish the truth of the 'theorem’ P  Q  P may be a conjunction of other hypotheses.  P  Q is a conjecture until a proof is produced.

35. Proofs35 Types of proof:  Vacuous Proof of P  Q The truth value of P  Q is true if P is false. If P can be shown false, then P  Q holds. Thus prove P  Q by showing P is false.  Example: If I am both rich and poor then hurricane Fran was a mild breeze.  This is of the form: (P  P)  Q and the hypotheses form a contradiction.

36. Proofs36 Trivial Proof of P  Q  If we know Q is true then P  Q is true!  Example: If it's raining today then the empty set is a subset of every set.  The assertion is trivially true independent of the truth of P.

37. Proofs37 Direct Proof of P  Q §Prove Q, using P as an assumption.  Thus prove P  Q by showing Q is true whenever P is true.  assume the hypotheses are true  use the rules of inference, axioms and any logical equivalences to establish the truth of the conclusion.  Example: the Cows don’t eat artichokes proof above

38. Proofs38 Direct Proof of P  Q  Theorem: If 6x + 9y = 101, then x or y is not an integer.  Proof: (Direct) Assume 6x + 9y = 101 is true. Then from the rules of algebra 3(2x + 3y) = 101. But 101/3 is not an integer so it must be the case that one of 2x or 3y is not an integer (maybe both). Therefore, one of x or y must not be an integer. Q.E.D.

39. Proofs39 Indirect Proof of P  Q  Prove the contrapositive, e.g.  Q   P is true, using a direct proof methods.

40. Proofs40 Indirect Proof of P  Q  Example: A perfect number is one which is the sum of all its divisors except itself. For example, 6 is perfect since 1 + 2 + 3 = 6. So is 28. Theorem: A perfect number is not a prime. Proof: (Indirect). We assume the number p is a prime and show it is not perfect. But the only divisors of a prime are 1 and itself. Hence the sum of the divisors less than p is 1 which is not equal to p. Hence p cannot be perfect. Q. E. D..

41. Proofs41 Proof by contradiction ( reductio ad absurdum) §Assume the negation of the proposition is true, then derive a contradiction.  Thus to prove of P  Q, assume P   Q is true, then derive a contradiction.

42. Proofs42 Proof by contradiction  Theorem :  2 is irrational. § Proof: Let P be the proposition “  2 is irrational” §Assume  P or “  2 is rational” §Then  2 = a/b, where a and b are integers and have no common factors (lowest terms). § Then (  2) 2 = (a/b) 2 is 2 = a 2 /b 2. § Thus 2b 2 = a 2. Thus a 2 is even, implying a is even. Since a is even, a = 2c for some integer c. §Thus 2b 2 = 4c 2, so b 2 = 2c 2. Hence b is even. §Contradiction! A and b are both even, so divisible by 2!

43. Proofs43 Proof by cases of P  Q  To prove P  Q, find a set of propositions P 1, P 2,..., P n, n  2, in which at least one P j must be true for P to be true. P  P 1  P 2 ...  P n  Then prove the n propositions P 1  Q, P 2  Q,..., P n  Q.

44. Proofs44 Proof by cases of P  Q  Let  be the operation 'max' on the set of integers:  if a  b then a  b = max{a, b} = a = b  a.  Theorem: The operation  is associative.  For all a, b, c  (a  b)  c = a  (b  c).

45. Proofs45 Proof by cases: The operation  is associative  Let a, b, c be arbitrary integers.  Then one of the following 6 cases must hold (are exhaustive): 1. a  b  c 2. a  c  b 3. b  a  c 4. b  c  a 5. c  a  b 6. c  b  a

46. Proofs46 Proof by cases: The operation  is associative  Case 1: a  b = a, a  c = a, and b  c = b.  Hence (a  b)  c = a = a  (b  c).  Therefore the equality holds for the first case.  The proofs of the remaining cases are similar.  Q. E. D.

47. Proofs47 Vacuous Proof §Consider the proposition: If you your grandfather dies as a baby then you will get an A in this class. §Proof of this statement: Your grandfather didn’t die, thus thus the premise must be false. Thus P  Q must be true.

48. Proofs48 Trivial Proof §Consider the proposition: l If 3n 2 + 5n -2  2n 2 + 7n - 16 then n = n 2. P(n). §Proof of P(0): l 0 = 0 2, thus P(0) is trivially true. QED.

49. Proofs49 Direct Proof §Consider: The sum of two even numbers is even. §Restate as:  x:  y: (x is even and y is even)  x + y is even §Proof: 1. Remember: x is even   a:x = 2a (definition) 2. Assume x is even and y is even (assume hypothesis) 3. x + y = 2a + 2b (from 1 and 2) 4. 2a + 2b = 2·(a+b) 5. By 1, 2·(a+b) is even - QED.

50. Proofs50 Direct Proof §Consider: Every multiple of 6 is also a multiple of 3.  Rewrite:  x  z  y:(6·x = y  3·z = y) §Proof: 1. Assume 6x = y (hypothesis) 2. 6x = y can be rewritten as 3 · 2x = y 3. Let z = 2x, then 3·z = y holds. QED.

51. Proofs51 Indirect Proofs §Prove the contrapositive, e.g. Prove that:  Q   P is true

52. Proofs52 Indirect Proofs §Prove: If x 2 is even, then x is even.  Rewrite:  x : (EVEN(x 2 )  EVEN(x))

53. Proofs53 Indirect Proofs §Prove: If x 2 is even, then x is even 1.  x : (ODD(x)  ODD(x 2 )) (contrapositive) 2. Assume 1 ODD(n) true for some n (hypothesis) 3. x is odd   a:x = 2a + 1 (definition) 4. n = 2a + 1 for some a (2 & 3) 5. n 2 = (2a + 1) 2 (substitution) 6. (2a + 1) 2 = (2a + 1)(2a + 1) = 4a 2 + 4a + 1 = 2 (2a 2 + 2a) + 1 7. 2 (2a 2 + 2a) + 1 is odd(3 & 6) QED

54. Proofs54 Proof by contradiction  To prove of P  Q, assume  P  Q , derive a contradiction.  Recall that: P  Q   P  Q  Then:  P  Q  P  Q)  P   Q (Demorgan’s)  Thus to prove P  Q we assume P   Q and show a contradiction.

55. Proofs55 Proof by contradiction §Consider Theorem: There is no largest prime number. §This can be stated as "If x is a prime number, then there exists another prime y which is greater"  Formally:  x  y: (PRIME(x)  PRIME(y)  x < y)

56. Proofs56 Proof by contradiction §There is no largest prime number l Assume largest prime number does exist. Call this number p. l Restate implication as p is prime, and there does not exist a prime which is greater. 1. Form a product r = 2 · 3 · 5 ·... p) (e.g. r is the product of all primes) 2. If we divide r+1 by any prime, it will have remainder 1 3. r+1 is prime, since any number not divisible by any prime which is less must be prime. 4. but r+1 > p, which contradicts that p is the greatest prime number.QED.

57. Proofs57 Proof by cases  To prove P  Q, find a set of propositions P 1, P 2,..., P n, n  2, in which at least one P j must be true for P to be true. P  P 1  P 2 ...  P n Then prove the n propositions P 1  Q, P 2  Q,..., P n  Q.  Thus: P  (P 1  P 2 ...  P n ) and (P 1  Q)  (P 2  Q) ...  (P n  Q)  (P  Q)

58. Proofs58 Proof by cases §Consider: For every nonzero integer x,x 2 > 0. §Let: P = "x is a nonzero integer” Q = x 2 > 0  We want to prove P  Q

59. Proofs59 Proof by cases §If: P = "x is a nonzero integer” Q = x 2 > 0  Prove P  Q §P can be broken up into two cases: P 1 = x > 0 P 2 = x < 0  Note that P  (P 1  P 2 ).

60. Proofs60 Proof by cases §For every nonzero integer x,x 2 > 0. §Prove each case - Prove P 1  Q: If x > 0, then x 2 > 0, since the product of two positive numbers is always positive. Prove P 2  Q: If x 0, since the product of two negative numbers is always positive. QED.

61. Proofs61 Existence Proofs  We wish to establish the truth of  xP( x). Constructive existence proof:  Establish P(c) is true for some c in the universe.  Then  xP( x) is true by Existential Generalization (EG).

62. Proofs62 Constructive Existence Proofs  Theorem: There exists an integer solution to the equation x 2 + y 2 = z 2.  Proof: Choose x = 3, y = 4, z = 5.

63. Proofs63 Constructive Existence Proofs  Theorem: There exists a bijection from A= [0,1] to B= [0, 2]. Proof: We build two injections and conclude there must be a bijection without ever exhibiting the bijection. Let f be the identity map from A to B. Then f is an injection (and we conclude that | A |  | B | ). Define the function g from B to A as g(x) = x/4. Then g is an injection. Therefore, | B |  | A |. We now apply a previous theorem which states that if | A |  | B | and | B |  | A | then | A | = | B |. Hence, there must be a bijection from A to B. (Note that we could have chosen g(x) = x/2 and obtained a bijection directly). Q. E. D.

64. Proofs64 Nonconstructive existence proof  Assume no c exists which makes P(c) true and derive a contradiction.  Example: Theorem: There exists an irrational number. Proof: Assume there doesn’t exist an irrational number. Then all numbers must be rational. Then the set of all numbers must be countable. Then the real numbers in the interval [0, 1] is a countable set. But we have already shown this set is not countable. Hence, we have a contradiction (The set [0,1] is countable and not countable). Therefore, there must exist an irrational number. Q. E. D. Note: we have not produced such a number!

65. Proofs65  Disproof by Counterexample:  Recall that  x  P(x)  xP(x ).  To establish that  xP(x ) is true (or  xP(x) is false) construct a c such that  P(c) is true or P(c) is false.  In this case c is called a counterexample to the assertion  xP(x)

66. Proofs66 Nonexistence Proofs  We wish to establish the truth of  xP( x) (which is equivalent to  x  P(x) ).  Use a proof by contradiction by assuming there is a c which makes P(c) true.

67. Proofs67 The (infamous) Halting Problem  We wish to establish the nonexistence of a universal debugging program. Theorem: There does not exist a program which will always determine if an arbitrary program P halts. We say the Halting Problem is undecidable. UDP P1 P2 P3 Yes (Halts) No (Infinite Loop)

68. Proofs68 Halting Problem  Proof: Suppose there is such a program called HALT which will determine if any input-free program P halts. HALT(P) prints 'yes' and halts if P halts, otherwise, HALT(P) prints 'no' and halts. We now construct another procedure as follows: procedure ABSURD; if HALT(ABSURD) = 'yes' then while true do print 'ha' (Note that ABSURD is input-free.)

69. Proofs69 Halting Problem  If ABSURD halts then we execute the loop which prints unending gales of laughter and thus the procedure does not halt.  If ABSURD does not halt then we will exit the program and halt. Hence, ABSURD halts if it doesn't and doesn't halt if it does  which is an obvious contradiction.  Hence such a program does not exist.  Q. E. D.

70. Proofs70 Universally Quantified Assertions  We wish to establish the truth of  xP(x)  We assume that x is an arbitrary member of the universe and show P(x) must be true. Using UG it follows that  xP(x).

71. Proofs71 Universally Quantified Assertions  Example: Theorem: For the universe of integers, x is even iff x 2 is even. Proof: The quantified assertion is  x[x is even  x 2 is even]

72. Proofs72 Universally Quantified Assertions  Proof: We assume x is arbitrary. Recall that P  Q is equivalent to (P  Q)  (Q  P). Case 1. We show if x is even then x 2 is even using a direct proof (the only if part or necessity). If x is even then x = 2k for some integer k. Hence, x 2 = 4k 2 = 2(2k 2 ) which is even since it is an integer which is divisible by 2. This completes the proof of case 1.

73. Proofs73 Universally Quantified Assertions Case 2. We show that if x 2 is even then x must be even (the if part or sufficiency). We use an indirect proof: Assume x is not even and show x 2 is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 which is odd and hence not even. This completes the proof of the second case. Therefore we have shown x is even iff x 2 is even. Since x was arbitrary, the result follows by UG. Q.E.D.

74. Proofs74

75. Proofs75