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Proofs, Recursion and Analysis of
                       Algorithms



           Mathematical Structures
            for Computer Science
                   Chapter 2	





Copyright © 2006 W.H. Freeman & Co.   MSCS Slides   Proofs, Recursion and Analysis of Algorithms
Proof Techniques
     ●        Informal proof methods :	

               ■    Inductive reasoning	

               ■    Deductive reasoning	

               ■    Proof by exhaustion	

               ■    Direct proof	

               ■    Proof by contraposition	

               ■    Serendipity	

     ●        A few terms to remember:	

               ■    Axioms: Statements that are assumed true.	

                     •  Example: Given two distinct points, there is exactly one line that
                        contains them.	

               ■    Definitions: Used to create new concepts in terms of existing ones.	

               ■    Theorem: A proposition that has been proved to be true.	

                     •  Two special kinds of theorems: Lemma and Corollary.	

                     •  Lemma: A theorem that is usually not too interesting in its own right
                        but is useful in proving another theorem.	

                     •  Corollary: A theorem that follows quickly from another theorem.	


Section 2.1                                         Proof Techniques                            1
Deductive Reasoning: Counter Example
     ●        Inductive Reasoning: Drawing a conclusion from a hypothesis
              based on experience. Hence the more cases you find where P
              follows from Q, the more confident you are about the conjecture
              P → Q.	

     ●        Usually, deductive reasoning is also applied to the same
              conjecture to ensure that it is indeed valid. 	

     ●        Deductive reasoning either looks for a counterexample that
              disproves the conjecture (i.e. a case when P is true but Q is
              false) or a to construct a proof (theorem).	

     ●        Example: Prove that “For every positive integer n, n! ≤ n2.”	

               ■    Start testing some cases say, n = 1, 2, 3 etc.	

               ■    It might seem like it is true for some cases but how far do you test,
                    say n = 4.	

               ■    We get n! = 24 and n2 = 16 which is a counter example for this
                    theorem. Hence,even finding a single case that doesn’t satisfy the
                    condition is enough to disprove the theorem.	



Section 2.1                                      Proof Techniques                           2
Counterexample
     ●        More examples of counterexample:	

              ■    All animals living in the ocean are fish.	

              ■    Every integer less than 10 is bigger than 5.	


     ●        Counter example is not trivial for all cases, so we have
              to use other proof methods.	





Section 2.1                                  Proof Techniques            3
Exhaustive Proof
     ●        If dealing with a finite domain in which the proof is to be shown to be
              valid, then using the exhaustive proof technique, one can go over all
              the possible cases for each member of the finite domain. 	

     ●        Final result of this exercise: you prove or disprove the theorem but you
              could be definitely exhausted.	

     ●        Example: For any positive integer less than or equal to 5, the square of
              the integer is less than or equal to the sum of 10 plus 5 times the
              integer.	


                               n	

   n2	

   10+5n	

       n2 < 10+5n	

                               0	

    0	

     10	

              yes	

                               1	

    1	

     15	

              yes	

                               2	

    4	

     20	

              yes	

                               3	

    9	

     25	

              yes	

                               4	

   16	

     30	

              yes	

                               5	

   25	

     35	

              yes	


Section 2.1                                             Proof Techniques                 4
Example: Exhaustive Proof
     ●        Example 1: If an integer between 1 and 20 is divisible by 6, then
              it is also divisible by 3.	





Section 2.1                                 Proof Techniques                      5
Direct Proof
     ●           Direct Proof:	

     ●           Used when exhaustive proof doesn’t work. Using the rules of
                 propositional and predicate logic, prove P → Q.	

     ●           Hence, assume the hypothesis P and prove Q. Hence, a formal proof
                 would consist of a proof sequence to go from P to Q.	

     ●           Consider the conjecture 	

              	

x is an even integer Λ y is an even integer → the product xy is an even integer.	

     ●           A complete formal proof sequence might look like the following: 	

               1.    x is an even integer Λ y is an even integer           	

       	

hyp 	

               2.    (∀x)[x is even integer →(∃k)(k is an integer Λ x = 2k)]         	

number fact
                      	

       	

       	

        	

       	

(definition of even integer) 	

               3.    x is even integer →(∃k)(k is an integer Λ x = 2k) 	

           	

2,ui 	

               4.    y is even integer →(∃k)(k is an integer Λ y = 2k) 	

           	

2,ui 	

               5.    x is an even integer 	

        	

       	

         	

       	

1,sim 	

               6.    (∃k)(k is an integer Λ x = 2k) 	

        	

         	

       	

3,5,mp 	

               7.    m is an integer Λ x = 2m        	

       	

         	

       	

6, ei 	

               8.    y is an even integer 	

        	

       	

         	

       	

1,sim 	



Section 2.1                                           Proof Techniques                                 6
Direct Proof Example (contd.)
     9.       (∃k)(k is an integer Λ y = 2k)     	

       	

          	

            	

4,8,mp 	

     10.      n is an integer and y = 2n         	

       	

          	

            	

9, ei 	

     11.      x = 2m          	

      	

       	

       	

          	

            	

7,sim 	

     12.      y = 2n          	

      	

       	

       	

          	

            	

10, sim 	

     13.      xy = (2m)(2n) 	

        	

       	

       	

11, 12, substitution of equals 	

     14.      xy = 2(2mn) 	

          	

       	

       	

          	

13, multiplication fact 	

     15.      m is an integer          	

       	

       	

          	

            	

7,sim "
     16.      n is an integer 	

      	

       	

       	

          	

10, sim 	

     17.      2mn is an integer        	

       	

       	

          	

15, 16, number fact 	

     18.      xy = 2(2mn) Λ 2mn is an integer 	

          	

          	

            	

14, 17, con 	

     19.      (∃k)(k is an integer Λ xy = 2k)    	

       	

          	

            	

18,eg 	

     20.      (∀x)((∃k)(k is an integer Λ x = 2k) → x is even integer)                 	

number fact
                   	

        	

      	

       	

       	

(definition of even integer) 	

     21.      (∃k)(k an integer Λ xy = 2k) → xy is even integer         	

            	

20, ui 	

     22.      xy is an even integer       	

        	

                	

   	

       	

19, 21, mp



Section 2.1                                                Proof Techniques                                 7
Indirect Proof: Proof by Contradiction
     ●        Derived from the definition of contradiction which says 	

          	

      	

     	

     	

P Λ Qʹ′ → 0	

     ●        Hence, (P Λ Qʹ′ → 0) → (P → Q) is a tautology.	

     ●        In a proof of contradiction, one assumes that the hypothesis and
              the negation of the conclusion are true and then to deduce some
              contradiction from these assumptions.	

     ●        Example 1: Prove that “If a number added to itself gives itself,
              then the number is 0.”	

              ■    The hypothesis (P) is x + x = x and the conclusion (Q) is x = 0.
                   Hence, the hypotheses for the proof by contradiction are:	

              ■    x + x = x and x ≠0 (the negation of the conclusion)	

              ■    Then 2x = x and x ≠0, hence dividing both sides by x, the result is 2
                   = 1, which is a contradiction. Hence, (x + x = x) → (x = 0).	





Section 2.1                                     Proof Techniques                           8
Proof by Contradiction
     ●        Example 2: Prove “For all real numbers x and y, if x + y ≥ 2,
              then either x ≥1 or y ≥1.”	

               ■    Proof: Say the conclusion is false, i.e. x < 1 and y < 1. (Note: or
                    becomes and in negation.)	

               ■    Adding the two conditions, the result is x + y < 2. 	

               ■    At this point, this condition is Pʹ′ if P = x + y ≥ 2, hence, P Λ Pʹ′
                    which is a contradiction. Hence, the statement above is true.	


     ●        Example 3: The sum of even integers is even.	

               ■       Proof: Let x = 2m, y = 2n for integers m and n and assume that 	

                    	

x + y is odd.	

               ■       Then x + y = 2m + 2n = 2k + 1 for some integer k.	

               ■       Hence, 2*(m + k - n) = 1, where m + n - k is some integer.	

               ■       This is a contradiction since 1 is not even.	





Section 2.1                                        Proof Techniques                         9
Proof by Contraposition
     ●        Use the variants of P → Q to prove the conjecture. From the
              inference rules of propositional logic, we know a rule called
              contraposition (termed “cont” in short). 	

     ●        It states that (Qʹ′ → Pʹ′) → (P → Q) (a Tautology).	

     ●        Qʹ′ → Pʹ′ is the contrapositive of P → Q. 	

     ●        Hence, proving the contrapositive implies proving the
              conjecture.	


     ●        Example: Prove that if the square of an integer is odd, then the
              integer must be odd	

               ■    Hence, prove n2 odd → n odd	

               ■    To do a proof by contraposition prove n even → n2 even	

               ■    If n is even, then it can be written as 2m where m is an integer, i.e.
                    even/odd.	

               ■    Then, n2 = 4m2 which is always even no matter what m2 is because
                    of the factor 4.	


Section 2.1                                       Proof Techniques                           10
Serendipity
     ●        Serendipity: Fortuitous happening or something by chance or
              good luck.	

     ●        Not a formal proof technique.	

     ●        Interesting proofs provided by this method although other
              methods can be used as well.	


     ●        Example: 342 players in a tennis tournament. One winner in the
              end. Each match is between two players with exactly one winner
              and the loser gets eliminated. Prove the total number of matches
              played in the tournament are 341.	

               ■    Solution: Only one champion at the end of the tournament, hence
                    341 losers at the end, hence 341 matches should have been played
                    to have 341 losers.	





Section 2.1                                     Proof Techniques                       11
Summarizing Proof techniques


              Proof Technique	

 Approach to prove P → Q 	

                Remarks	


              Exhaustive Proof	

      Demonstrate P → Q 	

         May only be used for
                                          for all cases	

           finite number of cases	

                Direct Proof	

        Assume P, deduce Q	

         The standard approach-
                                                                     usually the thing to try	

                  Proof by             Assume Qʹ′, derive Pʹ′	

       Use this Qʹ′ if as a
                 Contraposition	

                                    hypothesis seems to
                                                                     give more ammunition
                                                                          then P would	

                 Proof by            Assume P Λ Qʹ′ , deduce a        Use this when Q says
               Contradiction	

           contradiction	

            something is not true	

                 Serendipity	

          Not really a proof                Fun to know	

                                            technique	





Section 2.1                                       Proof Techniques                                 12
Class Exercise

     1.       Product of any 2 consecutive integers is even.	

     2.       The sum of 3 consecutive integers is even.	

     3.       Product of 3 consecutive integers is even.	

     4.       The square of an odd integer equals 8k+1 for some
              integer k.	

     5.       The sum of two rational numbers is rational.	

     6.       For some positive integer x, x + 1/x ≥ 2.	





Section 2.1                           Proof Techniques            13

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CPSC 125 Ch 2 Sec 1

  • 1. Proofs, Recursion and Analysis of Algorithms Mathematical Structures for Computer Science Chapter 2 Copyright © 2006 W.H. Freeman & Co. MSCS Slides Proofs, Recursion and Analysis of Algorithms
  • 2. Proof Techniques ●  Informal proof methods : ■  Inductive reasoning ■  Deductive reasoning ■  Proof by exhaustion ■  Direct proof ■  Proof by contraposition ■  Serendipity ●  A few terms to remember: ■  Axioms: Statements that are assumed true. •  Example: Given two distinct points, there is exactly one line that contains them. ■  Definitions: Used to create new concepts in terms of existing ones. ■  Theorem: A proposition that has been proved to be true. •  Two special kinds of theorems: Lemma and Corollary. •  Lemma: A theorem that is usually not too interesting in its own right but is useful in proving another theorem. •  Corollary: A theorem that follows quickly from another theorem. Section 2.1 Proof Techniques 1
  • 3. Deductive Reasoning: Counter Example ●  Inductive Reasoning: Drawing a conclusion from a hypothesis based on experience. Hence the more cases you find where P follows from Q, the more confident you are about the conjecture P → Q. ●  Usually, deductive reasoning is also applied to the same conjecture to ensure that it is indeed valid. ●  Deductive reasoning either looks for a counterexample that disproves the conjecture (i.e. a case when P is true but Q is false) or a to construct a proof (theorem). ●  Example: Prove that “For every positive integer n, n! ≤ n2.” ■  Start testing some cases say, n = 1, 2, 3 etc. ■  It might seem like it is true for some cases but how far do you test, say n = 4. ■  We get n! = 24 and n2 = 16 which is a counter example for this theorem. Hence,even finding a single case that doesn’t satisfy the condition is enough to disprove the theorem. Section 2.1 Proof Techniques 2
  • 4. Counterexample ●  More examples of counterexample: ■  All animals living in the ocean are fish. ■  Every integer less than 10 is bigger than 5. ●  Counter example is not trivial for all cases, so we have to use other proof methods. Section 2.1 Proof Techniques 3
  • 5. Exhaustive Proof ●  If dealing with a finite domain in which the proof is to be shown to be valid, then using the exhaustive proof technique, one can go over all the possible cases for each member of the finite domain. ●  Final result of this exercise: you prove or disprove the theorem but you could be definitely exhausted. ●  Example: For any positive integer less than or equal to 5, the square of the integer is less than or equal to the sum of 10 plus 5 times the integer. n n2 10+5n n2 < 10+5n 0 0 10 yes 1 1 15 yes 2 4 20 yes 3 9 25 yes 4 16 30 yes 5 25 35 yes Section 2.1 Proof Techniques 4
  • 6. Example: Exhaustive Proof ●  Example 1: If an integer between 1 and 20 is divisible by 6, then it is also divisible by 3. Section 2.1 Proof Techniques 5
  • 7. Direct Proof ●  Direct Proof: ●  Used when exhaustive proof doesn’t work. Using the rules of propositional and predicate logic, prove P → Q. ●  Hence, assume the hypothesis P and prove Q. Hence, a formal proof would consist of a proof sequence to go from P to Q. ●  Consider the conjecture x is an even integer Λ y is an even integer → the product xy is an even integer. ●  A complete formal proof sequence might look like the following: 1.  x is an even integer Λ y is an even integer hyp 2.  (∀x)[x is even integer →(∃k)(k is an integer Λ x = 2k)] number fact (definition of even integer) 3.  x is even integer →(∃k)(k is an integer Λ x = 2k) 2,ui 4.  y is even integer →(∃k)(k is an integer Λ y = 2k) 2,ui 5.  x is an even integer 1,sim 6.  (∃k)(k is an integer Λ x = 2k) 3,5,mp 7.  m is an integer Λ x = 2m 6, ei 8.  y is an even integer 1,sim Section 2.1 Proof Techniques 6
  • 8. Direct Proof Example (contd.) 9.  (∃k)(k is an integer Λ y = 2k) 4,8,mp 10.  n is an integer and y = 2n 9, ei 11.  x = 2m 7,sim 12.  y = 2n 10, sim 13.  xy = (2m)(2n) 11, 12, substitution of equals 14.  xy = 2(2mn) 13, multiplication fact 15.  m is an integer 7,sim " 16.  n is an integer 10, sim 17.  2mn is an integer 15, 16, number fact 18.  xy = 2(2mn) Λ 2mn is an integer 14, 17, con 19.  (∃k)(k is an integer Λ xy = 2k) 18,eg 20.  (∀x)((∃k)(k is an integer Λ x = 2k) → x is even integer) number fact (definition of even integer) 21.  (∃k)(k an integer Λ xy = 2k) → xy is even integer 20, ui 22.  xy is an even integer 19, 21, mp Section 2.1 Proof Techniques 7
  • 9. Indirect Proof: Proof by Contradiction ●  Derived from the definition of contradiction which says P Λ Qʹ′ → 0 ●  Hence, (P Λ Qʹ′ → 0) → (P → Q) is a tautology. ●  In a proof of contradiction, one assumes that the hypothesis and the negation of the conclusion are true and then to deduce some contradiction from these assumptions. ●  Example 1: Prove that “If a number added to itself gives itself, then the number is 0.” ■  The hypothesis (P) is x + x = x and the conclusion (Q) is x = 0. Hence, the hypotheses for the proof by contradiction are: ■  x + x = x and x ≠0 (the negation of the conclusion) ■  Then 2x = x and x ≠0, hence dividing both sides by x, the result is 2 = 1, which is a contradiction. Hence, (x + x = x) → (x = 0). Section 2.1 Proof Techniques 8
  • 10. Proof by Contradiction ●  Example 2: Prove “For all real numbers x and y, if x + y ≥ 2, then either x ≥1 or y ≥1.” ■  Proof: Say the conclusion is false, i.e. x < 1 and y < 1. (Note: or becomes and in negation.) ■  Adding the two conditions, the result is x + y < 2. ■  At this point, this condition is Pʹ′ if P = x + y ≥ 2, hence, P Λ Pʹ′ which is a contradiction. Hence, the statement above is true. ●  Example 3: The sum of even integers is even. ■  Proof: Let x = 2m, y = 2n for integers m and n and assume that x + y is odd. ■  Then x + y = 2m + 2n = 2k + 1 for some integer k. ■  Hence, 2*(m + k - n) = 1, where m + n - k is some integer. ■  This is a contradiction since 1 is not even. Section 2.1 Proof Techniques 9
  • 11. Proof by Contraposition ●  Use the variants of P → Q to prove the conjecture. From the inference rules of propositional logic, we know a rule called contraposition (termed “cont” in short). ●  It states that (Qʹ′ → Pʹ′) → (P → Q) (a Tautology). ●  Qʹ′ → Pʹ′ is the contrapositive of P → Q. ●  Hence, proving the contrapositive implies proving the conjecture. ●  Example: Prove that if the square of an integer is odd, then the integer must be odd ■  Hence, prove n2 odd → n odd ■  To do a proof by contraposition prove n even → n2 even ■  If n is even, then it can be written as 2m where m is an integer, i.e. even/odd. ■  Then, n2 = 4m2 which is always even no matter what m2 is because of the factor 4. Section 2.1 Proof Techniques 10
  • 12. Serendipity ●  Serendipity: Fortuitous happening or something by chance or good luck. ●  Not a formal proof technique. ●  Interesting proofs provided by this method although other methods can be used as well. ●  Example: 342 players in a tennis tournament. One winner in the end. Each match is between two players with exactly one winner and the loser gets eliminated. Prove the total number of matches played in the tournament are 341. ■  Solution: Only one champion at the end of the tournament, hence 341 losers at the end, hence 341 matches should have been played to have 341 losers. Section 2.1 Proof Techniques 11
  • 13. Summarizing Proof techniques Proof Technique Approach to prove P → Q Remarks Exhaustive Proof Demonstrate P → Q May only be used for for all cases finite number of cases Direct Proof Assume P, deduce Q The standard approach- usually the thing to try Proof by Assume Qʹ′, derive Pʹ′ Use this Qʹ′ if as a Contraposition hypothesis seems to give more ammunition then P would Proof by Assume P Λ Qʹ′ , deduce a Use this when Q says Contradiction contradiction something is not true Serendipity Not really a proof Fun to know technique Section 2.1 Proof Techniques 12
  • 14. Class Exercise 1.  Product of any 2 consecutive integers is even. 2.  The sum of 3 consecutive integers is even. 3.  Product of 3 consecutive integers is even. 4.  The square of an odd integer equals 8k+1 for some integer k. 5.  The sum of two rational numbers is rational. 6.  For some positive integer x, x + 1/x ≥ 2. Section 2.1 Proof Techniques 13