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Introduction to Proofs Goals 1.Introduce notion of proof & basic proof methods. 2.Distinguish between correct & incorrect arguments 3.Understand & construct basic types of proofs.

Copyright © Peter Cappello2 Introduction Proof: A valid argument establishing the truth of a mathematical statement. In CS, we want to prove –the correctness of an algorithm or program –more generally, a property of an algorithm/data structure, program, or system (e.g., Java can be parsed efficiently?) –AI strategies, automated theorem proving –the consistency/feasibility of a specification – consequences of a mathematical model of computing –...

Copyright © Peter Cappello3 Methods for Proving Theorems A theorem typically is of the form  x ( P( x )  Q( x ) ). The proof method typically is of the form 1.Let c be an arbitrary element in the domain. 2.Show that P( c )  Q( c ) is true. (The essence.) 3.Therefore,  x ( P( x )  Q( x ) ).

Copyright © Peter Cappello4 Direct Proof of p  q 1.Assume that p is true. 2.Show that q is true. Example –Let the domain be the integers. –Prove:  x ( ( 3 | x – 2 )  ( 3 | x 2 – 1 ) ) Proof: WHAT WHY 1.Let x be an arbitrary integer such that 3 | x – 2. (premise) 2.Let integer q = ( x – 2 ) / 3. (from step 1 & definition of “|”) 3.x = 3q + 2. (algebraic manipulation of step 2) 4. x 2 = ( 3q + 2 ) 2 = 9q 2 + 12q + 4 = 3( 3q 2 + 4q + 1 ) + 1. 5. x 2 – 1 = 3( 3q 2 + 4q + 1 ). (subtract 1 from both sides of step 4) 6. 3 | x 2 – 1. (from step 5 & definition of the meaning of “|”) 7. Therefore,  x ( ( 3 | x – 2 )  ( 3 | x 2 – 1 ) ) (Universal Generalization)

Copyright © Peter Cappello5 Proof by Contraposition of p  q Direct proof of the contrapositive: ~q  ~p 1.Assume q is false. 2.Show that p is false.

Copyright © Peter Cappello6 Proof by Contraposition of p  q Example The domain is the set of integers. Prove:  a  b ( (ab is even)  (a is even  b is even) ) Proof WHAT WHY 1.Let a, b be arbitrary integers. 2.Assume  (a is even  b is even). 3.a is odd  b is odd. (De Morgan’s law) 4.  m ( a = 2m + 1 )   n ( b = 2n + 1 ). (Defn. of odd.) 5.ab = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1 which is odd. 6.  a  b ( (ab is even)  (a is even  b is even) ) (U.G.)

Show p  q is true by showing that  (p  q) ≡  (  p  q) ≡ (p   q) is false: 1.Assume p   q. 2.Show that this leads to a contradiction. If you need to show   x P( x ) You may assume  x P( x ), then show that this leads to a contradiction. Copyright © Peter Cappello7 Proof by Contradiction of p  q

Copyright © Peter Cappello8 Example If 1, 2, …, 10 are placed randomly in a circle then the sum of some 3 adjacent numbers  17. Proof: 1.Place 1, 2, …, 10 randomly in a circle. (Premise) 2.Assume that the sum of no 3 adjacent numbers  17. 3. Then, x 1 + x 2 + x 3  16 (explain with diagram) x 2 + x 3 + x 4  16 x 3 + x 4 + x 5  16 … x 10 + x 1 + x 2  16. 4. Sum the 10 inequalities: 3( x 1 + x 2 +...+ x 10 )  16  10 = 160  3(1 + 2 + … + 10)  160  3*55 = 165  160, a contradiction. 5. Our assumption is false: If 1, 2, …, 10 are placed randomly in a circle then the sum of some 3 adjacent numbers  17.

Copyright © Peter Cappello9 Proving p  q p  q ≡ (p  q)  (q  p) ≡ (p  q)  (  p   q) To prove p  q it is necessary & sufficient to show either: (p  q)  (q  p), or equivalently (p  q)  (  p   q)

Copyright © Peter Cappello10 Presentation Perhaps the clearest proof presentation: –Give a numbered list of statements, where the last statement is the statement to be proved –For each statement, explain why its truth follows from: previous statements premises previously established mathematical facts or definitions.

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