3.1. Introduction Assumptions.

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1 Sectio 3. Proofs 3.1. Itroductio. A roof is a carefully reasoed argumet which establishes that a give statemet is true. Logic is a tool for the aalysis of roofs. Each statemet withi a roof is a assumtio, a aiom, a reviously rove theorem, or follows from revious statemets i the roof by a mathematical or logical rules ad defiitios Assumtios. Assumtios are the statemets you assume to be true as you try to rove the result. Eamle: If you wat to rove: If ad is eve, the > 0 Your roof should start with the assumtios that ad is eve. Further, you ca use the defiitio of a eve atural umber, ad write the assumtios as follows: Let, ad be eve, that is,, =. WUCT11 Logic 85

2 Assumtios are ofte thought to be the give iformatio or iformatio we kow that ca be used i our roof. As i the eamle above, whe you are rovig statemets of the form P Q, the the assumtio is the statemet P. Eercise: Write the statemet to be rove i the revious eamle usig logical otatio: Aioms. Aioms are laws i Mathematics that hold true ad require o roof. Eamles: = + 0 =, y, z,[ ( = y) ( y = z)] ( = z) WUCT11 Logic 86

3 Mathematical Rules. Mathematical Rules are kow rules that are ofte used. Eamle:, y, z, ( = y) ( + z = y + z) Logical Rules. Logical Rules are rules of logic such as Substitutio ad Substitutio of Equivalece usig the laws itroduced earlier 3.. The Law of Syllogism If P Q ad Q R are both tautologies, the so is P R. Eercise: Write the Law of Syllogism usig logical otatio: WUCT11 Logic 87

4 Show, usig the quick method that the Law of Syllogism is a tautology. ((P Q) (Q R)) ( P R) Ste: 1 3 5* WUCT11 Logic 88

5 WUCT11 Logic 89 Eamles: s is a square s is a rectagle s is a rectagle s is a arallelogram s is a arallelogram s is a quadrilateral s is a square s is a quadrilateral ) 6 ( 0 9) 6 ( 0 3) ( 0 3) ( Eercise: Comlete the followig usig the Law of Syllogism: t is studyig WUCT11 t is erolled i a diloma t is studet at WCA..

6 Most results i Mathematics that require roofs are of the form P Q. The Law of Syllogism rovides the most commo method of erformig roofs of such statemets. The Law of Syllogism is a kid of trasitivity that ca aly to. To use the Law of Syllogism, we set u a sequece of statemets, P P1, P1 P, P P3, K, P Q. The, by successive alicatios of the law, we have P Q. Eamle. We wish to rove that for eve. I logic otatio, we wish to rove: is eve (P) This has the form icludes is eve (Q)., if is eve, the is P Qad we ote that our assumtio ad P: is eve. WUCT11 Logic 90

7 WUCT11 Logic 91 Proof: ) ( ) ( ) ( ) ( eve is ) ( ) ( 4 4,, eve is Q P P P P P P P = = = = = = K K K K Comletig the roof is simly a matter of alyig the Law of Syllogism three times to get is eve is eve. The revious roof ca be simlified to: eve is ), ( 4, eve is = = = The use of Law of Syllogism is a matter of commo sese. We shall use the Law of Syllogism without direct referece. Note. The use of the coective i the revious roof seems a little reetitive, albeit valid. For variety, the coective ca be relaced by words such as therefore, thus, so we have, ad hece.

8 3.3. Modus Poes Rule of Modus Poes: If P ad P Q are both tautologies, the so is Q. I other words, Modus Poes simly says that if we kow P to be true, ad we kow that P imlies Q, the Q must also be true. Eercise: Write the rule of Modus Poes usig logical otatio: WUCT11 Logic 9

9 Show, usig the quick method that the rule of Modus Poes is a tautology. (P (P Q)) Q Ste 1 3* WUCT11 Logic 93

10 Eamles: If Zak is a cheater, the Zak sits i the back row Zak is a cheater Therefore Zak sits i the back row. If = 3 the I will eat my hat = 3 Therefore I will eat my hat Eercise: Comlete the followig usig Modus Poes If Zeus is a God, the Zeus is a God Therefore Zeus is immortal. If it is suy the I will go to the beach It is suy Therefore If I study hard the I will ass Therefore I will ass WUCT11 Logic 94

11 3.3.. Uiversal Rule of Modus Poes: If P() ad Q() are redicates, the uiversal rule of Modus Poes is (( P( ) Q( )) P( a)) Q( a). This meas Modus Poes ca be alied to redicates usig secific values for the variables i the domai. Eamles: If is eve [P()], the = [P(a)] Therefore is eve. [Q(a)] is eve [Q()] The Pricile of Mathematical Iductio says that whe you have a statemet, Claim(), that cocers, Claim(1) If P : the Claim() is Claim( k) Claim( k + 1), k true for all (Q) Thus we have P Q. WUCT11 Logic 95

12 Eercise: Accordig to Modus Poes, what must we establish so we ca aly this ricile to the followig statemet ad be able to say Claim() is true for all Claim(): 4 1? is a multile of 3. WUCT11 Logic 96

13 3.4. Modus Tolles Rule of Modus Tolles: If ~Q ad P Q are both tautologies, the so is ~P. I other words, Modus Poes simly says that if we kow ~Q to be true, ad we kow that P imlies Q, the ~P must also be true. Similarly if we kow Q to be false, ad we kow that P imlies Q, the P must also be false Eercise: Write the rule of Modus Poes usig logical otatio: WUCT11 Logic 97

14 Show, usig the quick method that the rule of Modus Tolles is a tautology. 1. ((P Q) ~Q ~P Ste: 3 1 4* WUCT11 Logic 98

15 Eamles: If Zak is a cheater, the Zak sits i the back row Zak sits i the frot row Therefore Zak is ot a cheater. If > 3 the Earth is flat The Earth is ot flat Therefore u 3 Eercise: Comlete the followig usig Modus Tolles If Zeus is a God, the Zeus is ot immortal Therefore Zeus is ot a God. If I go to the beach the it is suy It is ot suy Therefore If I arrive o time the I will be marked reset Therefore I did ot arrive o time. WUCT11 Logic 99

16 3.4.. Uiversal Rule of Modus Tolles: If P() ad Q() are redicates, the uiversal rule of Modus Tolles is: (( P( ) Q( )) ~ Q( a)) ~ P( a). This meas Modus Tolles ca be alied to redicates usig secific values for the variables i the domai. Eamle: If, the = Therefore. a, b, b 0, = a b Eercise: Comlete the followig usig the uiversal rule of Modus Tolles If, the 1 Therefore 1. WUCT11 Logic 100

17 3.5. Provig Quatified Statemets Provig Eistetial Statemets A statemet of the form D, P( ) P ( ) is true for at least oe D. is true if ad oly if To rove this kid of statemet, we eed to fid oe that makes P ( ) true. D Eamles: Prove that there eists a eve iteger that ca be writte two ways as the sum of two rimes. The statemet is of the form D, P( ), where D is the set of eve itegers ad P() is the statemet ca be writte as the sum of two rimes Thus we eed fid oly oe eve iteger which satisfies P(). Essetially, to fid the aroriate umber, we have to guess. Cosider 14 = (7 is rime); ad 14 = (3 ad 11 are rime). Therefore, there eists a eve iteger that ca be writte two ways as the sum of two rimes. WUCT11 Logic 101

18 Let r, s. Prove k, r + 18s = k The statemet is of the form k D, P( k), where D is the set of itegers ad P(k) is the statemet: r + 18s = k. Thus we eed fid oly oe iteger which satisfies P(k) r + 18s Eercises: = (11r + 9s) = k where Prove, + 5 = 0. k = 11r + 9s Prove that if a, b,the 10 a + 8b is divisible by (i.e., is eve). WUCT11 Logic 10

19 3.5.. Provig Uiversal Statemets A statemet of the form D, P( ) P ( ) is true for at every D. is true if ad oly if To rove this kid of statemet, we eed rove that for every D, P ( ) is true. I order to rove this kid of statemet, there are two methods: Method 1: Method of Ehaustio. The method of ehaustio is used whe the domai is fiite. Ehaustio caot be used whe the domai is ifiite. To erform the method of ehaustio, every member of the domai is tested to determie if it satisfies the redicate. WUCT11 Logic 103

20 Eamle: Prove the followig statemet: Every eve umber betwee ad 16 ca be writte as a sum of two rime umbers. The statemet is of the form D, P( ), where D = {4,6,8,10,1,14},ad P() is the statemet ca be writte as the sum of two rime umbers. The domai D is fiite so the method of ehaustio ca be used. Thus we must test every umber i D to show they ca be writte as the sum of two rimes. 4 = + 6 = = = = = Thus by the method of ehaustio every eve umber betwee ad 16 ca be writte as a sum of two rime umbers. WUCT11 Logic 104

21 Eercise: Prove for each iteger with 1 10, + 11 is rime. The statemet is of the form D, P( ), where D = {1,,3,4,5,6,7,8,9,10},ad P() is the statemet + 11 is rime. Thus we must show all umbers i D satisfy P(). Thus by the method of ehaustio for each iteger with 1 10, + 11 is rime. WUCT11 Logic 105

22 Method : Geeralised Proof. The geeralised roof method is used whe the domai is ifiite. It is called the method of geeralizig from the geeric articular. I order to show that every elemet of the domai satisfies the redicate, a articular but arbitrary elemet of the domai is chose ad show to satisfy the redicate. The method to show the redicate is satisfied will vary deedig o the form of the redicate. Secific techiques of geeralized roof will be outlied later i this sectio. WUCT11 Logic 106

23 Eamle: Pick ay umber, add 3, multily by 4, subtract 6, divide by two ad subtract twice the origial. The result is 3. Proof: Choose a articular but arbitrary umber, say, ad the determie if it satisfies the statemet. Ste Pick a umber Result Add Multily by 4 ( + 3) 4 = Subtract = Divide by ( 4 + 6) = + 3 Subtract twice the origial + 3 = = 3 I this eamle, is articular i that it reresets a sigle quatity, but arbitrarily chose as it ca rereset ay umber. WUCT11 Logic 107

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special