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Discrete Maths Objectives to introduce proofs: assumptions, conclusions, rules of inference; different ways of proving if- then statements 242-213, Semester 2, 2014-2015 4. Proofs 1

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Overview 1. What is a Proof? 2. Rules of Inference 3. Using Rules to Build a Proof 4. More Inference Rules 5. More Examples 6. More Proof Jargon 7. Two Simple Ways to Prove (p q ) 8. More Advanced Proofs of (p q) 9. Proving a Biconditional 10. Proof by Cases 11. Existence Proofs 12. Counter-examples 13. More Information 2

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1. What is a Proof? Two assumptions /premises / statements: All men are mortal. and Socrates is a man. Prove the conclusion: Socrates is mortal. 3 assumptions conclusion proof the steps in going from assumptions to a conclusion is called a proof

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In Logic 4 Rules of inference are used to create a proof most rules work with both propositional and predicate logic assumptions conclusion how do we prove this? how do we prove this? Pred. Sec 4.1

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2. Rules of Inference For proofs using propositional logic: 1. Modus Ponens 2. Modus Tollens 3. Transitivity 4. Disjunctive Syllogism 5. Addition 6. Simplification 7. Conjunction 8. Resolution 5 rules involving (used a lot) used in the Prolog logic programming language used in the Prolog logic programming language

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2.1. Modus Ponens Example Let p be It is snowing. Let q be I study discrete math. p q: If it is snowing, then I study discrete math. p: It is snowing. q: I study discrete math. 6 assumptions conclusion

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Why does Modus Ponens work? Look at the truth table for 7 assumptions p q T p T q ? There is only one possible value for q, true.

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2.2. Modus Tollens Example Let p be it is snowing. Let q be I study discrete math. p q: If it is snowing, then I study discrete math. q:I do not study discrete math. p:It is not snowing. 8

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Why does Modus Tollens work? Look at the truth table for 9 assumptions p q T qq T ? There is only one possible value for p, false. This means p == T p (i.e. q == F)

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Example Let p be it snows. Let q be I study discrete math. Let r be I will get an A. p q:If it snows, then I study discrete math. q r:If I study discrete math, then I will get an A. p r:If it snows, then I will get an A. 10 2.3. Transitivity (chaining) pqr assumptions conclusion This may be why math people use the symbol This may be why math people use the symbol

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11 R R p q T q r T possible values for p, r p r ? This shows why using truth tables for proofs does not scale up.

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2.4. Disjunctive Syllogism Example Let p be I study discrete math. Let q be I study English literature. p q:I study discrete math or I study English literature. p:I do not study discrete math. q:I study English literature. 12 p q T pp T q ?? the get part of rule

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2.5. Addition Example Let p be I study discrete math. Let q be I will visit the Moon. p:I study discrete math. p q:I study discrete math or I will visit the Moon. 13 p q T p ?? the add anything rule

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2.6. Simplification Example Let p be I study discrete math. Let q be I study English literature. p q: I study discrete math and English literature q:I study discrete math. 14 p p q T ? q Two rules ? p the get part of rule

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2.7. Conjunction Example Let p be I study discrete math. Let q be I study English literature. p:I study discrete math. q:I study English literature. p q:I study discrete math and I study English literature. 15 p q T ? p T q the combine two statements with rule -- useful the combine two statements with rule -- useful

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2.8. Resolution Example Let p be I study discrete math. Let r be I study English. Let q be I study databases. p r:I do not study discrete math or I study English. p q:I study discrete math or I study databases. q r:I study databases or I study English. 16 Graphically: p r p q q r

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3. Using Rules to Build a Proof Each statement (line) is either an assumption or a new statement made by using a rule of inference. S 1 S 2. S n C 17 assumptions or new statements and conclusion this sequence of statements is the proof

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Example 1 Assumption: Show that q is a conclusion. 18 Simplification

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Example 2 Four Assumptions: It is not sunny and it is colder than yesterday. We will go swimming if it is sunny. If we do not go swimming, then we will take a canoe trip. If we take a canoe trip, then we will be home by night. Show the conclusion: We will be home by night. 19 continued a q if p example a q if p example

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1. Create propositional variables: p: It is sunny. r: We will go swimming. t: We will be home by night. q: It is colder than yesterday. s: We will take a canoe trip. 2. Translation problem into propositional logic: Four assumptions: p qr p r ss t Conclusion: t 20

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3. Construct the proof 21

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4. More Inference Rules Rules using variables and quantification: 1. Universal Instantiation (UI) 2. Universal Generalization (UG) 3. Existential Instantiation (EI) 4. Existential Generalization (EG) 5. Universal Modus Ponens (MP) 6. Universal Modus Tollens (MT) 22 for predicate logic proofs versions of same-named propositional rules involving involving

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4.1. Universal Instantiation (UI) Example The domain U is all dogs. x P(x):All dogs are cute. P(c):Dog c is cute. 23 U P a b c... all elements are in P; nothing outside the choose any element rule

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Universial instantiation can be used on any expression surrounded by . For example: x P(x) Q(x) y P(y) Q(y) P(c) Q(c) P(a) Q(a) 24

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4.2. Universal Generalization (UG) 25 If you always choose a dog that is cute. Therefore, all dogs are cute. any c U P a b c... all elements are in P; nothing outside

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4.3. Existential Instantiation (EI) 26 U P a b c... some elements are in P; others are outside d... the choose an element carefully rule

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Existential instantiation can be used on any expression surrounded by . But you must be careful to select an element that is covered by the logic (i.e. inside the set). For example: x P(x) Q(x) y P(y) Q(y) P(c) Q(c) P(a) Q(a) 27

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4.4. Existential Generalization (EG) Example: P(c):Student c got an A in the class. x P(x):Someone got an A in the class. 28 U P a b c... some elements are in P; others are outside d...

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4.5. Universal Modus Ponens (MP) 29 See the Socrates example, in the next few slides. x (P(x) Q(x) P(a) Q(a) U a must be in here according to the assumptions P Q a

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4.6. Universal Modus Tollens (MT) 30 x (P(x) Q(x) Q(a) P(a) U a must be in here according to the assumptions P Q a

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5. More Examples Example 1 : Prove the conclusion John Smith has two legs Assumptions: Every man has two legs. John Smith is a man. Convert to predicate logic: M(x) is x is a man L(x) is x has two legs John Smith (J) is in the domain. 31 continued x M(x) L(x) Pred. Sec 4.1 M(J) L(J)

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Proof: 32 U M L J

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Example 2 Prove the conclusion Someone who passed the first exam has not read the book. Assumptions: Some students in this class have not read the book. Everyone in this class passed the first exam. 33 continued

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Convert to predicate logic: C(x) is x is in this class B(x) is x has read the book P(x) is x passed the first exam. Translate assumptions and conclusion: 34 continued Pred. Sec 4.3 Pred. Sec 4.1

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Proof 35

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Socrates Example 36 Proof "All men are mortal" "Socrates is a man" "Socrates is mortal" Pred. Sec 4.1

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6. More Proof Jargon A theorem is a conclusion shown to be true by writing a proof. A lemma is a statement that helps to prove a theorem. A corollary of a theorem is another statement that you can prove using the theorem. A conjecture is a conclusion that might be true. Once you have proved it, then it becomes a theorem. 37

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7. Two Simple Ways to Prove (p q ) 1. Trivial Proof: if q is true, then p q is true. e.g. If it is raining then 1=1. 38 p q this statement (p q) is true look at the cases in the truth table p doesn't affect the truth of (p q) p doesn't affect the truth of (p q)

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2. Vacuous Proof: if p is false then p q is true e.g. If I am both rich and poor, then 2 + 2 = 5. 39 look at the cases in the truth table p q this statement (p q) is true q doesn't affect the truth of (p q) q doesn't affect the truth of (p q)

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8. More Advanced Proofs of (p q) There are three main advanced techniques: 1. direct proof 2. proof by contrapositive (contraposition) 3. proof by contradiction 40

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8.1. Direct Proof of p q Assume that p is true. Use rules of inference to show that q is true. If p and q are true then (p q) is true 41 look at the cases in the truth table

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Example DP-1 Prove If n is an odd integer, then n 2 is odd. Assume that n is odd (i.e. p is true). Then n = 2 k + 1 for some integer k. Squaring both sides: n 2 = ( 2 k + 1) 2 = 4 k 2 + 4k +1 = 2(2 k 2 + 2k) + 1= 2r + 1 So n 2 is an odd integer (i.e. q is true). This means that p q is true. 42 p q

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Example DP-2 Definition: A real number x is rational if there exists integers a and b where b 0 such that x = a/b, and a and b have no common factors. Prove that the sum of two rational numbers is rational. As an if-then: if r is rational and s is rational then (r+s) is rational. Assume that r and s are rational (i.e. p is true). 43 pq continued

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Since r and s are rational, we can use the definition on the previous slide: This shows that the sum (r +s) is rational (i.e. that q is true) This means that p q is true. 44 where v = pu + qt w = qu 0

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8.2. Prove by Contrapositive Instead of proving (p q), we prove its contrapositive, which is ( q p) (p q) and ( q p) are logically the same (look at their truth tables), but sometimes it is easier to build a proof for ( q p). 45

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Proof Steps for ( q p) Assume that q is true. Use rules of inference to show that p is true. If q and p are true (i.e. that q and p are false) then ( q p ) is true see the truth table on the previous slide 46

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Example CP-1 Prove that if n is an integer and 3 n + 2 is odd, then n is odd. Prove the contrapositive ( q p ) is true This is: if (n is not odd) then (3n + 2 is not odd) Simplify using the mathematical meaning of odd/even: if (n is even) then (3n + 2 is even) 47 continued pq qq pp qq pp

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Assume n is even (i.e. that q is true). So, n = 2 k for some integer k. Thus 3 n + 2 = 3(2k) + 2 =6k +2 = 2(3k + 1) = 2j Therefore 3 n + 2 is even (i.e. p is true). We have shown (q p ) is true, so ( p q) is true. 48

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Example CP-2 Prove that if n 2 is odd, then n is odd. Prove the contrapositive ( q p ) is true This is: if (n is not odd) then (n 2 is not odd) Simplify using the mathematical meaning of odd/even: if (n is even) then (n 2 is even) 49 p q qq pp qq pp continued

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Assume n is even (i.e. q is true). Therefore, there is an integer k such that n = 2 k. Square both sides to get n 2 = 4 k 2 = 2 ( 2 k 2 ) This means that n 2 is even (i.e. p is true) We have shown (q p ) is true, so ( p q) is true. 50

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8.3. Prove by Contradiction Instead of proving that (p q) is true, prove the negative is false: i.e. prove that (p q) is false Transform: prove ( p q) is false prove (p q) is false In words this means that we must show that p and q cannot both be true. 51

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Example CD-1 Prove that if you pick 22 days from the calendar, at least 4 must fall on the same day of the week. p q: if (pick 22 days) then ( 4 are same day) Show that (p q) is false i.e. show that (p q) is false q = ( 4 are same day) = ( 3 are same day) 52 p q continued

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p = pick 22 days = 3 weeks + 1 day If the pick includes 3 weeks then each day is repeated 3 times. "+ 1 day" means that one of the days occurs 4 times. This is not the same as q In other words, (p q) is false 53

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Example CD-2 Prove that 2 is irrational. This famous problem is not of the form (p q), but we can use a slight variation of proof by contradiction. We want to prove that p is true. Instead we will prove that the negative is false: i.e. prove that p is false Assume p: 2 is not irrational 2 is rational 54 p continued

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If 2 is rational then 2 = a/b, where b 0 and a and b have no common factors (see slide 43). Then: Therefore a 2 is even. If a 2 is even then a is even (we should really prove this). Since a is even, a = 2c for some integer c. Then: Therefore b 2 is even. So b is even. 55 continued

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Since a and b are both even, then a and b have a common factor of 2 This contradicts p which includes the assumption that a and b have no common factors. So p is false, which means that p is true. Therefore, 2 is irrational. 56

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9. Proving a Biconditional (p q) Prove (p q) is true in two steps (cases): show that (p q ) is true and show that (q p) is true Why? Because (p q) (p q) (q p) 57 pq pqpqqpqpp q q p p q TTTTTT TFFTFF FTTFFF FFTTTT As shown in the truth table

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For each of the two cases, you can use any proof technique you like: e.g. trivial, vacuous, direct, contrapositive, contradiction 58

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Example Prove n is odd if and only if n 2 is odd. Two cases: p q:if (n is odd) then (n 2 is odd) q p:if (n 2 is odd) then (n is odd) we proved this case in example CP-2 59 p q

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10. Proof by Cases We want to prove the complex implication: We can rewrite this as a conjunction (and) of simple implications: ((p 1 q) (p 2 q) ... (p n q)) Each of the implications is a case that needs to be proved: show (p 1 q) is true, show (p 2 q) is true, etc. 60

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How to rewrite: Remember that (p q) ( p q) (p 1 p 2 ... p n ) q (p 1 p 2 ... p n ) q ( p 1 p 2 ... p n ) q // De Morgan's ( p 1 q) ( p 2 q) ... ( p n q) // distribution (p 1 q) (p 2 q) ... (p n q) 61

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Example Define max(a, b) = a if a b, otherwise = b if (a, b, c are real numbers) then max( max(a,b), c) == max(a, max(b, c)) Where are the cases inside p (i.e. p 1, p 2,..p n )? They are all the different possible values for a, b, and c. 62 continued q p (complex)

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There are only 6 p cases that we need to look at: 1. a b c 2. a c b 3. b a c 4. b c a 5. c a b 6. c b a 63 cases p1, p2,..., p6

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In other words, we must prove: (p 1 q)if (a b c) then max( max(a,b), c) == max(a, max(b, c)) (p 2 q)if (a c b) then max( max(a,b), c) == max(a, max(b, c)) : (p 6 q)if (c b a) then max( max(a,b), c) == max(a, max(b, c)) 64 and

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Case 1 Proof I'll just look at case 1 (p 1 q), since the other cases are similar. Use direct proof: Assume that p 1 is true. Use rules of inference to show that q is true. If p 1 and q are true then (p 1 q) is true 65

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Assume a b c (i.e. p 1 is true) Plug this information into max(): max(a, b) = a max(a, c) = amax(b, c) = b This means that max(max(a, b), c) = max(a,c) = a and max(a, max(b, c)) = max(a, b) = a Therefore q is true. So p 1 q is true. 66

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11. Existence Proofs Prove statements like e.g. is there a British person in this classroom? There are two main approaches: 1. Constructive existence proofs 2. Nonconstructive existence proofs 67 Does an element exist?

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Constructive Existence Proof Find an element c, for which P(c) is true. Then is true by Existential Generalization (EG). 68 U P a b c... some elements are in P; others are outside d... Proof by Example

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Example Prove: There is a positive integer that can be written as the sum of cubes of positive integers in two different ways. Proof: 1729 because 1729 = 10 3 + 9 3 = 12 3 + 1 3 69

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Nonconstructive Existence Proof Instead of proving x p(x) is true, we show that x p(x) is false Translate: prove x p(x) is false In words, we are trying to show that p() is not empty. 70 U P a b c... show that P is not empty i.e. that is has some elements d...

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12. Counter-examples A common problem is to show that a x P(x) statement is false: e.g. prove All the planets in our solar system are made of cheese. is false Logically: prove x P(x) is false x P(x) is true x P(x) is true In words, found an example that shows that P() is false called a counter-example. 71 U P a b c... show that this area is not empty

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Examples All the planets in our solar system are made of cheese. Counter-example: the Earth. So the statement is false. Every positive integer is the sum of the squares of 3 integers. Counter-example: 7. So the statement is false. 72

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13. More Information Discrete Mathematics and its Applications Kenneth H. Rosen McGraw Hill, 2007, 7th edition chapter 1, sections 1.6 1.8 73