1. 1 Copyright © 2011 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 1 Mathematical Tools and Techniques

2. 22 Logic and Proofs Logic involves propositions which have truth values, either true or false –“0 = 1” is a proposition whose value is false –“peanut butter is a source of protein” is true A proposition containing a free variable can be true or false, depending on the value of the variable –“x - 1 is prime” is true for x = 8, false for x = 10 Introduction to Computation

3. 3 Logic and Proofs (cont’d.) Compound propositions are constructed using the logical connectives ∧, ⋁, ¬, , and ↔ Introduction to Computation3

4. 4 Logic and Proofs (cont’d.) The truth value of ¬p is the opposite of the truth value of p Truth tables show possible combinations of p and q pq p ⋀ qp ∨ qp ∨ q p → qp ↔ q TTTTTT TFFTFF FTFTTF FFFFTT Introduction to Computation4

5. 5 Logic and Proofs (cont’d.) For a proposition like (p ∨ q) ∧ ¬(p → q), fill in the table in the order shown. Column 3 is the negation of column 2, column 4 is the conjunction of columns 1 and 3 1432 pq (p ∨ q)(p ∨ q) ∧ ¬ (p → q) TTTFFT TFTTTF FTTFFT FFFFFT Introduction to Computation5

6. 6 Logic and Proofs (cont’d.) A tautology is a proposition that is always true –Example: p ∨ ¬ p A contradiction is a proposition that is always false –Example: p ∧ ¬ p Two propositions P and Q are said to be logically equivalent if they always have the same truth value –This is written as P ⇔ Q A proposition P is said to logically imply a proposition Q if, whenever P is true, Q is also true –This is written as P ⇒ Q Introduction to Computation6

7. 7 Logic and Proofs (cont’d.) P → Q and P ⇒ Q look similar; however, –P → Q is a proposition; it has a truth value –P ⇒ Q is a “meta-statement”, an assertion about the relationship between propositions P and Q –P ⇒ Q means that P → Q is a tautology –Similarly, P ⇔ Q means that P ↔ Q is a tautology Introduction to Computation7

8. 8 Logic and Proofs (cont’d.) Logical identities can be used to simplify compound propositions: –The commutative laws p ∨ q  q ∨ p p ∧ q  q ∧ p –The associative laws p ∨ (q ∨ r)  (p ∨ q) ∨ r p ∧ (q ∧ r)  (p ∧ q) ∧ r –The distributive laws p ∨ (q ∧ r)  (p ∨ q) ∧ (p ∨ r) p ∧ (q ∨ r)  (p ∧ q) ∨ (p ∧ r) Introduction to Computation8

9. 9 Logic and Proofs (cont’d.) Logical identities (cont’d.) –The De Morgan laws  (p ∨ q)   p ∧  q  (p ∧ q)   p ∨  q –An equivalent formulation of the conditional (p  q)  (  p ∨ q) –The contrapositive of a conditional (p  q)  (  q   p) –An equivalent formulation of the biconditional (p  q)  (p  q) ∧ (q  p) Introduction to Computation9

10. 10 Logic and Proofs (cont’d.) The truth value of a proposition like “x - 1 is prime” depends on the value of x –We can use logical quantifiers,  (for every) or  (for some) to obtain statements that are no longer statements about specific elements in the domain, but statements about the domain itself  x (x - 1 is prime)  x (x - 1 is prime) –The first proposition is true if x – 1 is prime for every value of x in the domain; the second is true if x – 1 is prime for some (at least one) value of x. Introduction to Computation10

11. 11 Logic and Proofs (cont’d.) In statements with more than one quantifier, order matters: –  x (  y (x < y)) and  y (  x (x < y)) are not logically equivalent –The first says that for every x there is some larger y (perhaps depending on x); the second says that there is a single value y that is larger than every x Here are two identities involving the negation of a quantified statement:  (  x (P(x))   x (  P(x))  (  x (P(x))   x (  P(x)) Introduction to Computation11

12. 12 Logic and Proofs (cont’d.) A proof is a series of statements, each of which is derived from: –Initial assumptions –Statements that have been derived previously –Generally accepted facts The derivations use principles of logical reasoning Introduction to Computation12

13. 13 Logic and Proofs (cont’d.) A direct, constructive proof –Prove that for every two integers a and b, if a and b are odd, then ab is odd –Use the definition of odd to restate this as follows: If there exist integers i and j so that a = 2i +1 and b = 2j +1, then there exists k such that ab = 2k +1 Proof: ab = (2i + 1)(2j + 1) = 4ij + 2i + 2j + 1 = 2(2ij + i + j) + 1 If we let k be (2ij + i + j), we have the result we want Introduction to Computation13

14. 14 Logic and Proofs (cont’d.) An indirect proof –Prove that for every three positive integers i, j, and n, if ij = n, then i ≤  n or j ≤  n –Prove the contrapositive: assume there exist i, j, and n such that  (i ≤  n  j ≤  n) –By De Morgan, this implies  (i ≤  n)   (j ≤  n), or (i >  n)  (j >  n) –Therefore i j >  n  n = n, which implies that i j  n –This is a direct proof of the contrapositive statement, and thus an indirect proof of the original statement Introduction to Computation14

15. 15 Logic and Proofs (cont’d.) A proof by contradiction that  2 is irrational –Assume that there exist positive integers m and n such that m/n =  2 –Then, by dividing m and n by all common factors, we get p and q with no common factors such that p/q =  2 –Then all these statements are true: p = q  2; p 2 = 2q 2 ; p 2 is even; p is even; p = 2r; p 2 = 4r 2 ; q 2 = 2r 2 ; q is even. –Therefore, p and q have the common factor 2. This is a contradiction, which shows that the assumption is false Introduction to Computation15

16. 16 Logic and Proofs (cont’d.) A proof by cases –If we can enumerate all of the possible cases, and prove that the statement is true in each case, then we have proven the statement –For example, if we want to prove that one proposition P implies another proposition Q, then looking at the truth tables for P and Q gives us one way of enumerating all the possible cases Introduction to Computation16

17. 17 Sets A finite set can be described by listing its elements A = {1, 2, 4, 8} Sometimes we use ellipses B = {0, 3, 6, 9, … } C = {13, 14, 15, …, 71} E = {0, 3, 5, 6, 8, 9, … } (what set is this? Look 2 slides ahead) More generally, we can use a defining property B = {x | x is a nonnegative integer multiple of 3} C = {x | x is an integer and 13 ≤ x ≤ 71} B = {3y | y is a nonnegative integer} Introduction to Computation17

18. 18 Sets (cont’d.) x  A means that x is an element of A –Similarly, x  A means that x is not an element of A A  B means that A is a subset of B –i.e., every element of A is also an element of B The empty set is denoted by  The order of elements when we write a set is not significant: for example, {0, 1} = {1, 0} Repetition has no effect: {0, 0, 1, 1, 1, 2} = {0, 1, 2} To show that A = B, we need to show that A  B and that B  A Introduction to Computation18

19. 19 Sets (cont’d.) A few sets will come up frequently: –ℕ is the set of natural numbers, or nonnegative integers –ℤ is the set of all integers –ℝ is the set of all real numbers –ℝ + is the set of nonnegative real numbers Now we can write B and E from above more concisely: B = {3y | y  ℕ } E = {3i + 5j | i, j  ℕ } Introduction to Computation19

20. 20 Sets (cont’d.) The union, intersection and difference of two sets are defined as follows: A ∪ B = {x | x  A  x  B} A ∩ B = {x | x  A  x  B} A - B = {x | x  A  x  B} Examples: {1, 2, 3, 5} ∪ {2, 4, 6} = {1, 2, 3, 4, 5, 6} {1, 2, 3, 5} ∩ {2, 4, 6} = {2} {1, 2, 3, 5} - {2, 4, 6} = {1, 3, 5} Introduction to Computation20

21. 21 Sets (cont’d.) The complement of a set –We assume that A is a subset of some universal set U –Then the complement of A, written A’, is U - A –We think of A’ as the set of “everything that is not in A” –What this means, however, can be very different depending on the choice of U ; for example, what {0, 1}’ means depends on whether {0,1} is thought of as a subset of ℕ, ℤ, or ℝ. Introduction to Computation21

22. 22 Sets (cont’d.) Many useful set identities are analogous to the logical identities –Union and intersection are commutative and associative, and distribute across each other Two sets are disjoint if their intersection is empty A collection of sets is pairwise disjoint if every two distinct sets in the collection are disjoint A partition of a set S is a collection of pairwise disjoint sets whose union is S Introduction to Computation22

23. 23 Sets (cont’d.) We can describe the union of a set of sets –∪ {A i | 0 ≤ i ≤ n} = {x | x  A i for some i with 0 ≤ i ≤ n} Similarly for intersection: –∩ {A i | 0 ≤ i ≤ n} = {x | x  A i for every i with 0 ≤ i ≤ n} The set of all subsets of a set A is called the power set of A and is written 2 A –2 A = { X | X  A } –Example: 2 {a,b,c} = { , {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}} –Note that the empty set and the set A itself are in A’s power set Introduction to Computation23

24. 24 Sets (cont’d.) The Cartesian product of two sets A and B, denoted A  B, is the set of all ordered pairs with first element from A and second element from B –A  B = {(a, b) | a  A and b  B} –We can generalize this to ordered k-tuples: A 1  A 2  …  A k = {(a 1, a 2, …, a k | a i  A i for each i} Introduction to Computation24

25. 25 Functions and Equivalence Relations f : A →B means that f is a function from A to B –To each element of A, one element of B is assigned Examples: f : ℕ →R defined by the formula f(x) =  x g : 2 ℕ → 2 ℕ defined by g(A) = A ∪ {0} A is the domain of the function and B the codomain f and g are equal if and only if they have the same domain and codomain and f (x) = g(x) for every x in the domain Partial functions from A to B may assign values to only some elements of A Introduction to Computation25

26. 26 Functions and Equivalence Relations (cont’d.) The range of a function is the set of elements of the codomain that are actually values of the function –{ f (x) | x  A} (a subset of the codomain B) Introduction to Computation26

27. 27 Functions and Equivalence Relations (cont’d.) If f : A →B is a bijection then we can define the inverse function f -1 from B to A by these two formulas: for every x  A and y  B, –f -1 ( f(x) ) = x –f (f -1 (y)) = y It is easy to check that f -1 is also a bijection Introduction to Computation27

28. 28 Functions and Equivalence Relations (cont’d.) An n-ary operation on a set A is a function that assigns to every ordered n-tuple of elements of A an element of A Unary and binary operations are of most interest –Binary operations on the integers include addition –For every set S, binary operations on 2 S include union and intersection –Unary operations include negation (on the set of integers, for example) and complementation (on the set 2 A ) Introduction to Computation28

29. 29 Functions and Equivalence Relations (cont’d.) For a unary or binary operation on a set A –We say that a subset A 1 of A is closed under the operation if the result of applying the operation to elements of A 1 is an element of A 1 If A = 2 ℕ and A 1 is the set of nonempty subsets of A, then A 1 is closed under union but not under intersection The set of even natural numbers is closed under addition and multiplication The set of odd natural numbers is closed under multiplication but not addition Introduction to Computation29

30. 30 Functions and Equivalence Relations (cont’d.) Relations We can express relationships several ways: If R is a relation on a set, we can write “a is related to b” as a R b or as (a, b)  R Introduction to Computation30

31. 31 Functions and Equivalence Relations (cont’d.) Examples: –The equality relation (a relation on any set) –The relation on A containing all ordered pairs –The relation of congruence mod n on the set ℕ Introduction to Computation31

32. 32 Functions and Equivalence Relations (cont’d.) We may drop the subscript and just say [x] if there is no opportunity for confusion Theorem: If R is an equivalence relation on A, the equivalence classes with respect to R form a partition of A, and two elements of A are equivalent if and only if they are elements of the same equivalence class Introduction to Computation32

33. 33 Languages An alphabet is a finite set of symbols usually denoted by  –Examples: {a,b}, {0,1}, {A, B, C, …, Z} A string over  is a finite sequence of symbols |x| stands for the length of the string x n a (x) is the number of occurrences of a in the string x The null string  is a string over any alphabet  |  | = 0 Introduction to Computation33

34. 34 Languages (cont’d.) The set of all strings over  is  * Example: {a,b}* = { , a, b, aa, ab, ba, bb, aaa, aab,…} A language over  is a subset of  * Examples: –The empty language  –{ , a, aab}, a finite language –The palindromes over {a,b} (strings like , a, and baabaab that read the same backwards as forwards) –{x  {a, b}* | n a (x) > n b (x)} –{x  {a, b}* | |x| ≥ 2 and x begins and ends with b} Introduction to Computation34

35. 35 Languages (cont’d.) xy is the concatenation of the two strings x and y; this is the basic operation on strings –If x = ab and y = bab then xy = abbab and yx = babab –For every string x, x  =  x = x –|xy| = |x| + |y| Concatenation is associative, i.e., (xy)z = x(yz), so we can write xyz without worrying about how terms are grouped If s = tuv then t is a prefix of s, v is a suffix, and u is a substring. Every string is a prefix (and suffix, and substring) of itself Introduction to Computation35

36. 36 Languages (cont’d.) For languages L 1, and L 2 over  –L 1 ∪ L 2, L 1 ∩ L 2, and L 1 − L 2 are also languages over  If L   *, the complement of L is a language,  * - L For languages L 1, L 2 over  –L 1 L 2 is the language {xy | x  L 1 and y  L 2 } We use exponential notation a k = aaa…a, where there are k occurrences of a This also applies to strings (x k = xxx…x) and languages (L k = LLL…L) a 0 = x 0 = , L 0 = {  } (for every a  , x   *, L   *) Introduction to Computation36

37. 37 Languages (cont’d.) If L is a language over , then L* denotes the language of all strings that can be obtained by concatenating zero or more strings in L –This operation is known as the Kleene star L* = ∪ {L k | k  ℕ }   L* for every language L, since L 0 = {  } Strings are finite, and languages may not be, but to use a language we need a finite description –L 1 = {ab,bab}* ∪ {b} {ba}*{ab}* –L 2 = {x  {a,b}* | n a (x) > n b (x)} Introduction to Computation37

38. 38 Recursive Definitions A recursive definition of a set has a basis statement that specifies at least one member of the set, and a recursive part that specifies how additional members of the set can be generated in terms of given members. The prototypical example is ℕ, the set of natural numbers. It can be defined as follows: –Basis statement: 0  ℕ –Recursive part: if n  ℕ then n+1  ℕ –Every element of ℕ can be obtained from the first two statements Introduction to Computation38

39. 39 Recursive Definitions (cont’d.) Summary: –The third statement in the definition of ℕ is what says that ℕ is the smallest set that contains 0 and is closed under the successor operation (addition by 1) –The statement that the set being defined is the smallest is frequently omitted but always understood Example: the subset B = { 2 i 5 j | i, j  ℕ } –1  B –For every n  B, 2n  B –For every n  B, 5n  B Introduction to Computation39

40. 40 Recursive Definitions (cont’d.) We denote by F the subset of 2 {a,b}* defined by: – , {  }, {a}, {b}  F –if L 1, L 2  F then L 1 ∪ L 2  F –if L 1, L 2  F then L 1 L 2  F F is the smallest set of languages that contains the languages , {  }, {a}, and {b} and is closed under union and concatenation It is easy to see that F is the set of all finite languages over {a,b} Introduction to Computation40

41. 41 Recursive Definitions (cont’d.) The set of cities reachable from city s –Suppose that C is a finite set of cities, and the relation R is defined on C so that c R d means there is a nonstop commercial flight from c to d –For a city s  C, we would like to describe r(s), the set of cities reachable from s by taking zero or more nonstop flights. We can define r(s) this way: s  r(s) if c  r(s) and c R d then d  r(s) Introduction to Computation41

42. 42 Structural Induction Consider again the language Expr: –a  Expr. –For every x and every y in Expr, x ◦ y and x y are in Expr (where x ◦ y and x y are the strings x+y, x*y) –For every x  Expr, ◊(x)  Expr (where ◊(x) is (x) ) Note: this seems like unnecessarily confusing notation (why do we need the operator symbols ◦,, and ◊, when we already have the “operators” + and *?) The reason is that + and * are operations on numbers, but we’re talking about strings, not numbers. In this discussion +, *, (, and ) are just symbols in the alphabet Introduction to Computation42

43. 43 Structural Induction (cont’d.) To prove that every string x  Expr satisfies a condition P(x), use structural induction: show that –P(a) is true –For every x and every y in Expr, if P(x) and P(y) are true, then P(x ◦ y) and P(x y) are true –For every x  Expr, if P(x) is true, then P(◊(x)) is true In other words, show that the set of elements x satisfying the property P contains a and is closed under ◦,, and ◊. This set must then contain every element in the smallest set that contains a and is closed under the operations; i.e., every element in Expr Introduction to Computation

44. 44 Structural Induction (cont’d.) The recursive definition of Expr consists of –A basis part ( a  Expr ) –Three recursive parts, the first of the form If x, y  Expr then x ◦ y  Expr The basis step of the proof is to show that P(a) is true The induction hypothesis is that x, y  Expr and that P(x) and P(y) are true The first case of the induction step is to show that P(x ◦ y) is true. (Similarly for the other 2 operations) Introduction to Computation

45. 45 Structural Induction (cont’d.) For example, suppose P(x) is “x has odd length” The basis step of the proof is to show that a has odd length, and this is clearly true: |a| = 1 The induction hypothesis is that x, y  Expr and that x and y have odd length (i.e., |x| and |y| are odd) The three cases in the induction step are to show that x ◦ y, x y, and ◊(x ) have odd length These are all true, because | x ◦ y| = | x+y | = |x| + |y| + 1, and odd + odd + 1 = odd | x y| = | x*y | = |x| + |y| + 1 | ◊(x)| = |(x)| = |x| + 2, and odd + 2 = odd Introduction to Computation

46. 46 Structural Induction (cont’d.) Mathematical induction is simply structural induction based on the recursive definition of ℕ given earlier This is used to prove statements of the form “for every integer n ≥ n 0, P(n)” –Basis step: prove the statement P(n) for n = n 0 –Induction hypothesis: k is an integer ≥ n 0 and P(k) is true –Induction step: show using the induction hypothesis that P(k+1) is true Introduction to Computation46

47. 47 Structural Induction (cont’d.) Prove: For every n  ℕ, every set A with n elements, 2 A has exactly 2 n elements –Basis: for every set A with 0 elements, 2 A has 2 0 elements; this is true because only  has zero elements, and 2  = {  }, which has one element –Induction hypothesis: k  ℕ, and for every set A with k elements, 2 A has 2 k elements –Induction step: to show that for every set A with k+1 elements, 2 A has 2 k+1 elements Introduction to Computation47

48. 48 Structural Induction (cont’d.) Prove: For every n  ℕ, and every set A with n elements, 2 A has exactly 2 n elements (cont’d.) –Proof of induction step Let A be a set with k + 1 elements, and let a be any element of A (there is one, since k + 1 ≥ 1) Then A - {a} has k elements, and 2 A-{a} has 2 k elements by the induction hypothesis; therefore, A has 2 k subsets that do not contain a and 2 k subsets that do contain a, for a total of 2 k+1 subsets Introduction to Computation48

49. 49 Structural Induction (cont’d.) Strong induction is another form of induction Example: To prove that for every n ≥ 2, n is either prime or a product of two or more primes –Let us strengthen the statement: for n ≥ 2, every number m such that 2 ≤ m ≤ n is either prime or a product of two or more primes. –The reason for this, as we’ll see, is that it gives us a stronger induction hypothesis but doesn’t actually require that we prove any more than we would have anyway. Introduction to Computation49

50. 50 Structural Induction (cont’d.) Proof: –Basis step: To show that every m satisfying 2 ≤ m ≤ 2 is either prime or a product of primes. This reduces to showing that 2 is, and 2 is prime –Induction hypothesis: k ≥ 2, and for every m satisfying 2 ≤ m ≤ k, m is either a prime or a product of primes –Statement to prove in induction step: For every m satisfying 2 ≤ m ≤ k + 1, m is either prime or a product of primes. Introduction to Computation50

51. 51 Structural Induction (cont’d.) Proof of induction step: For every m with 2 ≤ m ≤ k, we already have the conclusion we want, from the induction hypothesis. The only additional statement we need to prove is that k + 1 is either prime or a product of primes If k + 1 is prime, we’re done; if not, k + 1 is the product of two smaller numbers, both bigger than 1. By the (stronger) induction hypothesis, both are prime or the product of primes. Therefore (in either case), k + 1 is the product of primes Introduction to Computation

52. 52 Structural Induction (cont’d.) The language Balanced was defined as follows: –   Balanced –If x, y  Balanced, then xy  Balanced and (x)  Balanced Let’s prove that x  Balanced if and only if B(x) is true, where B(x) is “x contains equal numbers of left and right parentheses, and no prefix of x contains more right than left” The “only if” part is straightforward; use structural induction, based on the definition of Balanced Introduction to Computation52

53. 53 Structural Induction (cont’d.) The basis step is to show that B(  ) is true. This is clear, because  has no symbols and no nonnull prefixes The induction hypothesis is that x, y  Balanced and both B(x) and B(y) are true We must show that B(xy) and B( (x) ) are true. Both xy and (x) have equal numbers of left and right parentheses, because x and y do. A prefix of xy is either a prefix of x, or xz for some prefix z of y; a prefix of (x) other than (x) is  or (z, for some prefix z of x. In all these cases, the statement is true by the induction hypothesis Introduction to Computation

54. 54 Structural Induction (cont’d.) For the reverse (the “if” part), use induction on the length of the string Prove: For every n  ℕ, if x is a string of parentheses so that |x| = n and B(x) is true then x  Balanced –Basis: if |x| = 0 then x = , so x  Balanced –Induction hypothesis: k  ℕ, and for every string x of parentheses, if |x| ≤ k and B(x), then x  Balanced (since we say |x| ≤ k, we’re using strong induction) –To prove: if |x| = k + 1 and B(x), then x  Balanced –Proof: the details get a little involved; see book Introduction to Computation54

55. 55 Structural Induction (cont’d.) Sometimes making a statement stronger makes it easier to prove. Let’s prove that no string in Expr contains ++ as a substring The basis step is easy to prove, and so are the induction steps for x*y and (x) In the case of x+y, what if x ends with + or y begins with +? Strengthen the statement: prove that no string contains ++ as a substring, or begins or ends with + –Now the stronger induction hypothesis makes it possible to complete the proof Introduction to Computation55

56. 56 Structural Induction (cont’d.) Recursive definitions of sets lead naturally to ways of defining functions on those sets Example: if we define a function f at 0, and then define f (n+1) assuming that f (n) is defined, then f is effectively defined over ℕ The factorial function is defined this way: f(0) = 1; f(n+1) = (n+1) * f(n) Definitions like these are particularly well suited to induction proofs of properties of the corresponding functions Introduction to Computation56