Discrete Mathematics ? Transparency No. 2-0 Chapter 3 Induction and Recursion Transparency No. 3-1 formal logic mathematical preliminaries

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No. 3-2 Contents 4.1 Mathematical Inductions 4.2 Strong Induction and Well-ordering 4.3 Recursive definitions & Structural Induction 4.4 Recursive algorithms 4.5 Program correctness

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Mathematical Induction (MI)  Principle of MI : To show that a property p hold for all nonnegative integer n, it suffices to show that 1. Basis step: P(0) is true 2. Inductive step: P(n)  P(n+1) is true for all nonnegative integer n. P(n) in 2. is called the inductive hypothesis. Notes: 1. Math. Ind. is exactly the inference rule: P(0),  n P(n)  P(n+1)  n P(n) for any property P 2. If the intended domain is all positive integers, then the basis step should be changed to: 1.Basis step: P(1) is true.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No. 3-4 Examples 1.Show that for all positive integers n, … + n = n (n+1) /2. Pf: Let P(n) denote the proposition: … + n = n (n+1) /2. The proof is by induction on n. Basis step: P(1) is true since 1 = 1 x (1+1) /2. Ind. step: Assume p(k) holds for arbitrary integer k > 0, i.e., … + k = k(k+1)/2. Then 1 + … + k + (k+1) = k(k+1)/2 + (k+1) = k(k+1)/2 + 2(k+1)/2 = (k+1)[(k+1)+1] /2. Hence p(k+1) is also true. This completes the proof of basis step and inductive step of MI, and hence by MI, p(n) holds for all positive integers n.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No. 3-5 Examples : 2:  i=1,n 2i-1 = n 2 3. n < 2n 4. 3 | n 3 - n if n > 0  i=1,n 2 i = 2 (n+1)  j=1,n ar j = ar n+1 - a / (r -1) 7. Let H k = 1 + 1/ /k => H 2 n  1 + n/2 8. |S| = n => |2 S | = 2 n. 9. If n > 3 => 2 n < n! 10. ~(S 1 ...  S n ) = ~S 1 U... U ~S n.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Let H k = 1 + 1/ /k. Then H 2 n  1 + n/2 for all non-negative integers n. pf: By induction on n. Let p(n) be the proposition H 2 n  1 + n/2 Basis Step: n = 0. Then H 2 0 = H 1 = 1  1 + 0/2. Hence p(0) is true. Ind. Step: Assume p(n) holds for any n  I.e., H 2 n  1 + n/2 holds for any n  0. Then H 2 n+1 = 1+… + 1/2 n + 1/(2 n +1) + … 1/(2 n +2 n )  H 2 n + 2 n x 1/(2 n +2 n )  1 + n/2 + ½ = 1 + (n+1)/2. This establishes the ind. step of MI. As a result p(n), i.e., H 2 n  1 + n/2, holds for all nonnegative integers n.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No. 3-7 More examples: 14. for every k  12, there are m,n  0 s.t. k = 4m + 5n. pf: By induction on k. Basis: If k < 12, then we are done. If k = 12, then = 4 x x 0. Inductive step: k = t + 1 > 12 ( Hence t  12 ) By Ind. Hyp., t = 4m + 5n. Then k = t + 1 = 4m + 5n + 1. case 1: m > 0 => k = 4(m-1) + 5 (n+1) case 2: m = 0 => t = 5n > 11 => n  3. hence t+1 = 5(n-3) = 4 x (n-3). Q.E.D.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No. 3-8 Correctness of MI.  Correctness of MI: Let p(.) be a property about positive integers. If p(1) holds and p(n) implies p(n+1) for all n, then it is true that p(n) holds for all n. Pf: Assume MI is incorrect. i.e. the set NP = {k | p(k) is false} is not empty. Let m be the least number of NP ---existence by well-order theorem Since p(0), 0  NP and m >0. => m-1 exists and p(m-1) is true => P(m) holds [by the inductive step of MI ] => m  NP => a contradiction. Q.E.D.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No. 3-9 Strong Induction and Well-Orering  [A problem MI is hard to prove: ] If n is a positive number > 1, then n can be written as a product of primes.  To prove this theorem using induction, we needs a stronger form of MI.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Strong Induction  [The 2 nd form of MI(Strong Induction; complete Induction)] To prove that p(n) holds for all non-negative integer n, where p(n) is a propositional function of n, It suffices to show that Basis step: P(0) holds Inductive step: P(0) /\ P(1) /\...,/\p(k-1)  P(k) holds for all k  0. I.e., Assume p(0),…p(k-1) hold for arbitrary k, and then show that p(k) is true as well.  P(0) /\ P(1) /\...,/\p(k-1) (or  t t<k  P(k)) is the called the induction hypothesis of the proof.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Example Ex2 : If n is a positive number > 1, then n can be written as a product of primes. Pf: Let p(n) be the proposition : if n> 1 then it can be written as a product of primes. Basis step: p(1) holds since n  1, not > 1. Ind. step: Let k be arbitrary positive number and assume p(t) holds for all t < k. There are two cases to consider: case 1: k is a prime number, then p(k) holds since k = k is the product of itself. case 2: k is a composite number. The by definition, there are two numbers 1< a, b < k such that k = ab. By ind. hyp., p(a) and p(b) hold and since a, b > 1, a and b can be written as a product of primes. let a = a 1,…,a i and b = b 1,…b j, then k = a 1 …a i x b 1 …b j is a product of primes.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Correctness of Strong Induction and well-ordering.  Correctness of SI: Let p(.) be a property about positive integers. If p(1) holds and p(1) /\ p(n) …/\p(n) implies p(n+1) for all n, then it is true that p(n) holds for all n. Pf: Assume SI is incorrect. i.e. the set NP = {k | p(k) is false} is not empty. Let m be the least number of NP ---existence by well-order property of positive integers Since p(1), 1  NP and m >1. => m-1 exists and p(m-1) is true => P(m) holds [by the inductive step of SI ] => m  NP => a contradiction. Q.E.D.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Well-ordered Property  [Well-ordered property of natural numbers]Every non-empty subset of non-negative integers has a least element.  The property can be used directly in the proof (in place of MI or SI). Ex: In round-robin tournament, every player plays every other exactly once and each match has a winner and a looser. We say p 1,p 2,…,p m form a cycle if p 1 beats p 2, p 2 beats p 3,…,p m beats p 1. Show that if there is a cycle of length m  3, then there must exist a cycle of 3.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Ex 6 pf: Let C be the set { k | there is a cycle of length k } in the tournament. Obviously, m  C and C is a subset of non-negative integers. So by well- ordering property, C has a least element, say k. Let p 1,p 2,…p k be such cycle. since there is no cycle of 1 or 2, k must  3. If k = 3, then we are done. O/w, k > 3 and consider p 1 and p 3. If p 3 beats p 1, then p 1,p 2, p 3 is a cycle of length 3 < k. a contradiction. If p 1 beats p 3, then p 1, p 3,…,p k form a cycle of length < k. This violates the fact that k is the least element of C. As a result, k must = 3.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Recursive definition of functions  Different ways of defining a functions Explicit listing  Suitable for finite functions only. Define by giving an explicit expression  Ex: F(n) = 2 n recursive (or inductive ) definition  Define value of objects (sequences, functions, sets,...) in terms of values of smaller similar ones.  Ex: the sequence 1,2,4,... (a n = 2 n ) can be defined recursively as follows: 1. a 0 = 1; 2. a n+1 = 2 x a n for n > 0.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Recursively defined functions  To define a function over natural numbers: specify the value of f at 0 (i.e., f(0)) Given a rule for finding f(n) from f(n-1),..., f(0).  i.e., f(n) = some expression in terms of n, f(n),..., f(0).  Ex1: f(n) = 3 if n = 0 = 2f(n-1) +3 if n >0 => f(0) = 3, f(1) = 2f(0) +3 = 9 f(2) = 2f(1)+3 = 21,... This guarantees f be defined for all numbers.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No More examples functions  Ex2: The factorial function f(n) = n! f(0) = 1 f(n) = n f(n-1) for all n > 0.  Recursively defined functions (over N) are well defined Pf: Let P(n) = "there is at least one value assigned to f(n)". Q(n) = "there are at most one value assigned to f(n)". We show P(n) hold for all n by MI.. basis: P(0) holds. Ind. : assume p(k) holds for all k ≤ n => since f(n+1) can be assigned a value by evaluating the expr(n,f(0),..,f(n)), where by ind. hyp. all f(i)s (i<n) have been assigned a value. The fact that Q(n) holds for all n is trivial, since each f(k) appear at the left hand side of the definition exactly once. QED

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No More examples: Ex5: The Fibonacci number: f(0) = 0; f(1) = 1; f(n) = f(n-1) + f(n-2) for n > 1. ==> 0,1,1,2,3,5,8,...

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Ex6: Show that f(n) >  n-2 whenever n ≥ 3, where  = (1+ sqrt(5))/2 = is the golden ratio  Properties of  :  2 = (1 +  ). Pf: (by MI). Let P(n) = "f(n) >  n-2 ". Basis: P(3) holds since f(3) = 2 >  3-2. Ind.step: (for n ≥ 4) If n = 4 =>f(4) = 3 >  4-2 = If n > 4 => by ind. hyp., f(n-1) >  n-3, f(n-2) >  n-4 Hence f(n) = f(n)+f(n-1) >  n-3 +  n-4 =(1+  n-4 =  n-2. QED

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Lame's theorem  a,b: positive integer with a  b. => #divisions used by the Euclidean algorithm to find gcd(a,b)  5 x #decimal digits in b. Pf: seq of equations used for finding gcd(a,b) where r 0 = a, r 1 = b. r 2 = r o mod r 1  0, r 3 = r 1 mod r 2  0... … r n = r n-2 mod r n-1  0, r n+1 = r n-1 mod r n = 0 i.e., until r n | r n-1. Then gcd(a,b) = r n. and #division used = n. Note: r n  1 = f 2 ; r n-1  2r n  2f 2 = f 3 ; r n-2  r n +r n-1 = f 2 + f 3 = f 4... r 2  r 3 + r 4  f n-1 +f n-2 =f n ; b = r 1  r 2  r 3  f n +f n-1 = f n+1.>  n-1. logb > (n-1) log  ~ (n-1) > (n-1)/5 n -1 n  5#digit(b).

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Recursively defined sets  Given a universal set U, a subset V of U and a set of operations OP on U, we often define a subset D of U as follows: 1. Init: Every element of V is an element of D. 2. Closure: For each operation f in OP, if f:U n ->U and t 1,..,t n are objects already known to be in the set D, then f(t 1,..,t n ) is also an object of D.  Example: The set S = {3n | n >0} N can be defined recursively as follows: 1. Init: 3 ∈ S (i.e., V = { 3 } ) 2. closure: S is closed under +. i.e., If a,b ∈ S then so are a+b. (OP = {+})

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Notes about recursively defined sets 1. The definition of D is not complete (in the sense that there are multiple subsets of U satisfying both conditions. Ex: the universe U satisfies (1) and (2), but it is not Our intended D. 2. In fact the intended defined set 3': D is the least of all subsets of U satisfying 1 & 2, or 3'': D is the intersection of all subsets of U satisfying 1 & 2 or 3''': Only objects obtained by a finite number of applications of rule 1 & 2 are elements of D. 3. It can be proven that 3',3'',and 3''' are equivalent. 4. Hence, to be complete, one of 3',3'' or 3''' should be appended to condition 1 & 2, though it can always be omitted(or replaced by the adv. inductively, recursively) with such understanding in mind.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Proof of the equivalence of 3',3'' and 3'''  D 1 : the set obtained by 1,2,3' D 1 satisfies 1&2 and any S satisfies 1&2 is a superset of D 1.  D 2 : the set obtained by 1,2,3''. D2 = the intersection of all subsets Sk of U satisfying 1&2.  D 3 : the set obtained by 1,2,3'''. For any x ∈ U, x ∈ D 3 iff there is a (proof) sequence x 1,...,x m = x, such that for each x i (i = 1..m) either  (init: ) x i ∈ V or  (closure:) there are f in OP and t 1,...t n in {x 1,..,x i-1 } s.t.  x i = f(t 1,..,t n ).

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No notes for the proof 1.D 2 satisfies 1&2 and is the least of all sets satisfying 1&2, Hence D 1 exists and equals to D 2. 2 (2.1) D 3 satisfies 1 & 2. (2.2) D 3 is contained in all sets satisfying 1 & 2. Hence D 3 = D 2. pf: 1.1: Let C = { T 1,…,T m,…} be the collection of all sets satisfying 1&2, and D 2, by definition, is ∩C. Hence V ∈ T k for all T k ∈ C and as a result V ∈ D (1) Suppose t 1,…,t n ∈ D 2, then t 1,…,t n ∈ T k for each T k in C, Hence f(t 1,…,t n ) ∈ T k and as a result f(t 1,..,t n ) ∈ D (2). 1.2: Since D 2 = ∩C, D 2 is a subset of all T k ’s which satisfies 1&2, D 2 is the least among these sets. Hence D 1 exists and equals to D 2.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No D 3 satisfies 1 & 2.[ by ind.] 2.2 D 3 is contained in all sets satisfying 1 & 2 [by ind.] Hence D 3 = D 2. pf: 2.1: two propositions need to be proved: V ⊆ D3 ---(1) and {t1,..,tn} ⊆ D3 => f(t1,…,tn) ∈ D3 ---(2). (1) is easy to show, since for each x in V, the singleton sequence x is a proof. Hence x ∈ D3. As to (2), since {t1,..,tn} ⊆ D3, by definition, there exist proof sequences S1,S2,…,Sn for t1,…,tn, respectively. We can thus join them together to form a new sequence S = S1,S2,…,Sn. We can then safely append f(t1,…,tn) to the end of S to form a new sequence for f(t1,…,tn), since all t1,…,tn have appeared in S. As a result f(t1,…,tn) ∈ D3. (2) thus is proved.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No D 3 is contained in all sets satisfying 1 & 2 [by ind.] pf: Let D be any set satisfying 1&2. We need to show that for all x, x ∈ D 3 =>x ∈ D. The proof is by ind. on the length of the m of the proof sequence : x 1,…,x m = x of x. If m = 1 then x=x 1 ∈ V, and hence x ∈ D. If m = k+1 > 1, then either x m ∈ V (and x m ∈ D) or ∃ j 1,j 2,…j n < m and x m = f(x j1,…,x jn ) for some f ∈ OP. For the latter case, by ind. hyp., x j1,…x jn ∈ D. Since D satisfies closure rule, f(x j1,…,x jn ) = x m ∈ D. Q.E.D

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Example:  Ex 7': The set of natural numbers can be defined inductively as follows: Init: 0 in N. closure: If x in N, then x' in N.  => 0, 0',0'',0''',... are natural numbers  (unary representation of natural numbers)

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Induction principles III (structural induction)  D: a recursively defined set  P; a property about objects of D.  To show that P(t) holds for all t in D, it suffices to show that 1. basis step: P(t) holds for all t in V. 2. Ind. step: For each f in OP and t 1,..,t n in D, if P(t1),...,P(tn) holds, then P(f(t1,..,tn)) holds, too.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Correctness of SI  Show the correctness of structural induction. Pf: Assume not correct. => NP = {t ∈ D | P(t) does not hold} is not empty. => ∃ x ∈ NP s.t. ∃ a derivation x 1,..x n of x and all x i (i<n) ∉ NP. => If n =1, then x 1 = x ∈ V (impossible) Else either n > 1 and x ∈ V (impossible, like n=1) or n > 1, and x=f(t 1,.,t n ) for some {t 1,..,t n } in {x 1,..x n-1 } and P holds for all t k ’s => P(x) holds too => x ∉ NP, a contradiction. QED.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No MI is a specialization of SI  Rephrase the SI to the domain N, we have: To show P(t) holds for all t ∈ N, it suffices to show that Init: P(0) holds Ind. step: [OP={ ‘ }] for any x in N, If P(x) holds than P(x') holds.  Notes: 1. The above is just MI. 2. MI is only suitable for proving properties of natural numbers; whereas SI is suitable for proving properties of all recursively defined sets. 3. The common variant of MI starting from a value c ≠ 0,1 is also a special case of SI with the domain D = {c, c+1, c + 2, … }

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No well-formed arithmetic expressions Ex: (2 +x), (x + (y/3)),... (ok) x2+, xy*/3... (no) Let Vr = {x,y,..,} be the set of variables, M = numerals = finite representations of numbers OP = {+,-,x,/,^} U = the set of all finite strings over Vr U M U OP U {(,)}. The set of all well-formed arithmetic expressions (wfe) can be defined inductively as follows: 1. Init: every variable x in Vr and every numeral n in M is a wfe. 2. closure: If A, B are wfe, then so are (x+y), (x-y), (x * y), (x / y) and (x ^ y). Note: "1 + x " is not a wfe. Why ?

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No More examples:  Ex9: Wff (well-formed propositional formulas) PV: {p 1,p 2,.. } a set of propositional symbols. OP = {/\, \/, ~, -> } U = the set of all finite strings over PV U OP U {(,)} Init: every p i in PV is a wff closure: If A and B are wffs, then so are (A/\B), (A \/B), (A->B), ~A.  Ex10: [strings]  : an alphabet  *: the set of finite strings over  is defined inductively as follows: 1. Init:  is a string. 2. closure: If x is a string and a a symbol in , then a·x is a string.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No  Ex11: Recursively define two functions on  *. len :  * -> N s.t. len(x) = the length of the string x. basis: len(  ) = 0 Ind. step: for any x in   and a in , len(ax) = len(x) + 1. · :  * x  *   * s.t. x · y = the concatenation of x and y. Basis:  · y = y for all string y. recursive step: (a · z) · y = a · (z · y) for all symbols a and strings z,y.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No  Prove properties of len(-) on  *: Ex12: show that len(x · y) = len(x) +len(y) for any x,y ∈  *. By SI on x. Let P(x) = "len(xy) = len(x) +len(y)". Basis: x = . => x · y = y => len(x · y) = len(y) = len(  ) + len(y). Ind. step: x = az len(x · y) = len((a · z) · y) = len((a · (z · y)) = 1 + len(zy) = 1+ len(z) + len(y) =len(x) +len(y).

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Where we use Recursion  Define a domain numbers, lists, trees, formulas, strings,...  Define functions on recursively defined domains  Prove properties of functions or domains by structural induction.  compute recursive functions --> recursive algorithm  Ex: len (x){ // x : a string if x =  then return(0) else return(1+ l(tl(x))) }

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Recursive algorithm  Definition: an algorithm is recursive if it solve a problem by reducing it to an instance of the same problem with smaller inputs.  Ex1: compute a n where a ∈ R and n ∈ N.  Ex2: gcd(a,b) a, b ∈ N, a > b  gcd(a,b) = def if b = 0 then a else gcd(b, a mod b).  Ex: show that gcd(a,b) will always terminate.  Comparison b/t recursion and iteration Recursion: easy to read, understand and devise. Iteration: use much less computation time. Result: programmer --> recursive program --> compiler --> iterative program --> machine.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Program correctness  After designing a program to solve a problem, how can we assure that the program always produce correct output?  Types of errors in a program: syntax error --> easy to detect by the help of compiler semantic error --> test or verify  Program testing can only increase our confidence about the correctness of a program; it can never guarantee that the program passing test always produce correct output.  A program is said to be correct if it produces the correct output for every possible input.  Correctness proof generally consists of two steps: Termination proof : Partial correctness: whenever the program terminates, it will produce the correct output.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Program verification  Problem: what does it mean that a program produce the correct output (or results)?  By specifying assertions (or descriptions) about the expected outcome of the program.  Input to program verifications: Pr : the program to be verified. Q : final assertions (postconditions), giving the properties that the output of the program should have P : initial assertions(preconditions), giving the properties that the initial input values are required to have.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Hoare triple:  P,Q; assertions  S: a program or program segment.  P {S} Q is called a Hoare triple, meaning that S is partially correct (p.c.) w.r.t P,Q,i.e., whenever P is true for I/P value of S and terminates, then Q is true for the O/P values of S. Ex1: x=1 {y := 2; z := x+ y} z = 3 is true. Why ? Ex 2: x = 1 { while x > 0 x++ } x = 0 is true. why?

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Typical program constructs: 1. assignment: x := expr x := x+y-3 2. composition: S1;S2 Execute S1 first, after termination, then execute S2. 3. Conditional: 3.1 If then S 3.2 If then S1 else S2. 4. Loop: 4.1 while do S 4.2 repeat S until // 4.3 do S while …  Other constructs possible, But it can be shown that any program can be converted into an equivalent one using only 1,2,3.1 and 4.1

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Assignment rule  P[x/expr] {x := expr } P P[x/expr] is the result of replacing every x in P by the expression expr. ex: P = "y P[x/3] = “y < 3 /\ 3+z = 5". Why correct? consider the variable spaces (...,x,...) == x := expr ==> (..., expr,...) |= P Hence if P[x/expr] holds before execution, P will hold after execution. Example: Q {y := x+y} x > 2y + 1 => Q = ? (x b,y b ) ==>{y a := x b +y b } ==>(x b,x b +y b ) = (x a,y a ) |= P(x a,y a ) = def ‘’x a > 2y a +1’’ => (x b,y b ) |= Q = P(x a,y a )[x a /x b ;y a /x b +y b ] = P(x b,x b +y b )  “x b > 2(x b +y b ) +1”

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Composition rules:  Splitting programs into subprograms and then show that each subprogram is correct.  The composition rule: P {S1} Q x = 0 { x:= x+2} ? Q {S2} R ? { x := x-1} x > P {S1;S2} R x=0 {x:= x+2; x:= x -1} x > 0  Meaning: Forward reading: Backward reading: to prove P{S1;S2}Q, it suffices to find an assertion Q s.t. P{S1}Q and Q {S2}R.  Problem: How to find Q ?

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Example:  Show that x =1 {y := 2; z := x +y} z = 3  x = 1 {y := 2; z := x+y} z = 3   x=1 {y := 2} ? ? {z := x+y} z = 3

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Classical rules Classical rules: P => P1 P {S} Q1 P => P1 P1 {S} Q Q1 => Q P1 {S} Q Q1 => Q P {S} Q P{S} Q P {S} Q Examples: x = 1 => x+1>1 x+1>0 {x := x + 1} x > 0 x+1>1 { x := x + 1 } x > 1 x > 0 => x ≠ x = 1 { x := x + 1} x > 1 x+1 > 0 {x := x+1 } x ≠ 0

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Conditional rules P /\ {S1} Q P /\~ {S2} Q P {if then S1 else S2 } Q T /\ x > y => x  x x  x {y:=x} y  x P /\ {S} Q T /\ x>y {y := x} y  x P /\~ => Q ~ x > y => y  x P {if then S} Q T {if x > y then y := x} y  x

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No While-loop rules  Loop invariant: A statement P is said to be a loop invariant of a while program: While do S, if it remains true after each iteration of the loop body S. I.e., P /\ {S} P is true.  While rule: P /\ {S} P P {while do S} P /\ ~  Issues: How to find loop invariant P? Most difficulty of program verification lies in the finding of appropriate loop invariants.

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No While loop example Show that n>0 { i:= 1; f := 1; while i < n do (i := i+1 ; f := f x i ) } f = n! To prove the program terminates with f = n!, a loop invariant is needed. Let p = "i ≤ n /\ f = i!" First show that p is a loop invariant of the while program i.e., i  n /\ f = i! /\ i < n { i:= i+1; f:= f x i} i  n /\ f=i!

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No while loop example(cont'd) n > i:= 1; i ≤ n f := 1; p = "i ≤ n /\ f = i! “ while i < n do (i := i+1 ; f := f x i ) p /\ ~ i i=n /\ f = i! ==> f = n!

Discrete Mathematics Ch 3 Mathematical reasoning Transparency No Another example: Ex5:Show that the following program is correct: Procedure prod(m,n: integer) : integer 1. If n < 0 then a := -n else a := n ; a = |n| 2. k := 0 ; x := 0 3. while k < a do --- p = "x = mk /\ k ≤ a" is a loop x := x + m; invariant. k := k+1 enddo --- x = mk /\ k ≤ a /\ ~k k=a /\ x=ma => x = m |n| 4. If n prod = - m |n| = mn else prod := x => Prod = m |n| = mn ---- prod = mn. Hence the program is [partially] correct ! Note: to be really correct, we need to show that the program will eventually terminates.