1 Software Correctness Proofs CIS 376 Bruce R. Maxim UM-Dearborn

2 Formal Analysis Refers to tool-based methods used to explore, debug, and verify formal specifications Methods –Theorem proving –Proof checking –Model checking –Animation and simulation

3 Formal Proof - part 1 Use deductive reasoning Proofs are based on a formal system that includes –set of primitives finite strings from a fixed alphabet – set of axioms specifying the rules of behavior for the primitives –set of inference rules allow deduction of additional true statements (known a theorems) within the system

4 Formal Proof - part 2 Deductive system –axioms and inference rules for a formal system Theory –axioms and derived theorems in a formal system Proof of theorem –sequence of statement transformations that adheres to the system’s inference rules s 1, s 2, s 3, …, s n |- T –theorem T is provable following the sequence s i

5 Formal System Properties Consistent –not possible to derive a statement and its contradiction form the same set of initial statements Complete –every true statement is provable Decidable –there is an algorithm for determining whether any legal statement is true Note: consistency must be present, completeness and decidability would be nice

6 Proof Construction Forward argument (deductive calculus) –starting with axioms and proven results the inference rules are used to prove the desired consequent Backward argument (test calculus) –starting with the desired result and applying the inference rules to derive a known result, axiom, or theorem

7 Program Verification Similar to writing a mathematical proof You must present a valid argument that is believable to the reader The argument must demonstrate using evidence that the algorithm is correct Algorithm is correct if code correctly transforms initial state to final state

8 State of Computation Most programming algorithms are based on the notion of transforming the inputs to outputs The state of computation may be defined by examining the contents of key variables before and after the execution of each statement

9 Assertions Assertions are facts about the state of the program variables It is wasteful to spend your time looking at variables that are not effected by a particular statement Default assertion –any variable not mentioned in the assertion for a statement do not affect the state of computation

10 Use of Assertions Pre-condition –assertion describing the state of computation before statement is executed Post condition –assertion describing the state of computation after a statement is executed Careful use of assertions as program comments can help control side effects

11 Code Verification Example Let’s assume we have an array with a search operation that corresponds to the English specifications below –The procedure searches and array A of length N for a value X –If X is found than the value of Y is set to the array index where X was found on exit –If no array element in A matches X then Y will be set to zero on exit

12 Example - Formal Specifications Pre-condition –N > 0 Post condition –{X = A[Y] and (1 <= Y <= N)} or –{(Y = 0) and (  k : (1 X)} Proof would need to show that algorithm transformed state of computation from pre- condition to the post condition

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15 Simple Algorithm Model {P} A {Q} –P = pre-condition –A = Algorithm –Q = post condition Sum algorithm {pre: x = x 0 and y = y 0 } z = x + y {post: z = x 0 + y 0 }

16 Sequence Algorithm Model if {P} A 1 {Q 1 } and {Q 1 } A 2 {Q} then {P} A 1 ; A 2 {Q} is true Swap algorithm {pre: x = x 0 and y = y 0 } temp = x x = y y = temp {post: temp = x 0 and x = y 0 and y = x 0 }

17 Intermediate Assertions Swap algorithm {pre: x = x 0 and y = y 0 } temp = x {temp = x 0 and x = x 0 and y = y 0 } x = y {temp = x 0 and x = y 0 and y = y 0 } y = temp {post: temp = x 0 and x = y 0 and y = x 0 }

18 Conditional Statements Absolute value {pre: x = x 0 } if x < 0 then y = - x else y = x {post: y = | x 0 |}

19 Intermediate Assertions if x < 0 then {x = x 0 and x 0 < 0} y = - x {y = | x 0 |} else {x = x 0 and x 0 >= 0} y = x {y = | x 0 |}

20 Understanding the While Loop Every well designed while loop must –make progress to ensure eventual termination –must maintain the loop invariant to ensure that it is valid at each loop exit point //invariant holds here while (condition) do //invariant holds here make_progress restore_invariant //invariant and not(condition) hold here

21 Loop Invariant Type of assertion that describes the variables which remain unchanged during the execution of a loop In general the stopping condition should remain unchanged during the execution of the loop Some people show the loop invariant as a statement which becomes false when loop execution is complete

22 Loop Invariant Example k = 0; //invariant: A[1], …, A[k] equal 0 //assert: N >= 0 while k < N do k = k +1;//make progress A[k] = 0;//restore invariant //assert: A[1], …, A[k] equal 0 and k >= N Note: say k=N will require an induction argument

23 While Loop Example Algorithm while y <> 0 do z = z + x y = y – 1 ans = z What does it do?

24 While Loop Assertions {pre: x = x 0 and z = z 0 and y = y 0 and y 0 >= 0} while y <> 0 do begin {0 < n <= y 0 and z’ = z 0 + n * x 0 } z = z + x y = y – 1 end {y = 0 and z’ = z 0 + y 0 * x 0 } ans = z {post: ans = z 0 + y 0 * x 0 }

25 Proof -1 If y = 0 loop does not execute and no variables change so z = z + 0 * x = ans If we assume that for n = k if program begins loop with y = k it will exit with ans = z + k * x

26 Proof - 2 We must prove that when program begins loop with y = k + 1 it will exit loop with ans = z + (k + 1) * x Suppose y = k + 1 at top of loop and the body of the loop executes one time x = x ‘ = x y = y’ = (k + 1) – 1 = k z = z’ = z + x

27 Proof - 3 Since we are at the top of the loop with y = k, we can use our induction hypothesis to get ans = z’ + k * x’ Substituting we get ans = (z + x) + k * x = z + (x + k * x) = z + (1 + k) * x = z + (k + 1) * x

28 Second While Loop Example Prod = 0 I = 0 {X and N are initialized and Prod = I * X and I <= N} while (I < N) do begin {Prod = I * X and I < N} Prod = Prod + X {Prod = (I + 1) * X and I < N} I = I + 1 {Prod = I * X and I <= N} end {loop exited with Prod = I * X and I >= N} {Prod = N * X}

29 Proof - 1 Check I = 0 Verify that invariant is true Prod = 0 Prod = 0 * X = 0 If N = 0 condition is false and loop terminates If N > 0 condition is true and loop execution continues

30 Proof - 2 Assume correctness of the invariant at the top of the loop when I = K Prod = K * X If K < N condition is true and the loop execution continues If K = N condition is false and loop terminates So post condition is true So we are assuming that X 0 + X 0 + X 0 + … + X 0 + X 0 = K * X 0 = Prod k times

31 Proof - 3 If we are at the top of the loop and execution continues the body adds X 0 to both sides Using our induction hypothesis we get (X 0 + X 0 + X 0 + … + X 0 + X 0 ) + X 0 = K * X 0 + X 0 k times Which gives us X 0 + X 0 + X 0 + … + X 0 + X 0 + X 0 = K * X 0 + X 0 k + 1 times Which can be written as X 0 + X 0 + X 0 + … + X 0 + X 0 + X 0 = (K + 1) *X 0 = Prod

32 Proof - 4 Since assuming the invariant is true for K we showed that it held for K + 1 If the loop condition is true execution continues otherwise is halts with I = N + 1 Prod = X 0 * N

33 Counting Loop Example This loop stores the sum of the first I array elements in position C[I] {pre: max >= M >= 1 and C initialized} for I = 1 to M C[I] = C[I] + C[I – 1] {post: I = M + 1 and for each J = 1 to I – 1 : C’[J] = C[0] + … + C[J]}

34 Proof - 1 Show for the case M = 1 that the program will exit with C[1] = C[1] + C[0] and C’[0] = C[0] If M = 1 loop executes with I = 1 and C’[1] = C[0] + C[1] Loop is satisfied and result is I = 2 and C[1] = C[1] + C[0] and C[0] = C[0] This gives us our intended result

35 Proof - 2 For our induction hypothesis we assume if loop reaches top with M = K the loop exits C’[K] = C[0] + … + C[K] … C’[1] = C[0] + C[1] C’[0] = C[0]

36 Proof - 3 We need to show that if loop reaches the top with M = K + 1 the loop ends with C’[K + 1] = C[0] + … + C[K + 1] C’[K] = C[0] + … + C[K] … C’[1] = C[0] + C[1] C’[0] = C[0]

37 Proof - 4 C’[K + 1] = C[K + 1] + C[(K + 1) - 1] C’[K + 1] = C[K + 1] + C[K] By our induction hypothesis C[K] now contains C’[K] so C’[K + 1] = C[K + 1] + C’[K] C’[K + 1] = C[K + 1] + C[K] + … + C[0]

38 Proof - 5 Loop condition is satisfied and I = (K + 1) + 1 = K + 2 The array should now contain C’[K + 1] = C[0] + … + C[K + 1] C’[K] = C[0] + … + C[K] … C’[1] = C[0] + C[1] C’[0] = C[0]

39 These proof examples were posted on the World Wide Web by Ken Abernathy Furman University

40 Proof Technique Sequent Calculus One type of deductive calculus A sequent is written  |- , which means /\  implies \/ , where  is a (possibly empty) list of formulas {A 1, …, A n } and  is a (possibly empty) list of formulas {B 1, …, B n } –the formulas in  are called the antecedents –the formulas in  are called the consequents To restate,  |-  means A 1 /\ … /\ A n implies B 1 \/ … \/ B n

41 Sequent Calculus A sequent calculus proof is a tree of sequents whose root is a sequent of the form |- T where T is the formula to be proved and the antecedent is empty The proof tree is then generated by applying inference rules of the form:  1 |-  1 …  n |-  n   |-  Intuitively, this rule replaces a leaf node in the proof tree of form  |-  with the n new leaves specified in the rule. If n is zero, that branch of the proof tree terminates. Rule N

42 Sequent Calculus - Example Rule 1 The Propositional Axiom (Prop_Axiom) is one of the rules of inference in sequent calculus. It has the following form form : , A  |- (A,  Intuitively, this rule indicates that a proof branch is complete when the sequent above is derived. Note that the consequent means the following:  /\ A implies A \/  which is obviously true. Prop_Axiom

43 Sequent Calculus - Example Rule 2 The Rule for Conjunction on the Right (And_Right) is another of the rules of inference in the sequent calculus. It has the following form:  |- A,  |- B,   |- (A /\ B,  This rule is typical of many sequent calculus inference rules which divide, but simplify, a branch of the proof tree. Note that the consequent is replaced by two simpler formulas which will be easier for a mechanized theorem prover to deal with. And_Right

44 Sequent Calculus - Example Rule 3 The Rule for Conjunction on the Left (And_Left) is another of the rules of inference in the sequent calculus. It has the following simple (non-branching) form:  A, B,  |-   (A /\ B,  |-  This rule is typical of several sequent calculus inference rules which simply restate the “obvious,” thereby providing a form easier for a mechanized theorem prover to deal with. And_Left

45 Sequent Calculus - Example Rule 4 The Rule for Implication on the Left (Implies_Left) is another of the rules of inference in the sequent calculus. It has the following form:  |- A,  B,  |-   (A => B,  |-  Similar to the And_Right rule, this rule again splits the proof into two cases, each of which will be easier for the mechanical prover to deal with. Implies_Left

46 Sequent Calculus - Example Rule 5 The Rule for Implication on the Right (Implies_Right) is another of the rules of inference in the sequent calculus. It has the following form:  A |- B,   |- (A => B,  This rule does not branch, but provides a form easier for a mechanized theorem prover to deal with. Implies_Right

47 Sequent Calculus Proof Example Theorem 1: ( P => (Q => R) ) => ( (P /\ Q) => R ) We begin the proof by forming the requisite sequent: Antecedents: none Consequents: Formula 1: ( P => (Q => R) ) => ( (P /\ Q) => R )

48 Proof Example - Step 1 As our first step we apply the rule Implies_Right. This rule will decompose the entire formula. Remember there is an implied “implies” in the sequent. In other words this sequent could be written |- (P => (Q => R)) => ((P /\ Q) => R). Antecedents: Formula 1: P => (Q => R) Consequents: Formula 1: (P /\ Q) => R

49 Proof Example Step 2 A second application of the rule Implies_Right will decompose the formula below the line in a similar way. Remember that rules applying to the “left” part of the sequent work on formulas above the bar; rules applying to the “right” part of the sequent work below the bar. Antecedents: Formula 1: P => (Q => R) Formula 2: P /\ Q Consequents: Formula 1: R

50 Proof Example Step 3 We next apply the rule And_Left -- this rule will modify (rewrite) Formula 2 above the line. Remember that all formulas above the line are connected by AND’s; formulas below the line are connected by OR’s. Antecedents: Formula 1: P => (Q => R) Formula 2: P Formula 3: Q Consequents: Formula 1: R

51 Proof Example Step 4 We next apply the rule Implies_Left -- this rule will modify Formula 1 above the line. Remember that Implies_Left splits the proof tree into two branches. We show and deal with Case 1 first, then move to Case 2 later. Case 1: Antecedents: Formula 1: Q => R Formula 2: P Formula 3: Q Consequents: Formula 1: R

52 Proof Example Step 5 To modify Formula 1 above the line, we next apply the rule Implies_Left again. Again this splits the proof tree into two branches. We show and deal with Case 1.1 first, then move to Case 1.2 later. Case 1.1: Antecedents: Formula 1: R Formula 2: P Formula 3: Q Consequents: Formula 1: R Case 1.1 will now yield to an application of Prop_Axiom

53 Proof Example Steps 5 & 6 As noted, an application of Prop_Axiom (Step 5) completes Case 1.1. We now move to Case 1.2. This is the second case resulting from the application of Implies_Left on the Case 1 sequent. Another application of Prop_Axiom (Step 6) completes Case 1.2 (and in turn Case 1 itself). Case 1.2: Antecedents: Formula 1: P Formula 2: Q Consequents: Formula 1: Q Formula 2: R Case 1.2 will also yield to an application of Prop_Axiom

54 Proof Example Step 7 Having completed the proof for Case 1, we now move to Case 2. Recall that this is the second case resulting from our first application of Implies_Left. Another application of Prop_Axiom (Step 7) completes Case 2 (and in turn the entire proof). Case 2: Antecedents: Formula 1: P Formula 2: Q Consequents: Formula 1: P Formula 2: R Case 2 will also yield to an application of Prop_Axiom

55 Mechanical Theorem Provers Many mechanical theorem provers require human interaction Users typically are required to choose the rule of inference to be applied at each step The theorem prover may be able to discover (by heuristic search) some rules to apply on its own The user needs to translate the statements to some normal form prior to beginning

56 Proving Algorithm Correctness Suppose you have a software component that accepts as input an array T of size N As output the component produces an T` which contains the elements of T arranged in ascending order How would we convert the code to its logical counterpart and prove its correctness?

57 Bubble Sort Algorithm T` = T more = true lab1: i = 0 if (more ~= true) then go to end not(more) = true //** assertion needed lab2:i = i + 1 if i >= N then go to lab1 if T`(i) < T`(i+1) then go to lab2 //* assertion needed exchange T`(i) with T`(i+1) more = true go to lab2

58 Step 1 Write assertions to describe the components input and output conditions input assertion A 1 : (T is an array) & (T is of size N) output assertion A end : (T` is an array) & (  i if i < N then (T`(i) <= T`(i+1)) & (  i if i < N then  j(T`(i) = T(j)) & (T` is of size N)

59 Step 2 Draw a flow diagram to represent the logical flow through the component Indicate points where data transformations will occur and write assertions Ex. Assuming a bubble sort is used two assertions might be –[(not(more) = true)) & (i T`(i+1))]  [T`(i) is exchanged with T`(i+1)] //* –[(not(more) = true)) & (I >= N)]  [T`(i) sorted] //**

60 Step 3 From the assertions a we generate a series of theorems to be proved If your first assertion is A 1 and the first transformation point has assertion A 2 associated with it the theorem to be proved is A 1  A 2 Once the list of theorems is established each must be proved individually (order does not matter)

61 Steps 4 and 5 We need to locate each loop in the flow diagram and write an if-then assertion for each loop condition To prove correctness, each logic path beginning with A 1 and ending with A end Following each of these paths allows us to demonstrate that the code shows that the truth of the input condition will lead to the truth of the output condition

62 Steps 6 & 7 After identifying each logic paths the truth of each path is proved rigorously (showing the the input assertion implies the output assertion according to the logic transformations found on that path) Finally you need to prove the program terminates (which may mean an induction argument if loops are involved)

63 Cost of Correctness Proofs - part 1 Advantages –You can discover algorithmic faults in the code –Gives you a formal understanding of the logical structures of the program –Regular use of proofs forces you to be more precise in specifying data, data structures and algorithmic rules Disadvantages –Code is often smaller size than its proof –It may take less effort to create code than to prove its correctness

64 Cost of Correctness Proofs - part 2 Disadvantages (continued) –Large programs require complex diagrams and contain many transformations to prove –Nonnumeric algorithms are hard to represent logically –Parallel processing is hard to represent –Complex data structures require complex transformations –Mathematical proofs have occasionally been found to be incorrect after years of use

65 Symbolic Execution Involves simulated execution of the program code using symbols rather than data variables The test program is viewed as having an input state defined by the input data and preconditions As each line of code is executed the program statement is checked for state changes Each logical path in the program corresponds to an ordered sequence of state changes The final state of each path must be a proper output state A program is correct if each input state generates the appropriate output state

66 Symbolic Execution Example Consider the following lines of code a = b + c; if (a > d) call_task1( ); else call_task2( );

67 Symbolic Execution Steps A symbolic execution tool would decide that (a>d) can be true or false, without worrying about the values assigned to a and d This gives us two states –(a > d) is false –(a > d) is true All data values are presumed to fall into one of the equivalence classes defined by the two states (so only a small number of test cases need be considered in a proof) This technique has many of the same costs and disadvantages of using logical correctness proofs

68 Structural Induction Induction can be used to show loop termination and correctness of list processing algorithms To show the f(list) is true for every list you must prove that –f(list) is true for an empty list//the base case –whenever f(list`) is true, so is f(x :: list) in other words adding an element to the list preserves truth no matter how the list is //induction step –f([x 1, …,x n ]) is true after n steps

69 Generalizing Induction Suppose you want to show that –  k addlen(k. list) = k + nlength(list) Give the following definitions nlength([ ])=0 nlength(x :: xs)= 1 + nlength(xs) addlen(k, [ ])=k addlen(k, x::xs)=addlen(k + 1, xs)

70 Induction Correctness Proof  k addlen(k. list) = k + nlength(list) Base case: addlen(k, [ ]) = k = k + 0 = k + nlength([ ]) Induction step - assume that addlen(k, x : :list` )= addlen(k + 1, list`) = k nlength(list`) //IHOP = k + nlength(x :: list`) // must prove