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Software Correctness Proofs CIS 376 Bruce R. Maxim UM-Dearborn 1 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 2 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 3 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 • s1, s2, s3, … , sn |- T – theorem T is provable following the sequence si 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 Example - Formal Specifications • Pre-condition –N>0 • Post condition – {X = A[Y] and (1 <= Y <= N)} or – {(Y = 0) and (k : (1 <= k <= N) A[k] <> X)} • Proof would need to show that algorithm transformed state of computation from pre- condition to the post condition 12 13 14 Simple Algorithm • Model {P} A {Q} – P = pre-condition – A = Algorithm – Q = post condition • Sum algorithm {pre: x = x0 and y = y0} z=x+y {post: z = x0 + y0} 15 Sequence Algorithm • Model if {P} A1 {Q1} and {Q1} A2 {Q} then {P} A1 ; A2 {Q} is true • Swap algorithm {pre: x = x0 and y = y0} temp = x x=y y = temp {post: temp = x0 and x = y0 and y = x0} 16 Intermediate Assertions • Swap algorithm {pre: x = x0 and y = y0} temp = x {temp = x0 and x = x0 and y = y0} x=y {temp = x0 and x = y0 and y = y0} y = temp {post: temp = x0 and x = y0 and y = x0} 17 Conditional Statements • Absolute value {pre: x = x0} if x < 0 then y=-x else y=x {post: y = | x0 |} 18 Intermediate Assertions if x < 0 then {x = x0 and x0< 0} y=-x {y = | x0 |} else {x = x0 and x0>= 0} y=x {y = | x0 |} 19 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 20 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 21 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 22 While Loop Example • Algorithm while y <> 0 do z=z+x y=y–1 ans = z • What does it do? 23 While Loop Assertions {pre: x = x0 and z = z0 and y = y0 and y0 >= 0} while y <> 0 do begin {0 < n <= y0 and z’ = z0 + n * x0} z=z+x y=y–1 end {y = 0 and z’ = z0 + y0 * x0} ans = z {post: ans = z0 + y0 * x0} 24 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 25 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 26 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 27 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} 28 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 29 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 X0 + X0 + X0 + … + X0 + X0 = K * X0 = Prod k times 30 Proof - 3 • If we are at the top of the loop and execution continues the body adds X0 to both sides • Using our induction hypothesis we get (X0 + X0 + X0 + … + X0 + X0) + X0 = K * X0 + X0 k times • Which gives us X0 + X0 + X0 + … + X0 + X0 + X0 = K * X0 + X0 k + 1 times • Which can be written as X0 + X0 + X0 + … + X0 + X0 + X0 = (K + 1) *X0 = Prod 31 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 = X0 * N 32 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]} 33 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 34 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] 35 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] 36 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] 37 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] 38 These proof examples were posted on the World Wide Web by Ken Abernathy Furman University 39 Proof Technique Sequent Calculus • One type of deductive calculus • A sequent is written |- , which means /\ implies \/ , where is a (possibly empty) list of formulas {A1, …, An} and is a (possibly empty) list of formulas {B1, …, Bn} – the formulas in are called the antecedents – the formulas in are called the consequents • To restate, |- means A1 /\ … /\ An implies B1 \/ … \/ Bn 40 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 Rule 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. 41 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: Prop_Axiom , 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. 42 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, And_Right |- (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. 43 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, |- And_Left (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. 44 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, |- Implies_Left (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. 45 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, Implies_Right |- (A => B, • This rule does not branch, but provides a form easier for a mechanized theorem prover to deal with. 46 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) 47 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 48 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 49 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 50 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 51 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 Case 1.1 will now Formula 2: P yield to an Formula 3: Q application of Prop_Axiom Consequents: Formula 1: R 52 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 Case 1.2 will also Formula 2: Q yield to an application of Consequents: Prop_Axiom Formula 1: Q Formula 2: R 53 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 Case 2 will also Formula 2: Q yield to an application of Consequents: Prop_Axiom Formula 1: P Formula 2: R 54 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 55 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? 56 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 57 Step 1 • Write assertions to describe the components input and output conditions • input assertion A1: (T is an array) & (T is of size N) • output assertion Aend: (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) 58 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 < N) & T`(i) > T`(i+1))] [T`(i) is exchanged with T`(i+1)] //* – [(not(more) = true)) & (I >= N)] [T`(i) sorted] //** 59 Step 3 • From the assertions a we generate a series of theorems to be proved • If your first assertion is A1 and the first transformation point has assertion A2 associated with it the theorem to be proved is A1 A2 • Once the list of theorems is established each must be proved individually (order does not matter) 60 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 A1 and ending with Aend 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 61 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) 62 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 63 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 64 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 65 Symbolic Execution Example • Consider the following lines of code a = b + c; if (a > d) call_task1( ); else call_task2( ); 66 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 67 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([x1, … ,xn]) is true after n steps 68 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) 69 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 + 1 + nlength(list`) //IHOP = k + nlength(x :: list`) // must prove 70

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