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Automated Theorem Proving Lecture 1. Program verification is undecidable! Given program P and specification S, does P satisfy S?

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Presentation on theme: "Automated Theorem Proving Lecture 1. Program verification is undecidable! Given program P and specification S, does P satisfy S?"— Presentation transcript:

1 Automated Theorem Proving Lecture 1

2 Program verification is undecidable! Given program P and specification S, does P satisfy S?

3 The model checking approach Create a model of the program in a decidable formalism Verify the model algorithmically Difficulties –Model creation is burden on programmer –The model might be incorrect. If verification fails, is the problem in the model or the program?

4 The axiomatic approach Add auxiliary specifications to the program to decompose the verification task into a set of local verification tasks Verify each local verification problem Difficulties –Auxiliary spec is burden on programmer –Auxiliary spec might be incorrect. If verification fails, is the problem with the auxiliary specification or the program?

5 Theorem Proving and Software Meets spec/Found Bug Theorem in a logic Program Specification Semantics VC generation Validity Provability (theorem proving) Soundness: –If the theorem is valid then the program meets specification –If the theorem is provable then it is valid

6 Overview of the Next Few Lectures From programs to theorems –Verification condition generation From theorems to proofs –Theorem provers –Decision procedures

7 Programs ! Theorems = Axiomatic Semantics Consists of: –A language for making assertions about programs –Rules for establishing when assertions hold Typical assertions: –During the execution, only non-null pointers are dereferenced –This program terminates with x = 0 Partial vs. total correctness assertions –Safety vs. liveness properties –Usually focus on safety (partial correctness)

8 Hoare Triples Partial correctness: { A } s { B} –When you start s in any state that satisfies A –If the execution of s terminates –It does so in a state that satisfies B Total correctness: [ A ] s [ B ] –When you start s in any state that satisfies A –The execution of s terminates and –It does so in a state that satisfies B Defined inductively on the structure of statements

9 Hoare Rules {A} s 1 {C} {C} s 2 {B} { A } s 1 ; s 2 {B} {A Æ E} s 1 {B} {A Æ : E} s 2 {B} {A} if E then s 1 else s 2 {B} {I Æ E} s {I} I Æ : E ) B { I} while E do s {B}

10 Hoare Rules: Assignment Example: { A } x := x + 2 {x >= 5 }. What is A? General rule: Surprising how simple the rule is ! The key is to compute backwards the precondition from the postcondition Forward rule is more complicated: { B[E/x] } x := E {B} { A } x := E { 9 x. A[x/x] Æ x = E[x/x] }

11 Weakest preconditions Dijkstras idea: To verify that { A } s { B} a) Let Pre(s, B) = {A | { A} s { B} } b) (Pre(s,B), )) is a lattice - false Pre(s,B) - if Pre(s,B) and Pre(s,B), then Pre(s,B) c) WP(s, B) = lub (Pre(s, B)) d) Compute WP(s, B) and prove A WP(s, B)

12 Predicate lattice falsetrue ) strong weak Pre(s, B) weakest precondition: WP(s, B) A

13 Weakest preconditions WP(x := E, B) = B[E/x] WP(s 1 ; s 2, B) = WP(s 1, WP(s 2, B)) WP(if E then s 1 else s 2, B) = E ) WP(s 1, B) Æ : E ) WP(s 2, B) WP(assert E, B) = E B

14 Example returns c requires true ensures c = a b bool or(bool a, bool b) { if (a) c := true else c := b } Conjecture to be proved: true (a true = a b) ( a b = a b) WP(S, c = a b) = (a true = a b) ( a b = a b) S

15 Weakest preconditions (Contd.) What about loops? Define a family of weakest preconditions –WP k (while E do S, B) = weakest precondition from which if the loop terminates in k or fewer iterations, then it terminates in B WP 0 = : E ) B WP i+1 = WP i (E ) WP(s, WP i )) WP(while E do S, B) = Æ k ¸ 0 WP k = glb {WP k | k ¸ 0} –Hard to compute –Can we find something easier yet sufficient ?

16 Not quite weakest preconditions Recall what we are trying to do: falsetrue ) strong weak Pre(s, B) weakest precondition: WP(s, B) A verification condition: VC(s, B) We shall construct a verification condition: VC(s, B) –The loops are annotated with loop invariants ! –VC is guaranteed stronger than WP –But hopefully still weaker than A: A ) VC(s, B) ) WP(s, B)

17 Verification condition generation Computed in a manner similar to WP Except the rule for while: VC(while I,T E do s, B) = I Æ ( I ((E ) VC(s, I)) Æ ( : E ) B)) )[T 0 /T] I is the loop invariant (provided by the programmer) T is the set of loop targets (can be approximated by scanning the loop body) I holds on entry I is preserved in an arbitrary iteration B holds when the loop terminates

18 returns c requires b >= 0 ensures c = a + b int add(int a, int b) { int t; t := b c := a invariant t 0 c = a + b - t while (t > 0) { c := c + 1 t := t - 1 } Example Conjecture to be proved: b 0 VC(A, c = a + b) VC(B, t 0 c = a + b - t) t - 1 0 c + 1 = a + b – (t – 1) B L A VC(L, c = a + b) t 0 c = a + b – t (t 0 c = a + b – t t > 0 t - 1 0 c + 1 = a + b – (t - 1) t 0 c = a + b)[c 0 /c,t 0 /t] VC(L, c = a + b) t 0 c = a + b – t (t 0 0 c 0 = a + b – t 0 t 0 > 0 t 0 - 1 0 c 0 + 1 = a + b – (t 0 - 1) t 0 0 c 0 = a + b) VC(A, c = a + b) b 0 a = a + b – b (t 0 0 c 0 = a + b – t 0 t 0 > 0 t 0 - 1 0 c 0 + 1 = a + b – (t 0 - 1) t 0 0 c 0 = a + b)

19 Invariants Are Not Easy Consider the following code from QuickSort int partition(int *a, int L 0, int H 0, int pivot) { int L = L 0, H = H 0 ; while(L < H) { while(a[L] < pivot) L ++; while(a[H] > pivot) H --; if(L < H) { swap a[L] and a[H] } L++; } return L } Consider verifying only memory safety What is the loop invariant for the outer loop ?

20 Verification conditions (Contd.) What about procedure calls? –Annotate each procedure with a precondition pre and a postcondition post –VC(f(), B) = pre (post B)

21 Program verification Annotate each procedure with a precondition and a postcondition and each loop with an invariant Verify each procedure separately requires pre ensures post f() { S; } Verify that the formula VC(f) pre VC(S, post) is valid

22 Proving verification conditions What is the decision procedure for proving validity of VC(f)? Depends on the logic in which VC(f) is expressed VC(f) pre VC(S, post)

23 Verification condition logic Atoms connected by boolean operators –,,, Atoms depend on the program variables and operations on them –boolean, integer, memory Atoms depend on the language of assertions, i.e., program assertions, loop invariants, preconditions and postconditions –quantification, reachability predicate

24 Assume each assertion is a quantifier-free boolean combination of expressions over program variables. VC(f) is a boolean combination of atoms –Each atom is a relation over terms –Each term is built using functions and logical constants Logical constants are different from program variables –program variables change over time –logical constants are fixed The logical constants in VC(f) refer to the values of program variables at the beginning of f.

25 Case I: Boolean programs Boolean-valued variables and boolean operations Formula := A | | A Atom := b b SymBoolConst

26 Example returns c requires true ensures c = a b bool or(bool a, bool b) { if (a) c := true else c := b } Conjecture to be proved: true (a true = a b) ( a b = a b) VC(S, c = a b) = (a true = a b) ( a b = a b) S

27 Case II: Arithmetic programs In addition, integer-valued variables with affine operations Formula := A | | A Atom := b | t = 0 | t < 0 | t 0 t Term := c | x | t + t | t – t | ct b SymBoolConst x SymIntConst c {…,-1,0,1,…}

28 returns c requires b >= 0 ensures c = a + b int add(int a, int b) { int t; t := b c := a invariant t 0 c = a + b - t while (t > 0) { c := c + 1 t := t - 1 } Example Conjecture to be proved: b 0 VC(A, c = a + b) VC(B, t 0 c = a + b - t) t - 1 0 c + 1 = a + b – (t – 1) B L A VC(L, c = a + b) t 0 c = a + b – t (t 0 c = a + b – t t > 0 t - 1 0 c + 1 = a + b – (t - 1) t 0 c = a + b)[c 0 /c,t 0 /t] VC(L, c = a + b) t 0 c = a + b – t (t 0 0 c 0 = a + b – t 0 t 0 > 0 t 0 - 1 0 c 0 + 1 = a + b – (t 0 - 1) t 0 0 c 0 = a + b) VC(A, c = a + b) b 0 a = a + b – b (t 0 0 c 0 = a + b – t 0 t 0 > 0 t 0 - 1 0 c 0 + 1 = a + b – (t 0 - 1) t 0 0 c 0 = a + b)

29 Case III: Memory programs In addition, a memory with read and write operations –an unbounded set of objects –a finite set of fields in each object –each field contains a boolean value, an integer value, or a reference to an object For each field f, two operations Select and Update –Select(f,o) is the content of the memory at object o and field f –Update(f,o,v) is a new memory obtained by updating field f of object o to v

30 Memory axioms for all objects o and o, and memories m: o = o Select(Update(m,o,v),o) = v o o Select(Update(m,o,v),o) = Select(m,o)

31 Modeling memory operations Treat each field f as a map variable: a = b.f a = Select(f,b) a.f = b f = Update(f,a,b) { ? } a.f = 5 { a.f + b.f = 10 } WP(a.f = 5, a.f + b.f = 10) WP(f = Update(f,a,5), Select(f,a) + Select(f,b) = 10) Select(Update(f,a,5),a) + Select(Update(f,a,5),b) = 10

32 Simplify using memory axiom Select(Update(f,a,5),a) + Select(Update(f,a,5),b) = 10 iff 5 + Select(Update(f,a,5),b) = 10 iff Select(Update(f,a,5),b) = 5 iff a = b 5 = 5 a b Select(f,b) = 5 iff a b Select(f,b) = 5

33 Formula := A | | A Atom := b | t = 0 | t < 0 | t 0 t Term := c | x | t + t | t – t | ct | Select(m,t) m MemTerm := f | Update(m,t,t) b SymBoolConst x SymIntConst c {…,-1,0,1,…}

34 Decision procedures Boolean programs –Propositional satisfiability Arithmetic programs –Propositional satisfiability modulo theory of linear arithmetic Memory programs –Propositional satisfiability modulo theory of linear arithmetic + arrays


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