Spring 2015 Program Analysis and Verification. Lecture 6: Axiomatic Semantics III. Roman Manevich Ben-Gurion University

Transcription

1 Spring 2015 Program Analysis and Verification Lecture 6: Axiomatic Semantics III Roman Manevich Ben-Gurion University

2 Tentative syllabus Semantics Static Analysis Abstract Interpretation fundamentals Analysis Techniques Crafting your own Natural Semantics Automating Hoare Logic Lattices Numerical Domains Soot Structural semantics Control Flow Graphs Fixed-Points Alias analysis From proofs to abstractions Axiomatic Verification Equation Systems Chaotic Iteration Interprocedural Analysis Systematically developing transformers Collecting Semantics Galois Connections Shape Analysis Domain constructors CEGAR Widening/ Narrowing 2

3 Previously Hoare logic Inference system Annotated programs Soundness and completeness Weakest precondition calculus Strongest postcondition calculus 3

4 Weakest (liberal) precondition rules 1. wlp(skip, Q) = Q 2. wlp(x := a, Q) = Q[a/x] 3. wlp(s 1 ; S 2, Q) = wlp(s 1, wlp(s 2, Q)) 4. wlp(if b then S 1 else S 2, Q) = (b wlp(s 1, Q)) ( b wlp(s 2, Q)) 5. wlp(while b do { } S, Q) = where {b } S { } and b Q Use consequence rule here 4

5 Strongest postcondition rules 1. sp(skip, P) = P 2. sp(x := a, P) = v. x=a[v/x] P[v/x] 3. sp(s 1 ; S 2, P) = sp(s 2, sp(s 1, P)) 4. sp(if b then S 1 else S 2, P) = sp(s 1, b P) sp(s 2, b P) 5. sp(while b do { } S, P) = b where {b } S { } and P b Use consequence rule here 5

6 Warm-up 6

7 Proof of swap by WP { y=b x=a } t := x { y=b t=a } x := y { x=b t=a } y := t { x=b y=a } 7

8 Prove swap via SP { y=b x=a } t := x { } x := y { } y := t { x=b y=a } 8

9 Prove swap via SP Quantifier elimination (a very trivial one) Quantifier elimination Quantifier elimination { y=b x=a } t := x { v. t=x y=b x=a } x := y { v. x=y t=v y=b v=a } { x=y y=b v. t=v v=a } { x=y y=b t=a } y := t { v. y=t x=v v=b t=a } { y=t t=a v. x=v v=b } { y=t t=a x=b } { x=b y=a } 9

10 Proof of absolute value via WP { x=v } { (-x= v x<0) (x= v x 0) } if x<0 then { -x= v } x := -x { x= v } else { x= v } skip { x= v } { x= v } 10

11 Prove absolute value by SP { x=v } { } if x<0 then { } x := -x { } else { } skip { } { } { x= v } 11

12 Proof of absolute value via WP { x=v } if x<0 then { x=v x<0 } x := -x { w. x=-w w=v w<0 } { x=-v v<0 } else { x=v x 0 } skip { x=v x 0 } { x=-v v<0 x=v x 0 } { x= v } 12

13 Agenda Some useful rules Extension for memory Proving termination 13

14 Making the proof system more practical 14

15 Conjunction rule [conj p ] { P } S { Q } { P } S { Q } { P P } S {Q Q } Allows breaking up proofs into smaller, easier to manage, sub-proofs 15

16 Breaks if C is nondeterministic [disj p ] More useful rules { P } C { Q } { P } C { Q } { P P } C {Q Q } [exist p ] { P } C { Q } { v. P } C { v. Q } v FV(C) { P } C { Q } [univ p ] { v. P } C { v. Q } v FV(C) [Inv p ] { F } C { F } Mod(C) FV(F)={} Mod(C) = set of variables assigned to in sub-statements of C FV(F) = free variables of F 16

17 Invariance + Conjunction = Constancy [constancy p ] { P } C { Q } { F P } C { F Q } Mod(C) FV(F)={} Mod(C) = set of variables assigned to in sub-statements of C FV(F) = free variables of F 17

18 Strongest postcondition calculus practice By Vadim Plessky ( [see page for license], via Wikimedia Commons 18

19 Floyd s strongest postcondition rule [ass Floyd ] { P } x := a { v. x=a[v/x] P[v/x] } where v is a fresh variable The value of x in the pre-state Example { z=x } x:=x+1 {? } 19

20 Floyd s strongest postcondition rule [ass Floyd ] { P } x := a { v. x=a[v/x] P[v/x] } where v is a fresh variable meaning: {x=z+1} Example { z=x } x:=x+1 { v. x=v+1 z=v } This rule is often considered problematic because it introduces a quantifier needs to be eliminated further on We will now see a variant of this rule 20

21 Small assignment axiom Create an explicit Skolem variable in precondition Then assign the resulting value to x First evaluate a in the precondition state (as a may access x) [ass floyd ] { x=v } x:=a { x=a[v/x] } where v FV(a) Examples: {x=n} x:=5*y {x=5*y} {x=n} x:=x+1 {x=n+1} {x=n} x:=y+1 {x=y+1} [exist p ] { n. x=n} x:=y+1 { n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancy p ] {z=9} x:=y+1 {z=9 x=y+1} 21

22 Small assignment axiom [ass { x=v } x:=a { x=a[v/x] } floyd ] Examples: {x=n} x:=5*y {x=5*y} {x=n} x:=x+1 {x=n+1} where v FV(a) {x=n} x:=y+1 {x=y+1} [exist p ] { n. x=n} x:=y+1 { n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancy p ] {z=9} x:=y+1 {z=9 x=y+1} 22

23 Small assignment axiom [ass { x=v } x:=a { x=a[v/x] } floyd ] Examples: {x=n} x:=5*y {x=5*y} {x=n} x:=x+1 {x=n+1} where v FV(a) {x=n} x:=y+1 {x=y+1} [exist p ] { n. x=n} x:=y+1 { n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancy p ] {z=9} x:=y+1 {z=9 x=y+1} 23

24 Small assignment axiom [ass { x=v } x:=a { x=a[v/x] } floyd ] Examples: {x=n} x:=5*y {x=5*y} {x=n} x:=x+1 {x=n+1} where v FV(a) {x=n} x:=y+1 {x=y+1} [exist p ] { n. x=n} x:=y+1 { n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancy p ] {z=9} x:=y+1 {z=9 x=y+1} 24

25 Prove using strongest postcondition { x=a y=b } t := x x := y y := t { x=b y=a } 25

26 Prove using strongest postcondition { x=a y=b } t := x { x=a y=b t=a } x := y y := t { x=b y=a } 26

27 Prove using strongest postcondition { x=a y=b } t := x { x=a y=b t=a } x := y { x=b y=b t=a } y := t { x=b y=a } 27

28 Prove using strongest postcondition { x=a y=b } t := x { x=a y=b t=a } x := y { x=b y=b t=a } y := t { x=b y=a t=a } { x=b y=a } // cons 28

29 Prove using strongest postcondition { x=v } if x<0 then { x=v x<0 } x := -x { x=-v x>0 } else { x=v x 0 } skip { x=v x 0 } { v<0 x=-v v 0 x=v } { x= v } 29

30 Prove using strongest postcondition { x=v } if x<0 then { x=v x<0 } x := -x { x=-v x>0 } else { x=v x 0 } skip { x=v x 0 } { v<0 x=-v v 0 x=v } { x= v } 30

31 Sum program specify Define Sum(0, n) = n {? } x := 0 res := 0 while (x<y) do res := res+x x := x+1 {? } { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } Background axiom 31

32 Sum program specify Define Sum(0, n) = n { y 0 } x := 0 res := 0 while (x<y) do res := res+x x := x+1 { res = Sum(0, y) } { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } Background axiom 32

33 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 res := 0 Inv = while (x<y) do res := res+x x := x+1 { res = Sum(0, y) } 33

34 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 Inv = while (x<y) do res := res+x x := x+1 { res = Sum(0, y) } 34

35 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = while (x<y) do res := res+x x := x+1 { res = Sum(0, y) } 35

36 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = { y 0 res=sum(0, x) x y } while (x<y) do res := res+x x := x+1 { res = Sum(0, y) } 36

37 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = { y 0 res=sum(0, x) x y } while (x<y) do { y 0 res=m x=n n y m=sum(0, n) x<y } { y 0 res=m x=n m=sum(0, n) n<y } res := res+x x := x+1 { res = Sum(0, y) } 37

38 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = { y 0 res=sum(0, x) x y } while (x<y) do { y 0 res=m x=n n y m=sum(0, n) x<y } { y 0 res=m x=n m=sum(0, n) n<y } res := res+x { y 0 res=m+x x=n m=sum(0, n) n<y } x := x+1 { res = Sum(0, y) } 38

39 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = { y 0 res=sum(0, x) x y } while (x<y) do { y 0 res=m x=n n y m=sum(0, n) x<y } { y 0 res=m x=n m=sum(0, n) n<y } res := res+x { y 0 res=m+x x=n m=sum(0, n) n<y } x := x+1 { y 0 res=m+x x=n+1 m=sum(0, n) n<y } { y 0 res=sum(0, x) x=n+1 n<y } // sum axiom { y 0 res=sum(0, x) x y } // cons { res = Sum(0, y) } 39

40 Sum program prove Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = { y 0 res=sum(0, x) x y } while (x<y) do { y 0 res=m x=n n y m=sum(0, n) x<y } { y 0 res=m x=n m=sum(0, n) n<y } x := x+1 { y 0 res=m x=n+1 m=sum(0, n) n<y } res := res+x { y 0 res=m+x x=n+1 m=sum(0, n) n<y } { y 0 res=sum(0, x) x=n+1 n<y } // sum axiom { y 0 res=sum(0, x) x y } // cons { y 0 res=sum(0, x) x y x y } { y 0 res=sum(0, y) x=y } { res = Sum(0, y) } 40

41 Buggy sum program 1 Define Sum(0, n) = n { x=sum(0, n) } { y=n+1 } { x+y=sum(0, n+1) } { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = { y 0 res=sum(0, x) x y } while (x<y) do { y 0 res=m x=n n y m=sum(0, n) x<y } { y 0 res=m x=n m=sum(0, n) n<y } res := res+x { y 0 res=m+x x=n m=sum(0, n) n<y } x := x+1 { y 0 res=m+n x=n+1 m=sum(0, n) n<y } { y 0 res=sum(0, n)+n x=n+1 n<y } { y 0 res=sum(0, x) x y } // cons { y 0 res=sum(0, x) x y x y } { y 0 res=sum(0, y) x=y } { res = Sum(0, y) } 41

42 Buggy sum program 2 { y 0 } x := 0 { y 0 x=0 } res := 0 { y 0 x=0 res=0 } Inv = { y 0 res=sum(0, x) } = { y 0 res=m x=n m=sum(0, n) } while (x y) do { y 0 res=m x=n m=sum(0, n) x y n y } x := x+1 { y 0 res=m x=n+1 m=sum(0, n) n y} res := res+x { y 0 res=m+x x=n+1 m=sum(0, n) n y} { y 0 res-x=sum(0, x-1) n y} { y 0 res=sum(0, x) } { y 0 res=sum(0, x) x>y } {res = Sum(0, y) } 42

43 Handling data structures 43

44 Problems with Hoare logic and heaps { P[a/x] } x := a { P } Consider the annotated program { y=5 } y := 5; { y=5 } x := &y; { y=5 } *x := 7; { y=5 } Is it correct? The rule works on a syntactic level unaware of possible aliasing between different terms (y and *x in our case) 44

45 Problems with Hoare logic and heaps { P[a/x] } x := a { P } {(x=&y z=5) (x &y y=5)} *x = z; { y=5 } What should the precondition be? 45

46 Problems with Hoare logic and heaps { P[a/x] } x := a { P } {(x=&y z=5) (x &y y=5)} *x = z; { y=5 } We split into cases depending on possible aliasing 46

47 Problems with Hoare logic and heaps { P[a/x] } x := a { P } {(x=&y z=5) (x &y y=5)} *x = z; { y=5 } What should the precondition be? Joseph M. Morris: A General Axiom of Assignment A different approach: heaps as arrays Really successful approach for heaps is based on Separation Logic 47

48 Axiomatizing data types S ::= x := a x := y[a] y[a] := x skip S 1 ; S 2 if b then S 1 else S 2 while b do S We added a new type of variables array variables Model array variable as a function y : Z Z Re-define program states State = Define operational semantics x := y[a], y[a] := x, 48

49 Axiomatizing data types S ::= x := a x := y[a] y[a] := x skip S 1 ; S 2 if b then S 1 else S 2 while b do S We added a new type of variables array variables Model array variable as a function y : Z Z We need the two following axioms: { y[x a](x) = a } { z x y[x a](z) = y(z) } 49

50 Array update rules (wlp) S ::= x := a x := y[a] y[a] := x skip S 1 ; S 2 if b then S 1 else S 2 while b do S Treat an array assignment y[a] := x as an update to the array function y y := y[a x] meaning y = v. v=a? x : y(v) A very general approach allows handling many data types [array-update] { P[y[a x]/y] } y[a] := x { P } [array-load] { P[y(a)/x] } x := y[a] { P } 50

51 Array update rules (wlp) example Treat an array assignment y[a] := x as an update to the array function y y := y[a x] meaning y = v. v=a? x : y(v) [array-update] { P[y[a x]/y] } y[a] := x { P } {x=y[i 7](i)} y[i]:=7 {x=y(i)} {x=7} y[i]:=7 {x=y(i)} [array-load] { P[y(a)/x] } x := y[a] { P } {y(a)=7} x:=y[a] {x=7} 51

52 Array update rules (sp) In both rules v, g, and b are fresh [array-update F ] { x=v y=g a=b } y[a] := x { y=g[b v] } [array-load F ] { y=g a=b } x := y[a] { x=g(b) } 52

53 practice proving programs with arrays 53

54 Array-max program specify nums : array N : int // N stands for num s length { N 0 nums=orig_nums } x := 0 res := nums[0] while x < N if nums[x] > res then res := nums[x] x := x { x=n } 2. { m. (m 0 m<n) nums(m) res } 3. { m. m 0 m<n nums(m)=res } 4. { nums=orig_nums } 54

55 Array-max program nums : array N : int // N stands for num s length { N 0 nums=orig_nums } x := 0 res := nums[0] while x < N if nums[x] > res then res := nums[x] x := x + 1 Post 1 : { x=n } Post 2 : { nums=orig_nums } Post 3 : { m. 0 m<n nums(m) res } Post 4 : { m. 0 m<n nums(m)=res } 55

56 Proof strategy Prove each goal 1, 2, 3, 4 separately and use conjunction rule to prove them all After proving {N 0} C {x=n} {nums=orig_nums} C {nums=orig_nums} We have proved {N 0 nums=orig_nums} C {x=n nums=orig_nums} We can refer to assertions from earlier proofs in writing new proofs 56

57 Array-max example: Post 1 nums : array N : int // N stands for num s length { N 0 } x := 0 { N 0 x=0 } res := nums[0] { x=0 } Inv = { x N } while x < N { x=k k<n } if nums[x] > res then { x=k k<n } res := nums[x] { x=k k<n } { x=k k<n } x := x + 1 { x=k+1 k<n } { x N x N } { x=n } 57

58 Array-max example: Post 2 nums : array N : int // N stands for num s length { nums=orig_nums } x := 0 { nums=orig_nums } res := nums[0] { nums=orig_nums } Inv = { nums=orig_nums } while x < N { nums=orig_nums x < N } if nums[x] > res then { nums=orig_nums } res := nums[x] { nums=orig_nums } { nums=orig_nums } x := x + 1 { nums=orig_nums } { nums=orig_nums x N } { nums=orig_nums } 58

59 Array-max example: Post 3 nums : array { N 0 0 m<n } // N stands for num s length x := 0 { x=0 } res := nums[0] { x=0 res=nums(0) } Inv = { 0 m<x nums(m) res } while x < N { x=k res=ores 0 m<k nums(m) ores } if nums[x] > res then { nums(x)>ores res=ores x=k 0 m<k nums(m) ores } res := nums[x] { res=nums(x) nums(x)>ores x=k 0 m<k nums(m) ores } { x=k 0 m k nums(m) res } { (x=k 0 m k nums(m) res) (ores nums(x) res=ores x=k res=ores 0 m<k nums(m) ores)} { x=k 0 m k nums(m) res } x := x + 1 { x=k+1 0 m k nums(m) res } { 0 m<x nums(m) res } { x=n 0 m<x nums(m) res} [univ p ]{ m. 0 m<n nums(m) res } 59

60 Proving termination By Noble0 (Own work) [CC BY-SA 3.0 ( via Wikimedia Commons 60

61 Total correctness semantics for While [ass p ] [skip p ] [comp p ] [if p ] [while p ] [ P[a/x] ] x := a [ P ] [ P ] skip [ P ] [ P ] S 1 [ Q ], [ Q ] S 2 [ R ] [ P ] S 1 ; S 2 [ R ] [ b P ] S 1 [ Q ], [ b P ] S 2 [ Q ] [ P ] if b then S 1 else S 2 [ Q ] [ P(z+1) ] S [ P(z) ] [ z. P(z) ] while b do S [ P(0) ] P(z+1) b P(0) b [cons p ] [ P ] S [ Q ] [ P ] S [ Q ] if P P and Q Q 61

62 Total correctness semantics for While Rank, or Loop variant [ass p ] [skip p ] [comp p ] [if p ] [while p ] [ P[a/x] ] x := a [ P ] [ P ] skip [ P ] [ P ] S 1 [ Q ], [ Q ] S 2 [ R ] [ P ] S 1 ; S 2 [ R ] [ b P ] S 1 [ Q ], [ b P ] S 2 [ Q ] [ P ] if b then S 1 else S 2 [ Q ] [ b P t=k ] S [ P t<k ] [ P ] while b do S [ b P ] P t 0 [cons p ] [ P ] S [ Q ] [ P ] S [ Q ] if P P and Q Q 62

63 Proving termination There is a more general rule based on wellfounded relations Partial orders with no infinite strictly decreasing chains Exercise: write a rule that proves only that a program S, started with precondition P terminates [ ] S [ ] 63

64 Proving termination There is a more general rule based on wellfounded relations Partial orders with no infinite strictly decreasing chains Exercise: write a rule that proves only that a program S, started with precondition P terminates [ P ] S [ true ] 64

65 Array-max specify termination nums : array N : int // N stands for num s length x := 0 res := nums[0] Variant = [? ] while x < N if nums[x] > res then res := nums[x] x := x + 1 [? ] 65

66 Array-max specify termination nums : array N : int // N stands for num s length x := 0 res := nums[0] Variant = [ N-x ] while x < N [? ] if nums[x] > res then res := nums[x] x := x + 1 [? ] [ true ] 66

67 Array-max prove loop variant nums : array N : int // N stands for num s length x := 0 res := nums[0] Variant = [ t=n-x ] while x < N [ x<n N-x=k N-x 0 ] if nums[x] > res then res := nums[x] x := x + 1 // [ N-x<k N-x 0 ] [ true ] 67

68 Array-max prove loop variant nums : array N : int // N stands for num s length x := 0 res := nums[0] Variant = [ t=n-x ] while x < N Capture initial value of x, since it changes in the loop [ x=x0 x0<n N-x0=k N-x0 0 ] if nums[x] > res then res := nums[x] x := x + 1 // [ N-x<k N-x 0 ] [ true ] 68

69 Array-max prove loop variant nums : array N : int // N stands for num s length x := 0 res := nums[0] Variant = [ t=n-x ] while x < N [ x=x0 x0<n N-x0=k N-x0 0 ] if nums[x] > res then res := nums[x] [ x=x0 x0<n N-x0=k N-x0 0 ] // Frame x := x + 1 // [ N-x<k N-x 0 ] [ true ] 69

70 Array-max prove loop variant nums : array N : int // N stands for num s length x := 0 res := nums[0] Variant = [ t=n-x ] while x < N [ x=x0 x0<n N-x0=k N-x0 0 ] if nums[x] > res then res := nums[x] [ x=x0 x0<n N-x0=k N-x0 0 ] // Frame x := x + 1 [ x=x0+1 x0<n N-x0=k N-x0 0 ] // [ N-x<k N-x 0 ] [ true ] 70

71 Array-max prove loop variant nums : array N : int // N stands for num s length x := 0 res := nums[0] Variant = [ t=n-x ] while x < N [ x=x0 x0<n N-x0=k N-x0 0 ] if nums[x] > res then res := nums[x] [ x=x0 x0<n N-x0=k N-x0 0 ] // Frame x := x + 1 [ x=x0+1 x0<n N-x0=k N-x0 0 ] [ N-x<k N-x 0 ] // cons [ true ] 71

72 Zune calendar bug while (days > 365) { if (IsLeapYear(year)) { if (days > 366) { days -= 366; year += 1; } } } else { days -= 365; year += 1; } 72

73 Fixed code while (days > 365) { if (IsLeapYear(year)) { if (days > 366) { days -= 366; year += 1; } else { break; } } else { days -= 365; year += 1; } } 73

74 Fixed code specify termination [? ] while (days > 365) { if (IsLeapYear(year)) { if (days > 366) { days -= 366; year += 1; } else { break; } } else { days -= 365; year += 1; } } [? ] 74

75 Fixed code specify variant [ true ] Variant = [? ] while (days > 365) { if (IsLeapYear(year)) { if (days > 366) { days -= 366; year += 1; } else { break; } } else { days -= 365; year += 1; } [? ] } [ true ] 75

76 Fixed code proving termination [ true ] Variant = [ t=days ] while (days > 365) { [ days>365 days=k days 0 ] if (IsLeapYear(year)) { if (days > 366) { days -= 366; year += 1; } else { break; [ false ] } } else { days -= 365; year += 1; } // [ days 0 days<k ] } [ true ] Let s model break by a small cheat assume execution never gets past it 76

77 Fixed code proving termination [ true ] Variant = [ t=days ] while (days > 365) { [ days 0 k=days days>365 ] -> [ days 0 k=days days>365 ] if (IsLeapYear(year)) { } [ k=days days>365 ] if (days > 366) { [ k=days days>365 days>366 ] -> [ k=days days>366 ] days -= 366; [ days=k-366 days>0 ] year += 1; [ days=k-366 days>0 ] } else { [ k=days days>365 days 366 ] break; [ false ] } [ (days=k-366 days>0) false ] -> [ days<k days>0 ] } else { [ k=days days>365 ] days -= 365; [ k-365=days days-365>365 ] -> [ k-365=days days 0 ] -> [ days<k days 0 ] year += 1; [ days<k days 0 ] [ days<k days 0 ] } [ true ] 77

78 Challenge: proving non-termination Write a rule for proving that a program does not terminate when started with a precondition P Prove that the buggy Zune calendar program does not terminate [while-nt p ] { b P } S { b } { P } while b do S { false } 78

79 conclusion 79

80 Extensions to axiomatic semantics Assertions for execution time Exact time Order of magnitude time Assertions for dynamic memory Separation Logic Assertions for parallelism Owicki-Gries Concurrent Separation Logic Rely-guarantee 80

81 Axiomatic verification conclusion Very powerful technique for reasoning about programs Sound and complete Extendible Static analysis can be used to automatically synthesize assertions and loop invariants 81

82 Next lecture: static analysis