Program Verification Using Hoare’s Logic Book: Chapter 7

1. Program VerificationUsing Hoare’s LogicBook: Chapter 7

2. While programs • Assignments y:=t • Composition S1; S2 • If-then-else if e the S1 else S2 fi • While while e do S od

3. Greatest common divisor {x1>0/\x2>0} y1:=x1; y2:=x2; while ¬(y1=y2) do if y1>y2 then y1:=y1-y2 else y2:=y2-y1 fi od {y1=gcd(x1,x2)}

4. Why it works? Suppose that y1,y2 are both positive integers. • If y1>y2 then gcd(y1,y2)=gcd(y1-y2,y2) • If y2>y1 then gcd(y1,y2)=gcd(y1,y2-y1) • If y1-y2 then gcd(y1,y2)=y1=y2

5. Assignment axiom {p[t/y]} y:=t {p} For example: {y+5=10} y:=y+5 {y=10} {y+y<z} x:=y {x+y<z} {2*(y+5)>20} y:=2*(y+5) {y>20} Justification: write p with y’ instead of y, and add the conjunct y’=t. Next, eliminate y’ by replacing y’ by t.

6. Why axiom works backwards? {p} y:=t {?} Strategy: write p and the conjunct y=t, where y’ replaces y in both p and t. Eliminate y’. {y>5} y:=2*(y+5) {?} {p} y:=t {$y’ (p[y’/y] /\ t[y’/y]=y)} y’>5 /\ y=2*(y’+5) y>20

7. Composition rule {p} S1 {r}, {r} S2 {q} {p} S1;S2 {q} For example: if the antecedents are 1. {x+1=y+2} x:=x+1 {x=y+2} 2. {x=y+2} y:=y+2 {x=y} Then the consequent is {x+1=y+2} x:=x+1; y:=y+2 {x=y}

8. More examples {p} S1 {r}, {r} S2 {q} {p} S1;S2 {q} {x1>0/\x2>0} y1:=x1 {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} {gcd(y1,x2)=gcd(y1,x2)/\y1>0/\x2>0} y2:=x2 {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0} {x1>0/\x2>0} y1:=x1 ; y2:=x2 {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0}

9. If-then-else rule {p/\e} S1 {q}, {p/\¬e} S2 {q} {p} if e then S1 else S2 fi {q} For example: p is gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\¬(y1=y2) e is y1>y2 S1 is y1:=y1-y2 S2 is y2:=y2-y1 q is gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0

10. While rule {p/\e} S {p} {p} while e do S od {p/\¬e} Example: p is {gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0} e is (y1=y2) S is if y1>y2 then y1:=y1-y2 else y2:=y2-y1 fi

11. Consequence rules • Strengthen a precondition rp, {p} S {q} {r} S {q} • Weaken a postscondition {p} S {q}, qr {p} S {r}

12. Use of first consequence rule Want to prove {x1>0/\x2>0} y1:=x1 {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} By assignment rule: {gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0} y1:=x1 {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} x1>0/\x2>0 gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0

13. Combining program {x1>0 /\ x2>0} y1:=x1; y2:=x1; {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0} while S do if e then S1 else S2 fi od {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0} Combine the above using concatenation rule!

14. Not completely finished {x1>0/\x2>0} y1:=x1; y2:=x1; while ~(y1=y2) do if e then S1 else S2 fi od {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2} But we wanted to prove: {x1>0/\x1>0} Prog {y1=gcd(x1,x2)}

15. Use of secend consequence rule {x1>0/\x2>0} Prog {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2} And the implication {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2} y1=gcd(x1,x2) Thus, {x1>0/\x2>0} Prog {y1=gcd(x1,x2)

16. {x1>0/\x2>0} y1:=x1; {gcd(x1,x2)=gcd(y1,x2) /\y1>0/\x2>0} y2:=x2; {gcd(x1,x2)=gcd(y1,y2) /\y1>0/\y2>0} while ¬(y1=y2) do {gcd(x1,x2)=gcd(y1,y2)/\ y1>0/\y2>0/\¬(y1=y2)} if y1>y2 then y1:=y1-y2 else y2:=y2-y1 fi od {y1=gcd(x1,x2)} Annotating a while program

17. {x>=0 /\ y>=0} a:=0; b:=x; while b>=y do b:=b-y; a:=a+1 od. {x=a*y+b/\b>=0/\b<y} Invariant: x=a*y+b /\ b>=0 Another example

18. Invariant • How to start the proof? Heuristics: Find invariant for each loop. For this example: x=a*y+b/\x>=0 Note: total correctness does not hold for y=0. Total correctness (with y>0) to be proved separately.

19. (1) {x=a*y+x/\x>=0} b:=x {x=a*y+b/\b>=0} (Assignment) (2) {x=0*y+x/\x>=0} a:=0 {x=a*y+x/\x>=0} (Assignment) (3){x=0*y+x/\x>=0} a:=0;b:=x{x=a*y+b/\x>=0} (Composition (2), (1)) Proof {p[t/y]} y:=t {p} {p}S1{r}, {r} S2{q} {p} S1;S2 {q}

20. (4){x=(a+1)*y+b/\b>=0} a:=a+1{x=a*y+b/\b>=0} (Assignment) (5){x=(a+1)*y+b-y/\b-y>=0} b:=b-y{x=(a+1)*y+b/\b>=0} (Assignment) (6){x=(a+1)*y+b-y/\b-y>=0} b:=b-y;a:=a+1{x=a*y+b/\b>=0} (Composition (5), (4)) Proof (cont.) {p[t/y]} y:=t {p} {p}S1{r}, {r} S2{q} {p} S1;S2 {q}

21. While rule {p/\e} S {p} {p} while e do S od {p/\¬e}

22. Consequence rules • Strengthen a precondition rp, {p} S {q} {r} S {q} • Weaken a postcondition {p} S {q}, qr {p} S {r}

23. Proof (cont.) (7) x=a*y+b/\b>=0/\b>=y x=(a+1)*y+b-y/\b-y>=0 (Logic) (8) {x=a*y+b/\b>=0/\b>=y} b:=b-y; a:=a+1 {x=a*y+b/\b>=0} (Consequence (6), (7)) (9) {x=a*y+b/\b>=0}while b>=y do b:=b-y; a:=a+1 od {x=a*y+b/\b>=0/\b<y} (while (8))

24. Proof (cont.) (10) {x=0*y+x/\x>=0} Prog {x=a*y+b/\b>=0/\b<y} (Composition (3), (9)) (11) x>=0/\y>=0 x=0*y+x/\x>=0 (Logic) (12) {x>=0/\y>=0} Prog {x=a*y+b/\b>=0/\b<y} (Consequence)

25. Soundness Hoare logic is sound in the sense that everything that can be proved is correct! This follows from the fact that each axiom and proof rule preserves soundness.

26. Completeness A proof system is called complete if every correct assertion can be proved. • Propositional logic is complete. • No deductive system for the standard arithmetic can be complete (Godel).

27. And for Hoare’s logic? Let S be a program and p its precondition. Then {p} S {false} means that S never terminates when started from p. This is undecideable. Thus, Hoare’s logic cannot be complete.

28. Weakest prendition, Strongest postcondition • For an assertion p and code S, let post(p,S) be the strongest assertion such that {p}S{post(p,S)} That is, if {p}S{q} then post(p,S)q. • For an assertion q and code S, let pre(S,q) be the weakest assertion such that {pre(S,q)}S{q} That is, if {p}S{q} then ppre(S,q).

29. Relative completeness • Suppose that either • post(p,S) exists for each p, S, or • pre(S,q) exists for each S, q. • Some oracle decides on pure implications. Then each correct Hoare triple can be proved. What does that mean? The weakness of the proof system stem from the weakness of the (FO) logic, not of Hoare’s proof system.

30. Extensions Many extensions for Hoare’s proof rules: • Total correctness • Arrays • Subroutines • Concurrent programs • Fairness

31. Proof rule for total correctness {p/\e/\t=z} S {p/\t<z}, pt>=0 {p} while e do S od {p/\¬e} where z - an int. variable, not appearing in p,t,e,S. t - an int. expression.