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Software Verification 1 Deductive Verification. Prof. Dr. Holger Schlingloff Institut für Informatik der Humboldt Universität und Fraunhofer Institut für Rechnerarchitektur und Softwaretechnik. Termination proof rule. Let (M,<) be a well-founded order and (z) be a formula involving z M
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Software Verification 1Deductive Verification Prof. Dr. Holger Schlingloff Institut für Informatik der Humboldt Universität und Fraunhofer Institut für Rechnerarchitektur und Softwaretechnik
Termination proof rule • Let (M,<) be a well-founded order and (z) be a formula involving zM • if ⊢ (z0) for some z0M and ⊢ (z)b (z’) ¬b for some z’<z, then ⊢ while (b) ¬b • (z) is called variant of the loop(special case: (z) = (z=t(x)), here t(x) is called the variant)
Termination - a more intricate example = {b=1; while (a<=100 | b!=1) if (a<=100) {a+=11; b++;} else {a-=10; b--;} a-=10; } Show: ⊢0<a<=100 a==91
We do the termination part only. • Hint for the invariant: • (0<b<=11 & 0<a<=111 & (a<=101 | b!=1)) • wfo: N0; Variant: (z) = (z==1111+111b-11a-1); • if 0<a<=100 & b==1, we have zN0 • Assume within the while-loop (z) & (a<=100 | b!=1)) • Case a<=100: {a+=11; b++} gives z-10==1111+111(b+1)-11(a+11)-1 • Case a>100: {a-=10; b--;} gives z-1==1111+111(b-1)-11(a-10)-1 • Thus, in both cases there exists z’<z such that (z’) holds
Finding Variants is Hard • Try this one: Mersenne = {n=0; k=0; while (k<48) {n++; if (isprim((2**n)-1)) k++}} • ... and apply for the Fields-medal if successful
Proof of Termination Proof Rule • if ⊢ (z) for some zM and ⊢ (z) (z’) ¬b for some z’<z then program while (b) terminates • Assume not. • Then there is an infinite execution ; ; ; ... such that b holds before and after each • Then there is an infinite descending chain z0, z1, z2, ... such that z0=z and zi+1<zi • Thus, M is not a wfo.
Binary Search Program : i=0; k=n; while (i<k) { s=i+(k-i-1)/2; //integer division if (a>x[s]) i=s+1 else k=s } Show n>=0 i(0<i<n (x[i-1]<x[i]) 0<=i<=n j(0<=j<i x[j]<a j(i<=j<n x[j]>=a
Variant (z)? • while (i<k) ... suggest (z) = (z=k-i) • ⊢ (z)b (z’) ¬b for some z’<z • what is a well-founded order for z?can we guarantee that zN0 ? • Example: (assume k>0, j>0) • {i=k; while (i!=0) i-=j} terminates iff k%j==0 • Assume k%j==0; wfo: (z) = (z=i/j); zN0 • {i=k; while (i>=0) i-=j} terminates always. Proof?
Transforming Variants We have to show: ⊢ (z) (z’) ¬b Most important case: ⊢ z=t(x) x=f(x) z’=t(x) ¬b Let z’=t(f(t-1(z))) ⊢ z=t(x) t-1(z)=x since t-1(t(x))=x ⊢ t-1(z)=x t(f(t-1(z)))=t(f(x)) ⊢ t(f(t-1(z)))=t(f(x)) x=f(x) t(f(t-1(z)))=t(x) (ass) Therefore, ⊢ z=t(x) x=f(x) t(f(t-1(z)))=t(x) • Ex.: ⊢ z=i+k i=i-j z’=i+k for z’=z-j
Proof for Binary Search Termination • Solution for binary search: z=(k-i)N0 ? • Show 0<=i<=k<=n is invariant (omitted) Let (z)= (k-i=z) • k-i=zi=i+(k-i-1)/2+1 k-i=z’ for z’ = (z-1)/2 - 1 < z Proof: let t(i) = k-i t(z) = k-z t-1(z)= (k-z) f(i) = i+(k-i-1)/2+1 t(f(t-1(z)))= k-((k-z) +(k- (k-z) -1)/2+1) = (z-1)/2-1 • k-i=zk=i+(k-i-1)/2 k-i=z’ for z’= i+((z+i)-i-1)/2-i=(z-1)/2 <z
Pre- and Postconditions • Dijkstra: wp-calculus (weakest precondition) • characterize the “weakest” formula which makes a Hoare-triple valid • =wp(.) iff ⊢ and⊢(') for every ’ for which ⊢’ • =wlp(.) iff ⊢{}{} and⊢(') for every ’ for which ⊢{’} {} • Example: wp(x++, x==7) = (x==6) • Dijkstra gives a set of rules for wp which can be seen as notational variant of Hoare logic
