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1 Program verification: flowchart programs (Book: chapter 7)

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2 History Verification of flowchart programs: Floyd, 1967 Hoares logic: Hoare, 1969 Linear Temporal Logic: Pnueli, Krueger, 1977 Model Checking: Clarke & Emerson, 1981

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3 Program Verification Predicate (first order) logic. Partial correctness, Total correctness Flowchart programs Invariants, annotated programs Well founded ordering (for termination) Hoares logic

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4 Predicate (first order logic) Variables, functions, predicates Terms Formulas (assertions)

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5 Signature Variables: v1, x, y18 Each variable represents a value of some given domain (int, real, string, …). Function symbols: f(_,_), g2(_), h(_,_,_). Each function has an arity (number of paramenters), a domain for each parameter, and a range. f:int*int->int (e.g., addition), g:real->real (e.g., square root) A constant is a predicate with arity 0. Relation symbols: R(_,_), Q(_). Each relation has an arity, and a domain for each parameter. R : real*real (e.g., greater than). Q : int (e.g., is a prime).

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6 Terms Terms are objects that have values. Each variable is a term. Applying a function with arity n to n terms results in a new term. Examples: v1, 5.0, f(v1,5.0), g2(f(v1,5.0)) More familiar notation: sqr(v1+5.0)

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7 Formulas Applying predicates to terms results in a formula. R(v1,5.0), Q(x) More familiar notation: v1>5.0 One can combine formulas with the boolean operators (and, or, not, implies). R(v1,5.0)->Q(x) x>1 -> x*x>x One can apply existentail and universal quantification to formulas. x Q(X) x1 R(x1,5.0) x y R(x,y)

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8 A model, A proofs A model gives a meaning (semantics) to a first order formula: A relation for each relation symbol. A function for each function symbol. A value for each variable. An important concept in first order logic is that of a proof. We assume the ability to prove that a formula holds for a given model. Example proof rule (MP) :

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9 Flowchart programs Input variables: X=x1,x2,…,xl Program variables: Y=y1,y2,…,ym Output variables: Z=z1,z2,…,zn start halt Y=f(X) Z=h(X,Y)

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10 Assignments and tests Y=g(X,Y)t(X,Y) FT

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11 start halt (y1,y2)=(0,x1) y2>=x2 (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) Initial condition Initial condition: the values for the input variables for which the program must work. x1>=0 /\ x2>0 F T

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12 start halt (y1,y2)=(0,x1) y2>=x2 (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) The input-output claim The relation between the values of the input and the output variables at termination. x1=z1*x2+z2 /\ 0<=z2

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13 Partial correctness, Termination, Total correctness Partial correctness: if the initial condition holds and the program terminates then the input-output claim holds. Termination: if the initial condition holds, the program terminates. Total correctness: if the initial condition holds, the program terminates and the input-output claim holds.

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14 Subtle point: The program is partially correct with respect to x1>=0/\x2>=0 and totally correct with respect to x1>=0/\x2>0 start halt (y1,y2)=(0,x1) y2>=x2 (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) T F

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15 start halt (y1,y2)=(0,x1) y2>=x2 (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) Annotating a scheme Assign an assertion for each pair of nodes. The assertion expresses the relation between the variable when the program counter is located between these nodes. A B CD E FT

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16 Invariants Invariants are assertions that hold at each state throughout the execution of the program. One can attach an assertion to a particular location in the code: e.g., at(B) (B). This is also an invariant; in other locations, at(B) does not hold hence the implication holds. If there is an assertion attached to each location, (A), (B), (C), (D), (E), then their disjunction is also an invariant: (A)\/ (B)\/ (C)\/ (D)\/ (E) (since location is always at one of these locations).

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17 Annotating a scheme with invariants A): x1>=0 /\ x2>=0 B): x1=y1*x2+y2 /\ y2>=0 C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 D):x1=y1*x2+y2 /\ y2>=0 /\ y2

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18 Preliminary: Relativizing assertions (B) : x1= y1 * x2 + y2 /\ y2 >= 0 Relativize B) w.r.t. the assignment, obtaining B) [Y\g(X,Y)] e (B) expressed w.r.t. variables at A.) (B) A = x1=0 * x2 + x1 /\ x1>=0 Think about two sets of variables, before={x, y, z, …} after={x,y,z…}. Rewrite (B) using after, and the assignment as a relation between the set of variables. Then eliminate after by substitution. Here: x1=y1 * x2 + y2 /\ y2>=0 /\ x1=x1 /\ x2=x2 /\ y1=0 /\ y2=x1 now eliminate x1, x2, y1, y2. (y1,y2)=(0,x1) A B A B Y=g(X,Y)

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19 Preliminary: Relativizing assertions (y1,y2)=(0,x1) A B A B A): (B) A (B) Y=g(X,Y)

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20 Verification conditions: assignment A) B) A where B) A = B)[Y\g(X,Y)] A): x1>=0 /\ x2>=0 B): x1=y1*x2+y2 /\ y2>=0 B) A = x1=0*x2+x1 /\ x1>=0 (y1,y2)=(0,x1) A B A B Y=g(X,Y)

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21 (y1,y2)=(y1+1,y2-x2) Second assignment C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 B): x1=y1*x2+y2 /\ y2>=0 B) C : x1=(y1+1)*x2+y2- x2 /\ y2-x2>=0 C B

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22 (z1,z2)=(y1,y2) Third assignment D):x1=y1*x2+y2 /\ y2>=0 /\ y2

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23 Verification conditions: tests B) /\ t(X,Y) C) B) /\¬t(X,Y) D) B): x1=y1*x2+y2 /\y2>=0 C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 D):x1=y1*x2+y2 /\ y2>=0 /\ y2

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24 Verification conditions: tests y2>=x2 B C D B C D t(X,Y) F T FT ¬t(X,Y) B) C)

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25 Partial correctness proof: An induction on length of execution B) D) C) Initially, states satisfy the initial conditions. Then, passing from one set of states to another, we preserve the invariants at the appropriate location. We prove: starting with a state satisfying the initial conditions, if are at a point in the execution, the invariant there holds. Not a proof of termination! start halt (y1,y2)=(0,x1) y2>=x2 (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) A B CD E A) no yes TF

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26 Exercise: prove partial correctness Initial condition: x>=0 Input-output claim: z=x! start halt (y1,y2)=(0,1) y1=x (y1,y2)=(y1+1,(y1+1)*y2)z=y2 TF

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27 What have we achieved? For each statement S that appears between points X and Y we showed that if the control is in X when (X) holds (the precondition of S) and S is executed, then (Y) (the postcondition of S) holds. Initially, we know that (A) holds. The above two conditions can be combined into an induction on the number of statements that were executed: If after n steps we are at point X, then (X) holds.

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28 Another example (A) : x>=0 (F) : z^2<=x<(z+1)^2 z is the biggest number that is not greater than sqrt x. start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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29 Some insight …+(2n+1)=(n+1)^2 y2 accumulates the above sum, until it is bigger than x. y3 ranges over odd numbers 1,3,5,… y1 is n-1. start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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30 Invariants It is sufficient to have one invariant for every loop (cycle in the programs graph). We will have (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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31 Obtaining (B) By backwards substitution in (C). (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 (B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1 start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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32 Check assignment condition (A)=x>=0 (B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1 (B) relativized is 0^2<=x /\ 0+1=(0+1)^2 /\ 1=2*0+1 Simplified: x>=0 start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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33 Obtaining (D) By backwards substitution in (B). (B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1 (D)=(y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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34 Checking (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 (C)/\y2<=x) (D) (D)=(y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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35 y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x (y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x (y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x (y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1

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36 Not finished! Still needs to: Calculate (E) by substituting backwards from (F). Check that (C)/\y2>x (E) start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2)z=y1 B C D F truefalse E y2=y2+y3

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37 Exercise: prove partial correctness. Initially: x1>0/\x2>0. At termination: z1=gcd(x1,x2). halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 FT y1>y2 y2=y2-y1 y1=y1-y2 TF

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38 Annotation of program with invariants halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 F T y1>y2 y2=y2-y1 y1=y1-y2 TF z1=gcd(x1,x2) x1>0 /\ x2>0 gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0 gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\y1 y2 gcd(y1,y2)=gcd(x1,x2)/\ y1>0/\y2>0/\y1

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39 Part 1 halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 F T y1>y2 y2=y2-y1 y1=y1-y2 TF (A)= x1>0 /\ x2>0 (B)=gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0 A B D E F G H (B)rel= gcd(x1,x2)=gcd(x1,x2)/\ x1>0/\x2>0 (A) (B)rel

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40 Part 2a halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 F T y1>y2 y2=y2-y1 y1=y1-y2 TF (B)= gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0 (D)=gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\y1 y2 A B D E F G H (B)/\ ¬(y1=y2) (D)

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41 Part 2b halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 F T y1>y2 y2=y2-y1 y1=y1-y2 TF (G)= y1=gcd(x1,x2) A B D E F G H (B)= gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0 (B)/\ (y1=y2) (G)

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42 Part 3 halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 F T y1>y2 y2=y2-y1 y1=y1-y2 TF (D)= gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\y1 y2 (E)=gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\y1

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43 Part 4 halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 F T y1>y2 y2=y2-y1 y1=y1-y2 TF x1>0 /\ x2>0 (B)= gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0 (E)= gcd(y1,y2)=gcd(x1,x2)/\ y1>0/\y2>0/\y1

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44 Annotation of program with invariants halt start (y1,y2)=(x1,x2) z1=y1 y1=y2 F T y1>y2 y2=y2-y1 y1=y1-y2 TF (H)= z1=gcd(x1,x2) (G)= y1=gcd(x1,x2) A B D E F G H (H)rel= y1=gcd(x1,x2) (G) (H)rel2

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45 Proving termination

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46 Well-founded sets Partially ordered set (W,<): If a**a2>a3>…
**

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47 Examples for well founded sets Natural numbers with the bigger than relation. Finite sets with the set inclusion relation. Strings with the substring relation. Tuples with alphabetic order: (a1,b1)>(a2,b2) iff a1>a2 or [a1=a2 and b1>b2]. (a1,b1,c1)>(a2,b2,c2) iff a1>a2 or [a1=a2 and b1>b2] or [a1=a2 and b1=b2 and c1>c2].

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48 Why does the program terminate y2 starts as x1. Each time the loop is executed, y2 is decremented. y2 is natural number The loop cannot be entered again when y2

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49 Proving termination Choose a well-founded set (W,<). Attach a function u(N) to each point N. Annotate the flowchart with invariants, and prove their consistency conditions. Prove that (N) (u(N) in W).

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50 How not to stay in a loop? Show that u(M)>=u(N)rel. At least once in each loop, show that u(M)>u(N). S M N T N M

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51 How not to stay in a loop? For stmt: (M) (u(M)>=u(N)rel) Relativize since we need to compare values not syntactic expressions. For test (true side): ( (M)/\test) (u(M)>=u(N)) For test (false side): ( (M)/\¬test) (u(M)>=u(L)) stmt M N test N M true L false

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52 What did we achieve? There are finitely many control points. The value of the function u cannot increase. If we return to the same control point, the value of u must decrease (its a loop!). The value of u can decrease only a finite number of times.

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53 Why does the program terminate u(A)=x1 u(B)=y2 u(C)=y2 u(D)=y2 u(E)=z2 W: naturals > : greater than start halt (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) (y1,y2)=(0,x1) A B D E false y2>=x2 C true

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54 Recall partial correctness annotation A): x1>=0 /\ x2>=0 B): x1=y1*x2+y2 /\ y2>=0 C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 D):x1=y1*x2+y2 /\ y2>=0 /\ y2

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55 Strengthen for termination A): x1>=0 /\ x2>0 B): x1=y1*x2+y2 /\ y2>=0/\x2>0 C): x1=y1*x2+y2 /\ y2>=0/\y2>=x2/\x2>0 D):x1=y1*x2+y2 /\ y2>=0 /\ y2 0 E):x1=z1*x2+z2 /\ 0<=z2

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56 Strengthen for termination A): x1>=0 /\ x2>0 u(A)>=0 B): x1=y1*x2+y2 /\ y2>=0/\x2>0 u(B)>=0 C): x1=y1*x2+y2 /\y2>=0 /\y2>=x2/\x2>0 u(c)>=0 D):x1=y1*x2+y2 /\ y2>=0 /\ y2 0 u(D)>=0 E):x1=z1*x2+z2 /\ 0 =0 This proves that u(M) is natural for each point M. u(A)=x1 u(B)=y2 u(C)=y2 u(D)=y2 u(E)=z2

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57 We shall show: u(A)=x1 u(B)=y2 u(C)=y2 u(D)=y2 u(E)=z2 A) u(A)>=u(B)rel B) u(B)>=u(C) C) u(C)>u(B)rel B) u(B)>=u(D) D) u(D)>=u(E)rel start halt (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) (y1,y2)=(0,x1) A B D E false y2>=x2 C true

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58 Proving decrement C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2/\x2>0 u(C)=y2 u(B)=y2 u(B)rel=y2-x2 C) y2>y2-x2 (notice that C) x2>0) start halt (y1,y2)=(y1+1,y2-x2)(z1,z2)=(y1,y2) (y1,y2)=(0,x1) A B D E false y2>=x2 C true

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59 Integer square prog. (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 (B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\y3=2*y1+1 start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F truefalse E y2=y2+y3

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60 u(A)=x+1 u(B)=x-y2+1 u(C)=max(0,x-y2+1) u(D)=x-y2+1 u(E)=u(F)=0 u(A)>=u(B)rel u(B)>u(C)rel u(C)>=u(D) u(C)>=u(E) u(D)>=u(B)rel Need some invariants, i.e., y2 0 at points B and D, and y3>0 at point C. start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F truefalse E y2=y2+y3

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61 Program Verification Using Hoares Logic Hoare triple is of the form {Precondition} Prog-segment {Postcondition} It expresses partial correctness: if the segment starts with a state satisfying the precondition and it terminates, the final state satisfies the postscondition. The idea is that one can decompose the proof of the program into smaller and smaller segments, depending on its structure.

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

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63 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)}

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64 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

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65 Assignment axiom {p[e/y] } y:=e {p} For example: {y+5=10} y:=y+5 {y=10} {y+y

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66 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. This y represents value of y before the assignment. {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

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67 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}

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68 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(x1,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}

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

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

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71 Consequence rules Strengthen a precondition r p, {p } S {q } {r } S {q } Weaken a postcondition {p } S {q }, q r {p } S {r }

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72 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

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73 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/\y1=y2} Combine the above using concatenation rule!

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74 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)}

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75 Use of second 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)}

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76 Annotating a while program {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)}

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

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78 Consequence rules Strengthen a precondition r p, {p} S {q} {r} S {q} Weaken a postcondition {p} S {q}, q r {p} S {r}

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79 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.

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80 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).

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81 And for Hoares 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, Hoares logic cannot be complete.

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82 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).

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83 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 Hoares proof system.

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84 Extensions Many extensions for Hoares proof rules: Total correctness Arrays Subroutines Concurrent programs Fairness

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85 Proof rule for total correctness Similar idea to Floyds termination: Well foundedness {p/\t/\f=z} S {p/\f =0) {p} while t do S od {p/\¬t} where z - an int. variable, not appearing in p,t,e,S. f - an int. expression.

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