program verification: flowchart programs
DESCRIPTION
Program verification: flowchart programs. Book: chapter 7. History. Verification of flowchart programs: Floyd, 1967 Hoare’s logic: Hoare, 1969 Linear Temporal Logic: Pnueli, Krueger, 1977 Model Checking: Clarke & Emerson, 1981. Program Verification. Predicate (first order) logic. - PowerPoint PPT PresentationTRANSCRIPT
Program verification: flowchart programs
Book: chapter 7
History Verification of flowchart programs: Floyd, 1967 Hoare’s logic: Hoare, 1969 Linear Temporal Logic: Pnueli, Krueger, 1977 Model Checking: Clarke & Emerson, 1981
Program Verification Predicate (first order) logic. Partial correctness, Total correctness Flowchart programs Invariants, annotated programs Well founded ordering (for
termination) Hoare’s logic
Predicate (first order logic)
Variables, functions, predicates
Terms
Formulas (assertions)
Signature Variables: v1, x, y18Each 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).
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)
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)
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) :
Flowchart programs
Input variables: X=x1,x2,…,xlProgram variables: Y=y1,y2,…,ymOutput variables: Z=z1,z2,…,zn
start
haltY=f(X)
Z=h(X,Y)
Assignments and tests
Y=g(X,Y) t(X,Y)FT
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
Initial conditionInitial condition: the
values for the input variables for which the program must work.
x1>=0 /\ x2>0
FT
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<x2
FT
Partial correctness, Termination, Total correctness Patial 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.
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
Subtle point:
The program ispartially correct
withrespect tox1>=0/\x2>=0and totally correctwith respect tox1>=0/\x2>0
T F
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
C D
E
FT
Annotating a scheme with invariants
A): x1>=0 /\ x2>=0B): x1=y1*x2+y2 /\
y2>=0C): x1=y1*x2+y2 /\
y2>=0 /\ y2>=x2D):x1=y1*x2+y2 /\
y2>=0 /\ y2<x2E):x1=z1*x2+z2 /\ 0<=z2<x2Notice: (A) is the initial
condition, is the input-output condition.
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
A
B
C D
E
FT
Verification conditions: assignment
A) B) [Y\g(X,Y)]
A): x1>=0 /\ x2>=0
B): x1=y1*x2+y2 /\ y2>=0
B) [Y\g(X,Y)] =x1=0*x2+x1 /\
x1>=0
(y1,y2)=(0,x1)
A
B
A
B
(y1,y2)=(0,x1)
Y=g(X,Y)
(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)[Y\g(X,Y]: x1=(y1+1)*x2+y2-x2 /\ y2-x2>=0
C
B
(z1,z2)=(y1,y2)
Third assignment
D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2
E):x1=z1*x2+z2 /\ 0<=z2<x2
E)[Z\g(X,Y]: x1=y1*x2+y2 /\ 0<=y2<x2
E
D
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<x2
y2>=x2
B
C
D
B
C
Dt(X,Y)
FT
FT
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
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.
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
A
B
C D
E
falsetrue
Annotating a scheme with invariants
A): x1>=0 /\ x2>=0B): x1=y1*x2+y2 /\
y2>=0C): x1=y1*x2+y2 /\
y2>=0 /\ y2>=x2D):x1=y1*x2+y2 /\
y2>=0 /\ y2<x2E):x1=z1*x2+z2 /\ 0<=z2<x2Notice: (A) is the initial
condition, Eis the input-output condition.
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
A
B
C D
E
falsetrue
Verification conditions: assignment
A) B) [Y\g(X,Y)]
A): x1>=0 /\ x2>=0B): x1=y1*x2+y2 /\
y2>=0
B) [Y\g(X,Y)] =x1=0*x2+x1 /\
x1>=0
A
B
(y1,y2)=(0,x1)
(y1,y2)=(0,x1)
A
B
Y=g(X,Y)
Assignment condition
(y1,y2)=(0,x1)
A
B
y1=2
y1=x1
2=x1
Another way to understand condition
(y1,y2)=(0,x1)
A
B
y1=2
y1=x1
Use two versions of variables: before assignment and after. E.g., y1 and y1’, respectively.
postcondition: y1’=x1assignment: y1’=2precondition: 2=x1
2=x1
Assignment condition
(y1,y2)=(0,x1)
A
B
y1=y1+5
y1=10
y1=5
Assignment condition
(y1,y2)=(0,x1)
A
B
y1=y1+5
y1=10
y1=5Postcondition: y1’=10
Assignment: y1’=y1+5
Precondition: y1+5=10, I.e., y1=5
Verification conditions: assignment
B): x1=y1’*x2+y2’ /\ y2’ >=0
Assignment: y1’=0 /\ y2’=x1
B) [Y\g(X,Y)] =x1=0*x2+x1 /\ x1>=0(or simply x1>=0)
A
B
(y1,y2)=(0,x1)
A): x1>=0 /\ x2>=0
Second assignment
Precondition:B): x1=y1*x2+y2 /\
y2>=0
Assignment:y1’=y1+1/\y2’=y2-x2
Postcondition:B)[Y\g(X,Y)]:
x1=(y1+1)*x2+y2-x2 /\ y2-x2>=0
(y1,y2)=(y1+1,y2-x2)
C
B
(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)[Y\g(X,Y)]: x1=(y1+1)*x2+y2-x2 /\ y2-x2>=0
C
B
(z1,z2)=(y1,y2)
Third assignment
D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2
E):x1=z1*x2+z2 /\ 0<=z2<x2
E)[Z\g(X,Y]: x1=y1*x2+y2 /\ 0<=y2<x2
E
D
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<x2
y2>=x2
B
C
D
B
C
Dt(X,Y)
falsetrue
falsetrue
Exercize: prove partial correctness
Initial condition: x>=0
Input-output claim: z=x!
start
halt
(y1,y2)=(0,1)
y1=x
(y1,y2)=(y1+1,(y2+1)y2) z=y2
truefalse
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 and S is executed, then (Y) 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.
Another example
(A) : x>=0
(F) : z^2<=x<(z+1)^2
z is the biggest numberthat is not greaterthan 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
Some insight
1+3+5+…+(2n+1)=(n+1)^2
y2 accumulates theabove sum, untilit is bigger than x.
y3 ranges over oddnumbers 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
Invariants
It is sufficient to have one invariant for every loop(cycle in the program’sgraph).
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
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
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+1Simplified: 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
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
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
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)+1y1^2<=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
Not finished!
Still needs to:
Calculate (E) bysubstituting backwardsfrom (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
Proving termination
Well-founded sets Partially ordered set (W,<):
If a<b and b<c then a<c (transitivity). If a<b then not b<a (asymmetry). Not a<a (irreflexivity).
Well-founded set (W,<): Partially ordered. No infinite decreasing chain a1>a2>a3>…
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].
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<x2.
start
halt
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
(y1,y2)=(0,x1)
A
B
D
E
falsey2>=x2
C
true
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).
How not to stay in a loop? Show that
u(M)>=u(N).
At least once in each loop, show that u(M)>u(N).
S
M
N
TN
M
How not to stay in a loop? For stmt: (M)(u(M)>=u(N)’rel)
For test (true side):((M)/\test)(u(M)>=u(N))
For test (false side):((M)/\
¬test)(u(M)>=u(L))
stmt
M
N
testN
M
true
L
false
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.
Why does the program terminate
u(A)=x1u(B)=y2u(C)=y2u(D)=y2u(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
falsey2>=x2
C
true
Recall partial correctness annotation
A): x1>=0 /\ x2>=0B): x1=y1*x2+y2 /\
y2>=0C): x1=y1*x2+y2 /\
y2>=0 /\ y2>=x2D):x1=y1*x2+y2 /\
y2>=0 /\ y2<x2E):x1=z1*x2+z2 /\ 0<=z2<x2
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
A
B
C D
E
falsetrue
Strengthen for termination
A): x1>=0 /\ x2>0B): x1=y1*x2+y2 /\
y2>=0/\x2>0C): x1=y1*x2+y2 /\
y2>=0 /\ y2>=x2/\x2>0D):x1=y1*x2+y2 /\
y2>=0 /\ y2<x2/\x2>0E):x1=z1*x2+z2 /\ 0<=z2<x2This proves that u(M) is
natural for each point M.
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
A
B
C D
E
falsetrue
We shall show:u(A)=x1u(B)=y2u(C)=y2u(D)=y2u(E)=z2u(A)>=u(B)u(B)>=u(C)u(C)>u(B)u(B)>=u(D)u(D)>=u(E)
start
halt
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
(y1,y2)=(0,x1)
A
B
D
E
falsey2>=x2
C
true
Proving decrement
C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2/\x2>0
u(C)=y2u(B)=y2u(B)’rel=y2-x2
C) y2>y2-x2(notice that C) x2>0)
start
halt
(y1,y2)=(0,x1)
y2>=x2
(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)
A
B
C D
E
falsetrue
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
u(A)=x+1u(B)=x-y2+1u(C)=max(0,x-y2)u(D)=x-y2+1u(E)=u(F)=0u(A)>=u(B)u(B)>u(C)u(C)>=u(D)u(D)>=u(B)Need some invariants,i.e., y2<=x/\y3>0at 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