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Functional Verification I
Software Testing and Verification Lecture Notes 21 Prepared by Stephen M. Thebaut, Ph.D. University of Florida
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Overview of Functional Verification Topics
Lecture Notes #21 - Functional Verification I Introduction Verifying correctness in program reading, writing, and validation Complete and sufficient correctness Compound programs and the Axiom of Replacement Lecture Notes #22 - Functional Verification II Correctness conditions and working correctness questions: sequencing and decision statements
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Overview of Functional Verification Topics
Lecture Notes #23 - Functional Verification III Iteration Recursion Lemma (IRL) (Very Cool!) Termination predicate Correctness conditions for while_do statement Correctness conditions for repeat_until statement Subgoal Induction Lecture Notes #24 – Functional Verification IV Invariant Status Theorem (EXTREMELY Cool!) While Loop Initialization
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Today’s Topics: Introduction
Verifying correctness in program reading, writing, and validation Complete and sufficient correctness Compound programs and the Axiom of Replacement
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Introduction What is functional verification?
A methodology originally developed by Mills for verifying program correctness with respect to an intended function specification. It represents a viable alternative to the axiomatic verification method developed by Hoare and Floyd.
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Introduction (cont’d)
References: Linger, Mills, & Witt, Structured Programming: Theory and Practice, Addison-Wesley, 1979. Dunlop & Basili, “A Comparative Analysis of Functional Correctness,” Computing Surveys, Vol. 14, No. 2, June 1982.† Linger, “Cleanroom Software Engineering for Zero- Defect Software,” Proceedings, 15th Int. Conf. on Soft. Eng. (1993), IEEE Computer Society Press.† † Required readings.
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Tasks in Program Reading, Writing, and Verification
Abstract a given program construct (e.g., an if_then_ else statement) into a hypothesized function f. To confirm that your understanding of the program is correct, show:
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Tasks in Program Reading, Writing, and Verification
Abstract a given program construct (e.g., an if_then_ else statement) into a hypothesized function f. To confirm that your understanding of the program is correct, show: f = [if p then G else H]
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Tasks in Program Reading, Writing, and Verification (cont’d)
Program Writing: Expand a given function f into a hypothesized program construct (e.g., an if_then_else statement). To confirm that your expansion of f into a program is correct, show:
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Tasks in Program Reading, Writing, and Verification (cont’d)
Program Writing: Expand a given function f into a hypothesized program construct (e.g., an if_then_else statement). To confirm that your expansion of f into a program is correct, show: f = [if p then G else H]
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Tasks in Program Reading, Writing, and Verification (cont’d)
Program Verification: You are given both function f and its hypothesized program expansion (e.g., an if_then_ else statement). To confirm the correctness of the hypothesized program expansion with respect to f, show:
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Tasks in Program Reading, Writing, and Verification (cont’d)
Program Verification: You are given both function f and its hypothesized program expansion (e.g., an if_then_ else statement). To confirm the correctness of the hypothesized program expansion with respect to f, show: f = [if p then G else H]
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Tasks in Program Reading, Writing, and Verification (cont’d)
In all three cases, the final task is to confirm the equivalence (or more typically, subset relationship) of two expressions, each representing the function of a program, i.e., confirm that f = [P] or f [P].
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Complete and Sufficient Correctness
Given a function f and a program P (claimed to implement f), correctness is concerned with one of two questions:
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Complete and Sufficient Correctness
Given a function f and a program P (claimed to implement f), correctness is concerned with one of two questions: Is f = [P] ? (“Is f equivalent to the function computed by P ?”) – A question of complete correctness.
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Complete and Sufficient Correctness
Given a function f and a program P (claimed to implement f), correctness is concerned with one of two questions: Is f = [P] ? (“Is f equivalent to the function computed by P ?”) – A question of complete correctness. Is f [P] ? (“Is f a subset of the function computed by P ?”) – A question of sufficient correctness.
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Complete and Sufficient Correctness (cont’d)
In the case of complete correctness, P computes the correct values of f for arguments in D(f) only; [P] is undefined (P does not terminate) for arguments outside D(f). In the case of sufficient correctness, P may compute values from arguments not in D(f). Note that, by definition, f = [P] implies f [P]
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] Sufficient (only) (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] Sufficient (only) (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] Sufficient (only) (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f Complete/Sufficient
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] Sufficient (only) (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f Complete/Sufficient
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] Sufficient (only) Neither (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f Complete/Sufficient
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] Sufficient (only) Neither (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f Complete/Sufficient
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Correctness Relationships
(X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P] f f [P] Sufficient (only) Neither (X,Y)f (X,Y)[P] (X,Y)f (X,Y)[P] [P], f [P] f Complete/Sufficient Neither
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Example For integers x,y consider the function:
f = (y≥0 x,y := x+y,0) and the programs: P1 = while y>0 do x,y := x+1,y-1 P2 = while y<>0 do x,y := x+1,y-1
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Example For integers x,y consider the function:
f = (y≥0 x,y := x+y,0) and the programs: P1 = while y>0 do x,y := x+1,y-1 P2 = while y<>0 do x,y := x+1,y-1 Use heuristics to hypothesize functions for P1 and P2 and compare these to f.
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 f = (y≥0 x,y := x+y,0) √
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := ?,? f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y := x+y,0 (since y=0) f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y f = (y≥0 x,y := x+y,0) √
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 x,y := ?,? f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 x,y := x,y f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 x,y := x,y f = (y≥0 x,y := x+y,0) So, is f = [P1]?
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 x,y := x,y f = (y≥0 x,y := x+y,0) So, is f = [P1]? no
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 x,y := x,y f = (y≥0 x,y := x+y,0) Is f [P1]?
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Example (cont’d) Consider P1 = while y>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 x,y := x,y f = (y≥0 x,y := x+y,0) Is f [P1]? yes
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 f = (y≥0 x,y := x+y,0) √
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y f = (y≥0 x,y := x+y,0) √
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 x,y := ?,? f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 infinite loop (i.e., “undefined”) f = (y≥0 x,y := x+y,0)
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 infinite loop f = (y≥0 x,y := x+y,0) So, is f = [P2]?
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 infinite loop f = (y≥0 x,y := x+y,0) So, is f = [P2]? yes
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 infinite loop f = (y≥0 x,y := x+y,0) Is f [P2]?
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Example (cont’d) Consider P2 = while y<>0 do x,y := x+1,y-1
y>0 x,y := x+(1)y,0 := x+y,0 y=0 x,y := x,y y<0 infinite loop f = (y≥0 x,y := x+y,0) Is f [P2]? yes
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Example (cont’d) Both programs satisfy sufficient correctness. (Both correctly compute f(x,y) for y≥0.) Only P2 satisfies complete correctness. (P1 terminates for negative y.)
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Example (cont’d) Both programs satisfy sufficient correctness. (Both correctly compute f(x,y) for y≥0.) Only P2 satisfies complete correctness. (P1 terminates for negative y.) But would YOUR MOM approve of writing “misbehaving” programs like P2 ?
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Defensive Programming: Handling Invalid Inputs
f and P can be redefined to handle invalid inputs: f’ = (y≥0 x,y,z := x+y,0,z | true x,y,z := x,y,‘error’) P’ = if y<0 then z := ‘error’ else while y>0 do x,y := x+1,y-1 end_while end_if_then_else Does f’ = [P’] ?
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Exercise Given P = if x>=y then x,y := y,x
“Identity” function: x,y := x,y Given P = if x>=y then x,y := y,x f1 = (x>y x,y := y,x | true I) f2 = (x>y x,y := y,x | x<y I) f3 = (x≠y x,y := y,x) Fill in the following “correctness table”: P C=Complete (and Sufficient) S=Sufficient (only) N=Neither f1 f2 f3
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Conventional meaning of “P computes f”
Henceforth, we will follow the lead of Dunlop and Basili (“A Comparative Analysis of Functional Correctness”) and adopt the convention of asserting “P computes f” if and only if f is a subset of [P]. That is, for “P to compute f,” no explicit require-ment is made concerning the behavior of P on inputs outside the domain of f. Thus, “P computes f,” which we will now write informally as “f=[P],” should henceforth be inter-preted as an assertion of sufficient correctness.
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Compound Programs and the Axiom of Replacement
The algebraic structure of compound program P permits decomposition into a hierarchy of abstractions. The proof of correctness of P is thereby decomposed into a proof of correctness of each such abstraction.
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Compound Programs and the Axiom of Replacement (cont’d)
For example, to show that compound program F implements function f, where F = if p then G else H and G, H are themselves programs: hypothesize functions g, h and attempt to prove g = [G] and h = [H]
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Compound Programs and the Axiom of Replacement (cont’d)
If successful, use the Axiom of Replacement to reduce the problem to proving f = if p then g else h If successful again, you will have proved f = [F]
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Compound Programs and the Axiom of Replacement (cont’d)
Thus, the Axiom of Replacement allows one to prove the correctness of complex programs in a bottom-up, incremental fashion. In the next lecture, we consider correctness conditions for sequencing and decision statements.
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Functional Verification I
Software Testing and Verification Lecture Notes 21 Prepared by Stephen M. Thebaut, Ph.D. University of Florida
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