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Functional Verification III Prepared by Stephen M. Thebaut, Ph.D. University of Florida Software Testing and Verification Lecture Notes 23

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Previously… Correctness conditions and working correctness questions: –sequencing –decision statements

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Today’s Topics Iteration Recursion Lemma (IRL) Termination predicate: term(f,P) Correctness conditions for while_do statement Sufficient correctness conditions Correctness conditions for repeat_until statement Subgoal Induction

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Iteration Recursion Lemma (IRL) The IRL reduces the verification of programs with loops to a question of termination and the verification of loop- free programs by converting iteration to recursion. For while loops, the Lemma states: f = [while p do g] = [if p then g;f end_if] (note recursion)

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Iteration Recursion Lemma (cont’d) p g T F p g T F p g T F p g T F f = = = f p g;f T F =

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Iteration Recursion Lemma (cont’d) Rather than verify directly that f is the program function of K = while p do g which can be very difficult, it is sufficient to prove that 1. K terminates for all X D(f), and that 2. f is the program function of Q = if p then g;f end_if because [K] = [Q].

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An important implication of the IRL Suppose for “input” X 0 the while loop term- inates after n iterations with “output” X n. Furthermore, let X 1, X 2,..., X n-1 be the in- termediate states generated by the loop. Then 0≤i

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An important implication of the IRL (cont’d) As f = [while p do g] = [if p then g;f end_if], it follows that f(X 0 ) = f(X 1 ) =... = f(X n ) = X n More generally, after each iteration of the loop, the function value of the current state, X, must be the same as the function value of the initial state, X 0. That is: f(X) = f(X 0 ) We will revisit this observation in connection with Mill’s Invariant Status Theorem later.

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Illustrative Example of IRL To further illustrate the fact that [while p do g] = [if p then g;f end_if] consider a concrete example... Let K = while y>0 do x,y := x+1,y−1 Claim: K is function equivalent to Q = if y>0 then x,y := x+1,y−1;k end_if where, by definition, k = [K]. p g p k o g

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Illustrative Example of IRL (cont’d) Case (y>0): For K = while y>0 do x,y := x+1,y−1, the loop body executes y times before the predicate y>0 becomes false. By observation, then, the final value of x is x 0 +(1)y 0 = x 0 +y 0 and the final value of y is 0. Thus, (y>0) => k = (x,y := x+y,0) Also, note that when y=0 initially, k = I = (x,y := x,y) = (x,y := x+0,y) = (x,y := x+y,0) Therefore, (y≥0) => k = (x,y := x+y,0)

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Illustrative Example of IRL (cont’d) Case (y>0): (cont’d) [Q] is a composition of two functions, i.e., k o g, and may be determined by direct substitution. For y>0 initially, y will be greater than OR EQUAL to 0 after executing the loop body, but since we know (y≥0) => k = (x,y := x+y,0), we have [Q] = (x,y := x+y,0) o (x,y := x+1,y−1) = (x,y := (x+1)+(y−1),0) = (x,y := x+y,0) = k (the function computed by K) Thus, [Q] = [K] when y>0.

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Illustrative Example of IRL (cont’d) Case (y≤0): Since the predicate (y>0) fails, both K and Q do nothing, and are therefore equivalent. Thus, [Q] = I = [K] when y≤0. Therefore, K is function equivalent to Q.

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Termination Predicate The correctness of a looping program P depends, in part, on termination. Consideration is limited to programs whose termination can be established and the following predicate is defined: term(f,P) ‘‘P terminates for every initial state X D(f)’’

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Before we continue… Take out a piece of paper and a pen/pencil. Without looking back in the lecture notes, write down the complete correctness con- ditions for: f = [if p then g]

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if_then Correctness Conditions Complete correctness conditions for f = [if p then g]: Prove: p (f = g) Л ¬p (f = I) So, aside from proving termination over the domain of f, what are the two corresponding conditions for: f = [while p do g] = [if p then fog] ?

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while_do Correctness Conditions Complete correctness conditions for f = [K] = [while p do G] (where g = [G] has already been shown): Prove: term(f,K) Л p (f = f o g) Л ¬p (f = I)

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while_do Correctness Conditions (cont’d) Working correctness questions: –Is loop termination guaranteed for any argument of f ? –When p is true does f equal f composed with g? –When p is false does f equal Identity?

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while_do Example Prove f = [T] where, for integers x, y, and z: f = (y≥0 z,y := z+xy,0) and T is: while y<>0 do z := z+x y := y−1 end_while p G

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while_do Example (cont’d) Proof: g = [G] = (z,y := z+x,y−1) by observation –term(f,T)? f = (y≥0 z,y := z+xy,0) and T is: while y<>0 do z := z+x y := y−1 end_while So, does y≥0 initially T will terminate?

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while_do Example (cont’d) Proof: g = [G] = (z,y := z+x,y−1) by observation –term(f,T)? √ (Prove this…)

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while_do Example (cont’d) Proof: g = [G] = (z,y := z+x,y−1) by observation –term(f,T)? √ (Prove this…) –Does (y=0) ( f = I )? ¬p¬p ( Recall: f = (y≥0 z,y := z+xy,0) )

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while_do Example (cont’d) Proof: g = [G] = (z,y := z+x,y−1) by observation –term(f,T)? √ (Prove this…) –Does (y=0) ( f = I )? √ (y=0) ( f = (z,y := z+x(0),0) = (z,y := z,0) ) (y=0) ( I = (z,y := z,0) )

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? p

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? case a: Does (y<0) ( f = f o g )? (y<0) ( f = undefined ) (y<0) ( f o g = f o (z,y := z+x,y−1) What is f when applied after g decrements the initially negative value of y? ( Recall: f = (y≥0 z,y := z+xy,0) )

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? case a: Does (y<0) ( f = f o g )? (y<0) ( f = undefined ) (y<0) ( f o g = undefined o (z,y := z+x,y−1) since y<0 g y (y<0)<0 ( Recall: f = (y≥0 z,y := z+xy,0) )

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? case a: Does (y<0) ( f = f o g )? (y<0) ( f = undefined ) (y<0) ( f o g = undefined o (z,y := z+x,y−1) = undefined )

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? case a: Does (y<0) ( f = f o g )? √ (y<0) ( f = undefined ) (y<0) ( f o g = undefined o (z,y := z+x,y−1) = undefined )

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? case b: Does (y>0) ( f = f o g )? ( Recall: f = (y≥0 z,y := z+xy,0) )

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–Does (y 0) ( f = f o g )? case b: Does (y>0) ( f = f o g )? (y>0) ( f = (z,y := z+xy,0) ) (y>0) ( f o g = f o (z,y := z+x,y−1) Again, what is f when applied after g decrements the initially positive value of y? ( Recall: f = (y≥0 z,y := z+xy,0) ) while_do Example (cont’d)

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–Does (y 0) ( f = f o g )? case b: Does (y>0) ( f = f o g )? (y>0) ( f = (z,y := z+xy,0) ) (y>0) ( f o g = (z,y := z+xy,0) o (z,y := z+x,y−1) since y>0 g y (y>0)≥0 ( Recall: f = (y≥0 z,y := z+xy,0) )

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? case b: Does (y>0) ( f = f o g )? (y>0) ( f = (z,y := z+xy,0) ) (y>0) ( f o g = (z,y := z+xy,0) o (z,y := z+x,y−1) = (z,y := (z+x)+x(y−1),0) = (z,y := z+xy,0) )

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? case b: Does (y>0) ( f = f o g )? (y>0) ( f = (z,y := z+xy,0) ) (y>0) ( f o g = (z,y := z+xy,0) o (z,y := z+x,y−1) = (z,y := (z+x)+x(y−1),0) = (z,y := z+xy,0) ) We could have also composed the full, conditional definition of f with g, i.e. (y≥0 z,y := z+xy,0) o (z,y := z+x,y−1) to yield (y≥1 z,y := z+xy,0) which is just (z,y := z+xy,0) when y>0.

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while_do Example (cont’d) –Does (y 0) ( f = f o g )? √ case b: Does (y>0) ( f = f o g )? √ (y>0) ( f = (z,y := z+xy,0) ) (y>0) ( f o g = (z,y := z+xy,0) o (z,y := z+x,y−1) = (z,y := (z+x)+x(y−1),0) = (z,y := z+xy,0) ) Therefore, f = [T].

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Exercise 1 For program M below, where all variables are integers, hypothesize a function f for [M] and prove f = [M]. while i

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Sufficient Correctness Conditions Given the complete correctness conditions for f = [H] = [while p do g]: Prove: term(f,H) Л p (f = f o g) Л ¬p (f = I)

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Sufficient Correctness Conditions (cont’d) What are the sufficient correctness conditions for f [H] = [while p do g]? Prove: f’ = [H] for some f’ Л f f’

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Sufficient Correctness Conditions (cont’d) What are the sufficient correctness conditions for f [H] (for ANY program, H)? Prove: f’ = [H] for some f’ Л f f’

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repeat_until Statement What are the complete correctness conditions for f = [R] = [repeat g until p]? p g F T f =

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repeat_until Statement (cont’d) An IRL for repeat_until statements: f = [repeat g until p] = [g; if ¬p then f]

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“Proof” by Picture p g F T f = p g F T = p g T F = f p g T F = f ¬p¬p g F T

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repeat_until Statement (cont’d) Therefore, it is sufficient to verify that 1. R terminates for all X D(f), and that 2. f is the program function of Q = g; if ¬p then f end_if because [R] = [Q].

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repeat_until Correctness Conditions Complete correctness conditions for f = [R] = [repeat G until p] (where g = [G] has already been shown): Prove: term(f,R) Л (p o g) (f = g) Л ¬(p o g) (f = f o g)

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repeat_until Correctness Conditions (cont’d) Working correctness questions: –Is loop termination guaranteed for any argument of f ? –When p o g is true does f equal g? –When p o g is false does f equal f o g?

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Exercise 2 For program R below, where all variables are integers, hypothesize a function r for [R] and prove r = [R]. repeat: x := x−1 y := y+2 until x=0

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Subgoal Induction “Subgoal induction” is a proof method pro- posed by Morris and Wegbreit † that can be viewed as a generalization of (while loop) functional verification. It uses a variation of the Iteration Recursion Lemma (IRL) to identify relatively simple correctness conditions for a while loop surrounded by pre- and post-processing code. † Morris, James & Ben Wegbreit, “Subgoal Induction,” CACM, Volume 20, No. 4, April 1977.

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Subgoal Induction (cont’d) The key observation underlying the method is: v = [while p do g end_while; t] ≡ [if p then g;v else t end_if_else] The function equivalence of these programs, like that asserted in the IRL, is perhaps best illustrated graphically...

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Subgoal Induction (cont’d) p g T F p T F p T F p T F = = v p g;v T F = t t g t g t g t v =

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Subgoal Induction (cont’d) Suppose, now, that compound program K is: h; while p do g end_while; t and that v = [while p do g end_while; t]. From the functional equivalence illustrated above and the fact that K = h;v, it therefore follows that: [K] = v o h = [if p then g;v else t end_if_else] o h

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Subgoal Induction (cont’d) Recall the complete correctness conditions for r = [if p then g else t]: (1) p (r=g) and (2) ¬p (r=t). Thus, the complete correctness conditions for f = [K] = [h; while p do g end_while; t] are: (1) term(f,K), (2) p (v=v o g), (3) ¬p (v=t), and (4) f=v o h where v = [while p do g end_while; t].

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Subgoal induction vs. functional verification How does subgoal induction differ from the program decomposition strategy employed in functional verification? To show f = [h; while p do g end_while; t] using functional verification, an intermediate hypothesis and “sub-proof” for the loop is required, whereas t is part of the intermediate hypothesis in the subgoal induction case. Note that if t is the identify function, the two strategies are identical.

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Subgoal induction vs. functional verification (cont’d) But, if h is the identify function, then subgoal induction has an advantage since intended function f (if given) can then be used as the intermediate hypothesis. (In this case, treating the loop and t as a whole results in a more efficient proof.)

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Subgoal Induction Example Use subgoal induction to prove f = [K] where, for integers x, y, and z: f = (x≥0 x,y,z := 0,2 x,2 x ) and K is: y := 1 H while x<>0 do y := y2 x := x-1 end_while z := y T G

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Subgoal Induction Example (cont’d) We need to show: (1) term(f,K), (2) p (v=v o g), (3) ¬p (v=t), and (4) f=v o h But first, we must hypothesize a function for v (our “intermediate hypothesis”): v = [while x<>0 do g end_while; z := y]

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Subgoal Induction Example (cont’d) What is the function, v, of this program? while x<>0 do y := y2 x := x-1 end_while z := y

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Subgoal Induction Example (cont’d) What is the function, v, of this program? while x<>0 do y := y2 x := x-1 end_while z := y x>0 x,y,z := 0, y2 x, y2 x x=0 x,y,z := x, y, y := 0, y2 x, y2 x x<0 undefined Therefore, v is hypothesized to be: (x≥0 x,y,z := 0, y2 x, y2 x )

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh (1) Does K terminate for all x≥0? YES y := 1 while x<>0 do y := y2 x := x-1 end_while z := y (Prove this using the Method of Well-Founded Sets.)

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh (2) Does (x 0) ( v = v o g )? p

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh (2) Does (x 0) ( v = v o g )? case a: Does (x<0) ( v = v o g )? YES (x<0) ( v = undefined ) (x<0) ( v o g = undefined o (x,y := x-1,2y) = undefined ) ( Recall: hypoth. v = (x≥0 x,y,z := 0, y2 x, y2 x ) )

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh (2) Does (x 0) ( v = v o g )? YES case b: Does (x>0) ( v = v o g )? YES (x>0) ( v = (x,y,z := 0, y2 x, y2 x ) ) (x>0) ( v o g = (x,y,z := 0, y2 x, y2 x ) o (x,y := x-1,2y) = (x,y,z := 0, 2y2 x-1, 2y2 x-1 ) = (x,y,z := 0, y2 x, y2 x ) ) ( Recall: hypoth. v = (x≥0 x,y,z := 0, y2 x, y2 x ) )

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh (3) Does (x=0) ( v = t )? YES (x=0) ( v = (x,y,z := 0, y2 0, y2 0 ) ) = (x,y,z := 0, y, y) ) (x=0) ( t = (x,y,z := 0, y, y) ) ( Recall: hypoth. v = (x≥0 x,y,z := 0, y2 x, y2 x ) )

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh (4) Does f = v o h ?

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh (4) Does f = v o h ? YES f = (x≥0 x,y,z := 0,2 x,2 x ) voh = (x≥0 x,y,z := 0, y2 x, y2 x ) o (x,y,z := x,1,z) = (x≥0 x,y,z := 0, (1)2 x, (1)2 x ) = (x≥0 x,y,z := 0, 2 x, 2 x ) ( Recall: hypoth. v = (x≥0 x,y,z := 0, y2 x, y2 x ) )

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Subgoal Induction Example (cont’d) Returning to the four correctness conditions: (1) term(f,K), (2) p (v=vog), (3) ¬p (v=t), and (4) f=voh Therefore, for f = (x≥0 x,y,z := 0,2 x,2 x ) and K: y := 1 while x<>0 do y := y2 x := x-1 end_while z := y we conclude, by subgoal induction, that f = [K].

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Summary Iteration Recursion Lemma (IRL) Termination predicate: term(f,P) Correctness conditions for while_do statement Sufficient correctness conditions Correctness conditions for repeat_until statement Subgoal Induction

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Coming up next… Thinking about invariants again Invariant Status Theorem (IST) While Loop Initialization Utility of IST

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Functional Verification III Prepared by Stephen M. Thebaut, Ph.D. University of Florida Software Testing and Verification Lecture Notes 23

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