1. Predicate Transforms II
Software Testing and Verification
Lecture Notes 20
Prepared by
Stephen M. Thebaut, Ph.D.
University of Florida

2. Predicate Transforms II
Transform rules for while loops:
Weakest pre-conditions (wp’s)
Weakest liberal pre-conditions (wlp’s)
Relationships between wp’s and wlp’s with loop invariants
Strongest post-conditions (sp’s)
On the power of axiomatic verification and the relative usefulness of predicate transforms (when dealing with loops)

3. Predicate Transforms II
Transform rules for while loops:
Weakest pre-conditions (wp’s)
Weakest liberal pre-conditions (wlp’s)
Relationships between wp’s and wlp’s with loop invariants
Strongest post-conditions (sp’s)
On the power of axiomatic verification and the relative usefulness of predicate transforms (when dealing with loops)

4. wp Rule for while_do Statement
In order for the program while b do S to terminate in state Q, it is necessary that:

5. wp Rule for while_do Statement
In order for the program while b do S to terminate in state Q, it is necessary that:
0. b is initially false and Q holds, OR

6. wp Rule for while_do Statement
In order for the program while b do S to terminate in state Q, it is necessary that:
0. b is initially false and Q holds, OR
1. b is initially true and after executing S, ¬b and Q hold, OR

7. wp Rule for while_do Statement
In order for the program while b do S to terminate in state Q, it is necessary that:
0. b is initially false and Q holds, OR
1. b is initially true and after executing S, ¬b and Q hold, OR
2. b is initially true and after executing S, b is still true, and after executing S a second time, ¬b and Q hold, OR

8. wp Rule for while_do Statement
In order for the program while b do S to terminate in state Q, it is necessary that:
0. b is initially false and Q holds, OR
1. b is initially true and after executing S, ¬b and Q hold, OR
2. b is initially true and after executing S, b is still true, and after executing S a second time, ¬b and Q hold, OR .

9. wp Rule for while_do Statement (cont’d)
Thus, we can write
wp(while b do S, Q)  H0 V H1 V H2 V…
where
H0  ¬b Л Q
H1  b Л wp(S, ¬b Л Q)
H2  b Л wp(S, b Л wp(S, ¬b Л Q)) .

10. wp Rule for while_do Statement (cont’d)
Thus, we can write
wp(while b do S, Q)  H0 V H1 V H2 V…
where
H0  ¬b Л Q
H1  b Л wp(S, ¬b Л Q)
H2  b Л wp(S, b Л wp(S, ¬b Л Q)) .

11. wp Rule for while_do Statement (cont’d)
Thus, we can write
wp(while b do S, Q)  H0 V H1 V H2 V…
where
H0  ¬b Л Q
H1  b Л wp(S, ¬b Л Q)
H2  b Л wp(S, b Л wp(S, ¬b Л Q)) .

12. wp Rule for while_do Statement (cont’d)
Equivalently, we can write
wp(while b do S, Q)  H0 V H1 V H2 V…
where
H0  ¬b Л Q
H1  b Л wp(S, H0)
H2  b Л wp(S, H1)
Hi  b Л wp(S, Hi-1) … …

13. wp Rule for while_do Statement (cont’d)
Equivalently, we can write
wp(while b do S, Q)  H0 V H1 V H2 V…
where
H0  ¬b Л Q
H1  b Л wp(S, H0)
H2  b Л wp(S, H1)
Hi  b Л wp(S, Hi-1) … …

14. wp Rule for while_do Statement (cont’d)
Equivalently, we can write
wp(while b do S, Q)  H0 V H1 V H2 V…
where
H0  ¬b Л Q
H1  b Л wp(S, H0)
H2  b Л wp(S, H1)
Hi  b Л wp(S, Hi-1) … …

15. Something to think about…
How do these terms compare to the (infinite) set of necessary conditions derived for the while_do ROI?

16. FLASHBACK to Lecture Notes #18…
Something to think about… (cont'd)
FLASHBACK to Lecture Notes #18…
So, we know that {P} while b do S {Q} will hold if the following conditions hold:
Case 0: (P Л b)  Q
Case 1: {P Л b} S {K1}, (K1 Л b)  Q
Case 2: {K1 Л b} S {K2}, (K2 Л b)  Q …
Case N: {KN-1 Л b} S {KN}, (KN Л b)  Q …

17. Something to think about… (cont'd)
What is the relationship between
wp(while b do S, Q)
and an invariant, I, for which initialization, preservation, and finalization hold?

18. Something to think about… (cont'd)
What is the relationship between
wp(while b do S, Q)
and an invariant, I, for which initialization, preservation, and finalization hold?
We'll come back to this question later...

19. Example
For what initial values of i, n, and t will the following program terminate with t=xn?
while i <= n do
t := t*x
i := i+1
end_while
How about i=1, t=1, and n=2? Can you think of any others? For example...
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?

20. Example (cont’d)
Find the wp of this program with respect to the post-condition {t=xn}. (Attempt to find a regularity in terms that allows a closed-form expression.)

21. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x = i := i+1
H1  b Л wp(S, H0)
H2  b Л wp(S, H1)
while i <= n do
t := t*x
i := i+1
end_while

22. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0) =
H2  b Л wp(S, H1)
while i <= n do
t := t*x
i := i+1
end_while

23. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn) =
H2  b Л wp(S, H1)
while i <= n do
t := t*x
i := i+1
end_while

24. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn) =
H2  b Л wp(S, H1)
while i <= n do
t := t*x
i := i+1
end_while

25. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn)
= i≤n Л i+1>n Л tx=xn =
H2  b Л wp(S, H1)
while i <= n do
t := t*x
i := i+1
end_while

26. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn)
= i≤n Л i+1>n Л tx=xn
= i=n Л t=xn-1
H2  b Л wp(S, H1) =
while i <= n do
t := t*x
i := i+1
end_while

27. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn)
= i≤n Л i+1>n Л tx=xn
= i=n Л t=xn-1
H2  b Л wp(S, H1)
= i≤n Л wp(S, i=n Л t=xn-1) =
while i <= n do
t := t*x
i := i+1
end_while

28. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn)
= i≤n Л i+1>n Л tx=xn
= i=n Л t=xn-1
H2  b Л wp(S, H1)
= i≤n Л wp(S, i=n Л t=xn-1) =
while i <= n do
t := t*x
i := i+1
end_while

29. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn)
= i≤n Л i+1>n Л tx=xn
= i=n Л t=xn-1
H2  b Л wp(S, H1)
= i≤n Л wp(S, i=n Л t=xn-1)
= i≤n Л i+1=n Л tx=xn-1 =
while i <= n do
t := t*x
i := i+1
end_while

30. Example (cont’d) while i <= n do H0  ¬b Л Q t := t*x
= i>n Л t=xn
H1  b Л wp(S, H0)
= i≤n Л wp(S, i>n Л t=xn)
= i≤n Л i+1>n Л tx=xn
= i=n Л t=xn-1
H2  b Л wp(S, H1)
= i≤n Л wp(S, i=n Л t=xn-1)
= i≤n Л i+1=n Л tx=xn-1
= i=n-1 Л t=xn-2
while i <= n do
t := t*x
i := i+1
end_while

31. Example (cont’d) H3  b Л wp(S, H2) while i <= n do = t := t*x.
Hk  b Л wp(S, Hk-1)
while i <= n do
t := t*x
i := i+1
end_while

32. Example (cont’d) H3  b Л wp(S, H2) while i <= n do
= i≤n Л wp(S, i=n-1 Л t=xn-2) = .
Hk  b Л wp(S, Hk-1)
while i <= n do
t := t*x
i := i+1
end_while

33. Example (cont’d) H3  b Л wp(S, H2) while i <= n do
= i≤n Л wp(S, i=n-1 Л t=xn-2) = .
Hk  b Л wp(S, Hk-1)
while i <= n do
t := t*x
i := i+1
end_while

34. Example (cont’d) H3  b Л wp(S, H2) while i <= n do
= i≤n Л wp(S, i=n-1 Л t=xn-2)
= i≤n Л i+1=n-1 Л tx=xn-2) = .
Hk  b Л wp(S, Hk-1)
while i <= n do
t := t*x
i := i+1
end_while

35. Example (cont’d) H3  b Л wp(S, H2) while i <= n do
= i≤n Л wp(S, i=n-1 Л t=xn-2)
= i≤n Л i+1=n-1 Л tx=xn-2)
= i=n-2 Л t=xn-3 .
Hk  b Л wp(S, Hk-1) =
while i <= n do
t := t*x
i := i+1
end_while

36. Example (cont’d) H3  b Л wp(S, H2) while i <= n do
= i≤n Л wp(S, i=n-1 Л t=xn-2)
= i≤n Л i+1=n-1 Л tx=xn-2)
= i=n-2 Л t=xn-3 .
Hk  b Л wp(S, Hk-1)
= i=n-(k-1) Л t=xn-k =
while i <= n do
t := t*x
i := i+1
end_while

37. Example (cont’d) H3  b Л wp(S, H2) while i <= n do
= i≤n Л wp(S, i=n-1 Л t=xn-2)
= i≤n Л i+1=n-1 Л tx=xn-2)
= i=n-2 Л t=xn-3 .
Hk  b Л wp(S, Hk-1)
= i=n-(k-1) Л t=xn-k
= i=n-k+1 Л t=xn-k
while i <= n do
t := t*x
i := i+1
end_while

38. Example (cont’d) Thus, we have: H0 = i>n Л t=xn
Hk = i=n-k+1 Л t=xn-k (for all k>0)

39. Example (cont’d) Thus, we have: H0 = i>n Л t=xn
Hk = i=n-k+1 Л t=xn-k (for all k>0)

40. Example (cont’d) Thus, we have: H0 = i>n Л t=xn
Hk = i=n-k+1 Л t=xn-k (for all k>0)
and since i=n-k+1  n-k=i-1

41. Example (cont’d) Thus, we have: H0 = i>n Л t=xn
Hk = i=n-k+1 Л t=xn-k (for all k>0)
and since i=n-k+1  n-k=i-1
= i≤n Л t=xi-1 (where i≤n for all k>0)

42. Example (cont’d) Thus, we have: H0 = i>n Л t=xn
Hk = i=n-k+1 Л t=xn-k (for all k>0)
and since i=n-k+1  n-k=i-1
= i≤n Л t=xi-1 (where i≤n for all k>0)
Therefore, wp  H0 V H1 V H2 V ...
= (i>n Л t=xn) V (i≤n Л t=xi-1)

43. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?

44. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?

45. Example (cont’d) So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?
(1>(1,2,…) Л 1=x(1,2,…)) V (1≤(1,2,…) Л 1=x1-1)

46. Example (cont’d) So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?
(1>(1,2,…) Л 1=x(1,2,…)) V (1≤(1,2,…) Л 1=x1-1) √

47. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}? √

48. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}? √

49. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?
(3>1 Л x=x1) V (3≤1 Л x=x3-1) √

50. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?
(3>1 Л x=x1) V (3≤1 Л x=x3-1) √ √

51. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}? √ √

52. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}? √ √

53. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?
(2>5 Л x=x5) V (2≤5 Л x=x2-1) √ √

54. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}?
(2>5 Л x=x5) V (2≤5 Л x=x2-1) √ √ √

55. (i>n Л t=xn) V (i≤n Л t=xi-1)
Example (cont’d)
So, given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
which of the following initial states will result in the program terminating with t=xn?
{i=1 Л t=1 Л n≥1}?
{i=3 Л t=x Л n=1}?
{i=2 Л t=x Л n=5}? √ √ √

56. Addendum (based on a question raised in class)
Another example…given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
will the following initial state values result in the program terminating with t=xn?
{i=1 Л t=1 Л n=0}

57. Addendum (based on a question raised in class)
Another example…given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
will the following initial state values result in the program terminating with t=xn?
{i=1 Л t=1 Л n=0}
(1>0 Л 1=x0) V (1≤0 Л 1=x1-1)

58. Addendum (based on a question raised in class)
Another example…given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
will the following initial state values result in the program terminating with t=xn?
{i=1 Л t=1 Л n=0}
(1>0 Л 1=x0) V (1≤0 Л 1=x1-1) √

59. Addendum (based on a question raised in class)
Another example…given that the wp is
(i>n Л t=xn) V (i≤n Л t=xi-1)
will the following initial state values result in the program terminating with t=xn?
{i=1 Л t=1 Л n=0}
(1>0 Л 1=x0) V (1≤0 Л 1=x1-1) √ √

60. Predicate Transforms II
Transform rules for while loops:
Weakest pre-conditions (wp’s)
Weakest liberal pre-conditions (wlp’s)
Relationships between wp’s and wlp’s with loop invariants
Strongest post-conditions (sp’s)
On the power of axiomatic verification and the relative usefulness of predicate transforms (when dealing with loops)

61. wlp Rule for while_do Statement
In order for the program while b do S to either terminate in state Q, or not term- inate at all, it is necessary that:
Q will hold on program termination, OR
the program will not terminate.
Therefore, wlp(while b do S, Q) ≡
wp(while b do S, Q) V ¬wp(while b do S, true)

62. wlp Rule for while_do Statement
In order for the program while b do S to either terminate in state Q, or not term- inate at all, it is necessary that:
Q will hold on program termination, OR
the program will not terminate.
Therefore, wlp(while b do S, Q) ≡
wp(while b do S, Q) V ¬wp(while b do S, true)
(Note: wp(M, true) is the weakest pre-condition ensuring termination of program M.)

63. Example
Use the wlp rule for while_do statements to determine the weakest liberal pre-condition for the following program with respect to post-condition t=x5.
while i<>3 do
t := t*x
i := i+1
end_while

64. Step 1: determine wp with respect to Q
H0  ¬b Л Q =
H1  b Л wp(S, H0) .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

65. Step 1: determine wp with respect to Q
H0  ¬b Л Q
= i=3 Л t=x5
H1  b Л wp(S, H0) = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

66. Step 1: determine wp with respect to Q
H0  ¬b Л Q
= i=3 Л t=x5
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3 Л t=x5) = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

67. Step 1: determine wp with respect to Q
H0  ¬b Л Q
= i=3 Л t=x5
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3 Л t=x5) = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

68. Step 1: determine wp with respect to Q
H0  ¬b Л Q
= i=3 Л t=x5
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3 Л t=x5)
= i≠3 Л i+1=3 Л tx=x5 = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

69. Step 1: determine wp with respect to Q
H0  ¬b Л Q
= i=3 Л t=x5
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3 Л t=x5)
= i≠3 Л i+1=3 Л tx=x5
= i=2 Л t=x4 .
Hk  b Л wp(S, Hk-1) =
while i<>3 do
t := t*x
i := i+1
end_while

70. Step 1: determine wp with respect to Q
H0  ¬b Л Q
= i=3 Л t=x5
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3 Л t=x5)
= i≠3 Л i+1=3 Л tx=x5
= i=2 Л t=x4 .
Hk  b Л wp(S, Hk-1)
= i=3-k Л t=x5-k
while i<>3 do
t := t*x
i := i+1
end_while

71. Step 1: determine wp with respect to Q
Thus, we have:
H0 = i=3 Л t=x5
Hk = i=3-k Л t=x5-k (for all k>0)

72. Step 1: determine wp with respect to Q
Thus, we have:
H0 = i=3 Л t=x5
Hk = i=3-k Л t=x5-k (for all k>0)

73. Step 1: determine wp with respect to Q
Thus, we have:
H0 = i=3 Л t=x5
Hk = i=3-k Л t=x5-k (for all k>0)
and since i=3-k => 5-k=i+2
= i<3 Л t=xi+2 (where i<3 for all k>0)

74. Step 1: determine wp with respect to Q
Thus, we have:
H0 = i=3 Л t=x5
Hk = i=3-k Л t=x5-k (for all k>0)
and since i=3-k => 5-k=i+2
= i<3 Л t=xi+2 (where i<3 for all k>0)
Therefore, the wp w.r.t. Q, H0 V H1 V H2 V... is:

75. Step 1: determine wp with respect to Q
Thus, we have:
H0 = i=3 Л t=x5
Hk = i=3-k Л t=x5-k (for all k>0)
and since i=3-k => 5-k=i+2
= i<3 Л t=xi+2 (where i<3 for all k>0)
Therefore, the wp w.r.t. Q, H0 V H1 V H2 V... is:
i≤3 Л t=xi+2

76. Step 2: determine wp with respect to true
H0  ¬b Л true =
H1  b Л wp(S, H0) .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

77. Step 2: determine wp with respect to true
H0  ¬b Л true
= i=3
H1  b Л wp(S, H0) = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

78. Step 2: determine wp with respect to true
H0  ¬b Л true
= i=3
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3) = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

79. Step 2: determine wp with respect to true
H0  ¬b Л true
= i=3
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3) = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

80. Step 2: determine wp with respect to true
H0  ¬b Л true
= i=3
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3)
= i≠3 Л i+1=3 = .
Hk  b Л wp(S, Hk-1)
while i<>3 do
t := t*x
i := i+1
end_while

81. Step 2: determine wp with respect to true
H0  ¬b Л true
= i=3
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3)
= i≠3 Л i+1=3
= i=2 .
Hk  b Л wp(S, Hk-1) =
while i<>3 do
t := t*x
i := i+1
end_while

82. Step 2: determine wp with respect to true
H0  ¬b Л true
= i=3
H1  b Л wp(S, H0)
= i≠3 Л wp(S, i=3)
= i≠3 Л i+1=3
= i=2 .
Hk  b Л wp(S, Hk-1)
= i=3-k
while i<>3 do
t := t*x
i := i+1
end_while

83. Step 2: determine wp with respect to true
Thus, we have:
H0 = i=3
Hk = i=3-k (for all k>0)

84. Step 2: determine wp with respect to true
Thus, we have:
H0 = i=3
Hk = i=3-k (for all k>0)
= i<3

85. Step 2: determine wp with respect to true
Thus, we have:
H0 = i=3
Hk = i=3-k (for all k>0)
= i<3
Therefore, the wp w.r.t. true, H0 V H1 V H2 V... is:
i≤3

86. Step 3: combine wp’s into one disjunct
Thus, wlp(while i<>3 do t := t*x; i := i+1, t=x5) =
(i≤3 Л t=xi+2) V i>3
Exercise: In light of this, for which of the following
initial states is the program weakly correct with re-
spect to t=x5?
{i=1 Л t=1 Л x=1}?
{i=2 Л t=x Л x=2}?
{i=5 Л t=8 Л x=3}?

87. Step 3: combine wp’s into one disjunct
Thus, wlp(while i<>3 do t := t*x; i := i+1, t=x5) =
(i≤3 Л t=xi+2) V i>3
Exercise: In light of this, for which of the following
initial states is the program weakly correct with re-
spect to t=x5?
{i=1 Л t=1 Л x=1}?
{i=2 Л t=x Л x=2}?
{i=5 Л t=8 Л x=3}?

88. Predicate Transforms II
Transform rules for while loops:
Weakest pre-conditions (wp’s)
Weakest liberal pre-conditions (wlp’s)
Relationships between wp’s and wlp’s with loop invariants
Strongest post-conditions (sp’s)
On the power of axiomatic verification and the relative usefulness of predicate transforms (when dealing with loops)

89. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? P  wlp ?
For while loops, does {w(l)p Л b} S {w(l)p} ?
Does (w(l)p Л ¬b)  Q ?

90. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? P  wlp ?
For while loops, does {w(l)p Л b} S {w(l)p} ?
Does (w(l)p Л ¬b)  Q ?

91. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? yes P  wlp ? no
For while loops, does {w(l)p Л b} S {w(l)p} ?
Does (w(l)p Л ¬b)  Q ?

92. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? yes P  wlp ? no
For while loops, does {w(l)p Л b} S {w(l)p} ?
Does (w(l)p Л ¬b)  Q ?

93. {wp Л b} S
{wp} ???

94. {wp Л b} S

95. {wp Л b}
= {[H0 V H1 V …] Л b} S

96. {wp Л b}
= {[H0 V H1 V …] Л b}
= {[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л b} S

97. {wp Л b}
= {[H0 V H1 V …] Л b}
= {[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л b} S

98. {wp Л b}
= {[H0 V H1 V …] Л b}
= {[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л b}
= {(b Л wp(S, H0)) V (b Л wp(S, H1)) V …} S

99. {wp Л b}
= {[H0 V H1 V …] Л b}
= {[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л b}
= {(b Л wp(S, H0)) V (b Л wp(S, H1)) V …}
= {H1 V H2 V …} S

100. {wp Л b}
= {[H0 V H1 V …] Л b}
= {[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л b}
= {(b Л wp(S, H0)) V (b Л wp(S, H1)) V …}
= {H1 V H2 V …} S
{H0 V H1 V …}

101. {wp Л b}
= {[H0 V H1 V …] Л b}
= {[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л b}
= {(b Л wp(S, H0)) V (b Л wp(S, H1)) V …}
= {H1 V H2 V …} S
{H0 V H1 V …} = {wp}

102. Similarly, it can be shown that {wlp Л b} S {wlp}.
{wp Л b}
= {[H0 V H1 V …] Л b}
= {[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л b}
= {(b Л wp(S, H0)) V (b Л wp(S, H1)) V …}
= {H1 V H2 V …} S
{H0 V H1 V …} = {wp}
Similarly, it can be shown that {wlp Л b} S {wlp}.

103. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? yes P  wlp ? no
For while loops, does {w(l)p Л b} S {w(l)p} ? yes
Does (w(l)p Л ¬b)  Q ?

104. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? yes P  wlp ? no
For while loops, does {w(l)p Л b} S {w(l)p} ? yes
Does (w(l)p Л ¬b)  Q ?

105. {wp Л ¬b}

106. {wp Л ¬b}
{[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л ¬b}

107. {wp Л ¬b}
{[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л ¬b}
{(¬b Л Q)}  Q

108. {wp Л ¬b}
{[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л ¬b}
{(¬b Л Q)}  Q

109. Similarly, it is easy to show that {wlp Л ¬b}  Q.
{wp Л ¬b}
{[(¬b Л Q) V (b Л wp(S, H0)) V (b Л wp(S, H1)) V …] Л ¬b}
{(¬b Л Q)}  Q
Similarly, it is easy to show that {wlp Л ¬b}  Q.

110. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? yes P  wlp ? no
For while loops, does {w(l)p Л b} S {w(l)p} ? yes
Does (w(l)p Л ¬b)  Q ? yes

111. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? yes P  wlp ? no
For while loops, does {w(l)p Л b} S {w(l)p} ? yes
Does (w(l)p Л ¬b)  Q ? yes
_________________________
wp  weakest while loop invariant which guarantees termination!

112. Loop Invariants and w(l)p’s
In general, are loops guaranteed to terminate when: P  wp ? yes P  wlp ? no
For while loops, does {w(l)p Л b} S {w(l)p} ? yes
Does (w(l)p Л ¬b)  Q ? yes
_________________________
wp  weakest while loop invariant which guarantees termination!
wlp  weakest while loop invariant which does NOT guarantee termination!

113. Predicate Transforms II
Transform rules for while loops:
Weakest pre-conditions (wp’s)
Weakest liberal pre-conditions (wlp’s)
Relationships between wp’s and wlp’s with loop invariants
Strongest post-conditions (sp’s)
On the power of axiomatic verification and the relative usefulness of predicate transforms (when dealing with loops)

114. sp Rule for while_do Statement
What is the strongest condition on the final state of program while b do S given that P holds initially? (Note that the post-condition is undefined when the program does not terminate.)
Recall our derivation of the while loop Rule of Inference from Lecture Notes #18 (Axiomatic Verification II).
(flashback follows...)

115. Necessary Conditions: while_do
So, we know that {P} while b do S {Q} will hold if the following conditions hold:
Case 0: (P Л b)  Q
Case 1: {P Л b} S {K1}, (K1 Л b)  Q
Case 2: {K1 Л b} S {K2}, (K2 Л b)  Q …
Case N: {KN-1 Л b} S {KN}, (KN Л b)  Q …
Great! But who has the time to show that an infinite number of conditions hold?

116. sp Rule for while_do Statement
In order to eliminate the infinite sequence of necessary conditions, we replaced each Ki with I (a loop invariant.)
But for i≥1, Ki is just the strongest post- condition of S with respect to (Ki-1 Л b), where K0 = P.

117. sp Rule for while_do Statement
Thus, if the loop terminates,
sp(while b do S, P) = ¬b Л (K0 V K1 V K2 V ...)
where
K0  P
K1  sp(S, b Л P)
K2  sp(S, b Л sp(S, b Л P))
K3  sp(S, b Л sp(S, b Л sp(S, b Л P))) .

118. sp Rule for while_do Statement
Thus, if the loop terminates,
sp(while b do S, P) = ¬b Л (K0 V K1 V K2 V ...)
where
K0  P
K1  sp(S, b Л P)
K2  sp(S, b Л sp(S, b Л P))
K3  sp(S, b Л sp(S, b Л sp(S, b Л P))) .

119. sp Rule for while_do Statement
Thus, if the loop terminates,
sp(while b do S, P) = ¬b Л (K0 V K1 V K2 V ...)
where
K0  P
K1  sp(S, b Л P)
K2  sp(S, b Л sp(S, b Л P))
K3  sp(S, b Л sp(S, b Л sp(S, b Л P))) .

120. sp Rule for while_do Statement
Thus, if the loop terminates,
sp(while b do S, P) = ¬b Л (K0 V K1 V K2 V ...)
where
K0  P
K1  sp(S, b Л P)
K2  sp(S, b Л sp(S, b Л P))
K3  sp(S, b Л sp(S, b Л sp(S, b Л P))) .

121. sp Rule for while_do Statement
Equivalently, we can write: on termination,
sp(while b do S, P) = ¬b Л (K0 V K1 V K2 V ...)
where
K0  P
K1  sp(S, b Л K0)
K2  sp(S, b Л K1)
KN  sp(S, b Л KN-1) … …

122. sp Rule for while_do Statement
Equivalently, we can write: on termination,
sp(while b do S, P) = ¬b Л (K0 V K1 V K2 V ...)
where
K0  P
K1  sp(S, b Л K0)
K2  sp(S, b Л K1)
KN  sp(S, b Л KN-1) … …

123. sp Rule for while_do Statement
Equivalently, we can write: on termination,
sp(while b do S, P) = ¬b Л (K0 V K1 V K2 V ...)
where
K0  P
K1  sp(S, b Л K0)
K2  sp(S, b Л K1)
KN  sp(S, b Л KN-1) … …

124. Example Use the Strongest Post-condition ROI to prove: {true} Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

125. Example Use the Strongest Post-condition ROI to prove: T {true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}
We need to show:
sp(T, Z=X Л J=1)  Z=XY
where T is: while J<>Y do
Z := Z+X
J := J+1
end_while
if T terminates. T

126. Example (cont’d) K0  P = {true} K1  sp(S, b Л K0) Z := X J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

127. Example (cont’d) K0  P = Z=X Л J=1 {true} K1  sp(S, b Л K0) Z := X =
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

128. Example (cont’d) K0  P = Z=X Л J=1 {true} K1  sp(S, b Л K0) Z := X =
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

129. Example (cont’d) K0  P = Z=X Л J=1 {true} K1  sp(S, b Л K0) Z := X
= Z=Z’+X Л J=J’+1 Л J’≠Y Л
Z’=X Л J’=1 =
K2  sp(S, b Л K1)
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

130. Example (cont’d) K0  P = Z=X Л J=1 {true} K1  sp(S, b Л K0) Z := X
= Z=Z’+X Л J=J’+1 Л J’≠Y Л
Z’=X Л J’=1
= Z=2X Л J=2 Л Y≠1
K2  sp(S, b Л K1) =
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

131. Example (cont’d) K0  P = Z=X Л J=1 {true} K1  sp(S, b Л K0) Z := X
= Z=Z’+X Л J=J’+1 Л J’≠Y Л
Z’=X Л J’=1
= Z=2X Л J=2 Л Y≠1
K2  sp(S, b Л K1) =
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

132. Example (cont’d) K0  P = Z=X Л J=1 {true} K1  sp(S, b Л K0) Z := X
= Z=Z’+X Л J=J’+1 Л J’≠Y Л
Z’=X Л J’=1
= Z=2X Л J=2 Л Y≠1
K2  sp(S, b Л K1)
Z’=2X Л J’=2 Л Y≠1 =
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

133. Example (cont’d) K0  P = Z=X Л J=1 {true} K1  sp(S, b Л K0) Z := X
= Z=Z’+X Л J=J’+1 Л J’≠Y Л
Z’=X Л J’=1
= Z=2X Л J=2 Л Y≠1
K2  sp(S, b Л K1)
Z’=2X Л J’=2 Л Y≠1
= Z=3X Л J=3 Л Y≠1 Л Y≠2
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

134. Example (cont’d) . K3  sp(S, b Л K2) = {true} Z := X J := 1
KN  sp(S, b Л KN-1)
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

135. Example (cont’d) . K3  sp(S, b Л K2) = Z=4X Л J=4 Л Y≠1 Л {true}
Y≠2 Л Y≠3 .
KN  sp(S, b Л KN-1) =
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

136. Example (cont’d) . K3  sp(S, b Л K2) = Z=4X Л J=4 Л Y≠1 Л {true}
Y≠2 Л Y≠3 .
KN  sp(S, b Л KN-1)
= Z=(N+1)X Л J=N+1 Л
Y≠1 Л Y≠2 Л ... Л Y≠N .
{true}
Z := X
J := 1
while J<>Y do
Z := Z+X
J := J+1
end_while
{Z=XY}

137. Example (cont’d) Thus, when T terminates (i.e., when Y≥1),
sp(T, Z=X Л J=1) =
J=Y Л [(Z=X Л J=1) V (Z=2X Л J=2 Л Y≠1) V
(Z=3X Л J=3 Л Y≠1 Л Y≠2) V ...]

138. Example (cont’d) Thus, when T terminates (i.e., when Y≥1),
sp(T, Z=X Л J=1) =
J=Y Л [(Z=X Л J=1) V (Z=2X Л J=2 Л Y≠1) V
(Z=3X Л J=3 Л Y≠1 Л Y≠2) V ...]
=> [(Z=XY Л Y=1) V (Z=XY Л Y=2) V ...]

139. Example (cont’d) Thus, when T terminates (i.e., when Y≥1),
sp(T, Z=X Л J=1) =
J=Y Л [(Z=X Л J=1) V (Z=2X Л J=2 Л Y≠1) V
(Z=3X Л J=3 Л Y≠1 Л Y≠2) V ...]
=> [(Z=XY Л Y=1) V (Z=XY Л Y=2) V ...]
=> (Z=XY Л Y≥1)
=> Q (i.e., Z=XY)

140. sp(T, Z=X Л J=1) is undefined
Example (cont’d)
When T does NOT terminate (i.e., when Y<1),
sp(T, Z=X Л J=1) is undefined

141. sp(T, Z=X Л J=1) is undefined
Example (cont’d)
When T does NOT terminate (i.e., when Y<1),
sp(T, Z=X Л J=1) is undefined
Therefore, by the Strongest Post-Condition ROI, the
assertion of weak correctness holds.

142. Predicate Transforms II
Transform rules for while loops:
Weakest pre-conditions (wp’s)
Weakest liberal pre-conditions (wlp’s)
Relationships between wp’s and wlp’s with loop invariants
Strongest post-conditions (sp’s)
On the power of axiomatic verification and the relative usefulness of predicate transforms (when dealing with loops)

143. On the power of axiomatic verification and the relative usefulness of predicate transforms
Hoare Logic is a deductive system that is both sound and relatively complete (i.e., complete to the extent that we can decide the validity of assertions in ROI’s) for deriving proofs of Hoare triples.
Predicate transforms operationalize this system by providing a way to produce valid correctness specifications.
Weakest pre-conditions (wp’s) are typically easier to use in this respect than either wlp’s or sp’s when dealing with loops.

144. Problem Set 6: Predicate Transforms
Note especially Problem 6: deriving and using the weakest pre-condition for the repeat_until construct.

145. Predicate Transforms II
Software Testing and Verification
Lecture Notes 20
Prepared by
Stephen M. Thebaut, Ph.D.
University of Florida