1. Predicate Transforms I
Software Testing and Verification
Lecture Notes 19
Prepared by
Stephen M. Thebaut, Ph.D.
University of Florida

2. Predicate Transforms I
Introduction:
weakest pre-conditions (wp’s)
weakest liberal pre-conditions (wlp’s)
strongest post-conditions (sp’s)
Proving {P} S {Q} using predicate transforms
Transform rules for:
assignment statements
sequencing
selection statements

3. Predicate Transforms I
Introduction:
weakest pre-conditions (wp’s)
weakest liberal pre-conditions (wlp’s)
strongest post-conditions (sp’s)
Proving {P} S {Q} using predicate transforms
Transform rules for:
assignment statements
sequencing
selection statements

4. Introduction What are Predicate Transforms†?
Rules for transforming post-conditions into pre-conditions or vice-versa.
They provide algorithms to reduce the problem of verifying Hoare triples to proving predicate calculus formulas.
Thus, predicate transforms operationalize Hoare Logic.
† Also known as “Predicate Transformers”

5. Introduction (cont’d)
What is a “weakest pre-condition”?
It is the weakest condition on the initial state of program S ensuring termination in state Q.
It is denoted wp(S,Q) and read, “the weakest pre-condition of S with respect to Q.”

6. Introduction (cont’d)
What is a “weakest liberal pre-condition”?
It is the weakest condition on the initial state of program S ensuring state Q on termination if S terminates.
It is denoted wlp(S,Q) and read, “the weakest liberal pre-condition of S with respect to Q.”

7. Introduction (cont’d)
What is a “strongest post-condition”?
It is the strongest condition on the final state of program S given that P holds initially and given that S terminates.
It is denoted sp(S,P) and read, “the strongest post-condition of S with respect to P.”

8. Predicate Transforms I
Introduction:
weakest pre-conditions (wp’s)
weakest liberal pre-conditions (wlp’s)
strongest post-conditions (sp’s)
Proving {P} S {Q} using predicate transforms
Transform rules for:
assignment statements
sequencing
selection statements

9. ROI’s (algorithms) for proving program correctness using predicate transforms
P  wp(S,Q)
{P} S {Q} strongly
P  wlp(S,Q)
{P} S {Q}
sp(S,P)  Q

10. ROI’s (algorithms) for proving program correctness (cont’d)
Note the relationship between weakest liberal pre-conditions and strongest post-conditions:
P  wlp(S,Q) ≡ sp(S,P)  Q
We now consider rules for computing predi-cate transforms for structured programs comprised of assignment statements, if-then (-else) statements, and (in part II) while loops.

11. ROI’s (algorithms) for proving program correctness (cont’d)
Note the relationship between weakest liberal pre-conditions and strongest post-conditions:
P  wlp(S,Q) ≡ sp(S,P)  Q
We now consider rules for computing predi-cate transforms for structured programs comprised of assignment statements, if-then (-else) statements, and (in part II) while loops.

12. Predicate Transforms I
Introduction:
weakest pre-conditions (wp’s)
weakest liberal pre-conditions (wlp’s)
strongest post-conditions (sp’s)
Proving {P} S {Q} using predicate transforms
Transform rules for:
assignment statements
sequencing
selection statements

13. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)

14. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) =

15. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0

16. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0
= y>-3

17. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0
= y>-3
w(l)p(x:=x+1, xn+1) =

18. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0
= y>-3
w(l)p(x:=x+1, xn+1) = x+1n+1

19. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0
= y>-3
w(l)p(x:=x+1, xn+1) = x+1n+1
= xn

20. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0
= y>-3
w(l)p(x:=x+1, xn+1) = x+1n+1
= xn
w(l)p(x:=7, x=7) =

21. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0
= y>-3
w(l)p(x:=x+1, xn+1) = x+1n+1
= xn
w(l)p(x:=7, x=7) = (7=7)

22. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples:
w(l)p(x:=y+3, x>0) = y+3>0
= y>-3
w(l)p(x:=x+1, xn+1) = x+1n+1
= xn
w(l)p(x:=7, x=7) = (7=7)
= true
(cont’d)

23. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) =

24. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) = (7=6)

25. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) = (7=6)
= false

26. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) = (7=6)
= false
w(l)p(x:=7, y=7) =

27. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) = (7=6)
= false
w(l)p(x:=7, y=7) = y=7

28. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) = (7=6)
= false
w(l)p(x:=7, y=7) = y=7
w(l)p(y:=-x, y=|x|) =

29. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) = (7=6)
= false
w(l)p(x:=7, y=7) = y=7
w(l)p(y:=-x, y=|x|) = -x=|x|

30. wp and wlp Rule for Assignment Statements
w(l)p(x:=E, Q(x,y,z))  Q(E,y,z)
Examples: (cont’d)
w(l)p(x:=7, x=6) = (7=6)
= false
w(l)p(x:=7, y=7) = y=7
w(l)p(y:=-x, y=|x|) = -x=|x|
= x0

31. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)

32. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) =

33. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3

34. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3
(=> x>0)

35. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3
(=> x>0)
sp(x:=x+1, x<n) =

36. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3
(=> x>0)
sp(x:=x+1, x<n) = x=x’+1 Л x’<n

37. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3
(=> x>0)
sp(x:=x+1, x<n) = x=x’+1 Л x’<n
(=> x-1<n)

38. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3
(=> x>0)
sp(x:=x+1, x<n) = x=x’+1 Л x’<n
(=> x-1<n)
sp(x:=7, true) =

39. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3
(=> x>0)
sp(x:=x+1, x<n) = x=x’+1 Л x’<n
(=> x-1<n)
sp(x:=7, true) = x=7 Л true

40. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples:
sp(x:=y+3, y>-3) = x=y+3 Л y>-3
(=> x>0)
sp(x:=x+1, x<n) = x=x’+1 Л x’<n
(=> x-1<n)
sp(x:=7, true) = x=7 Л true
= x=7
(cont’d)

41. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples: (cont’d)
sp(x:=7, false) =

42. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples: (cont’d)
sp(x:=7, false) = x=7 Л false

43. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples: (cont’d)
sp(x:=7, false) = x=7 Л false
= false

44. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples: (cont’d)
sp(x:=7, false) = x=7 Л false
= false
sp(x:=7, y=7) =

45. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples: (cont’d)
sp(x:=7, false) = x=7 Л false
= false
sp(x:=7, y=7) = x=7 Л y=7

46. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples: (cont’d)
sp(x:=7, false) = x=7 Л false
= false
sp(x:=7, y=7) = x=7 Л y=7
sp(y:=-x, y=|x|) =

47. sp Rule for Assignment Statements
sp(x:=E, P)  x=E(x’,y,z) Л P(x’,y,z)
Examples: (cont’d)
sp(x:=7, false) = x=7 Л false
= false
sp(x:=7, y=7) = x=7 Л y=7
sp(y:=-x, y=|x|) = y=-x Л y’=|x|

48. Predicate Transforms I
Introduction:
weakest pre-conditions (wp’s)
weakest liberal pre-conditions (wlp’s)
strongest post-conditions (sp’s)
Proving {P} S {Q} using predicate transforms
Transform rules for:
assignment statements
sequencing
selection statements

49. wp and wlp Rule for Sequencing
w(l)p(S1;S2;...;Sn-1;Sn, Q) 
w(l)p(S1, w(l)p(S2,...w(l)p(Sn-1, w(l)p(Sn, Q))…))

50. wp and wlp Rule for Sequencing
w(l)p(S1;S2;...;Sn-1;Sn, Q) 
w(l)p(S1, w(l)p(S2,...w(l)p(Sn-1, w(l)p(Sn, Q))…))

51. wp and wlp Rule for Sequencing
w(l)p(S1;S2;...;Sn-1;Sn, Q) 
w(l)p(S1, w(l)p(S2,...w(l)p(Sn-1, w(l)p(Sn, Q))…))

52. wp and wlp Rule for Sequencing
w(l)p(S1;S2;...;Sn-1;Sn, Q) 
w(l)p(S1, w(l)p(S2,...w(l)p(Sn-1, w(l)p(Sn, Q))…))

53. wp and wlp Rule for Sequencing
w(l)p(S1;S2;...;Sn-1;Sn, Q) 
w(l)p(S1, w(l)p(S2,...w(l)p(Sn-1, w(l)p(Sn, Q))…))

54. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)

55. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
C:=D+1
B:=C2
A:=B2
{ A=36 }

56. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
C:=D+1
B:=C2
{ B2 = 36 }
A:=B2
{ A=36 }

57. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
C:=D+1
{ (2C)2 = 36 }
B:=C2
{ B2 = 36 }
A:=B2
{ A=36 }

58. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
{ (2(D+1))2 = 36 }
C:=D+1
{ (2C)2 = 36 }
B:=C2
{ B2 = 36 }
A:=B2
{ A=36 }

59. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
{ (2(D+1))2 = 36 }
C:=D+1
{ (2C)2 = 36 }
B:=C2
{ B2 = 36 }
A:=B2
{ A=36 }
C:=D+1
B:=C2
A:=B2
{ A=36 }

60. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
{ (2(D+1))2 = 36 }
C:=D+1
{ (2C)2 = 36 }
B:=C2
{ B2 = 36 }
A:=B2
{ A=36 }
C:=D+1
B:=C2
{ B=6 V B=-6 }
A:=B2
{ A=36 }

61. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
{ (2(D+1))2 = 36 }
C:=D+1
{ (2C)2 = 36 }
B:=C2
{ B2 = 36 }
A:=B2
{ A=36 }
C:=D+1
{ C=3 V C=-3 }
B:=C2
{ B=6 V B=-6 }
A:=B2
{ A=36 }

62. wp and wlp Rule for Sequencing (cont’d)
Example:
w(l)p(C:=D+1; B:=C2; A:=B2, A=36)
{ (2(D+1))2 = 36 }
C:=D+1
{ (2C)2 = 36 }
B:=C2
{ B2 = 36 }
A:=B2
{ A=36 }
{ D=2 V D=-4 }
C:=D+1
{ C=3 V C=-3 }
B:=C2
{ B=6 V B=-6 }
A:=B2
{ A=36 }

63. sp Rule for Sequencing Rule: sp(S1;S2;...;Sn-1;Sn, P) 
sp(Sn, sp(Sn-1,...sp(S2, sp(S1, P))…))

64. sp Rule for Sequencing Rule: sp(S1;S2;...;Sn-1;Sn, P) 
sp(Sn, sp(Sn-1,...sp(S2, sp(S1, P))…))

65. sp Rule for Sequencing Rule: sp(S1;S2;...;Sn-1;Sn, P) 
sp(Sn, sp(Sn-1,...sp(S2, sp(S1, P))…))

66. sp Rule for Sequencing Rule: sp(S1;S2;...;Sn-1;Sn, P) 
sp(Sn, sp(Sn-1,...sp(S2, sp(S1, P))…))

67. sp Rule for Sequencing Rule: sp(S1;S2;...;Sn-1;Sn, P) 
sp(Sn, sp(Sn-1,...sp(S2, sp(S1, P))…))

68. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)

69. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
B:=C2
A:=B2

70. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=D+1 Л D=1 }
B:=C2
A:=B2

71. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=D+1 Л D=1 }
B:=C2
{ B=2C Л C=D+1 Л D=1 }
A:=B2

72. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=D+1 Л D=1 }
B:=C2
{ B=2C Л C=D+1 Л D=1 }
A:=B2
{ A=B2 Л B=2C Л C=D+1 Л D=1 }

73. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
B:=C2
A:=B2

74. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=D+1 Л D=1 }
B:=C2
A:=B2

75. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=2 Л D=1 }
B:=C2
A:=B2

76. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=2 Л D=1 }
B:=C2
{ B=2C Л C=2 Л D=1 }
A:=B2

77. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=2 Л D=1 }
B:=C2
{ B=4 Л C=2 Л D=1 }
A:=B2

78. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=2 Л D=1 }
B:=C2
{ B=4 Л C=2 Л D=1 }
A:=B2
{ A=B2 Л B=4 Л C=2 Л D=1 }

79. sp Rule for Sequencing (cont’d)
Example 1:
sp(C:=D+1; B:=C2; A:=B2, D=1)
{ D=1 }
C:=D+1
{ C=2 Л D=1 }
B:=C2
{ B=4 Л C=2 Л D=1 }
A:=B2
{ A=16 Л B=4 Л C=2 Л D=1 }

80. sp Rule for Sequencing (cont’d)
Compound programs often include multiple assignments to the same variable, e.g.,
X := X+1; ...; X := Y-X; ...; X := ZY; ...
It is sometimes useful to “anchor” the initial values of such variables using some suitable notation such as X0 when applying the sp Rule for Assignment Statements.
Consider the following example...

81. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1

82. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1
{ true }
(1) X:=X+1
(2) X:=YX
(3) X:=X-1

83. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1
{ true }
(1) X:=X+1
{ X=X’+1 Л true } = { X=X0+1 }
(2) X:=YX
(3) X:=X-1

84. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1
{ true }
(1) X:=X+1
{ X=X’+1 Л true } = { X=X0+1 }
(2) X:=YX
(3) X:=X-1
“anchoring” initial value of X to X0:

85. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1
{ true }
(1) X:=X+1
{ X=X’+1 Л true } = { X=X0+1 }
(2) X:=YX
{ X=YX’ Л X’=X0+1 } = { X=Y(X0+1) }
(3) X:=X-1
“anchoring” initial value of X to X0:

86. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1
{ true }
(1) X:=X+1
{ X=X’+1 Л true } = { X=X0+1 }
(2) X:=YX
{ X=YX’ Л X’=X0+1 } = { X=Y(X0+1) }
(3) X:=X-1
{ X=X’-1 Л X’=Y(X0+1) } = { X=Y(X0+1)-1 }
“anchoring” initial value of X to X0:

87. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1
{ true }
(1) X:=X+1
{ X=X’+1 Л true } = { X=X0+1 }
(2) X:=YX
{ X=YX’ Л X’=X0+1 } = { X=Y(X0+1) }
(3) X:=X-1
{ X=X’-1 Л X’=Y(X0+1) } = { X=Y(X0+1)-1 }
Therefore, sp(S, true) is X=YX’+Y-1
“anchoring” initial value of X to X0:

88. sp Rule for Sequencing (cont’d)
Example 2:
sp(S, true) where S is X:=X+1; X:=YX; X:=X-1
{ true }
(1) X:=X+1
{ X=X’+1 Л true } = { X=X0+1 }
(2) X:=YX
{ X=YX’ Л X’=X0+1 } = { X=Y(X0+1) }
(3) X:=X-1
{ X=X’-1 Л X’=Y(X0+1) } = { X=Y(X0+1)-1 }
Therefore, sp(S, true) is X=YX’+Y-1
“anchoring” initial value of X to X0:
reverting to standard X’ notation:

89. Predicate Transforms I
Introduction:
weakest pre-conditions (wp’s)
weakest liberal pre-conditions (wlp’s)
strongest post-conditions (sp’s)
Proving {P} S {Q} using predicate transforms
Transform rules for:
assignment statements
sequencing
selection statements

90. wp and wlp Rule for if-then-else Statement
w(l)p(if b then S1 else S2, Q) 
(b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q))

91. wp and wlp Rule for if-then-else Statement
w(l)p(if b then S1 else S2, Q) 
(b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q)) T F b S1 S2
{Q}

92. wp and wlp Rule for if-then-else Statement
w(l)p(if b then S1 else S2, Q) 
(b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q)) T F b
b Л w(l)p(S1, Q) S1 S2
{Q}

93. wp and wlp Rule for if-then-else Statement
w(l)p(if b then S1 else S2, Q) 
(b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q)) T F b
b Л w(l)p(S1, Q)
¬b Л w(l)p(S2, Q) S1 S2
{Q}

94. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|)

95. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|) b S1 S2 Q

96. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|)
= (b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q)) b S1 S2 Q

97. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|)
= (b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л w(l)p(y:=x, y=|x|) b S1 S2 Q

98. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|)
= (b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л w(l)p(y:=x, y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л x=|x|) b S1 S2 Q

99. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|)
= (b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л w(l)p(y:=x, y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л x=|x|)
= (x<0 Л x≤0) V (x≥0 Л x≥0) b S1 S2 Q

100. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|)
= (b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л w(l)p(y:=x, y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л x=|x|)
= (x<0 Л x≤0) V (x≥0 Л x≥0)
= (x<0 V x≥0) b S1 S2 Q

101. wp and wlp Rule for if-then-else Statement
Example:
w(l)p(if x<0 then y:=-x else y:=x, y=|x|)
= (b Л w(l)p(S1, Q)) V (¬b Л w(l)p(S2, Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л w(l)p(y:=x, y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л x=|x|)
= (x<0 Л x≤0) V (x≥0 Л x≥0)
= (x<0 V x≥0)
= true b S1 S2 Q

102. wp and wlp Rule for if-then Statement
w(l)p(if b then S, Q) 
(b Л w(l)p(S, Q)) V (¬b Л Q)

103. wp and wlp Rule for if-then Statement
w(l)p(if b then S, Q) 
(b Л w(l)p(S, Q)) V (¬b Л Q) T b F S
{Q}

104. wp and wlp Rule for if-then Statement
w(l)p(if b then S, Q) 
(b Л w(l)p(S, Q)) V (¬b Л Q) T b
b Л w(l)p(S, Q) F S
{Q}

105. wp and wlp Rule for if-then Statement
w(l)p(if b then S, Q) 
(b Л w(l)p(S, Q)) V (¬b Л Q) T b
b Л w(l)p(S, Q) F S
¬b Л Q
{Q}

106. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)

107. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|) b S Q

108. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)
= (b Л w(l)p(S, Q)) V (¬b Л Q)) b S Q

109. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)
= (b Л w(l)p(S, Q)) V (¬b Л Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л y=|x|) b S Q

110. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)
= (b Л w(l)p(S, Q)) V (¬b Л Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л y=|x|) b S Q

111. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)
= (b Л w(l)p(S, Q)) V (¬b Л Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л y=|x|)
= (x<0 Л x≤0) V (x≥0 Л y=|x|) b S Q

112. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)
= (b Л w(l)p(S, Q)) V (¬b Л Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л y=|x|)
= (x<0 Л x≤0) V (x≥0 Л y=|x|)
= (x<0 V (x≥0 Л y=|x|)) b S Q

113. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)
= (b Л w(l)p(S, Q)) V (¬b Л Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л y=|x|)
= (x<0 Л x≤0) V (x≥0 Л y=|x|)
= (x<0 V (x≥0 Л y=|x|))
= (x<0 V (x≥0 Л y=x)) b S Q

114. wp and wlp Rule for if-then Statement (cont’d)
Example:
w(l)p(if x<0 then y:=-x, y=|x|)
= (b Л w(l)p(S, Q)) V (¬b Л Q))
= (x<0 Л w(l)p(y:=-x, y=|x|) V
(x≥0 Л y=|x|)
= (x<0 Л -x=|x|) V (x≥0 Л y=|x|)
= (x<0 Л x≤0) V (x≥0 Л y=|x|)
= (x<0 V (x≥0 Л y=|x|))
= (x<0 V (x≥0 Л y=x))
= (x<0 V y=x) b S Q

115. Exercise Prove the assertion below using the wlp ROI.
{Z=B} if A>B then Z := A {Z=Max(A,B)}

116. Exercise Prove the assertion below using the wlp ROI.
{Z=B} if A>B then Z := A {Z=Max(A,B)} P S Q

117. Exercise Prove the assertion below using the wlp ROI.
{Z=B} if A>B then Z := A {Z=Max(A,B)}
Recall the wlp ROI:
P  wlp(S,Q)
{P} S {Q} P S Q

118. sp Rule for if-then-else Statement
sp(if b then S1 else S2, P) 
sp(S1, b Л P) V sp(S2, ¬b Л P)

119. sp Rule for if-then-else Statement
sp(if b then S1 else S2, P) 
sp(S1, b Л P) V sp(S2, ¬b Л P)
{P} T F b S1 S2

120. sp Rule for if-then-else Statement
sp(if b then S1 else S2, P) 
sp(S1, b Л P) V sp(S2, ¬b Л P)
{P} T F b S1 S2
sp(S1, b Л P)

121. sp Rule for if-then-else Statement
sp(if b then S1 else S2, P) 
sp(S1, b Л P) V sp(S2, ¬b Л P)
{P} T F b S1 S2
sp(S1, b Л P)
sp(S2, ¬b Л P)

122. sp Rule for if-then Statement
sp(if b then S, P) 
sp(S, b Л P) V (¬b Л P)

123. sp Rule for if-then Statement
sp(if b then S, P) 
sp(S, b Л P) V (¬b Л P)
{P} T b F S

124. sp Rule for if-then Statement
sp(if b then S, P) 
sp(S, b Л P) V (¬b Л P)
{P} T b F S
sp(S, b Л P)

125. sp Rule for if-then Statement
sp(if b then S, P) 
sp(S, b Л P) V (¬b Л P)
{P} T b F S
¬b Л P
sp(S, b Л P)

126. Example Prove the assertion: {y=x} if x<0 then y:=-x {y=|x|}
using the strongest post-condition (sp) ROI.

127. Example Prove the assertion: {y=x} if x<0 then y:=-x {y=|x|}
using the strongest post-condition (sp) ROI.
Recall the sp ROI:
sp(S,P)  Q
{P} S {Q}

128. Example Prove the assertion: {y=x} if x<0 then y:=-x {y=|x|}
using the strongest post-condition (sp) ROI.
(1) sp(if x<0 then y:=-x, y=x) =
sp(y:=-x, x<0 Л y=x) V (x≥0 Л y=x) =
(y=-x Л x<0 Л y’=x) V (x≥0 Л y=x) P

129. Example Prove the assertion: {y=x} if x<0 then y:=-x {y=|x|}
using the strongest post-condition (sp) ROI.
(1) sp(if x<0 then y:=-x, y=x) =
sp(y:=-x, x<0 Л y=x) V (x≥0 Л y=x) =
(y=-x Л x<0 Л y’=x) V (x≥0 Л y=x) P

130. Example Prove the assertion: {y=x} if x<0 then y:=-x {y=|x|}
using the strongest post-condition (sp) ROI.
(1) sp(if x<0 then y:=-x, y=x) =
sp(y:=-x, x<0 Л y=x) V (x≥0 Л y=x) =
(y=-x Л x<0 Л y’=x) V (x≥0 Л y=x) P

131. Example Prove the assertion: {y=x} if x<0 then y:=-x {y=|x|}
using the strongest post-condition (sp) ROI.
(1) sp(if x<0 then y:=-x, y=x) =
sp(y:=-x, x<0 Л y=x) V (x≥0 Л y=x) =
(y=-x Л x<0 Л y’=x) V (x≥0 Л y=x)
(2) (y=-x Л x<0 Л y’=x) V (x≥0 Л y=x) =>
(x<0 Л y=-x) V (x≥0 Л y=x) => y=|x| √ P Q

132. Example Prove the assertion: {y=x} if x<0 then y:=-x {y=|x|}
using the strongest post-condition (sp) ROI.
(1) sp(if x<0 then y:=-x, y=x) =
sp(y:=-x, x<0 Л y=x) V (x≥0 Л y=x) =
(y=-x Л x<0 Л y’=x) V (x≥0 Л y=x)
(2) (y=-x Л x<0 Л y’=x) V (x≥0 Л y=x) =>
(x<0 Л y=-x) V (x≥0 Л y=x) => y=|x| √ P Q

133. Coming Up Next…
Transform rules for while loops

134. Predicate Transforms I
Software Testing and Verification
Lecture Notes 19
Prepared by
Stephen M. Thebaut, Ph.D.
University of Florida