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# Pre and Post Condition Rules - PowerPoint PPT Presentation

Pre and Post Condition Rules. Definition : If R and S are two assertions, then R is said to be stronger than S if R -> S (R implies S). Example : the assertion i < 0 is stronger than the assertion i < 3 because i < 0 implies that i < 3 ( i < 0 is true implies that i < 3 is true)

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• Definition : If R and S are two assertions, then R is said to be stronger than S if R -> S (R implies S).

• Example :

• the assertion i < 0 is stronger than the assertion i < 3 because i < 0 implies that i < 3 ( i < 0 is true implies that i < 3 is true)

• draw the set and subset diagram to get a better “feel,” if needed

• Note that if R is stronger than S, then all states that satisfy R will satisfy S. But there is at least one state that satisfies S which will not satisfy R. So the number of states that satisfies S is larger than that of R. Thus one may view the notion of “stronger” as “more selective” because less states will satisfy the stronger condition.

• Of course, if R is strongerthan S, then S is weaker than R

• Rule 1: If P’ is stronger than P and if {P}C{Q} triple is correct, then with the strengthened precondition assertion, {P’}C{Q} triple is also correct

• Example :

• if {P}C{Q} is correct, for P which is (x>0), then for P’, which asserts (x>2), the triple {P’}C{Q} will also be correct.

• Rule 1 - More formally :

• P’ -> P (strengthening p to p’)

• {P}C{Q}

• {P’}C{Q}

• Example : if {x <5} x:= x+1 {x < 6} is correct, then strengthening {P} to {x < 3} should give us {x<3} x:=x+1 {x<6} as also correct because :

• {x<3} -> {x<5}

• {x<5} x := x+1 {x<6}

• {x<3} x := x+1 {x<6}

• Rule 2: If Q -> Q’ and {P}C{Q} triple is correct then {P}C{Q’} triple is correct

• Rule 2 : Formally we have:

• {P}C{Q}

• Q -> Q’

• {P}C{Q’}

• Example :

• if { } max := b {max=b} then show { } max:=b {max >= b}

{ } max := b {max =b}

{max = b} -> {max >= b}

{ } max:=b {max >=b }

• Rule 3: If C is a piece of code, {P}C{Q} AND {P’}C{Q’} are correct (note that both of the conditions have to be True simultaneously), then {P AND P’} C {Q AND Q’} is also correct

• Formally :

• {P} C {Q}

• {P’} C {Q’}

• {P AND P’} C {Q AND Q’}

• Rule 4:If C is a piece of code, {P}C{Q} AND {P’}C{Q’}, then {P OR P’} C {Q OR Q’} is also correct

• Formally :

• {P} C {Q}

• {P’} C {Q’}

• {P OR P’} C {Q OR Q’}

• Problem : given the following Hoare Triples:

• { } x:= x+1 { x = x+1 } AND [note that x:= x + 1 has multiple meanings here !]

• {x>0} x:= x +1 {x > 0}

• show that {x>0} x:=x+1 {x > -1}

• Proof : (a little more detailed than needed)

• a) using conjunction rule, we get {x>0} x:= x+ 1{x=x+1 AND x>0}

• b) using the weakening the post-condition rule, we have {x=x+1 and x>0} -> {x>0}; thus the conjunction triple {x>0} x:=x+1 {x=x+1 AND x>0} also imply {x>0} x:=x+1 {x>0} is correct by weakening the post condition

• Furthermore {x > 0 } -> { x > - 1} (or x > 0 is stronger than x > -1)

• Therefore we have {x>0} x := x+1 {x>-1} (by weakening the post condition)

• Alternatively : note that the weakening of post condition can be achieved through dropping x=x+1 and also directly weakening x > 0 to x > -1.