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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.

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Presentation on theme: "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."— Presentation transcript:

1 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) 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 stronger than S, then S is weaker than R

2 Precondition Strengthening 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 {x<5} {x<5} x := x+1 {x<6} {x<3} x := x+1 {x<6}

3 Post Condition Weakening 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 }

4 Conjunction (AND) and Disjunction (OR) Rules 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’}

5 Example using Conjunction Rule & more 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.


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