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[Frama-c-discuss] Jessie sort specification

You can also have a look at the sorting example in the tutorial:

I know, it is only given "as is" and it needs to be explained, but I 
hope it helps anyway.

- Claude

Virgile Prevosto wrote:
> Hello Kerstin,
> Le mer. 03 juin 2009 13:43:48 CEST,
> "Kerstin Hartig" <kerstin.hartig at> a ?crit :
>> the array are growing. There's always a problem with the loop 
>> invariants. The post conditions are valid.
> Just a remark here: the post-condition is valid modulo the assumption
> that the loop invariant is valid (if you have a look at the why gui,
> you'll see an hypothesis corresponding to the invariant) and the loop
> test is false (same thing). But this still leaves the issue of proving
> that the loop invariant holds.
>> Is it possible we miss something in the specification? Maybe another loop invariant?
>> It is possible to prove that \forall integer k; 0 <= k < size ==> a[k] ==  k;
> Not with your current loop invariant. Remember that when considering a
> loop, the only thing that jessie knows of all the steps preceding the
> current ones is the loop invariant. Conversely, everything which is not
> stated in the loop invariant is out of reach. In order to complete the
> proof, you'll thus have to add another invariant stating that:
> loop invariant \forall integer k; 0<=k<i ==> a[k] == k;
> (in fact, since this invariant implies that \forall k; 0<=k<i-1 ==>
> a[k]<=a[k+1], the latter becomes useless as a loop invariant).
> Best regards,

Claude March?                          | tel: +33 1 72 92 59 69
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