Frama-C-discuss mailing list archives

This page gathers the archives of the old Frama-C-discuss archives, that was hosted by Inria's gforge before its demise at the end of 2020. To search for mails newer than September 2020, please visit the page of the new mailing list on Renater.

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

[Frama-c-discuss] Jessie sort specification

Hello Claude,

thanks for the pointer to the jessie tutorial.
We were trying to prove the postcondition without "advanced algebraic  
The reason for this attempt is a customer that feels a bit overwhelmed  
by inductive definitions.

Do you think it is possible that Jessie can prove a sort algorithm  
that has not been "inductively specified"?

Regards Jens

Am 03.06.2009 um 16:42 schrieb Claude March?:

> 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,