Copyright  (c) Levent Erkok 

License  BSD3 
Maintainer  erkokl@gmail.com 
Stability  experimental 
Safe Haskell  None 
Language  Haskell2010 
Proof of correctness of an imperative summation algorithm, using weakest preconditions. We investigate a few different invariants and see how different versions lead to proofs and failures.
Synopsis
Program state
The state for the sum program, parameterized over a base type a
.
Instances
Functor SumS Source #  
Foldable SumS Source #  
Defined in Documentation.SBV.Examples.WeakestPreconditions.Sum fold :: Monoid m => SumS m > m # foldMap :: Monoid m => (a > m) > SumS a > m # foldr :: (a > b > b) > b > SumS a > b # foldr' :: (a > b > b) > b > SumS a > b # foldl :: (b > a > b) > b > SumS a > b # foldl' :: (b > a > b) > b > SumS a > b # foldr1 :: (a > a > a) > SumS a > a # foldl1 :: (a > a > a) > SumS a > a # elem :: Eq a => a > SumS a > Bool # maximum :: Ord a => SumS a > a #  
Traversable SumS Source #  
SymVal a => Fresh IO (SumS (SBV a)) Source # 

Show a => Show (SumS a) Source #  
(SymVal a, Show a) => Show (SumS (SBV a)) Source #  Show instance for 
Generic (SumS a) Source #  
Mergeable a => Mergeable (SumS a) Source #  
type Rep (SumS a) Source #  
Defined in Documentation.SBV.Examples.WeakestPreconditions.Sum type Rep (SumS a) = D1 (MetaData "SumS" "Documentation.SBV.Examples.WeakestPreconditions.Sum" "sbv8.7DbQHjiKtor73WzWR2O4MT3" False) (C1 (MetaCons "SumS" PrefixI True) (S1 (MetaSel (Just "n") NoSourceUnpackedness NoSourceStrictness DecidedLazy) (Rec0 a) :*: (S1 (MetaSel (Just "i") NoSourceUnpackedness NoSourceStrictness DecidedLazy) (Rec0 a) :*: S1 (MetaSel (Just "s") NoSourceUnpackedness NoSourceStrictness DecidedLazy) (Rec0 a)))) 
The algorithm
algorithm :: Invariant S > Maybe (Measure S) > Stmt S Source #
The imperative summation algorithm:
i = 0 s = 0 while i < n i = i+1 s = s+i
Note that we need to explicitly annotate each loop with its invariant and the termination measure. For convenience, we take those two as parameters, so we can experiment later.
Precondition for our program: n
must be nonnegative. Note that there is
an explicit call to abort
in our program to protect against this case, so
if we do not have this precondition, all programs will fail.
Postcondition for our program: s
must be the sum of all numbers up to
and including n
.
imperativeSum :: Invariant S > Maybe (Measure S) > Program S Source #
A program is the algorithm, together with its pre and postconditions.
Correctness
correctness :: Invariant S > Maybe (Measure S) > IO (ProofResult (SumS Integer)) Source #
Check that the program terminates and s
equals n*(n+1)/2
upon termination, i.e., the sum of all numbers upto n
. Note
that this only holds if n >= 0
to start with, as guaranteed
by the precondition of our program.
The correct termination measure is ni
: It goes down in each
iteration provided we start with n >= 0
and it always remains
nonnegative while the loop is executing. Note that we do not
need a lexicographic measure in this case, hence we simply return
a list of one element.
The correct invariant is a conjunction of two facts. First, s
is
equivalent to the sum of numbers 0
upto i
. This clearly holds at
the beginning when i = s = 0
, and is maintained in each iteration
of the body. Second, it always holds that i <= n
as long as the
loop executes, both before and after each execution of the body.
When the loop terminates, it holds that i = n
. Since the invariant says
s
is the sum of all numbers up to but not including i
, we
conclude that s
is the sum of all numbers up to and including n
,
as requested.
Note that coming up with this invariant is neither trivial, nor easy to automate by any means. What SBV provides is a way to check that your invariant and termination measures are correct, not a means of coming up with them in the first place.
We have:
>>>
:set XNamedFieldPuns
>>>
let invariant SumS{n, i, s} = s .== (i*(i+1)) `sDiv` 2 .&& i .<= n
>>>
let measure SumS{n, i} = [n  i]
>>>
correctness invariant (Just measure)
Total correctness is established. Q.E.D.
Example proof attempts
It is instructive to look at several proof attempts to see what can go wrong and how the weakestprecondition engine behaves.
Always false invariant
Let's see what happens if we have an always false invariant. Clearly, this will not
do the job, but it is instructive to see the output. For this exercise, we are only
interested in partial correctness (to see the impact of the invariant only), so we
will simply use Nothing
for the measures.
>>>
import Control.Monad (void)
>>>
let invariant _ = sFalse
>>>
void $ correctness invariant Nothing
Following proof obligation failed: ================================== Invariant for loop "i < n" fails upon entry: SumS {n = 0, i = 0, s = 0}
When the invariant is constant false, it fails upon entry to the loop, and thus the proof itself fails.
Always true invariant
The invariant must hold prior to entry to the loop, after the loopbody executes, and must be strong enough to establish the postcondition. The easiest thing to try would be the invariant that always returns true:
>>>
let invariant _ = sTrue
>>>
void $ correctness invariant Nothing
Following proof obligation failed: ================================== Postcondition fails: Start: SumS {n = 0, i = 0, s = 0} End : SumS {n = 0, i = 0, s = 1}
In this case, we are told that the end state does not establish the
postcondition. Indeed when n=0
, we would expect s=0
, not s=1
.
The natural question to ask is how did SBV come up with this unexpected
state at the end of the program run? If you think about the program execution, indeed this
state is unreachable: We know that s
represents the sum of all numbers up to i
,
so if i=0
, we would expect s
to be 0
. Our invariant is clearly an overapproximation
of the reachable space, and SBV is telling us that we need to constrain and outlaw
the state {n = 0, i = 0, s = 1}
. Clearly, the invariant has to state something
about the relationship between i
and s
, which we are missing in this case.
Failing to maintain the invariant
What happens if we pose an invariant that the loop actually does not maintain? Here is an example:
>>>
let invariant SumS{n, i, s} = s .<= i .&& s .== (i*(i+1)) `sDiv` 2 .&& i .<= n
>>>
void $ correctness invariant Nothing
Following proof obligation failed: ================================== Invariant for loop "i < n" is not maintaned by the body: Before: SumS {n = 2, i = 1, s = 1} After : SumS {n = 2, i = 2, s = 3}
Here, we posed the extra incorrect invariant that s <= i
must be maintained, and SBV found us a reachable state that violates the invariant. The
before state indeed satisfies s <= i
, but the after state does not. Note that the proof fails in this case not because the program
is incorrect, but the stipulated invariant is not valid.
Having a bad measure, Part I
The termination measure must always be nonnegative:
>>>
let invariant SumS{n, i, s} = s .== (i*(i+1)) `sDiv` 2 .&& i .<= n
>>>
let measure SumS{n, i} = [ i]
>>>
void $ correctness invariant (Just measure)
Following proof obligation failed: ================================== Measure for loop "i < n" is negative: State : SumS {n = 3, i = 2, s = 3} Measure: 2
The failure is pretty obvious in this case: Measure produces a negative value.
Having a bad measure, Part II
The other way we can have a bad measure is if it fails to decrease through the loop body:
>>>
let invariant SumS{n, i, s} = s .== (i*(i+1)) `sDiv` 2 .&& i .<= n
>>>
let measure SumS{n, i} = [n + i]
>>>
void $ correctness invariant (Just measure)
Following proof obligation failed: ================================== Measure for loop "i < n" does not decrease: Before : SumS {n = 1, i = 0, s = 0} Measure: 1 After : SumS {n = 1, i = 1, s = 1} Measure: 2
Clearly, as i
increases, so does our bogus measure n+i
. Note that a counterexample where i
is
negative is also possible for this failure, as the SMT solver will find a counterexample to induction, not
necessarily a reachable state. Obviously, all such failures need to be addressed for the full proof.