Dafny Sum and Max solution

A Dafny solution to the Sum and Max verification benchmark. I added some additional specification to verify that the result max is really the maximum and the sum is really the sum.

method SumMax(a: array<int>) returns (sum:int, max:int)
 requires a != null;
 requires a.Length >= 0;
 requires forall i :: 0 <= i < a.Length ==> a[i] >= 0;
 ensures sum <= a.Length*max;
 ensures forall i :: 0 <= i < a.Length ==> a[i] <= max;
 ensures a.Length == 0 || 
         (exists i :: 0 <= i < a.Length && a[i] == max);
 ensures sum == Sum(a);
{
 max := 0;
 sum := 0;
 
 var idx:int := 0;
 while idx<a.Length 
  invariant idx <= a.Length;
  invariant forall i :: 0 <= i < idx ==> a[i] <= max;
  invariant (idx==0 && max==0) || 
            exists i :: 0 <= i < idx && a[i] == max;
  invariant sum <= idx*max;
  invariant sum == Sum'(a, 0, idx);
 {
  if (max < a[idx])
  {
   max := a[idx];
  }
  sum := sum + a[idx];
  idx := idx + 1;
 }
}
 
function Sum(a: array<int>) : int
 reads a;
 requires a != null;
{
 Sum'(a, 0, a.Length)
}
 
function Sum'(a: array<int>, i:int, j:int) : int
 reads a;
 requires a != null;
 requires 0 <= i <= j <= a.Length;
{
 if j == i then 0 else Sum'(a, i, j-1) + a[j-1]
}
 
lemma SumOfIntervalOneIsElement(a: array<int>, i:int)
 requires a != null;
 requires 0 <= i < a.Length;
 ensures a.Length > 0;
 ensures Sum'(a, i, i+1) == a[i];
{ }
 
lemma SumOfIntervalIsSumOfSubIntervals
  (a: array<int>, i:int, j:int, k:int)
 requires a != null;
 requires 0 <= i <= j <= k <= a.Length;
 ensures Sum'(a, i, j) + Sum'(a, j, k) == Sum'(a, i, k);
{ }

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