(* Max

  Given an array `a` of integers with length `n` greater than `0`,
  return `max_idx`, the index of the maximum element of that array.
  
  E.g. `max_idx [5, 12, 34, 10] 4` will return `2`
  E.g. `max_idx [4, 3, 2] 3` will return `0`
  E.g. `max_idx [1, 2, 3, 4] 4` will return `3`
  
  Prove the below program indeed follows the given specification,
  by giving an appropriate invariant.
*)

module Max

  use int.Int
  use ref.Ref
  use array.Array

  let max_idx (a: array int) (n: int) : (max_idx: int)
    requires { length a > 0 }
    requires { n = length a }
    ensures { 0 <= max_idx <= n-1 }
    ensures { forall i. 0 <= i <= n-1 -> a[i] <= a[max_idx] }
  = 
    let ref max_idx = 0 in
    for i = 0 to n-1 do
      invariant { 0 <= max_idx <= n-1 }
      (* IMPORTANT: DON'T MODIFY THE ABOVE LINES *)
      invariant { true (* TODO: Replace `true` with your solution. Your solution MUST be a single line, at line number 30. DON'T add another line of codes. *) }
      (* IMPORTANT: DON'T MODIFY THE BELOW LINES *)
      if a[max_idx] < a[i] then max_idx <- i;
    done;
    max_idx

end