Complete "Δ is a permutation" is sufficient for simplest reindexing proof

In other words, if the loop is of the form

  for(i = 0; i < N; i++)
    A[i] = ...stuff not involving A...

and Δ is a permutation over 0..N-1, then the above is equivalent to

  for(i = 0; i < N; i++)
    A[Δ[i]] = ...stuff not involving A with any
                 occurrences of i replaced by Δ[i]...