What does "unspecified" mean in this context? If you mean "the value of a signed integer is unspecified if it overflows", then I can still construct cases in which you lose the optimization that widens a loop induction variable from 32-bit to 64-bit. For example, take this code:

    for (int i = 0; i != n; i++)
        sum += a[i];

If n is -2, will this function include the element at a[(uint64_t)INT_MAX + 1] in the sum? If signed overflow is UB, then the answer is "maybe", and thus i and n can legally be promoted to 64-bit. If signed overflow is unspecified, but the result must still fit in a 32-bit integer, then the answer is "no", and i and n can't legally be promoted.