The algorithm ADDN implements N-bit fixed-width binary addition for non-negative integers and ignores overflows. For example, ADD4((1101)2,(1100)2) = (1001)2 because (1101)2 + (1100)2 = (11001)2 but the leading bit can’t fit in the 4-bit register. A standard way for computers to represent negative integers is the "two’s complement" method TN (x). Non-negative integers from 0 to 2N−1 − 1 are represented using ordinary fixed-width binary (e. g. T4(3) = (0011)2), and a negative integer n with −2 N−1 ≤ n is represented using the binary expansion of the (positive) integer 2N + n (e. g. T4(−3) = (1101)2 because 24 + (−3) = 13 = (1101)2). This representation allows us to use ADDN unchanged for both positive and negative integers! To partially prove this claim, show that if a and b are negative integers with −2 N−1 ≤ a + b, then ADDN (TN (a), TN (b)) = TN (a + b). (Hint: In what situation does ADDN (x, y) not equal x + y, and then what does it equal instead?)