~x + ~y == ~(x + y) is always false?

Assume for the sake of contradiction that there exists some x and some y (mod 2ⁿ) such that

~(x+y) == ~x + ~y

By two’s complement*, we know that,

      -x == ~x + 1
<==>  -1 == ~x + x

Noting this result, we have,

      ~(x+y) == ~x + ~y
<==>  ~(x+y) + (x+y) == ~x + ~y + (x+y)
<==>  ~(x+y) + (x+y) == (~x + x) + (~y + y)
<==>  ~(x+y) + (x+y) == -1 + -1
<==>  ~(x+y) + (x+y) == -2
<==>  -1 == -2

Hence, a contradiction. Therefore, ~(x+y) != ~x + ~y for all x and y (mod 2ⁿ).

^{*It is interesting to note that on a machine with one’s complement arithmetic, the equality actually holds true for all x and y. This is because under one’s complement, ~x = -x. Thus, ~x + ~y == -x + -y == -(x+y) == ~(x+y).}