~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) == … Read more