The Cout function of a 3-bit adder is as follows:
AB + Cin (A ⊕ B) ----- i
It, being a majority function, can also be written as:
AB+BCin+CinA
which is equivalent to
AB + Cin (A+B) ------- ii
So, if we consider eqn i and ii, doesn't it show that (A ⊕ B) = (A+B). What am I missing here?
