X + y = (X & Y) <1) + (x ^ y) proof

Method 1:
We consider adding X and Y to bits in binary representation.
Where the I-th Digit
XI + yi = (XI & Yi) <1) + (Xi ^ Yi)
(XI & Yi) <1 indicates that when XI and Yi are both 1, carry 1 is required.
Xi ^ Yi indicates that carry is not considered. The value of the current BIT.
Add all the data, that is
X + y = sum {Xi * 2 ^ I} + sum {Yi * 2 ^ I} = sum {(XI + yi) * 2 ^ I}
= Sum {(XI & Yi) <1) + (Xi ^ Yi) * 2 ^ I}
= Sum {(XI & Yi) * 2) * 2 ^ I + (Xi ^ Yi) * 2 ^ I}
= Sum {(XI & Yi) * 2 ^ I} * 2 + sum {(Xi ^ Yi) * 2 ^ I}
= (X & Y) * 2 + (x ^ y)
= (X & Y) <1) + (x ^ y)

Method 2:

X = (X & Y) + (x ^ y) & X)
Y = (X & Y) + (x ^ y) & Y)
(X ^ y) & x) + (x ^ y) & Y) = x ^ y
----------------------------------
X + y = 2 * (X & Y) + (x ^ y)

Application:

Implement Addition of two integers using bitwise operations

Int add (int A, int B) {If (B = 0) return a; int sun, carry; sum = a ^ B; // complete Step 1 without carry addition caryy = (A & B) <1; // complete Step 1 with a carry addition, and carry return add (sum, carry ); // perform recursive addition}

X + y = (X & Y) <1) + (x ^ y) proof