Checking overlow after addition of 2 integers in 2's complement form

Question

Q) Let a_n−1a_n−2...a₀ and b_n−1b_n−2...b₀ denote the 2’s complement representation of two integers A and B respectively. Addition of A and B yields a sum S=s_n−1s_n−2...s₀.The outgoing carry generated at the most significant bit position, if any, is ignored. Show that an overflow (incorrect addition result) will occur only if the following Boolean condition holds:
(~s_n−1)⊕(a_n−1s_n−1)=b_n−1(s_n−1⊕a_n−1)

where ⊕ denotes the Boolean XOR operation and ~ denotes the boolean NOT operation. You may use the Boolean identity: X+Y=X⊕Y⊕(XY) to prove your result.

My Attempt :- When a_n−1b_n−1(~s_n−1) + (~a_n−1)∗(~b_n−1)∗s_n−1 is 1 then overflow OCCURS otherwise no overflow. Means when 2 positive numbers are added and answer is negative and when 2 negative numbers are added and answer is positive then overflow occurs otherwise it does not occur.
Here,
(~s_n−1)⊕(a_n−1s_n−1)=b_n−1(s_n−1⊕a_n−1)
(s_n−1)(a_n−1s_n−1) + (~s_n−1)(~(a_n−1s_n−1))= b_n−1(s_n−1(~a_n−1)+ a_n−1(~s_n−1))
After solving further, I am getting finally :-
s_n−1a_n−1 + (~s_n−1) = b_n−1s_n−1(~a_n−1) + b_n−1a_n−1(~s_n−1)

I don't know how to solve further to get the desired result and I also don't know whther I am going in right direction or not. Please help.

Answer 1

I'll describe this in words, hopefully it will be easy to understand.

Consider the msb as the sign. If the numbers are opposite sign, then no overflow is possible.

If the signs are the same, and the result has a different sign from either of the two numbers, then overflow has occurred. Otherwise it has not.