To my understanding, the desired DFA can be obtained by using a modified product automaton, as used for the intersection of L1 and L2, but the terminal states have to be defined differently. Instead of making a product state (q_1,q_2) a terminal state if and only if q_1 and q_2...

binary,overflow,twos-complement,complement

Short answer: if you are performing arithmetic on fixed-width binary numbers, using two's complement representation for negative numbers, then yes, you ignore the one-bit overflow. Long Answer: You can consider each ith bit in n-bit two's complement notation have place value 2^i, for 0 <= i < n - 1,...