Logic indeed
[(AB) + (B XOR A)] = (B + A) + [{~[~A(~B)]}(A + B)] = [(A + B) + (BA)] = (B + A)(A + B)(B + A)...(B + A)(A + B)(B + A)...(B + A)(A + B)[A + (BB)]]...(A + B)(B + A)...[(AA) + (BB)]...[(AAAAAA...) + (BBBBBB...)]...(AA + BB..)...(A + B)... (A + B)...(A + B)...(B + A)...(A + B)111111111111111111111111111111111111 and if A will never be equal to B in all Binary condition then calculate (A NN...NNOR B) provided that the number of times of Negation of "V" is an element of even number
The answer is 1.
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