$\overline{(A\cap \bar{B})} \cap \overline{(B \cap \bar{A})}$ = A∩B $(\bar{A}\cup B) \cap (\bar{B} \cup A)$ = A∩B Everything without A and having B part ∩ Everything without B and having A part = A∩B Common to Both is A∩B and Everything except A and B.$\neq$ A∩B Modification : A $\cup$ B is Universe. then equation hold.