How to prove P[(A∪B)∩(¯A∪¯B)] = P(A) + P(B) - 2P(A∩B) via probability manipulations?

Question

blockcolder · Answer

$$P((A\cup B)\cup(A'\cup B'))=P(A\cup B)+P(A' \cup B')-P((A \cup B)\cap(A'\cup B'))\
P((A\cup B)\cup(A'\cup B'))=P(A \cup A' \cup B\cup B')=P(U)=1\
P(A\cup B)=P(A)+P(B)-P(A\cap B)\
P(A'\cup B')=P((A\cap B)')=1-P(A\cap B)$$
Put these all together and you get
$$1=P(A)+P(B)-P(A\cap B)+1-P(A\cap B)-P((A \cup B)\cap(A'\cup B'))\
\Rightarrow P((A \cup B)\cap(A'\cup B'))=P(A)+P(B)-2P(A\cap B)$$