How to prove $P(A\cap B) - P(A)P(B) = P(A)P(B^c) - P(A\cap B^c)$? I'm trying to prove the identity in the title. I have tried all the results I'm aware of but I can't find the way out. I know that
$$P(A) = P(A\cap B)+ P(A\cap B^c),$$
and that
$$P(A\cup B) = P(A)+P(B)-P(A\cap B),$$
but I don't see how to combine these (or other properties I am missing) to get the desired result...
Any hint is highly appreciated.

$$P(A\cap B) - P(A)P(B) = P(A)P(B^c) - P(A\cap B^c)$$
is the same statement as:$$P(A\cap B)+P(A\cap B^c)=P(A)(P(B)+P(B^c))$$(just replace negative terms to the other side)
or shorter:$$P(A)=P(A)$$which is evidently a true statement.
