I am given the following equality: $$ P((A \cap B^c) \cup (A^c \cap B)) = P(A) + P(B) - 2P(A \cap B)$$ And, I am asked to arrive at the right-hand expression from the left one appealing to one of the three probability axioms at each step.
I had a go at it and simplified
$$ P((A \cap B^c) \cup (A^c \cap B)) $$
$$\Downarrow$$ Using the third axiom. This identity was in the textbook:
$$ P((A \cap B^c) + (A^c \cap B)) - P( (A \cap B^c) \cap (A^c \cap B) ) $$
$$\Downarrow$$ Using associativity and commutativity:
$$ P((A \cap B^c) + (A^c \cap B)) - P( (A \cap A^c) \cap (B \cap B^c) ) $$
$$\Downarrow$$ Now, since $ P( (A \cap A^c) \cap (B \cap B^c) ) = P( ( \varnothing ) \cap (\varnothing) ) = 0 $, I write:
$$\Downarrow$$ $$ P((A \cap B^c) + (A^c \cap B)) $$
And, now I am stuck... This is my first time posting in the math section, so correct my formatting if I made mistakes. Also, is how I reasoned correct so far?
Given sets $A$, $B$, we write $A\setminus B=A\cap B^c$. Note that $A\setminus B$ and $B\setminus A$ are disjoint.
Note that $$ \begin{align} P(A)&=P(A\setminus B)+P(A\cap B)\\ P(B)&=P(B\setminus A)+P(A\cap B) \end{align} $$ since $A$ is the disjoint union of $A\setminus B$ and $A\cap B$ and $B$ is the disjoint union of $B\setminus A$ and $A\cap B$ so $$ P(A)+P(B)-2P(A\cap B)=P(A\setminus B)+P(B\setminus A)= P((A\cap B^c)\cup (B\cap A^c)) $$