Give a mathematical derivation of the formula by appealing to probability axioms or set operation rules 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
$$
P(A)+P(B)-2P(A\cap B)=P(A\setminus B)+P(B\setminus A)= P((A\cap B^c)\cup (B\cap A^c))
$$
