Given two sets $A$ and $B$. How to write $A$ or $B$, but not both in set notation? My answer is $A \cup B  \setminus  \left(A \cap B\right)$. My textbook says $(A^c \cap B) \cup(A \cap B^c)$. I have drawn a Venn diagram for both sets. If i have not made any mistakes then it looks like that both a valid ways. I'm I correct?
Also, I'm not sure whether I have drawn the Venn diagrams correctly or not. I have just drawn a rectangle in which two circles intersect. Is this the correct way of doing it?
Last question: how to I mathematically prove that (if that is the case) $A \cup B  \setminus  \left(A \cap B\right) = (A^c \cap B) \cup(A \cap B^c)$ ? 
 A: Intuitively, you can read the left hand side as "the collection of elements such that they are in A or B but not in A and B" and you can read the right hand side as "the collection of elements where they are either in A but not B or in B but not A." This might give us some idea of why these are equal apriori.
More formally, let $A, B \subset X$, then we have $(A \cup B) \setminus (A \cap B) = (A \cup B) \cap (A \cap B)^c$. Use DeMorgan's to write $(A \cap B)^c = (A^c \cup B^c)$. Now use distributivity to get $(A \cup B) \cap (A^c \cup B^c) = (A \cap A^c) \cup (A \cap B^c) \cup(B \cap A^c) \cup (B \cap B^c)$. 
We now have, for any set $Y$, $Y \cap Y^c = \varnothing$, so we can rewrite the above to get $(A \cup B) \setminus (A \cap B) = (A \cap B^c) \cup (B \cap A^c)$.
A: For any $x$ in the universal set exactly on of the following is true
Either: 
1) $x \not \in A\cup B$.
or 
2) $x \in A\cup B$.
And if $x \in A\cup B$ then exactly one of the following is true.
2a) $x \in A\cup B$ and $x \in A\cap B$
or 
2b) $x\in A\cup B$ and $x \not \in A\cap B$
SO for every $x$ in the univers exactly one of $1,2a, 2b$ is true.
If case 1) $x \not \in A\cup B$ then $x \not \in (A\cup B)\setminus (A\cap B)$ and $x$ is in neither $A\cap B^c$ (because $x \not\in A$) nor in $A^c \cap B$ (because $x \not \in B$). So $x \not \in (A^c\cap B)\cup (A\cup B^c)$. 
So $(A\cup B)\setminus (A\cap B)$ and $(A^c\cap B)\cup (A\cup B^c)$ both exclude all elements not in $A\cup B$.
If case 2) $x \not \in (A\cup B)\setminus (A\cap B)$ (because $x$ is in $A\cap B$).  And $x\not \in A^c \cap B$ (because $x  \in A$ so $x\not \in A^c$) and $x \not \in A\cap B^c$ (because $x \in B$ so $x \not \in A^c$). So $x \not \in (A^c\cap B)\cup(A\cap B^c)$. 
So $(A\cup B)\setminus (A\cap B)$ and $(A^c\cap B)\cup (A\cup B^c)$ both exclude all elements  in $A\cup B$ but also in $A\cap B$.
If case 3) $x\in (A\cup B)\setminus (A\cap B)$ because .... $x \in A\cup B$ and $x\not \in A\cap B$.....  And $x\in A\cup B$ so either $x$ is in $A$ or $x$ is in $B$.  If $x \in A$ then  $x\not \in A\cap B$ so $x \not \in B$ and $x\in A\cap B^c$.  If $x \in B$ then $x \not \in A\cap B$ so $x \not \in A$ so $x \in A^c \cap B$.  Either way either $x \in (A\cap B^c)$ or $x \in (A^c \cap B)$ so $x \in (A^c \cap B)\cup (A\cap B^c)$.
So $(A\cup B)\setminus (A\cap B)$ and $(A^c\cap B)\cup (A\cup B^c)$ both include all elements  in $A\cup B$ but not in $A\cap B$.
....
So $(A\cup B)\setminus (A\cap B)$ and $(A^c\cap B)\cup (A\cup B^c)$ both contain and exclude the exact same elements.  SO they are equal.
