Prove $A+B= A \cup B$ if and only if $A \cap B = \emptyset$ using the definition of $A+B$ Let $A$ and $B$ be sets. Define the symmetric difference of $A$ and $B$, written $A+B$, by $A+B=(A \cup B) \backslash (A \cap B)$.
Prove $A+B= A \cup B$ if and only if $A \cap B = \emptyset$.
My attempt:
Part 1b. If $A+B = A \cup B$, then $ A \cap B = \emptyset$
$A+B=(A \cup B) \backslash (A \cap B)$.
There shouldn't be any elements inside $A \cap B$.
$A+B=(A \cup B) \backslash (\emptyset)$
Therefore, $A \cap B = \emptyset$. 
Edit: What if I put...
Definition: Let $(A \cup B)$ and $(A \cap B)$ be sets. The complement of $(A \cup B)$ relative to $(A \cap B)$, written $ (A \cup B)\backslash (A \cap B)$ is the set $(A \cup B) \backslash (A \cap B) = [x: x \in (A \cup B) \land x \notin (A \cap B)]$
While the elements are inside $(A \cup B)$, they don't belong in $(A \cap B)$
hmmmm they isn't any element in common...so it's an emptyset... wait a sec... emptyset means that there aren't any elements. 
Part 2b. If $A \cap B = \emptyset$, then $A+B = A \cup B$
$A+B=(A \cup B) \backslash (\emptyset)$
Definition: Let $(A \cup B)$ and $\emptyset$ be sets. The complement of $(A \cup B)$ relative to $\emptyset$, written $ (A \cup B)\backslash \emptyset$ is the set $(A \cup B) \backslash \emptyset = [x: x \in (A \cup B) \land x \notin \emptyset]$
There are elements inside $A \cup B$, but not in $A \cap B$.
As a result, $A+B = (A \cup B)$
Did I do this correctly?
 A: I will try to show you the result exploiting the definitions.
Define the symmetric difference of $A$ and $B$ (written $A+B$), by 

$A+B = (A \cup B) \backslash (A \cap B)$.

We have that 

$x \in A+B \quad$ iff $\quad ( x \in A \lor x \in B) \quad$ and $\quad \lnot (x \in A \land x \in B)$

But we have also that 

$x \in A \cup B \quad$ iff $\quad ( x \in A \lor x \in B)$

Now we want that $A + B$ must be equal to $A \cup B$.
If we "compare" the two above condition, we can write them as 

$P \land Q$ and $P$, and we want that $P \land Q \equiv P$.

By truth-table, we can check that $P \land Q \equiv P$ iff $Q$ is identically true.
So, we want that our condition $\lnot (x \in A \land x \in B)$ must be identically true.
This is simply :

$\forall x \lnot (x \in A \land x \in B)$.

By definition, $x \in A \cap B$ iff $x \in A \land x \in B$.
So, our condition is : $\forall x (x \notin A \cap B)$ and this, in turn, means that $A \cap B$ is $\emptyset$ (the empty set).
A: A + B := A\B U B\A. We will use this definition to prove the claim and it is equivalent to the definition given in the statement above. 
===> If A + B = A U B. Then we show the intersection of A and B to be empty. If this is not true then there is a y in A and B. So y in A U B and therefore y in A + B. So that means y is in A\B or B\A. But neither of these can happen since y is in both sets A and B. Thus we obtain a contradiction. So the intersection of A and B is empty. 
<=== If A intersects B is empty. We show A + B = A U B. We have A U B = A\B U B\A U C,
where C is the intersection of A and B. So if C = empty set, then A U B = A\B U B\A = A + B
