When is the set statement: (A⊕B) = (A ∪ B) true? "When is the set statement:

(A⊕B) = (A ∪ B)

a true statement? Is it true sometimes, never, or always? If it is sometimes, state the cases where it is."
How would you go about finding the answer to the question or ones like this one? 
Thanks for your time!
 A: If I've made the right assumptions in my comment above, a good way to approach this problem is by drawing a Venn diagram.
Here's $A\oplus B$:

Here's $A\cup B$:

So, the area that's filled in in $A\cup B$ but not $A\oplus B$ is $A\cap B$. What do I need to be true about $A\cap B$ to make the two Venn diagrams have the same area filled in?
A: If you define $A \bigoplus B$ as Kevin did we see that it is true when $A \cap B = \emptyset$. This is because $$(A \cap B^{c}) \cup (B \cap A^{c}) = A \cup B - (A \cap B) = A \cup B.$$ This tells us that for $A \cap B \neq \emptyset$ they are not equal. 
A: To add to the previous answers, you might try creating a truth table for or and xor with two bits, then three bits and seeing the results.
The Venn diagram, trying some cases and the other answers should get you going!
A: HINT: $A\oplus B$ is the set of objects that belong to exactly one of the sets $A$ and $B$. This should be fairly easy to see whether you define $A\oplus B$ as $(A\setminus B)\cup(B\setminus A)$ or as $(A\cup B)\setminus(A\cap B)$. $A\cup B$, on the other hand, is the set of things that are in at least one of the sets $A$ and $B$. Obviously these aren’t always going to be the same. What possibility do you have to rule out in order to get in exactly one of the sets to mean the same as in at least one of the sets?
A: I would calculate this by simplifying at the logic level, using the not-so-well-known definition
$$
x \in A \oplus B \;\equiv\; x \in A \not\equiv x \in B
$$
We can simplify $\;A \oplus B \;=\; A \cup B\;$ as follows:
\begin{align}
& A \oplus B \;=\; A \cup B \\
\equiv & \;\;\;\;\;\text{"set extensionality"} \\
& \langle \forall x :: x \in A \oplus B \;\equiv\; x \in A \cup B \rangle \\
\equiv & \;\;\;\;\;\text{"definitions of $\;\oplus, \cup\;$; drop parentheses since $\;\not\equiv, \equiv\;$ are mutually associative"} \\
& \langle \forall x :: x \in A \;\not\equiv\; x \in B \;\equiv\; x \in A \lor x \in B \rangle \\
\equiv & \;\;\;\;\;\text{"bring $\;\lnot\;$ to the outside -- to prepare for the next step"} \\
& \langle \forall x :: \lnot(x \in A \;\equiv\; x \in B \;\equiv\; x \in A \lor x \in B) \rangle \\
\equiv & \;\;\;\;\;\text{"apply (what Dijkstra c.s. call) the golden rule"} \\
& \langle \forall x :: \lnot(x \in A \land x \in B) \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$; definition of $\;\varnothing\;$"} \\
& A \cap B = \varnothing
\end{align}
