Set Operations Question (subtraction, union, intersection) I have a questions reguarding order of operations for sets:
$\forall A,B $ $(A-B) \cup (B - A) \cup (A \cap B) = A \cup B$
If I'm to understand this correctly, the first union $\big((A-B) \cup (B - A)\big)$ would be an empty set correct? so if I were to take that set and combine it with $(A \cap B)$ wouldn't I end up with the intersection of A and B and not the union?
I also have a second question:
$\forall A,B$ $ (A\triangle B) \triangle C = A \triangle (B \triangle C)$
Which has the same principles, and I believe once I figure out the first I can figure out the second.
EDIT: $\triangle$ is symmetric difference
 A: (A−B) gives you the slice of A with the B part punched out. It's like with a Venn Diagram For A and B except you remove the middle intersecting part AND the B part leaving the sliver that is ONLY A. The same goes for (B−A) which leaves you the sliver that is ONLY B.
((A−B)∪(B−A)) is the union (not intersection) of the two slivers. That is analogous to sort of adding them together to get a new set C. If you were to intersect them, then you would get the empty set.
The remaining (A∩B) is that center portion of your standard Venn Diagram, the part where the A and B both overlap. This is the intersection. 
So we see that ((A−B)∪(B−A)) gives all of A and B except for the intersecting part and (A∩B) gives ONLY the intersecting part. Putting them together, in union, produces the whole of A∪B .
Your second question looks like the associative property. See 'http://en.wikipedia.org/wiki/Associative_property'.
With the triangle symbols as symmetric difference you would have A△B = ((A−B)∪(B−A)) . The key is to use the fact that (A−B) = (A∩not(B)) adn DeMorgan's Laws. Then do:
(A△B)△C = ((A−B)∪(B−A))△C
= ((((A−B)∪(B−A))−C)∪(C−((A−B)∪(B−A))))
= ((((A−B)∪(B−A))−C)∪(C−((A−B)∪(B−A))))
= ((((A∩not(B))∪(B∩not(A)))−C)∪(C−((A∩not(B))∪(B∩not(A)))))
= ((((A∩not(B))∪(B∩not(A)))∩not(C))∪(C∩not((A∩not(B))∪(B∩not(A)))))
= ((((A∩not(B))∪(B∩not(A)))∩not(C))∪(C∩(not(A∩not(B))∩not(B∩not(A)))))
// use DeMorgan's Laws to get from the above line to the below line. 
= ((A∩not((B∩not(C))∪(C∩not(B))))∪(((B∩not(C))∪(C∩not(B)))∩not(A)))
= ((A−((B∩not(C))∪(C∩not(B))))∪(((B∩not(C))∪(C∩not(B)))−A))
= ((A−((B−C)∪(C−B)))∪(((B−C)∪(C−B))−A))
= A△((B−C)∪(C−B))
= A△(B△C)
DeMorgan's Laws can be found at 'http://mathworld.wolfram.com/deMorgansLaws.html' and they are to be applied in the middle step that was omitted (omitted mostly because I was losing track of all the parentheses).
