# 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

• $(A-B) \cup (B-A)$ is not empty. – Extremal Dec 6 '14 at 3:24
• Think about the first union more carefully. If it were an intersection, you would have the empty set but this is not the case. – mphy Dec 6 '14 at 3:24
• Also, could you define the triangle for the second part? I am not familiar with that notation. – mphy Dec 6 '14 at 3:26
• I think the first union will be $(A-B)\cup(B-A)=A\cup B-A\cap B$. Because the set $(A-B)$ will contain only elements of set A that doenst as on B and $(B-A)$ will contain elements of B that doenst as on A, then the union of the two set will be the union of set A and B except for the elements of intersection – cand Dec 6 '14 at 3:26
• $(A-B) \cup (B-A)= (A \cap B^c) \cup (B \cap A^c)=(A\cup B)-(A\cap B)$ – Extremal Dec 6 '14 at 3:26

(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).

• How would I go about proving it for the second question? would I simply expand and re-organize the first one to prove it's equal to the second? – dustyblade Dec 6 '14 at 4:04