Discrete Mathematics - (A∪B)-(A∩B)=(A-B)∪(B-A) - prove by cases?

[nigara]

Discrete Mathematics - (A∪B)-(A∩B)=(A-B)∪(B-A) - prove by cases??

Hi, I'm new to these forums so please redirect me if I've posted this in the wrong place.
I'm trying to graduate and this is my last class, but as I'm not a math major, I'm really struggling with this particular problem. I've been able to manage most other simple proofs pretty well, but this one has me stumped:

(A∪B)-(A∩B)=(A-B)∪(B-A)

The instructor hinted that this problem could/should be done with Proof by Cases. Any chance someone could walk me through this?? Like I said, I've managed to handle other similar problems but the elements of this one have me thrown... any help would be appreciated. I tend to learn well by example, but there's nothing in my notes/textbook that show a similarly structured problem...

Thanks!

 

[TylerH]

I don't see how to do it with with proof by cases. I would do it by expanding the set operations into set builder definitions. $\left(A \cup B \right) - \left( A \cup B \right) = \left\{ a : \left( a \in A \vee a \in B \right) \wedge \neg \left( a \in A \wedge a \in B \right) \right\}$ and $\left( A-B \right) \cup \left( B-A \right) = \left\{ a : \left( a \in A \wedge a \notin B \right) \vee \left( a \in B \wedge a \notin A \right) \right\}$. You could prove that $\left( a \in A \wedge a \notin B \right) \vee \left( a \in B \wedge a \notin A \right) \Leftrightarrow \left( a \in A \vee a \in B \right) \wedge \neg \left( a \in A \wedge a \in B \right)$, which would imply what you wish to show.

That would be the hard way to do it. Hopefully, someone will come along and explain the method your instructor was referring to.

 

[HallsofIvy]

If x is any member of the universal set, then one of four "cases' applies:
1) x is in A but not in B.
2) x is in B but not in A.
3) x is in both A and B.
4) x in in neither A nor B.

For each of those cases, determine if x is in $A\cup B- A\cap B$ and if it is in $(A- B)\cup(B- A)$. If, in every case, x is in one if and only if it is in the other, then the two sets are equal.

 

[TylerH]

If x is any member of the universal set, then one of four "cases' applies:
1) x is in A but not in B.
2) x is in B but not in A.
3) x is in both A and B.
4) x in in neither A nor B.

For each of those cases, determine if x is in $A\cup B- A\cap B$ and if it is in $(A- B)\cup(B- A)$. If, in every case, x is in one if and only if it is in the other, then the two sets are equal.

OH! That's a lot simpler.

 

[solakis]

If x is any member of the universal set, then one of four "cases' applies:
1) x is in A but not in B.
2) x is in B but not in A.
3) x is in both A and B.
4) x in in neither A nor B.

For each of those cases, determine if x is in $A\cup B- A\cap B$ and if it is in $(A- B)\cup(B- A)$. If, in every case, x is in one if and only if it is in the other, then the two sets are equal.

Could you please show me how the case: xεΑ and xεΒ,

Would make the sets: $A\cup B- A\cap B$ and $(A- B)\cup(B- A)$. equal??