# Homework Help: Abstract - Prove (A-B)union(B-A)=(AunionB)-(AintersectB)

1. Sep 5, 2010

### kathrynag

1. The problem statement, all variables and given/known data

Prove:
(A-B)$$\cup$$(B-A)=(A$$\cup$$B)-(A$$\cap$$B)

2. Relevant equations

3. The attempt at a solution
We need to show (A-B)$$\cup$$(B-A)$$\subseteq$$(A$$\cup$$B)-(A$$\cap$$B)
and (A$$\cup$$B)-(A$$\cap$$B)$$\supseteq$$(A-B)$$\cup$$(B-A).

We begin by showing the first:
Let x$$\in$$(A-B)$$\cup$$(B-A).
By definition of union, x$$\in$$A-B or x$$\in$$B-A.
If x$$\in$$A-B, we know x$$\in$$A .......

This is where I've begun to get stuck. Not sure where to go next.

2. Sep 5, 2010

### Office_Shredder

Staff Emeritus
We know x is in A, AND x is NOT in B.

What does it mean for an element to be in $$A\cup B- A\cap B$$? Is x inside of it in this case?

3. Sep 5, 2010

### kathrynag

An elelemnt is in A or B and not in A and B.

4. Sep 5, 2010

### Office_Shredder

Staff Emeritus
So does the x that we were looking at satisfy that requirement?

5. Sep 5, 2010

### kathrynag

I'm thinking yes, but I'm having trouble visualizing that just from the x an element of A and not B.

6. Sep 5, 2010

### vela

Staff Emeritus
Just break it down into simple pieces. If x is an element of A and not an element of B:

1. Is x in A or B?
2. Is x in A and B?

Therefore...

7. Sep 5, 2010

### kathrynag

x is in A or B, but not in A and B.
Therefore, we have the right side of the equation.

Ok, but what about if x is an element of B-A?
x is in B, bot not in A.
Then x is in B or A, but not in B and A.
So, therefore, we have the right side of the equation.

Ok, and then I just work the other way to prove equality?

8. Sep 5, 2010

### vela

Staff Emeritus
Yup.

9. Sep 6, 2010

### kathrynag

Thanks a lot. That makes a lot more sense now.