# Math Help - Symmetric Difference Proof

1. $\displaystyle \text{let} \ x \in (A \cap B^{c}) \cup (B \cap A^{c})$

$\displaystyle \text{then} \ x \in (A \cap B^{c}) \ \text{or} \ x \in (B \cap A^{c})$

$\text{so} \ (x \in A \ \text{and} \ x \in B^{c}) \ \text{or} \ (x \in B \ \text{and} \ x \in A^{c})$

$\Big((x \in A \ \text{and} \ x \in B^{c}) \ \text{or} \ x \in B \Big) \ \text{and} \ \Big((x \in A \ \text{and} \ x \in B^{c}) \ \text{or} \ x \in A^{c} \Big)$

$\Big((x \in A \ \text{or} \ x\in B) \ \text{and} \ (x \in B^{c} \ \text{or} \ x \in B) \Big) \ \text{and} \ \Big((x \in A \ \text{or} \ x \in A^{c}) \ \text{and} \ (x \in B^{c} \ \text{or} \ x \in A^{c}) \Big)$

$\Big((x \in (A \cup B) \cap U \Big) \ \text{and} \ \Big(x \in U \cap (B^{c} \cup A^{c}) \Big)$

$x \in (A \cup B) \ \text{and} \ x \in (A \cap B)^{c}$

$x \in (A \cup B) \cap (A \cap B)^{c}$

2. as this seems to be a great solution I am not supposed to be using anything besides the basic set theory laws

3. Random Variable's solution is actually very nice since the proof doesn't involve knowledge of any laws or theorems - just the basic definitions. It's probably worth it for you to review his argument. The argument using the laws is much quicker and simpler than this argument (as are most arguments that use already established results).

4. Originally Posted by dwsmith
$(A\cup B)- (A\cap B)=(A\cup B)\cap (A\cap B)^c=(A\cup B)\cap (A^c\cup B^c)=...$

Use distributive law here.

What will you get?
$[(A\cup B)\cap A^c]\cup [(A\cup B)\cap B^c]=(B\cap A^c)\cup (A\cap B^c)=(B-A)\cup (A-B)$

Here is the result. If you would have just tried to distribute, you may have had the answer by using the laws per your request.

5. I was having trouble with your notation versus mine, I just couldn't translate your solution/hints into my notation

6. You were using / and I was using -. I am just used to using A-B for what is in A but not in B.

7. I was talking more about the compliment c here
the variations of De Morgan I learned don't use c so I was having trouble switching from one to the other

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