# Thread: Symmetric Difference Proof

1. ## Symmetric Difference Proof

We know that

$A \triangle B = (A\backslash B) \cup (B \backslash A)$

and we also know that

$A \triangle B = (A \cup B) \backslash (A \cap B)$

show that

$(A\backslash B) \cup (B \backslash A) = (A \cup B) \backslash (A \cap B)$

using only the following laws: DeMorgan's, Complement, Distributive, Absorption... basically the main set theory laws.

I have attempted to do this but I'm running into some trouble trying to figure out how to move from the union of the two to the difference of the two.

2. [(a-b) u b] - [(a-b) u a] ......

3. not sure how you came to this..

This is my first experience with set theory so I might be a little slow

4. Here's most of the work:

$(A\cap B')\cup (B\cap A') = [(A\cap B')\cup B]\cap [(A\cap B']\cup A']=[(A\cup B)\cap (B'\cup B)]\cap[(A\cup A')\cap (B' \cup A')]$

See if you can finish it.

5. Or you can write $(A\cup B)- (A\cap B)=(A\cup B)\cap (A\cap B)^c=(A\cup B)\cap (A^c\cup B^c)=...$

6. Reply to dwsmith:

If you're in ZF set theory, set complements aren't allowed. In naive set theory, they are. Just a side note.

[EDIT]: Spelling of "complements", as per dwsmith's post below. What a nice set!

7. so how do I start from the first step of (A-B) union (B-A)?

8. complements

9. Originally Posted by Ackbeet
Reply to dwsmith:

If you're in ZF set theory, set complements aren't allowed. In naive set theory, they are. Just a side note.

[EDIT]: Spelling of "complements", as per dwsmith's post below. What a nice set!
I have to get something right every now and then.

10. the only complement laws I know are $A \cup (B \backslash A) = A \cup B$
and $A \cap (B \backslash A) = 0$

11. $A\setminus B$ is usually defined as $A\cap B'$ (I'm using ' for complement)

It literally translates to "the elements in A and not in B"

The first step in my argument above is just this translation.

12. Originally Posted by gutnedawg
the only complement laws I know are $A \cup (B \backslash A) = A \cup B$
and $A \cap (B \backslash A) = 0$
$\emptyset \ \mbox{or} \ \{\}$ not = 0

13. I understand this however this is not the notation that we are using. I was instructed that the absorption laws would prove to be very useful

14. Originally Posted by dwsmith
$\emptyset \ \mbox{or} \ \{\}$ not = 0
Yea I know I just didn't feel like looking up how to write the null set

15. $(A\cup B)- (A\cap B)=(A\cup B)\cap (A\cap B)^c=(A\cup B)\cap (A^c\cup B^c)=...$

Use distributive law here.

What will you get?

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