Symmetric Difference Proof

**gutnedawg** · January 6th 2011, 11:23 AM

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.

**dwsmith** · January 6th 2011, 11:26 AM

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

**gutnedawg** · January 6th 2011, 11:31 AM

not sure how you came to this..

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

**DrSteve** · January 6th 2011, 11:39 AM

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.

**dwsmith** · January 6th 2011, 11:40 AM

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

**Ackbeet** · January 6th 2011, 11:44 AM

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!

**gutnedawg** · January 6th 2011, 11:45 AM

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

**dwsmith** · January 6th 2011, 11:45 AM

complements

**dwsmith** · January 6th 2011, 11:53 AM

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.

**gutnedawg** · January 6th 2011, 11:55 AM

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

**DrSteve** · January 6th 2011, 11:57 AM

$A\setminus B$ 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.

**dwsmith** · January 6th 2011, 12:01 PM

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

**gutnedawg** · January 6th 2011, 12:02 PM

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

**gutnedawg** · January 6th 2011, 12:04 PM

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

**dwsmith** · January 6th 2011, 12:05 PM

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

Use distributive law here.

What will you get?

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