# Need confirmation on set identity problem

• Feb 2nd 2010, 01:24 PM
Runty
Need confirmation on set identity problem
I believe I may have solved this properly, but I have some nagging doubts about my process. Here is the question.

Show whether the set equality $\displaystyle (A-B) \cup (B-A) = (A \cup B) - (A \cap B)$ is true, using set identities.

My process has led me to believe it is false, but I can't shake the feeling that I made a mistake somewhere. Here is my work.

$\displaystyle (A-B) \cup (B-A) = (A \cup B) - (A \cap B)$
$\displaystyle (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A \cap B})$
$\displaystyle (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A} \cup \overline{B})$ De Morgan's Law
$\displaystyle (A \cap \overline{B}) \cup (\overline{A} \cap B) = (A \cup B) \cap (\overline{A} \cup \overline{B})$ Commutative Law
$\displaystyle (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cap \overline{A}) \cup (B \cap \overline{B})$ Distributive Laws (Twice) (this is where I think I screwed up)
$\displaystyle U \cap U = \emptyset \cup \emptyset$ Complement Laws (multiple times)
$\displaystyle U = \emptyset$ Identity Laws

I do not know whether or not my application of the Distributive Laws in that case was correct. Could someone please check this?
• Feb 2nd 2010, 01:36 PM
Drexel28
Quote:

Originally Posted by Runty
I believe I may have solved this properly, but I have some nagging doubts about my process. Here is the question.

Show whether the set equality $\displaystyle (A-B) \cup (B-A) = (A \cup B) - (A \cap B)$ is true, using set identities.

My process has led me to believe it is false, but I can't shake the feeling that I made a mistake somewhere. Here is my work.

$\displaystyle (A-B) \cup (B-A) = (A \cup B) - (A \cap B)$
$\displaystyle (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A \cap B})$
$\displaystyle (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A} \cup \overline{B})$ De Morgan's Law
$\displaystyle (A \cap \overline{B}) \cup (\overline{A} \cap B) = (A \cup B) \cap (\overline{A} \cup \overline{B})$ Commutative Law
$\displaystyle (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cap \overline{A}) \cup (B \cap \overline{B})$ Distributive Laws (Twice) (this is where I think I screwed up)
$\displaystyle U \cap U = \emptyset \cup \emptyset$ Complement Laws (multiple times)
$\displaystyle U = \emptyset$ Identity Laws

I do not know whether or not my application of the Distributive Laws in that case was correct. Could someone please check this?

Too much stuff going on there. Just show us what you did in the distributive step...show both distributions.
• Feb 2nd 2010, 01:38 PM
Runty
$\displaystyle (A \cap \overline{B}) \cup (\overline{A} \cap B) = (A \cup B) \cap (\overline{A} \cup \overline{B})$ Starting point
$\displaystyle (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cup B) \cap (\overline{A} \cup \overline{B})$ Distributive step #1
$\displaystyle (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cap \overline{A}) \cup (B \cap \overline{B})$ Distributive step #2
• Feb 2nd 2010, 01:52 PM
Plato
$\displaystyle \begin{gathered} \left( {A \cup B} \right) \cap \left( {\overline A \cup \overline B } \right) \hfill \\ \left[ {A \cap \left( {\overline A \cup \overline B } \right)} \right] \cup \left[ {B \cap \left( {\overline A \cup \overline B } \right)} \right] \hfill \\ \left[ {A \cap \overline B } \right] \cup \left[ {B \cap \overline A } \right] \hfill \\ \end{gathered}$
• Feb 2nd 2010, 02:08 PM
Runty
Quote:

Originally Posted by Plato
$\displaystyle \begin{gathered} \left( {A \cup B} \right) \cap \left( {\overline A \cup \overline B } \right) \hfill \\ \left[ {A \cap \left( {\overline A \cup \overline B } \right)} \right] \cup \left[ {B \cap \left( {\overline A \cup \overline B } \right)} \right] \hfill \\ \left[ {A \cap \overline B } \right] \cup \left[ {B \cap \overline A } \right] \hfill \\ \end{gathered}$

So I was wrong about it being false. Thank you.