# Set Theory problem

• February 24th 2008, 02:35 PM
Foink
Set Theory problem

For all sets A, B, C, prove
If $A-C = \emptyset$ then $(A \cup B) - (B \cap C) = (A-B) \cup (B-C)$

This is what I have done so far, but I seem to be stuck:

$(A \cup B) - (B \cap C)$ Thm $A-B=A \cap B'$
$(A \cup B) \cap (B \cap C)'$ DeMorgans
$(A \cup B) \cap (B' \cup C')$

I can't figure it out from there. I'm guessing it might have something to do with distribution but i'm not sure. Any help is appreciated
• February 24th 2008, 03:30 PM
Plato
$\left( {A \cup B} \right) \cap \left( {B' \cup C'} \right) = \left( {A \cap B'} \right) \cup \left[ {\left( {A \cap C'} \right) \cup \left( {B \cap C'} \right)} \right]
$
• February 24th 2008, 04:34 PM
Soroban
Hello, Foink!

You're right . . . the Distributive Property will finish it.

Quote:

For all sets $A, B, C,$ prove:
If $A-C = \emptyset$ then $(A \cup B) - (B \cap C) = (A-B) \cup (B-C)$

This is what I have done so far:

$\begin{array}{cc}(A \cup B) - (B \cap C) & \text{Given} \\
(A \cup B) \cap (B \cap C)' & \text{d{e}f. subtraction} \\
(A \cup B) \cap (B' \cup C\,') & \text{DeMorgan} \end{array}$

. . . Good!

We know: . $\begin{array}{cc}S \cap S' \:=\:\emptyset & {\color{blue}[1]} \\ S \cup \emptyset \:=\:S & {\color{blue}[2]}\end{array}$

We are also given: . $A - C \:=\:\emptyset\quad\Rightarrow\quad A \cap C\,' \:=\:\emptyset\;\;{\color{blue}[3]}$

$\begin{array}{cc}
[(A \cup B) \cap B'] \cup [(A \cup B) \cap C\,'] & \text{Distr.} \\ \\

[(A \cap B') \cup (B \cap B')] \cup [(A \cap C\,') \cup (B \cap C\,')] & \text{Distr.} \\ \\

[(A \cap B') \cup \emptyset\:] \cup[\:\emptyset \cup (B \cap C\,')] & [1], [3] \\ \\

(A \cap B') \cup (B \cap C\,') & [2] \\ \\

(A - B) \cup (B - C) & \text{d{e}f. subtraction}
\end{array}$

• February 26th 2008, 04:06 PM
CPR
What is B' or C' in your problem?
Just wondering if the B' meant elements not in set B or elements not in set C.
• February 26th 2008, 04:08 PM
Jhevon
Quote:

Originally Posted by CPR
Just wondering if the B' meant elements not in set B or elements not in set C.

correct. B' and C' means B-compliment and C-compliment respectively
• February 26th 2008, 04:33 PM
CPR
Question on A' U B
A= { 2, 4, 5}
B = {1,2,4,5,6}

so what will be A' U B

and let say that there was an U={ 1,2, 3, 4, 5, 6,7}

What would B'?
• February 26th 2008, 04:36 PM
Jhevon
Quote:

Originally Posted by CPR
A= { 2, 4, 5}
B = {1,2,4,5,6}

so what will be A' U B

and let say that there was an U={ 1,2, 3, 4, 5, 6,7}

What would B'?

given that U, A' = {1, 3, 6, 7} and B' = {3,7}

you can find A' U B from that
• February 26th 2008, 04:42 PM
CPR
Thanks
It would be {1,6}.
• February 26th 2008, 04:54 PM
Jhevon
Quote:

Originally Posted by CPR
It would be {1,6}.

yes