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 August 8th, 2012, 08:39 AM #1 Senior Member   Joined: Sep 2011 Posts: 140 Thanks: 0 Set theory How do you prove this? $A \cap (B \cap C'=(A\cap B) \cap (A\cap C)'" /> This is how i am approaching it but im stuck a one point: So lets have $x \in A \cap (B \cap C'" />, which means that $x \in A$ and$x \in (B \cap C'" />, this also means that $x \in B$ and $x \in C'$, Then we have: $x \in A$ and $x \in B$ so $x \in A \cap B$. But what do i do with the $C'$? Also my teacher said we have to show for both LHS and RHS. If i start with RHS this time it becomes weirder. let $x \in (A\cap B) \cap (A\cap C)'$, this means that $x \in (A\cap B)$ and $x \in (A\cap C)',$ which gives $x \in A ,$ $x \in B$, $x \in A'$ and $x \in C'$. How is this possible when $x \in A$ and we have $x \in A'$ . Im so confused with this proof. Similarly how do you prove: $(A \cup B) \cap (A' \cup B) = B$ Can anyone help me understand this clearly?? will appreciate any help.
 August 8th, 2012, 09:18 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Set theory Use De'Morgan's law.
August 8th, 2012, 09:26 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Set theory

Hello, gaussrelatz!

Quote:
 $\text{How do you prove this? }\:A\,\cap\,(B\,\cap\,C' ) \;=\; (A\,\cap\,B)\,\cap\,(A\,\cap\,C)#39;$

$\text{Start with the right side . . .}$

$\begin{array}{ccccccc}1. & (A\,\cap\,B)\,\cap\,(A\,\cap\,C)' &\;\;\;& 1. & \text{Given} \\ \\ \\
2. & (A\,\cap\,B)\,\cap\,(A' \,\cup\,C') && 2. & \text{DeMorgan} \\ \\ \\
3. & \big[(A\,\cap\,B)\,\cap\,A'\big]\,\cup\,\big[(A\,\cap\,B)\,\cap\,C'\big] && 3. & \text{Distributive} \\ \\ \\
4. & \big[(A\,\cap\,A')\,\cap\,B\big] \,\cup\,\big[A\,\cap\,(B\,\cap\,C')\big] && 4. & \text{Comm. Assoc.} \\ \\ \\
5. & \big[\emptyset \,\cap\,B\big] \,\cup\,\big[A\,\cap\,(B\,\cap\,C')\big] && 5. & S\,\cap\,S' \:=\:\emptyset \\ \\ \\
6. & \emptyset \,\cup\,\big[A\,\cap\,(B\,\cap\,C')\big] && 6. & \emptyset\,\cap\,S \:=\:\emptyset \\ \\ \\
7. & A\,\cap\,(B\,\cap\,C') && 7. & \emptyset\,\cup\,S \:=\:S \end{array}$

Quote:
 $\text{Similarly, how do you prove: }\:(A\,\cup\,B)\,\cap\,(A'\,\cup\,B) \;=\; B$

$\text{Start with the left side . . .}$

$\begin{array}{ccccccc}
1. & (A\,\cup\,B)\,\cap\,(A'\,\cup\,B) &\;\;\;& 1. & \text{Given} \\ \\ \\
2. & (A\,\cap\,A')\,\cup\,B && 2. & \text{Distr.} \\ \\ \\
3. & \emptyset \,\cup\,B && 3. & S\,\cap\,S' \:=\:\emptyset \\ \\ \\ \\
4. & B && 4. & \emptyset \,\cup\,S \:=\:S \end{array}$

 August 8th, 2012, 09:35 AM #4 Senior Member   Joined: Sep 2011 Posts: 140 Thanks: 0 Re: Set theory Thank you very much for the help. Ok i see the proof using the algebra of sets, could you also please give me a proof using the elements for example, by beginning with : let $x \in A$etc... and show from there. I tried to prove it set algebraically but my teacher said i should be using the proof as i mentioned like above in my post.

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