Hi using set algebra I have to simplify the following:

( (A n B n C) n (A u B u C') u (A u B') )

the answer I have been given is A n (B' n C) but I get A u B'

can anybody verify and break down the steps used?

Many thanks

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- Jan 3rd 2009, 07:02 AMbarc0deSet Algebra
Hi using set algebra I have to simplify the following:

( (A n B n C) n (A u B u C') u (A u B') )

the answer I have been given is A n (B' n C) but I get A u B'

can anybody verify and break down the steps used?

Many thanks - Jan 3rd 2009, 07:03 AMMush
- Jan 3rd 2009, 07:05 AMbarc0de
- Jan 3rd 2009, 07:22 AMPlato
- Jan 3rd 2009, 07:27 AMSoroban
Hello, barc0de!

Quote:

Simplify: .,$\displaystyle (A \cup B \cup C) \cap [A \cup (B \cap C)] $

. . . $\displaystyle \begin{array}{cccc}\bigg[A \cup B \cup C\bigg] \cap \bigg[A \cup (B \cap C)\bigg] & &\text{Given} \\ \bigg[ A \cup B \cup C \bigg] \cap \bigg[ (A \cup B) \cap (A \cup C)\bigg] & & \text{Distributive} \\ \\[-4mm] \bigg[(A \cup B \cup C) \cap (A \cup B)\bigg] \cap \bigg[A \cup C\bigg] && \text{Associative} \\ \bigg[A \cup B\bigg] \cap \bigg[A \cup C\bigg] && \text{Subset} \\ \\[-4mm] A \cup (B \cap C) && \text{Distributive}\end{array}$

- Jan 3rd 2009, 07:38 AMPlato
You need to edit the post by adding some grouping symbols.

Because $\displaystyle \left( {X \cap Y} \right) \cup Z \ne X \cap \left( {Y \cup Z} \right)$.

Which one is your problem? - Jan 3rd 2009, 07:43 AMPlato
**Your edit did nothing to clear away the confusion.** - Jan 3rd 2009, 07:48 AMbarc0de
- Jan 3rd 2009, 07:49 AMbarc0de
Thanks Soroban. Could you take a look at the following please:

(A n B n C) n (A u B u C') u (A u B')

the answer I have been given is A n (B' n C) but I get A u B'

can you verify and break down the steps used - Jan 3rd 2009, 07:55 AMPlato
Which way are they grouped?

$\displaystyle \begin{gathered}

\left[ {\left( {A \cap B \cap C} \right) \cap \left( {A \cup B \cup C'} \right)} \right] \cup \left( {A \cup B'} \right) \hfill \\

\left( {A \cap B \cap C} \right) \cap \left[ {\left( {A \cup B \cup C'} \right) \cup \left( {A \cup B'} \right)} \right] \hfill \\

\end{gathered}

$

It makes a huge difference. - Jan 3rd 2009, 07:57 AMbarc0de
This may be where the problem lies in getting 2 different answers. I have typed the problem in as I have been given it. Would it be too much trouble for you to do both?? I would really appreciate it.

Many thanks - Jan 3rd 2009, 08:10 AMPlato
$\displaystyle \begin{gathered}

\left[ {\left( {A \cap B \cap C} \right) \cap \left( {A \cup B \cup C'} \right)} \right] \cup \left( {A \cup B'} \right) = A \cup B' \hfill \\

\left( {A \cap B \cap C} \right) \cap \left[ {\left( {A \cup B \cup C'} \right) \cup \left( {A \cup B'} \right)} \right] = \left( {A \cap B \cap C} \right) \hfill \\

\end{gathered} $ - Jan 3rd 2009, 08:21 AMbarc0de
Could you please break down the steps used in finding the first solution?