Set Algebra

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January 3rd 2009, 07:02 AM
barc0de

Set 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
January 3rd 2009, 07:03 AM
Mush

Quote:

Originally Posted by barc0de

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

What is B'?
January 3rd 2009, 07:05 AM
barc0de

Quote:

Originally Posted by Mush

What is B'?

B' is the complement of B
January 3rd 2009, 07:22 AM
Plato

Quote:

Originally Posted by barc0de

I have to simplify the following: (A u B u C) n (A u (B n C))

Here is a start:
$\left[ {A \cup B \cup C} \right] \cap \left[ {A \cup \left( {B \cap C} \right)} \right] = A \cup \left[ {\left( {B \cup C} \right) \cap \left( {B \cap C} \right)} \right]$
January 3rd 2009, 07:27 AM
Soroban

Hello, barc0de!

Quote:

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

. . . $\begin{array}{cccc}\bigg[A \cup B \cup C\bigg] \cap \bigg[A \cup (B \cap C)\bigg] & &\text{Given} \\ \bigg[ A \cup B \cup C \bigg] \cap \bigg[ (A \cup B) \cap (A \cup C)\bigg] & & \text{Distributive} \\ \\[-4mm] \bigg[(A \cup B \cup C) \cap (A \cup B)\bigg] \cap \bigg[A \cup C\bigg] && \text{Associative} \\ \bigg[A \cup B\bigg] \cap \bigg[A \cup C\bigg] && \text{Subset} \\ \\[-4mm] A \cup (B \cap C) && \text{Distributive}\end{array}$
January 3rd 2009, 07:38 AM
Plato

You need to edit the post by adding some grouping symbols.
Because $\left( {X \cap Y} \right) \cup Z \ne X \cap \left( {Y \cup Z} \right)$ .
Which one is your problem?
January 3rd 2009, 07:43 AM
Plato

Your edit did nothing to clear away the confusion.
January 3rd 2009, 07:48 AM
barc0de

Quote:

Originally Posted by Plato

Your edit did nothing to clear away the confusion.

Sorry I'm not sure what you mean.

I have to algebraically simplify:

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

that is all I have
January 3rd 2009, 07:49 AM
barc0de

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
January 3rd 2009, 07:55 AM
Plato

Which way are they grouped?
$\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.
January 3rd 2009, 07:57 AM
barc0de

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
January 3rd 2009, 08:10 AM
Plato

Quote:

Originally Posted by barc0de

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.

$\begin{gathered} \left[ {\left( {A \cap B \cap C} \right) \cap \left( {A \cup B \cup C'} \right)} \right] \cup \left( {A \cup B'} \right) = A \cup B' \hfill \\ \left( {A \cap B \cap C} \right) \cap \left[ {\left( {A \cup B \cup C'} \right) \cup \left( {A \cup B'} \right)} \right] = \left( {A \cap B \cap C} \right) \hfill \\ \end{gathered}$
January 3rd 2009, 08:21 AM
barc0de

Could you please break down the steps used in finding the first solution?