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
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
Hello, barc0de!
Simplify: .,$\displaystyle (A \cup B \cup C) \cap [A \cup (B \cap C)] $
. . . $\displaystyle \begin{array}{cccc}\bigg[A \cup B \cup C\bigg] \cap \bigg[A \cup (B \cap C)\bigg] & &\text{Given} \\ \bigg[ A \cup B \cup C \bigg] \cap \bigg[ (A \cup B) \cap (A \cup C)\bigg] & & \text{Distributive} \\ \\[-4mm] \bigg[(A \cup B \cup C) \cap (A \cup B)\bigg] \cap \bigg[A \cup C\bigg] && \text{Associative} \\ \bigg[A \cup B\bigg] \cap \bigg[A \cup C\bigg] && \text{Subset} \\ \\[-4mm] A \cup (B \cap C) && \text{Distributive}\end{array}$
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.
$\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} $