91Ó°ÊÓ

For every two sets of A and B, \(A \cap B\)and \(A \cap {B^c}\)are disjoint. Moreover, \(A = \left( {A \cap B} \right) \cup \left( {A \cap {B^c}} \right)\).

In addition, B and \(A \cap {B^c}\) are disjoint. Moreover, \(A \cup B = B \cup \left( {A \cap {B^c}} \right)\).

Let an arbitrary element \(x \in \left( {A \cap B} \right) \cup \left( {A \cap {B^c}} \right)\).

Using the distributive law, the above set can be written as shown below.

\(\begin{aligned}{c}\left( {A \cap B} \right) \cup \left( {A \cap {B^c}} \right)\\ = \left\{ {\left( {A \cap B} \right) \cup A} \right\} \cap \left\{ {\left( {A \cap B} \right) \cup {B^c}} \right\}\end{aligned}\)

By applying the distributive law again, we get,

\(\begin{aligned}{c} &= A \cap \left\{ {\left( {A \cup {B^c}} \right) \cap \left( {B \cup {B^c}} \right)} \right\}\\ &= A \cap \left\{ {\left( {A \cup {B^c}} \right) \cap \Omega } \right\}\\ &= A \cap \left( {A \cup {B^c}} \right)\\ &= A\end{aligned}\)

Here, \(\Omega \) is the sample space.

Therefore, \(x \in A\) and \(A = \left( {A \cap B} \right) \cup \left( {A \cap {B^c}} \right)\).

Use the distributive law as shown below.

\(\begin{aligned}{c}B \cup \left( {A \cap {B^c}} \right)\\ &= \left( {B \cup A} \right) \cap \left( {B \cup {B^c}} \right)\\ &= \left( {B \cup A} \right) \cap \Omega \\ &= B \cup A\\ = A \cup B\end{aligned}\)

Here, \(\Omega \)is the sample space.

Thus, \(A \cup B = B \cup \left( {A \cap {B^c}} \right)\).