For any sets A and B show that i. (A B) (A-B)=A ii. A (B-A)=A B

Question

For any sets A and B show that
i. $(A \cap B) \cup(A-B)=A$
ii. $A \cup(B-A)=A \cup B$

✓

Answer

$\begin{array}{l}\text { i. }(A \cap B) \cup(A-B)=A \\ \text { L.H.S. }=(A \cap B) \cup(A-B)\end{array}$
$\begin{array}{l}=( A \cap B ) \cup\left(A \cap B^{\prime}\right)\left[\therefore( A - B )= A \cap B^{\prime}\right] \\ = A \cap\left(B \cup B^{\prime}\right)[ By \text { distributive law }] \\ = A \cap U \left[\left(B \cup B^{\prime}\right)= U =\text { Universal set }\right] \\ = A \\ =\text { R.H.S. }\end{array}$
$\begin{array}{r}\text { ii. } A \cup(B-A)=A \cup B \\ \text { L.H.S. }=A \cup(B-A)\end{array}$
$\begin{array}{l}= A \cup\left( B \cap A^{\prime}\right)\left[\therefore( B - A )= B \cap A^{\prime}\right] \\ =( A \cup B ) \cap\left(A \cap A^{\prime}\right)[ By \text { distributive law }] \\ =( A \cup B ) \cap u \left[\therefore A \cup A^{\prime}= u =\text { Universal set }\right] \\ = A \cup B \\ =\text { R.H.S. }\end{array}$

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