For any two sets A and B, prove the following:

Solution:

(i)

$\mathrm{LHS}=A \cap\left(A^{\prime} \cup B\right)$

$=\left(A \cap A^{\prime}\right) \cup(A \cap B)$

$=(\phi) \cup(A \cap B)$

$=A \cap B=\mathrm{RHS}$

Hence proved.

(ii)

LHS $=A-(A-B)$

$=A-\left(A \cap B^{\prime}\right)$

$=A \cap\left(A \cap B^{\prime}\right)$

$=A \cap\left(A \cap B^{\prime}\right)^{\prime}$

$=A \cap\left\{A^{\prime} \cup\left(B^{\prime}\right)^{\prime}\right\}$

$=\left(A \cap A^{\prime}\right) \cup(A \cup B)$

$=(\emptyset) \cup(A \cup B)$

$=(A \cup B)=$ RHSb

Hence proved.

(iii)

$\mathrm{LHS}=A \cap(A \cup B)$

$=A \cap\left(A^{\prime} \cap B^{\prime}\right)$

$=\left(A \cap A^{\prime}\right) \cap\left(A \cap B^{\prime}\right)$

$=(\phi) \cap\left(A \cap B^{\prime}\right)$

$=\phi=\mathrm{RHS} \quad[\phi \cap A=\phi]$

Hence proved.

(iv)

$\mathrm{LHS}=A \Delta(A \cap B)$

$=\{A-(A \cap B)\} \cup\{(A \cap B)-A\}$

$=\left\{A \cap(A \cap B)^{\prime}\right\} \cup\left\{(A \cap B) \cap A^{\prime}\right\}$

$=\left(A \cap B^{\prime}\right) \cup(\phi)$

$=\left(A \cap B^{\prime}\right)$

$=A-B=\mathrm{RHS}$

Hence proved.