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


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

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

(ii) $A-(A-B)=A \cap B$

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

(iv) $A-B=A \Delta(A \cap B)$.



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

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

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

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

Hence proved.


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.


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

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


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

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

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

Hence proved.


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

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


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

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

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



Hence proved.



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