For any two sets A and B, prove that

(i)

$(A \cup B)-B=(A \cup B) \cap B^{\prime} \quad\left(X-Y=X \cap Y^{\prime}\right)$

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

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

$=A \cap B$

$=A-B$

(ii)

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

$=A \cap\left(A^{\prime} \cup B^{\prime}\right) \quad($ De Morgan law $)$

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

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

$=A \cap B^{\prime}$

$=A-B$

(iii)

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

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

$=A \cap\left[A^{\prime} \cup\left(B^{\prime}\right)^{\prime}\right]$   (De Morgan law)

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

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

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

$=A \cap B$

(iv)

$A \cup(B-A)=A \cup\left(B \cap A^{\prime}\right) \quad\left(X-Y=X \cap Y^{\prime}\right)$

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

$=(A \cup B) \cap \cup \quad(\cup$ is the universal set $)$

$=A \cup B$

(v)

$(A-B) \cup(A \cap B)=\left(A \cap B^{\prime}\right) \cup(A \cap B) \quad\left(X-Y=X \cap Y^{\prime}\right)$

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

$=A \cap \cup$

$=A$