Question

Show that $A-(A-B)=$ $\varphi \cup (A\cap B)=A\cap B$

Answer 1

Consider, $A-(A-B)$

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

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

$=A\cap (A^{\prime }\cup (B^{\prime }{)}^{\prime })$ (Demorgan's law)

$=A\cap (A^{\prime }\cup B)$ [$(A^{\prime }{)}^{\prime }=A$]

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

$=\varphi \cup (A\cap B)$ (Intersection of set and its complement is always a null set)

$=A\cap B$