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Show that for any sets $A$ and $B$,
$A=(A \cap B) \cup(A-B)$
and $A \cup(B-A)=(A \cup B)$
$A=(A \cap B) \cup(A-B)$
and $A \cup(B-A)=(A \cup B)$
Solution:
1908 Upvotes
Verified Answer
$(A \cap B) \cup(A-B)=(A \cap B) \cup\left(A \cap B^{\prime}\right)$
$\left[\because A-B=A \cap B^{\prime}\right]$
$=A \cap\left(B \cup B^{\prime}\right) \quad[$ by distributive law $]$
$=A \cap X=A \quad[X$ is a universal set $]$
$A \cup(B-A)=A \cup\left(B \cap A^{\prime}\right)$
${\left[\because B-A=B \cap A^{\prime}\right] }$
$=(A \cup B) \cap\left(A \cup A^{\prime}\right)$
[by distributive law]
$=(A \cup B) \cap X$
$\quad\left[\because X=A \cup A^{\prime}\right.$ is universal set $]$
$=A \cup B$
$\left[\because A-B=A \cap B^{\prime}\right]$
$=A \cap\left(B \cup B^{\prime}\right) \quad[$ by distributive law $]$
$=A \cap X=A \quad[X$ is a universal set $]$
$A \cup(B-A)=A \cup\left(B \cap A^{\prime}\right)$
${\left[\because B-A=B \cap A^{\prime}\right] }$
$=(A \cup B) \cap\left(A \cup A^{\prime}\right)$
[by distributive law]
$=(A \cup B) \cap X$
$\quad\left[\because X=A \cup A^{\prime}\right.$ is universal set $]$
$=A \cup B$
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