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## Here's the Solution to this Question

a) $A-B=A\cap B^C$

We must show $A-B\sube A\cap B^C$ and $A\cap B^C\sube A-B$

Show that $A-B\sube A\cap B^C$

Let $x\in A-B.$ By definition of set difference, $x\in A$ and $x\notin B.$ By definition of complement, $x\notin B$ implies that $x\in B^C.$ Hence, it is true that both, $x\in A$ and $x\in B^C.$ By definition of intersection, $x\in A\cap B^C.$

Show that $A\cap B^C\sube A-B.$

Let $x\in A\cap B^C.$ By definition of intersection, $x\in A$ and $x\in B^C.$ By definition of complement, $x\in B^C$ implies that $x\notin B.$ Hence, $x\in A$ and $x\notin B.$ By definition of set difference, $x\in A-B.$

Thus, $A-B=A\cap B^C.$

b) $(A-B)\cup(A\cap B)=A$

Let $x\in A.$ There are two cases, $x\in A$ and $x\notin B$ or $x\in A$ and $x\in B.$

In the first case $x\in A, x\notin B,$ so by definition set difference, $x\in A-B.$

In the second case $x\in A$ and $x\in B,$ so by definition of intersection $x\in A\cap B.$

By definition of union $x\in (A-B)\cup(A\cap B).$

Thus if $x\in A,$ then $x\in (A-B)\cup(A\cap B).$

Let $x\in (A-B)\cup(A\cap B).$ This means that either $x\in A-B$ or $x\in A\cap B.$

In the first case $x\in A, x\notin B,$ in the second case $x\in A$ and $x\in B.$

Then in either case $x\in A.$

Two sets are equal, since they have the same elements.Therefore

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