a) A−B=A∩BC
We must show A−B⊆A∩BC and A∩BC⊆A−B
Show that A−B⊆A∩BC
Let x∈A−B. By definition of set difference, x∈A and x∈/B. By definition of complement, x∈/B implies that x∈BC. Hence, it is true that both, x∈A and x∈BC. By definition of intersection, x∈A∩BC.
Show that A∩BC⊆A−B.
Let x∈A∩BC. By definition of intersection, x∈A and x∈BC. By definition of complement, x∈BC implies that x∈/B. Hence, x∈A and x∈/B. By definition of set difference, x∈A−B.
Thus, A−B=A∩BC.
b) (A−B)∪(A∩B)=A
Let x∈A. There are two cases, x∈A and x∈/B or x∈A and x∈B.
In the first case x∈A,x∈/B, so by definition set difference, x∈A−B.
In the second case x∈A and x∈B, so by definition of intersection x∈A∩B.
By definition of union x∈(A−B)∪(A∩B).
Thus if x∈A, then x∈(A−B)∪(A∩B).
Let x∈(A−B)∪(A∩B). This means that either x∈A−B or x∈A∩B.
In the first case x∈A,x∈/B, in the second case x∈A and x∈B.
Then in either case x∈A.
Two sets are equal, since they have the same elements.Therefore
(A−B)∪(A∩B)=A
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