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Exercise 1.5 · Q6

Q.Let AA and BB be subsets of the universal set NN, the set of natural numbers. Then A′∪[(A∩B)∪B′]A'\cup[(A\cap B)\cup B'] is

(1) AA
(2) A′A'
(3) BB
(4) NN
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Step 1. Simplify the inner bracket using distributivity: (A∩B)∪B′=(A∪B′)∩(B∪B′)=(A∪B′)∩U=A∪B′(A\cap B)\cup B'=(A\cup B')\cap(B\cup B')=(A\cup B')\cap U=A\cup B'.

Step 2. So the full expression becomes A′∪(A∪B′)=(A′∪A)∪B′=U∪B′=UA'\cup(A\cup B')=(A'\cup A)\cup B'=U\cup B'=U (since A′∪A=UA'\cup A=U always, and U∪U\cup anything =U=U). …

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