Show that if A and B are sets then (A intersection B) union (A intersection B compliment) = A first by showing that each side is a subset of the other side and then by using a membership table.

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  1. If element x belongs to set B and set A, the intersection A∩B also contains x. Not B (complement) does not contain x, so the intersection with A does not contain x.
  2. If x does not belong to set B but does belong to set A, A∩B does not contain x, but since not B does, its intersection with A contains x.
  3. If x does not belong to set A, then neither intersection contains x.
  4. If we consider the union of the two intersections from 1 and 2 above, x belongs to the union in each case.
  5. For 3 above, the union does not contain x.

For all elements x in A, the final union consists of all such elements, therefore the union is set A itself.


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