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DeMorgan's Law extended to Union AND Intersection

  1. Jun 4, 2015 #1

    We all know that DeMorgan's Law is as follows:
    (A∪B)' = A'∩B'
    (A∩B)' = A'∪B'
    where ' refers to the complement of a set and A and B are both sets.

    We also know that this can be extended to more than two terms.

    My question is whether or not the following is true:
    (A∩B∪C)' = A'∪B'∩C'

    Here is my logic:
    (A∩B∪C)' = ((A∩B)∪C)' = (A∩B)'∩C' = A'∪B'∩C'
  2. jcsd
  3. Jun 5, 2015 #2


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    Yes that's right. You can reassure yourself by drawing a Venn diagram in which A and B are intersecting circles, dividing the Venn box into four regions counting the region outside both circles. Then draw C as a blob that intersects all four regions.

    By the way, I would write A∩B∪C as (A∩B)∪C because the leftmost precedence of operators is not universally understood or accepted for set operators, and
    (A∩B)∪C is not in general equal to A∩(B∪C).
  4. Jul 10, 2015 #3
    ((A∩B)∪C)' ≡ ((A∪C)∩(B∪C))'
    ≡ (A∪C)'∪(B∪C)'
    ≡ (A'∩C')∪(B'∩C')
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