DeMorgan's Law extended to Union AND Intersection

  • #1

Main Question or Discussion Point

Hello.

We all know that DeMorgan's Law is as follows:
(A∪B)' = A'∩B'
and
(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'
 

Answers and Replies

  • #2
andrewkirk
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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).
 
  • #3
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((A∩B)∪C)' ≡ ((A∪C)∩(B∪C))'
≡ (A∪C)'∪(B∪C)'
≡ (A'∩C')∪(B'∩C')
 

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