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Help understanding the proof of De Morgan's Law

  • Thread starter Isaac12
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  • #1
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I need some help on understanding the proof of De Morgan's Law.

(A intersection B)' = A' U B'
I know that it is proof is along the lines of

Let x be in (A intersection B)'.
Then x is NOT in A intersection B
<=> x not in A AND x not in B (use logical connectors, negations, etc)
<=> x in A' OR x in B'
<=> x in A' U B'

but my question is how does the AND become an OR?
 

Answers and Replies

  • #2
HallsofIvy
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The statement "I am going to work AND I am going to have pizza for lunch" is true only if BOTH are true. If I want to show "It is NOT true that I am going to work and I going to have pizza for lunch" I only need to show ONE of those two if false. That is, either I am NOT going to work or I am NOT going to have pizza for lunch. I don't have to show that they are both false, only that they are not both false. "Not A and B" is the same as "not A or not B".
 

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