Help understanding the proof of De Morgan's Law

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SUMMARY

The discussion centers on the proof of De Morgan's Law, specifically the expression (A ∩ B)' = A' ∪ B'. The proof demonstrates that if an element x is not in the intersection of sets A and B, then it must be in the union of their complements. The key logical transformation involves understanding that negating a conjunction (AND) results in a disjunction (OR), as illustrated by the example of negating the statement "I am going to work AND I am going to have pizza for lunch." This transformation is a fundamental aspect of propositional logic.

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  • Understanding of set theory concepts such as intersection and union
  • Familiarity with logical operators, specifically AND, OR, and NOT
  • Basic knowledge of propositional logic
  • Ability to manipulate logical expressions and proofs
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  • Study the principles of propositional logic and truth tables
  • Explore additional proofs of De Morgan's Laws in various contexts
  • Learn about logical equivalences and their applications in mathematics
  • Investigate the implications of De Morgan's Laws in computer science, particularly in Boolean algebra
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Isaac12
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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?
 
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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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