is there any proof for De Morgan's law?

Asked by ashwinivg | 25th Jun, 2010, 12:00: AM

Expert Answer:

Dear Student
De Morgan laws can be proved as follows
(i) (AUB)' = A' ∩ B'
Let x (AUB)'
So x(AUB)
So x A and x B
So x A' and x B
So xA'∩B'
So (AUB)' A'∩B'
Similarly reversing the steps we can show that A'∩B' (AUB)'
Hope it clarified your answer
Team Topperlearning

Answered by  | 14th Jul, 2010, 03:10: PM

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