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#1 2009-09-08 23:46:29
- serendipity
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- Registered: 2009-08-27
- Posts: 1
proof of De Morgan rules
Hello guys!(and sorry for my english..)
i am looking for a proof with induction at the above excercise...is at Propositional Logic...
the problem is this : (p1 ^p2...^pn) <-> ~(~p1 \/ ~p2 ...\/pn)...
thanx in advance for your help...
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#2 2009-09-09 15:14:57
- Ricky
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- Registered: 2005-12-04
- Posts: 3,791
Re: proof of De Morgan rules
If you understand the proof of De Morgan's rule for just two variables, then use induction to get the general statement.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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