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Jan 25th, 2008, 05:48 AM
#1
Thread Starter
New Member
Proving De Morgan's Law
Hi guys!
I need to prove De Morgan's law for a logic class. My friend and I were successful with the first theorem:
~(P^Q)<=>~Pv~Q
1. ~(PvQ) |P(1)
2. P |A(2)
3. PvQ |(vIntro) 2 (2)
4. falsum |(~Elim) 1,3 (1,2)
5. ~P |(~Intro) 2,4 (1,2)
6. Q |A(6)
7. PvQ |(vIntro) 6 (6)
8. falsum |(~Elim) 1,7 (1,6)
9. ~Q |(~Intro) 6,8 (1,6)
10. ~P^~Q |(^Intro) 5,9 (1)
I've been trying to solve the second theorem but I think that I'm stuck. Can you help me prove ~(pvq)<=>~p^~q?
Thanks!!!
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Jan 30th, 2008, 02:55 AM
#2
Re: Proving De Morgan's Law
Do you mean you want help proving the other direction (for some reason), that is ~Pv~Q => ~(P^Q)? Or do you want help proving the other theorem, ~(PvQ) <=> ~P^~Q? [that is, with "and" and "or" switched].
If it's the first one I think you just proceed by cases with v elimination. If it's the second it should be at least similar to this one... but maybe it's not in some subtle way? I proved these last semester but my formal logic is draining out of me....
The time you enjoy wasting is not wasted time.
Bertrand Russell
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