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How to proove De Morgan's Law for Logic? 
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#1
Apr1805, 04:25 AM

P: 2

Without using Truth table
thanks folks DeMorgan's Law: ~(P^Q)<=>~Pv~Q; ~(PvQ)<=>~P^~Q; 


#2
Apr1805, 06:30 AM

PF Gold
P: 2,330

Several laws bear his name which one do you want? Just write it out.



#3
Apr1805, 07:23 AM

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P: 3,684

I imagine he wants
[tex]\vee_{i=1}^nk_i=\wedge_{i=1}^nk_i[/tex] [tex]\wedge_{i=1}^nk_i=\vee_{i=1}^nk_i[/tex] (I used  for not; how do you get the proper symbol?) 


#4
Apr1805, 08:34 AM

P: 2

How to proove De Morgan's Law for Logic?
De Morgan's Law:
~(P^Q)<=>~Pv~Q; ~(PvQ)<=>~P^~Q; 


#5
Apr1805, 10:55 AM

P: 552

Well, here's a way to prove the first implication of the first law, ~(p&q) > ~p v ~q
Here the indented lines following an assumption indicate the scope of the assumption:
can be proved similarly. 


#6
Apr1805, 11:59 AM

P: 552

The key to that proof was the use of exportation to get line 7. Both halves of the second DeMorgan's law can also be proved by the same general idea, though it's slightly trickier.



#7
Apr1905, 08:03 AM

P: 13

Example: [tex] \neg \exists x . P(x) \Leftrightarrow \forall x. \neg P(x) [/tex] 


#8
Apr1905, 09:49 PM

P: 1,408

The way I learned this is that is follows from the semantic defintion of truth.
Say you you have an interpretation M and a sentence F by definition M= ~F iff not M=F M= (F&G) iff M= F and M=G M= (FvG) iff M= F or M=G Thus a consequence of the definition of truth for negation and of two junctions is the fact that (F&G) is true iff ~(~Fv~G) is true, and (FvG) is true iff ~(~F&~G) is true. You have ~(PvQ) iff ~P&~Q, which would be equivalent to ~(FvG) iff ~~(~F&~G), which is the same as saying ~(FvG) iff (~F&~G). 


#9
Apr2005, 10:41 AM

P: 552




#10
Oct2505, 12:22 PM

P: 5

Is there no easier way to prove DeMorgan's theorem without having to use EXPORTATION and DISJUNCTIVE SYLLOGISM rules? Is there a way to prove this Law by just using modus ponens, modus tollens, disjunctive argument, conjunctive argument, simplification, and so on?



#11
Oct2505, 12:47 PM

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#12
Oct2805, 01:13 AM

P: 97




#13
Oct2905, 11:14 PM

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P: 2,586




#14
Aug708, 12:57 AM

P: 1

Hi,
how to simplify (p^q)v(pvq) using the logical equivalence? 


#15
Dec1108, 01:16 PM

P: 2

How do you prove the following? I understand that it is part of demorgans law.. but how do i get rid of the double negative?
(A v B) therefore (A & B) 


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