Re: Play "Stump the Intuitionists!"
 From: Herbert Newman <nomail@invalid>
 Date: Sat, 10 Jan 2009 20:17:58 +0100
On 10 Jan 2009 05:40:31 0800, Daryl McCullough wrote:
I guess, you meant it implies __, right? With other words,You can let P == Av~A. Then ~(Av~A) is a contradiction.
I'm not mistaken, (~P > P) > P is not intvalid.
~(A v ~A) > __
or
~~(A v ~A)
is a theorem in int. logic. But how can we _prove_ that? (Well, Glivenko's
theorem comes to mind. Did you invoke it here?)
Anyway, then ~(Av~A) > __ and __ > Av~A would give ~(Av~A) > Av~A.
So ~(Av~A) > Av~A. So the [formula] (~P > P) > P impliesAh, nice.
excluded middle.
Herb
.
 FollowUps:
 Re: Play "Stump the Intuitionists!"
 From: Daryl McCullough
 Re: Play "Stump the Intuitionists!"
 From: Herbert Newman
 Re: Play "Stump the Intuitionists!"
 References:
 Play "Stump the Intuitionists!"
 From: MoeBlee
 Re: Play "Stump the Intuitionists!"
 From: MoeBlee
 Re: Play "Stump the Intuitionists!"
 From: Daryl McCullough
 Play "Stump the Intuitionists!"
 Prev by Date: hidden assumptions again
 Next by Date: Re: What is a logical contradiction?
 Previous by thread: Re: Play "Stump the Intuitionists!"
 Next by thread: Re: Play "Stump the Intuitionists!"
 Index(es):
Relevant Pages
