
Re: Two Finite Arithmetics
Posted:
Apr 9, 2014 9:27 PM


On Wed, 9 Apr 2014, Dan Christensen wrote: > On Wednesday, April 9, 2014 4:41:31 AM UTC4, William Elliot wrote:
> > That's your unworkable criteria. > You are free, of course, to concoct a system that meets some of the > requirements. Just don't expect anyone to take notice. > > > All of the difficultiest that you haven't been able > > to resolved are smmothly handled by allowing Sm. > > It is certainly a challenge to come up with a suitable formalism that meets > all the requirements. Your inelegant kluge certainly does not. Quit berating a most eligant resolution.and finish what you started without ending in mumblings by showing how you can pack all your baggage into complete and constitent redition of finite arithmatic
> > BTW, just as I told you, in naive finite arithmatic > > > > it's a theorem, ie provalbe, that for all x /= m, > > x /= Sx > > > > by induction over the equivalent statement > > P(x) when x = m or x /= Sx. > > You can wave your hands all you want, William, but, by definition, the > socalled largest number in finite arithmetic simply cannot have a > successor. As challenging as it may be to formalize, there is no way around > it. That's your hangup definition. With that, I'm hanging up this thread.

