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Topic: Two Finite Arithmetics
Replies: 19   Last Post: Apr 9, 2014 9:27 PM

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William Elliot

Posts: 1,489
Registered: 1/8/12
Re: Two Finite Arithmetics
Posted: Apr 9, 2014 9:27 PM
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On Wed, 9 Apr 2014, Dan Christensen wrote:
> On Wednesday, April 9, 2014 4:41:31 AM UTC-4, 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
> so-called 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.

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