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Topic: The axioms for ultrafinitist number theory?
Replies: 56   Last Post: Mar 30, 2014 10:20 AM

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 LudovicoVan Posts: 4,165 From: London Registered: 2/8/08
Re: The axioms for ultrafinitist number theory?
Posted: Mar 22, 2014 11:56 AM

"William Elliot" <marsh@panix.com> wrote in message
news:Pine.NEB.4.64.1403220138030.27476@panix3.panix.com...
> On Sat, 22 Mar 2014, William Elliot wrote:
>> On Sat, 22 Mar 2014, Julio Di Egidio wrote:
>> > "William Elliot" <marsh@panix.com> wrote in message
>
>> > > A10 needs a simple correction (using successor function S as it's
>> > > easier)
>> > > If P subset N, 0 in P, for all x in P, Sx in P, then N subset P.
>> > > A10 needs to be:
>> > > If P subset N, 0 in P, for all x in P\max, Sx in P, then N subset P.
>> > >
>> > > Thus if 0 in p and for all x in P\max, Sx in P, then max in P.

>> >
>> > Instead you have missed the truly salient part, which was not my
>> > ruminations
>> > but rather where I notice: "add an axiom to the effect that max has no
>> > predecessor, and you'll see that your proof [that all elements but zero
>> > have a
>> > predecessor] goes through just as well. "

>>
>> Ok, it get it and the patch is weak. Namely one can conclude
>> for all x in N, Sx in N, thus Smax in N
>> which can be discared as S is a partial funcition over N\max.
>>
>> Ok, let's define Smax = max. Now A10 can be stated:
>> If P subset N, 0 in P, for all x in P, Sx in P, then P = N.
>>
>> Whoops, I see a contradiciton. For all n, Sn /= n.
>> Proof by induction. S0 /= 0, by axiom.
>> If Sn /= n, then SSn /= Sn by contradiction as S is injective over N\max.
>> Whoops, I see problem.
>>
>> Ok, with Smax = max, iduction needs changing.
>> If P subset N, 0 in P, for all x in P\max, Sx in P, then P\max = N\max
>>
>> If P subset N, 0 in P, max in P, for all x in P\max, Sx in P, then P = N.
>>
>> How are those? Once that hammered out, then look out when it comes
>> to inductive definitions of addition and multiplication.
>>
>> 0 + n = 0; Sm + n = m + Sn. So addition would max out with max.
>> S0 + max = 0 = Smax = Smax = max. That seems consistent. So the
>> problem now becomes which of the axioms need changing to accomodate
>> max and can the expected theorems still be proved or will more axioms
>> be needed for max?
>>
>> Oh the other hand, simply by letting Smax = 0 and keeping the other
>> axioms (except no Smax), wouldn't we have integers modulus max for
>> a model. Ok, I think Smax = 0 is how to rssolve the problem you

The problem I have raised was about max having or not having a
*predecessor*, so you still manage to miss the point.

> It seems to me that those axioms may have an infinite model, namely
> omega_0 + 1 with max = omega_0.

Oh yeah, in a way or the other, you are catching up.

Julio

Date Subject Author
3/18/14 Dan Christensen
3/18/14 William Elliot
3/18/14 Dan Christensen
3/18/14 LudovicoVan
3/18/14 LudovicoVan
3/18/14 LudovicoVan
3/18/14 LudovicoVan
3/18/14 Dan Christensen
3/18/14 LudovicoVan
3/18/14 LudovicoVan
3/18/14 Dan Christensen
3/18/14 LudovicoVan
3/18/14 Dan Christensen
3/18/14 LudovicoVan
3/18/14 Dan Christensen
3/18/14 LudovicoVan
3/20/14 LudovicoVan
3/20/14 Dan Christensen
3/20/14 LudovicoVan
3/19/14 Wizard-Of-Oz
3/19/14 Dan Christensen
3/20/14 Wizard-Of-Oz
3/20/14 Dan Christensen
3/20/14 Shmuel (Seymour J.) Metz
3/20/14 Virgil
3/20/14 Shmuel (Seymour J.) Metz
3/20/14 William Elliot
3/20/14 Dan Christensen
3/20/14 William Elliot
3/20/14 Dan Christensen
3/20/14 Dan Christensen
3/20/14 Dan Christensen
3/20/14 William Elliot
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3/21/14 LudovicoVan
3/21/14 William Elliot
3/21/14 LudovicoVan
3/22/14 LudovicoVan
3/25/14 ross.finlayson@gmail.com
3/22/14 William Elliot
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3/22/14 LudovicoVan
3/22/14 Dan Christensen
3/23/14 Dan Christensen
3/23/14 LudovicoVan
3/23/14 Dan Christensen
3/23/14 LudovicoVan
3/23/14 Dan Christensen
3/23/14 Virgil
3/23/14 David Hartley
3/23/14 Dan Christensen
3/23/14 Dan Christensen
3/23/14 Dan Christensen
3/25/14 William Elliot
3/26/14 Dan Christensen
3/30/14 Dan Christensen
3/30/14 Dan Christensen