"William Elliot" <firstname.lastname@example.org> wrote in message news:Pine.NEB.email@example.com... > On Sat, 22 Mar 2014, William Elliot wrote: >> On Sat, 22 Mar 2014, Julio Di Egidio wrote: >> > "William Elliot" <firstname.lastname@example.org> wrote in message > >> > > A10 needs a simple correction (using successor function S as it's >> > > easier) >> > > Instead o: >> > > 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 >> raised. Comments? Has somebody already doen this?
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.