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Topic: A compact reformulation of MK-Foundation-Choice.
Replies: 2   Last Post: Mar 23, 2013 2:31 PM

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 Zaljohar@gmail.com Posts: 2,665 Registered: 6/29/07
Re: A compact reformulation of MK-Foundation-Choice.
Posted: Mar 23, 2013 2:31 PM

On Mar 23, 12:25 am, Charlie-Boo <shymath...@gmail.com> wrote:
> On Mar 19, 12:41 pm, Zuhair <zaljo...@gmail.com> wrote:
>
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> > This is also another reformulation of MK.
>
> > Define: Set(x) iff Exist y. x in y
>
> > Axioms:
>
> > Unique construction: if phi is a formula in which x is not free, then:
> > (Exist!x for all y ( y in x iff Set(y) & phi)) is an axiom.

>
> > Pairing: (For all y (y in x -> y=a or y=b)) -> Set(x)
>
> > Size limitation: Set(x) & |y| =< |H(TC(x))|  -> Set(y)
>
> > /
>
> > Definitions:
> > TC(x)={y| for all t (x subclass_of t & t is transitive -> y in t)}
> > t is transitive iff for all m,n (m in n & n in t -> m in t)
> > H(x)= {y| for all z(z in TC({y}) -> |z| =< |x|)}
> > |z| =< |x| iff Exist f (f:z -->x & f is injective).
> > /

>
>  > This is a more compact presentation of MK-Foundation-Choice. Which
> of
>  > course can interpret the whole MK and of course can construct a
> model
>  > of ZFC thus proving its consistency.
>
> But do you know that your system is consistent?  Or if it has a
> different set of theorems than ZFC (otherwise you are saying that if
> ZFC is consistent then ZFC is consistent)?
>
> C-B
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> > Zuhair

Yes you are right. All of what I'm saying here is that If this theory
is consistent then ZFC is. But of course we already know that this
theory cannot be consistent if ZFC is not. When I said it proves
Con(ZFC) I mean relative to it, it's just a relative consistency
proof. It is not a consistency proof in the sense of how Gentzen
proved PA.

Zuhair

Date Subject Author
3/19/13 Zaljohar@gmail.com
3/22/13 Charlie-Boo
3/23/13 Zaljohar@gmail.com