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Topic: A reformulation of MK-Foundation-Choice: Even more compact!
Replies: 3   Last Post: Mar 28, 2013 2:25 PM

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 Zaljohar@gmail.com Posts: 2,665 Registered: 6/29/07
Re: A reformulation of MK-Foundation-Choice: Even more compact!
Posted: Mar 28, 2013 2:25 PM
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On Mar 28, 2:07 pm, Zuhair <zaljo...@gmail.com> wrote:
> On Mar 23, 8:33 pm, Zuhair <zaljo...@gmail.com> wrote:
>

> > This is even more compact reformulation of MK-Foundation-Choice.
>
> > Unique Comprehension: 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.

>
> > Size limitation: Set({}) & [Set(x) & y =< H(TC(x)) -> Set(y)]
> > /

>
> It might be possible to further weaken that to the following
>
> Set({}) & [Set(x) & y=<H(x) -> Set(y)]

No this won't work we need H(TC(x)) as in the original formulation.

But for the sake of proving Con(ZC) yes we can use the weak axiom

Set({}) & [Set(x) & y =< H(x) -> Set(y)]

where =< is defined as:

y =< x iff Exist z (for all m. m in y & ~m in x -> m=z)

Zuhair
>
> I think this can interpret MK over the sub-domain of well founded
> sets, thus proving the consistency of ZFC relative to it.
>
> Also I do think that if we re-define =< to the following modified
> subset relation, then the resulting theory would prove the consistency
> of ZC relative to it.
>
> Def.) y =< x iff Exist z (for all m. m in y & ~m in x -> m=z)
>
> Zuhair
>

Date Subject Author
3/23/13 Zaljohar@gmail.com
3/23/13 Zaljohar@gmail.com
3/28/13 Zaljohar@gmail.com
3/28/13 Zaljohar@gmail.com

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