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Topic: Interpreting Z and ZF
Replies: 2   Last Post: Apr 21, 2013 7:52 AM

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 Zaljohar@gmail.com Posts: 2,665 Registered: 6/29/07
Re: Interpreting Z and ZF
Posted: Apr 21, 2013 7:52 AM

On Apr 20, 4:21 pm, Zuhair <zaljo...@gmail.com> wrote:
> On Apr 19, 5:35 pm, Zuhair <zaljo...@gmail.com> wrote:
>
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> > Using theories that *interpret* Z; ZF can greatly shrink the
> > axiomatics needed for proving the consistency of those theories, an
> > old observation of course. The following couple of first order
> > theories is a nice demonstration of that:

>
> > C shall be used to denote the known subset relation.
>
> > Theory 1 is the closure of all sentences following from:
>
> > 1. Separation If phi is a formula in which y is free but not x,
> > then all closures of:

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> > For all A Exist x For all y (y in x iff y C A & phi)
>
> > are axioms.
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> > 2. Infinity.
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> > This interprets Z.
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> > Theory 2 is the closure of all sentences following from:
>
> > (1) Replacement: If phi(x,y) is a formula in which x,y are free but
> > not B,
> > then all closures of:

>
> > For all A  ([For all x C A Exist z for all y (Phi(x,y) -> y C z)]
> > -> Exist B for all y (y in B iff Exist x C A (phi(x,y))))

>
> Or we can use:
>
> VA EB [(VyeB(ExeA.P)) & (VxeA (Ey.P -> EyeB.P))]
>
> Where V signifies "for all"
>
>

I meant:

VA EB [(VyeB(ExCA.P)) & (VxCA (Ey.P -> EyeB.P))]

where V signifies "for all" and C signifies subset relation.

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> > are axioms.
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> > (2) Infinity
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> > This interprets ZF.
>
> > If one seeks to prove the consistency of Z or ZF, then proving the
> > above theories would be sufficient, there is no need to try prove ALL
> > the known axioms of those theories.

>
> > Zuhair

Date Subject Author
4/19/13 Zaljohar@gmail.com
4/20/13 Zaljohar@gmail.com
4/21/13 Zaljohar@gmail.com