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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: > > > > > > > > > > > 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: > > > For all A Exist x For all y (y in x iff y C A & phi) > > > are axioms. > > > 2. Infinity. > > > This interprets Z. > > > 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.
> > > > > > > are axioms. > > > (2) Infinity > > > 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



