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Topic: An equivalent of MK-Foundation-Choice
Replies: 10   Last Post: Feb 23, 2013 11:20 PM

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Charlie-Boo

Posts: 1,585
Registered: 2/27/06
Re: An equivalent of MK-Foundation-Choice
Posted: Feb 23, 2013 7:28 AM
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On Feb 22, 11:54 pm, Zuhair <zaljo...@gmail.com> wrote:
> On Feb 23, 5:48 am, Charlie-Boo <shymath...@gmail.com> wrote:
>
>
>
>
>

> > On Feb 22, 3:38 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > > On Feb 22, 10:23 pm, Charlie-Boo <shymath...@gmail.com> wrote:
>
> > > > On Feb 22, 5:14 am, Zuhair <zaljo...@gmail.com> wrote:
>
> > > > > On Feb 21, 11:10 pm, Charlie-Boo <shymath...@gmail.com> wrote:
>
> > > > > > On Feb 20, 6:01 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > > > > > > This is just a cute result.
>
> > > > > > > The following theory is equal to MK-Foundation-Choice
>
> > > > > > > Language: FOL(=,e)
>
> > > > > > > Define: Set(x) iff Ey. x e y
>
> > > > > > > Axioms: ID axioms+
>
> > > > > > > 1.Extensionality: (Az. z e x <-> z e y) -> x=y
>
> > > > > > > 2. Construction: if phi is a formula in which x is not free,
> > > > > > > then (ExAy.y e x<->Set(y)&phi) is an axiom

>
> > > > > > > 3. Pairing: (Ay. y e x -> y=a or y=b) -> Set(x)
>
> > > > > > > 4. Size limitation
> > > > > > > Set(x) <-> Ey. y is set sized & Azex(Emey(z<<m))

>
> > > > > > > where y is set sized iff Es. Set(s) & y =< s
> > > > > > > and z<<m iff z =<m & AneTC(z).n =<m

>
> > > > > > > TC(z) stands for 'transitive closure of z' defined in the usual manner
> > > > > > > as the minimal transitive class having z as subclass of; transitive of
> > > > > > > course defined as a class having all its members as subsets of it.

>
> > > > > > > y =< s iff Exist f. f:y-->s & f is injective.
>
> > > > > > > /
>
> > > > > > > Of course this theory PROVES the consistency of ZFC.
> > > > > > > Proofs had all been worked up in detail. It is an enjoying experience
> > > > > > > to try figure them out.

>
> > > > > > Why don't you supply your proof of ZFC consistency in detail then?  If
> > > > > > that is too much, then why not give it for 2 potentially conflicting
> > > > > > axioms in detail, as I suggested recently?

>
> > > > > > Why waste space with unsubstantiated claims of grandeur?
>
> > > > > > C-B
>
> > > > > > > Zuhair
>
> > > > > Of course I'll supply them in DETAIL. There is no grandeur, nothing
> > > > > like that. That matter has been PROVED. I just wanted some to enjoy

>
> > > > What has been proved?
>
> > > > > figuring it out before I send the whole written proof.
>
> > > > What are you waiting for?
>
> > > > I'll bet you $25 to your $1 that you don't supply a proof, payable via
> > > > PayPal.  Are we on?  Only condition is you have to answer every
> > > > question - no obfuscation, please.

>
> > > > It'd be well worth $25 if you have a proof of ZF consistency and I was
> > > > one of the first to be able to give it.  Has it been proven before?
> > > > It seems people say "if ZF were consistent".  Who has tried?  Does
> > > > Gödel's 2nd Theorem mean you'll do math that ZF can't?

>
> > > > C-B
>
> > > > > Zuhair
>
> > > There is some confusion here. What I'm claiming is that IF we hold
> > > that theory presented here to be consistent then MK minus foundation
> > > minus choice would be consistent, and thus ZFC would be proved
> > > consistent. I'm speaking about a relative consistency proof here.

>
> > > Zuhair
>
> > "Of course this theory PROVES the consistency of ZFC.  Proofs had all
> > been worked up in detail."  This seems to be saying clearly that you
> > have developed a proof that ZFC is consistent.

>
> > Now you're saying if one system is consistent then another is?  Sure,
> > if system X plus a few more axioms is consistent, then by definition
> > system X is also consistent.  Knowing that connection between 2
> > systems might signify nothing, depending on the relationship between
> > the two systems.

>
> > C-B
>
> Yes that's all what I'm saying, it is a relative consistency proof,
> but it is cute


Is that a common thing, proving that one system being consistent
implies that another is? By any method other than proving each
axiom? And this is a new theorem or a new proof?

> in that it is not too trivial,

If we are proving each axiom, then is it not too trivial in one of
these proofs - then which one? - or in all of them?

C-B

> although simple really, try yourself to
> prove all axioms
> of MK minus foundation minus choice in the above theory.It's enjoying
>
> Zuhair- Hide quoted text -
>
> - Show quoted text -





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