
Re: An equivalent of MKFoundationChoice
Posted:
Feb 23, 2013 7:28 AM


On Feb 22, 11:54 pm, Zuhair <zaljo...@gmail.com> wrote: > On Feb 23, 5:48 am, CharlieBoo <shymath...@gmail.com> wrote: > > > > > > > On Feb 22, 3:38 pm, Zuhair <zaljo...@gmail.com> wrote: > > > > On Feb 22, 10:23 pm, CharlieBoo <shymath...@gmail.com> wrote: > > > > > On Feb 22, 5:14 am, Zuhair <zaljo...@gmail.com> wrote: > > > > > > On Feb 21, 11:10 pm, CharlieBoo <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 MKFoundationChoice > > > > > > > > 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? > > > > > > > CB > > > > > > > > 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? > > > > > CB > > > > > > 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. > > > CB > > 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?
CB
> 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 

