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Re: An equivalent of MK-Foundation-Choice
Posted:
Feb 22, 2013 2:23 PM
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On Feb 22, 5:14 am, Zuhair <zaljo...@gmail.com> wrote:
> On Feb 21, 11:10 pm, Charlie-Boo <shymath...@gmail.com> wrote:
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> > On Feb 20, 6:01 pm, Zuhair <zaljo...@gmail.com> wrote:
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> > > This is just a cute result.
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> > > The following theory is equal to MK-Foundation-Choice
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> > > Language: FOL(=,e)
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> > > Define: Set(x) iff Ey. x e y
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> > > Axioms: ID axioms+
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> > > 1.Extensionality: (Az. z e x <-> z e y) -> x=y
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> > > 2. Construction: if phi is a formula in which x is not free,
> > > then (ExAy.y e x<->Set(y)&phi) is an axiom
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> > > 3. Pairing: (Ay. y e x -> y=a or y=b) -> Set(x)
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> > > 4. Size limitation
> > > Set(x) <-> Ey. y is set sized & Azex(Emey(z<<m))
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> > > where y is set sized iff Es. Set(s) & y =< s
> > > and z<<m iff z =<m & AneTC(z).n =<m
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> > > 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.
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> > > y =< s iff Exist f. f:y-->s & f is injective.
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> > > /
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> > > 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.
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> > 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?
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> > Why waste space with unsubstantiated claims of grandeur?
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> > C-B
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> > > Zuhair
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> 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
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