Re: V
- From: zuhair <zaljohar@xxxxxxxxx>
- Date: 30 May 2007 20:40:05 -0700
On May 30, 10:22 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:
On May 31, 12:28 pm, zuhair <zaljo...@xxxxxxxxx> wrote:
On May 30, 9:06 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:
On May 30, 11:07 pm, zuhair <zaljo...@xxxxxxxxx> wrote:
On May 30, 12:43 am, Rupert <rupertmccal...@xxxxxxxxx> wrote:
On May 30, 6:13 am, zuhair <zaljo...@xxxxxxxxx> wrote:
Hi all,
I would like to discuss the following theory , which might serve as an
introductory discussion to the Multilayer theories that I presented in
another thread.
Theory T is the set of all sentences entailed( from first order logic
with identity and the primitive constant V) by the following non
logical axioms:
Primitives: e,=,V
1) Extensionality: AxAy(y=x<->Az(zey<->zex)).
2) Regularity:Ax( Ey(yex) -> Ey(yex & y disjoint x)).
3) Comprehension schema: if F is a formula in one free variable and in
which x is not free then all closures of
ExAy(yex<->(yeV & F(y)))
are axioms.
4) Pairing: ArAsExAy( (yex<->(y=r v y=s)) &
( (reV & seV) -> xeV ) ).
5) Union: ArExAy((yex<->Ez(yez&zer)) & (reV -> xeV)).
6) Power:ArExAy((yex<-> Az(zey->zer)) & (reV -> xeV)).
7) Infinity:Ex( xeV & 0ex & Ay(yex->yu{y}ex)).
8) Membership:Ax(xeV<->(Az(zex->zeV) & x subnumerous_to V)).
/
Theory definition finished.
This theory can be proved consistent in ZF.
Then ZF is inconsistent.
This theory proves ZF.
Since this theory proves Replacement within V
Remember the version I was talking about here didn't have any version
of Replacement. So no, it doesn't. Or if it does, then you're right,
quite a weak fragment of ZF is inconsistent. So show me your proof.
Theorem schema of Replacement:if F
is a formula in two variables and in which x is not
free then all closures of
AxE!y(F(x,y))Ay(F(x,y)->yeV) -> AreVEceVAy(yec<->ExerF)
are axoims.
Proof:from the left of the implication we have
Ax(xec -> xeV)
from the uniqueness on the left we have
~c supernumerous_to r.
But P(r) supernumerous_to r
Then P(r) supernumerous to c
But P(r)eV
Then ceV . (Membership).
theorem proved.
Anyhow Rupert, at the beginning of this post I posted two theories,
ONe that doesn't have replacement in it ( the MK like theory but not
MK itself) and the other one
is the ZFC like theory with the 10 axioms ( see the second message in
this thread) which has Replacement in it.
However even the first theory proves Replacement within V
but it doesn't prove it outside of V.
The ZFC like theory has replacement and separation as a axiom that
works within and outside V, and thus it is a stronger theory than the
first one,but not stronger than
the oo-Multilayer theory.
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Okay, I get it. You're right. Your theory proves the consistency of
ZF. It can be interpreted in ZF+"an inaccessible cardinal exists".
Ok, Rupert, I would like to know
the formula of 'x is inaccessible cardinal'.
can you please write it in FOL.
I think ZF+inaccessible cardinal is writtin in this manner.
ZF axioms
+ axiom of inacessible cardinal which is
Ex( x is inaccessible cardinal ).
Now my question to you , were do this inaccessible cardinal exist.
Does it exist in V
or outside V.
What happens if I add the following axiom to theory
I posted in this thread( the one with the 10 axioms,
what I called the ZFC like thoery )
Ex(xeV & x is inaccessible cardinal)
Would this addition of this axiom result in an inconsistent theory.
Or it will result in a theory that is stronger
than ZF+inaccessible cardinal exists.
Zuhair
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