Re: Russell paradox solved by infinite multi-layering

On Oct 9, 8:47 pm, Zaljo...@xxxxxxxxx wrote:
First order logic with identity:

e,=, Vi for each i for i=1,2,3,.......

1) Ey(yex) -> Ey(yex & ~Ec( cey & cex ) )

2) Az(zex<->zey) -> x=y

3) (xeVi & yex) -> yeVi is an axiom

4) E!y( Am(mex->mey) & y is transitive &
Am( (mey & ~mex) -> Ez( zey & mez ) ) ).

Definition:

y=Tc(x) <-> ( Am(mex->mey) & y is transitive &
Am( (mey & ~mex) -> Ez( zey & mez ) ) ).

5) z1,...,zn e Vi , F(y,z1,...,zn) is a formula
in which x is not free

Ex( xeVi+1 & Ay(yex<->(yeVi & F(y,z1,...,zn))) &
(xeVi <->Ay(xeTc(y)->~F(y,z1,...,zn))) ).

is an axiom.

Zuhair

I think that 5 can be replaced by two axiom schemas:

5) for every i , if P is a formula in which x is not free then all
closures of

Ex( xeVi+1 & Ay(yex<->(yeVi & P(y))))

are axioms.

6) for every i, if P is a formula in which x is not free then all
closures of

Ax( (Ay(yex<->(yeVi&P(y))) & Ay(xeTc(y)->~P(y))) -> xeVi )

are axioms.

Theorems:

ExAy ~yex

Proof: let P(y)<->~y=y

Then from 5 we have

Ex (xeV2 & Ay( ~yex) )

From extensionality this set is unique

Thus we can define x as

x=0 <-> (xeV2 & Ay( ~yex ))

Now from 6 we have the following

Let x=0 and P(y)<-> ~y=y and i=1 so we have:


(Ay(ye0<->~y=y)) & Ay(0eTc(y)->y=y)) -> 0eV1

From extensionality we know that Ay(0eTc(y) ->y=y)
is a true statement.
Also we have Ay(ye0<->~y=y) is a true statement.

Thus 0eV1.

So x=0 <-> (xeV1 & Ay(~yex))

Pairing can be easily prove, so is infinity.
HOwever I am not sure of separation and power.

Zuhair

.

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