Re: Ramification Schema.

On May 23, 9:07 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:
On May 24, 10:25 am, zuhair <zaljo...@xxxxxxxxx> wrote:





On May 23, 10:38 am, zuhair <zaljo...@xxxxxxxxx> wrote:


Not only differential pairing and infinity can be proved in this
theory, actually even Union can be proved, so actually one should drop
union from the list of axioms of this theory.

5) Ramification schema: if F is a formula in which x is not free then
all closures of
AyEu(Az(F(y,z)->zeu)) ->
AmEx(mex & Ay(yex -> Az(F(y,z)->zex))).
are axioms.

Now Let u=y

What makes you think that is allowed?

From first order logic with identity we know for sure that this is
allowed.

Proof:

Ay(Az(F(y,z)->zey)) -> AyEu(Az(F(y,z)->zeu))
Since AyEu(Az(F(y,z)->zeu)) -> AmEx(mex & Ay(yex -> Az(F(y,z)->zex))).

Then:Ay(Az(F(y,z)->zey))->
AmEx(mex & Ay(yex -> Az(F(y,z)->zex))).

Since from FOL we know that for formulas F P Q
(F->P->Q) -> (F->Q)

Zuhair




.

Relevant Pages

  • Re: ZF subtheory of T
    ... For all a in W,b in W (a union b in W). ... Of course a union b is the class of exactly all members of a and b. ... minus regularity pluse the axiom for all x: ...
    (sci.logic)
  • Re: Two results of set geometry
    ... if each is a proper subset of some other in the set. ... we PROVE from the axioms that w is a counterexample. ... Moreover, you say you do endorse that there exist infinite sets, so ... and the union of the set is infinite. ...
    (sci.math)
  • Re: ZFC in another shape.
    ... First order logic with identity axioms also. ... Anyhow I think it is the strong replacement, ... But you haven't proven union. ... Simply replace the members of a set X that is equinmerous to the ...
    (sci.math)
  • Re: Orlow cardinality question
    ... > over the ZFaxioms and the Peano construction etc... ... > that since you can construct all the finite sets then you ... > understanding of the ZF axioms is then that well, we can't quite union ... The existence of countable union S = U ...
    (sci.math)
  • Re: ZFC in another shape.
    ... First order logic with identity axioms also. ... countable language (and, I'm supposing that you have only a countable ... Anyhow I think it is the strong replacement, ... But you haven't proven union. ...
    (sci.math)