
Re: Interpreting ZFC
Posted:
Apr 30, 2013 3:45 PM


On Apr 29, 5:36 am, Graham Cooper <grahamcoop...@gmail.com> wrote: > On Apr 28, 3:58 am, Zuhair <zaljo...@gmail.com> wrote: > > > > > > > > > > > On Apr 27, 3:55 pm, Jan Burse <janbu...@fastmail.fm> wrote: > > > No > > > > Zuhair schrieb: > > > > > PreZFC is a first order theory with the following axioms: > > > > > (1) Powerful Boundedness: if phi is a formula in which x,y are free, > > > > then > > > > all closures of: > > > > > EB: (Vy in B(Ex C A:phi)) & (Vx C A ((Ey:phi) >(Ey in B:phi))) > > > > > are axioms. > > > > > C is subset relation. > > > > V;E signifies universal; existential quantification respectively. > > > > > 2) Infinity. > > > > > / > > > > > The whole of ZFC can be interpreted in PreZFC. > > > > > Zuhair > > > Hmmm,... you must have figured out some flaw somewhere, what is it? > > > Zuhair > > B is any set in the world of mathematics! > > EB: (Vy in B(Ex C A:phi)) & (Vx C A ((Ey:phi) >(Ey in B:phi))) > > is > > Exist B ALL y in B ... Exist X C A:phi > & > All X C A (Exist y:phi > Exist y in B:phi ) > > *************** > > 1st line: > > yeB <> SUBSET X OF A with elements that satisfy phi > > 3rd line: > > ALL subsets of A.. > y satisfies phi > y e B (that satisfy phi) > >  > > firstly, is the final phi in B:phi necessary
Yes. > since phi already designates members of B > yea but it doesn't enforce which of phi objects are members of B. The last phi is necessary to enforce one phi object for Eeach x subset of A, to be a member of B. > secondly, All subsets of A is a POWERSET operation > on all SETS in the THEORY which has huge complexity > > thirdly, this is starting to look like mereology where > on starting equation is given to derive the rest.. > > the problem with mereology is it uses ALL(S) quantifier > and C (subset) to codefine each other.. > > fourth, perhaps you could show LINE BY LINE how > phi(x) <> x ~e x > > is barred from inferring an existent set B. > > Herc >  > EARTH, WIND, FIRE, WATER... is my bet!

