[FOM] Generalization Axiom Scheme
Zuhair Abdul Ghafoor Al-Johar
zaljohar at yahoo.com
Sat Nov 19 13:49:14 EST 2011
I think this approach can be salvaged
The basic principle of the scheme can be modified
to the following concept:
Any formula phi(y) closed on hereditarily
finite classes defines a set, provided that
if it has parameters then no parameter free
subformula of it can hold for finite classes only.
In symbols:
for all n=0,1,2,3,...; if phi(y) is a formula
in which only z1...zn occur as parameters, in which
x is not free, and in which u doesn't occur, and
where Q1...Qm are all parameter free subformulas of it,
then:
(n>0 ->(not[(u).Q1(u) -> finite(u)]&...& not[(u).Qm(u) -> finite(u)])) and
((z1)...(zn) are HF. (y). phi(y) -> y is HF)
->
(z1)...(zn) are sets. Exist x. set(x) & (y). y e x iff phi(y)
is an axiom.
Qi(u) refers to the formula Qi but with all occurrences of the free
variable in Qi replaced by u.
/
This scheme is so powerful that when added to Axioms of
Extensionality and impredicative class comprehension then
it can prove the existence of inaccessible set cardinals.
To see this take the following formulas:
phi_1<-> (y is hereditarily subnumerous to z1 and y is ordinal)
phi_2 <-> (Exist k. k is hereditarily strictly subnumerous to y and
k equinumerous to z1 and y= (U(k)) U k and y is ordinal)
phi_1 enable us to define the successor set cardinal of every
set cardinal.
phi_2 enable us to define the successor inaccessible set cardinal
of any inaccessible set cardinal (0,1,Omega_0 are considered inaccessible)
So the successor inaccessible cardinal of Omega_0 which is
the first uncountable weak inaccessible cardinal is definable as a set here.
So this theory proves ZF.
The interesting thing is that this theory can make us speak about
inaccessibles in a manner that is derived from mimicking what
is happening in the hereditarily finite world, ZF and VGB axioms
can all be interpreted as a part of mimicking what is happening in
the hereditarily finite world. This theory however goes beyond ZF
and VGB and divulge the inaccessible realm.
To me seeing that ZF can be proved by a theory that basically mimic the
hereditarily finite world is a point in favor of consistency of ZF.
Zuhair
--- On Fri, 11 Nov 2011 23:00:35 -0800 (PST)
Zuhair Abdul Ghafoor Al-Johar <zaljohar at yahoo.com>
wrote:
> Dear Sirs.
>
> This theory of mine is inconsistent.
>
> Let phi(x) be
>
> (for all y. (y is finite and y supernumerous to z1) -> x
> hereditarily subnumerous to y)
>
> Let z1 be some infinite set, and we get the set of all sets
> and thus
> reproducing Russell's paradox.
>
> Best regards
>
> Zuhair
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