Re: ZFC in another shape.

On Feb 4, 6:38 pm, "zuhair" <zaljo...@xxxxxxxxx> wrote:
Hi All,

Is the following theory equivalent to ZFC?

Theory X.

Primitives: e,=

Definition Schema:

x is an ordinal <-> AmAn(((mex&nem)->nex)&Az(zen->zem)).

Axioms:

1)Extensionality
2)Foundation
3)Ordinal comprehension: Ex: x is an ordinal.


what is that? I don't believe I wrote this axiom.
this is a mistake.

Correction:

What I wanted to write is a kind of axiom schema. were the existance
of every ordinal as defined above is quarrenteed.

3)Ordinal comprehension Schema:

ExAy(yex<->((yed or y=d) & d is an ordinal)). For every individual
instance of the variable d.
(were d is a variable that rang over sets).

In reality it might be better writtin in the following manner.

For every individual instance of d the following

"ExAy(yex<->((yed or y=d) & d is an ordinal))"

is an axiom.

here Ordinal comprehension is meant to be a schema, and not a single
axiom. Something like how separation is a schema in ZFC.

To be more clear about this I mean the following.

when d is not an ordinal then we have

x ={ }. since we'll have the axiom

ExAy(yex<->false).

so the existance of the empty set is quarenteed.

Now { } is an ordinal. (definition schema 1)

then when d={ } we have the following Axiom.

ExAy(yex<->((ye{} or y={}) & {} is an ordinal))

and this is:

ExAy(yex<->(y={} & {} is an ordinal))

Now either x ={} or not.

if x={}, then yex is false

and we have Ay(false<->true), a contradiction.

Then ~x={}

Now for every y:~y=0 we also have

(yex<->false)

and therefore ~yex when ~y=0.

and therefore only when y=0, y is a member of x.

so x={0}. ( note 0={}).

Now {0} is and ordinal ( definition schema 1).

Therefore we have the following axiom,

when d={0}.

ExAy(yex<->((ye{0} or y={0}) & {0} is an ordinal))

and this leads to x={0,{0}}.

But {0,{0}} is an ordinal, and so we have the following Axiom :

when d={0,{0}}.

ExAy(yex<->((ye{0,{0}} or y={0,{0}}) & {0,{0}} is an ordinal)).

and here x={0,{0},{0,{0}}}, which is an ordinal,...

And so on, having an uncountablly infinite number of Axioms.(one axiom
per each value of d), since there are uncountably infinite ordinals.

Now I think my idea of the "Ordinal Comprehension" schema. is not
difficult to understand , and I assume it is not weard.

Of course The set of all ordinals, do not exist in this theory. Since
its existance violate foundation.

Now, Separation is implied by replacement.

Pairing, can be acheived from replacing members in {0,{0}} , by x and
y,using the appropriate replacement function.

Union of sets x and y can be simply acheived by replacing the members
of a set z that is
equinumerous to the members of x and y together, by the members of x
and y.

Example x= {a,b} , y= {c,d}

Let z={0,{0},{0,{0}},{0,{0},{0,{0}}}

Replacing 0 by a, {0} by b, {0,{0}} by c
and {0,{0},{0,{0}} by d. then according to Replacement we have
the set {a,b,c,d}. as an existing set.

The axiom of the empty set is not needed since it is proved from
ordinal comprehension schema when d is not an ordinal. Also it can be
proved from replacement.

Therefore: Axioms of Empty set,Union,Pairing
Separation are all theorums in this theory.

The axiom of Power set , is the one hardest to
prove as a theorum in this theory.

But I guess that a method similar to that proving union as a theorum
in this theory can be used to prove the theorum of power, of course
depending on choice, which states that every set is well ordered.

In nutshell, I think this theory is equivalent to ZFC.

Sorry for the mistake.

Zuhair





4)Replacement
5)Infinity
6)Choice

All axioms except ordinal comprehension are as in ZFC.

I think that this theory is equivalent to ZFC!

Is that right?

Zuhair


.

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