Re: ZFC in another shape.

On Feb 4, 9:02 pm, "Jesse F. Hughes" <j...@xxxxxxxxxxxxx> wrote:
"Jesse F. Hughes" <j...@xxxxxxxxxxxxx> writes:

"zuhair" <zaljo...@xxxxxxxxx> writes:

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)).

You sure that's the right definition? As consequence of this
definition, you get:

x is an ordinal -> AmAnAz(z e n -> z e m)

Thus, if there are any ordinals, then every set is equal (by
extensionality).

For the remainder, I will pretend you actually gave a correct
definition of ordinal.





Axioms:

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

What a strange name for that axiom.

4)Replacement
5)Infinity
6)Choice

All axioms except ordinal comprehension are as in ZFC.

I think that this theory is equivalent to ZFC!

Why would you think that? Seriously, what makes you think it is?

To put it another way, consider the following theory:

Axioms:

1')Extensionality
2')Foundation
3')The empty set exists: (Ex)(Ay)(~ y e x)
4')Replacement
5')Infinity
6')Choice

This theory is equivalent to yours, I'd say. Certainly it is at least
as strong as yours, since (3') implies (3). After all, the empty set
is an ordinal, and so if the empty set exists, an ordinal exists.

I'm almost sure that (3) implies (3'), but I don't see any reason to
check.

So your new hypothesis is much simpler: the existence of an empty set
(along with these other axioms) proves both power set and separation.

Doesn't seem too plausible.

--
Jesse F. Hughes

"Most of my research is irreducibly complex."
-- James S. Harris- Hide quoted text -

- Show quoted text -

Yea , true , I committed a mistake.
I corrected it.

Zuhair

.

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