Re: ZFC in another shape.

On Feb 6, 2:59 pm, "MoeBlee" <jazzm...@xxxxxxxxxxx> wrote:
On Feb 6, 11:20 am, "zuhair" <zaljo...@xxxxxxxxx> wrote:

Not only that. I discovered that The axiom of ordina succession I've
made , is in reality a theorum in a theory consisting of the other six
axioms. we can simply prove that using power and replacement alone.
Since every ordinal subsets its power set, and is a member of its
power set, then by replacement the the sucessor ordinal is a subset of
the power set.
what I want to say is that S(x) is a subset of P(x) for all x: x is an
ordinal.

Then the axiom of ordinal succession should be converted into the
theorum of ordinal succession.

You don't even need replacement. Power set with separation will do the
job.

Meanwhile, I think you can find proofs that union is not derivable
from the other axioms (if I'm not mistaken, in the usual treatments of
the independence proofs). So your new theory is just ZFCR (ZFC with
regularity) without union. So you've set up all this rigmarole just to
state ZFCR without union.

MoeBlee

From all of that I deduce the following.

ZFC is in reality the following axioms.

1) Extensionality 2) Replacement (strong version)
3) Power 4) Union 5) Infinity

+- 6) Regularity and 7) Choice.

I have a question , can replacement be modefied in such a manner as to
embrace union as well.

Example:

AxEy!(P(x,y))->AaEbAy(yeb<->Exea(P(x,y))).

Can't that be modified to:

AxEy!(P(x,y))->AaAcEbAy(yeb<->Ex((xea \/ xec)&P(x,y))).

I am not sure if this can cover union. But having choice, I think we
can always have a choice function that using this axiom can turn union
into a theorum in a theory that consists of
Extensionality,Replacement(my version),Power,Infinity,
Regularity ,Choice.

Zuhair

.

Relevant Pages

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  • 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: 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. ...
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  • Re: ZFC in another shape.
    ... we can simply prove that using power and replacement alone. ... Since every ordinal subsets its power set, and is a member of its ... from the other axioms (if I'm not mistaken, ... regularity) without union. ...
    (sci.math)
  • Re: ZFC in another shape.
    ... we can simply prove that using power and replacement alone. ... Since every ordinal subsets its power set, and is a member of its ... from the other axioms (if I'm not mistaken, ... regularity) without union. ...
    (sci.math)
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