Re: ZFC in another shape.

On Feb 6, 12:49 pm, "MoeBlee" <jazzm...@xxxxxxxxxxx> wrote:
On Feb 6, 8:29 am, "hagman" <goo...@xxxxxxxxxxxxx> wrote:

'0' and '{}' can be defined, for sure, now that I know you've chosen
the strong version of replacement. But you haven't proven union.

Is that right? Can replacement produce an empty set {F(x): x in A} if
A is not empty?

Certain formulations of replacement (such as found in Enderton's and
in Suppes's set theory texts) prove the existence of an empty set.
Meanwhile, the derivation you suggest is obtainable from separation,
which is also provable from those certain formulations of replacement.

One is tempted to try to prove the existence of the non-ordinal {{}}.
However that would in itself require {}.

As far as I can tell, it's all rather mixed up. However, we do get the
empty set from replacement, then from power set we get {0}, then from
power set we get {0 {0}}, then from separation (which we get from
replacement) we get {{0}}, which is not an ordinal. But then, of
course, we don't need his ordinal successor axiom for any of that
(there's no reason to prove there is a non-ordinal just to prove the
existence of the empty set when we already have the existence of the
empty set and indeed use it to prove the existence of a non-ordinal).

MoeBlee

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.

Zuhair

.

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