Re: Implementable Set Theory and Consistency of ZFC
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Thu, 01 Nov 2007 07:22:21 -0600
On Thu, 01 Nov 2007 10:03:34 +0100, Han de Bruijn
<Han.deBruijn@xxxxxxxxxxxxxx> wrote:
MoeBlee wrote:
On Oct 31, 1:59 am, Han de Bruijn <Han.deBru...@xxxxxxxxxxxxxx> wrote:
Jesse F. Hughes wrote:
I will say it once more. I am typing this slowly, since I don't want
you to miss anything I say. If you have a proof in a theory
consisting of axioms (1)-(4), it is also a proof in the theory
consisting of axioms (1)-(4)+(X).
How do you "know" that? Has the Pope told you, by dogma, that it is so?
No, you royal ignoramus, it's PROVEN as a basic property of the
deductive system. It's the property of monotonicity of deduction. And
we PROVE it.
Yeah, yeah. Then why are (5-9) required in ZFC, once Infinity has become
an axiom of it? Why is e.g. Choice provable in (ZFC-Infinity) and not in
common ZFC?
This has been explained several times. If by ZFC-Infinity you mean
ZFC without the axiom of infinity then Choice is _not_ provable
in that syatem. "ZFC without the axiom of infinity" is not the
same as "ZFC plus the negation of the axiom of infinity" - choice
_is_ provable in ZFC plus the negation of the axiom of infinity.
But ZFC does not consist of "ZFC plus the negation of Infinity"
plus more axioms (it _is_ equal to "ZFC withot Infinity" plus
more axioms).
Han de Bruijn
************************
David C. Ullrich
.
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