Re: Implementable Set Theory and Consistency of ZFC

David C. Ullrich wrote:

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.

Sure. (ZFC - Foo) is not the same as (ZFC - Foo + ~Foo). Quite clear.
Tip, hint: how can Foo be "denied" iff Foo is just plain nonsense in
the first place? But there is more.
Because (apples without pears) is not the same as (apples without pears
and (no pears)). And I am so stupid that I don't understand this? Wow!

But ZFC does not consist of "ZFC plus the negation of Infinity"
plus more axioms (it _is_ equal to "ZFC without Infinity" plus
more axioms).

Han de Bruijn

.

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