Re: Implementable Set Theory and Consistency of ZFC

On Nov 1, 4:52 am, Han de Bruijn <Han.deBru...@xxxxxxxxxxxxxx> wrote:
David C. Ullrich wrote:

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?

I am not very strong in this field but my understanding is that
your "Foo" above would be a sentence of first order logic.
So I'm not sure what it would mean to say it was "nonsense."
Denying a sentence in FOL is just coming up with a new
sentence that's the same as the old one but with a negation
symbol in front of it.


But there is more.
Because (apples without pears) is not the same as (apples without pears
and (no pears)).

I don't think that analogy will quite work.

Instead of physical objects, let's just use these two variables: A, P.
And let's try to prove A^P.

1) Given: A is true, P is true.

Can we prove A^P? Yes.

2) Given: A is true.

Can we prove A^P? No; it could be either true or false.

3) Given: A is true, P is false.

Can we prove A^P? No. In fact, we can prove that A^P is false.

My analogy relative to the issue being discussed is:

1) ZFC, including Infinity.
2) ZFC, without Infinity.
3) ZFC, with the negation of Infinity.

Is it clear why 2) and 3) are different?

Apologies for jumping in.


Marshall

.

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