From: Aatu Koskensilta on
Frederick Williams <frederick.williams2(a)tesco.net> writes:

> Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize
> it in ZFC.

This is a pretty silly way of proving the consistency of PA in set
theory. That PA is consistent is a triviality. The interest in Gentzen's
proof lies elsewhere.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on
Charlie-Boo <shymathguy(a)gmail.com> writes:

> ZFC was designed to avoid paradoxes by making explicit what can be a
> set. It doesn't do anything else except what the Peano Axioms give
> it.

Does PA give us Borel determinacy? Is Borel determinacy trivial?

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on
Charlie-Boo <shymathguy(a)gmail.com> writes:

> On Jun 28, 8:58�am, Frederick Williams <frederick.willia...(a)tesco.net>
> wrote:
>
>> Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize
>> it in ZFC.
>
> Give the slightest bit of details.

Read a logic book.

>> It has everything to do with V_omega.
>
> That's not a ZFC axiom.

Indeed. It is a set.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on
billh04 <hale(a)tulane.edu> writes:

> Are you saying that it is a theorem of ZFC that PA is consistent?

Sure. That is, the statement "PA is consistent" formalized in the
language of set theory as usual is formally derivable in ZFC (and
already in much weaker theories).

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on
MoeBlee <jazzmobe(a)hotmail.com> writes:

> By the way, here's Aatu's proof:
>
> Theorem (of Z-"ax regularity"-"ax infinity"+"~ ax infinity"):
>
> Ax x is finite.

You need an F up there, I believe.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus