https://en.wikipedia.org/wiki/%CE%A9-consistent_theory#Relat...
The Wikipedia article says T is omega-consistent if "T + RFN_T + the set of all true sentences is consistent", which should mean the same thing as "T + RFN_T is true".
> I think just PA+"PA is consistent" is enough?
It's not clear to me how. I believe PA+"PA is consistent" would allow a model where Goodstein's theorem is true for the standard natural numbers, but that also contains some nonstandard integer N for which Goodstein's theorem is false. I think that's exactly the case that's ruled out by the stronger statement of ω-consistency.
On the first question, how do you encode omega-consistency as a formula of PA? Just curious, it's not at all obvious to me.
I was also wondering about that, but going by the Wikipedia definition, it doesn't seem too complicated: you say, "For all encoded propositions P(x): If for every x there exists an encoded proof of P(x), then there does not exist an encoded proof of 'there exists an x such that ¬P(x)'." That is, if you can prove a proposition for each integer in the metatheory, then quantifiers within the target theory must respect that.
So I'm not convinced this is enough to conclude "forall x . P(x)" within PA itself, which is what you need to prove Goodstein's. We can merely conclude that there exists a (possibly non-standard) n coding for that statement.
ChatGPT tells me that PA+"PA is consistent" is not quite enough. I believe that it has digested enough logic textbooks that I'll believe that claim.