That’s sufficient to be a contradiction, since it’s possible for the speaker to say “X” and then say “X*”. You have to constrain the speaker to say true things, and thus for the speaker to begin enumerating axioms of Peano Arithmetic, you have to assume those axioms to be true, thus contradiction.
It’s sufficient to allow an adversarial (and dishonest) speaker to force a contradiction, sure. But the theorem is completely subjective. It says that even from the agent’s perspective there is a problem. IE, even if we think the speaker to be completely honest, we can’t (computably) have (even minimally) consistent beliefs. So it’s more surprising than simply saying that if we believe a speaker to be honest then that speaker can create a contradiction by lying to us. (At least, more surprising to me!)
If we think the speaker to be completely honest (incapable of saying false things), and also we have nonzero belief that the speaker will enumerate axioms of Peano Arithmetic, then we already believe a contradiction (because assuming that a proof of X implies X creates all the contradictions).
Adding to the axioms of PA, the statement “a proof of X from the axioms of PA implies X”, does not create any contradictions. This is just the belief that PA is sound.
What would be contradictory would be for PA itself to believe that PA is sound. It is fine for an agent to have the belief that PA is sound.
Believe the speaker to be honest.
Have minimally consistent beliefs.
That’s sufficient to be a contradiction, since it’s possible for the speaker to say “X” and then say “X*”. You have to constrain the speaker to say true things, and thus for the speaker to begin enumerating axioms of Peano Arithmetic, you have to assume those axioms to be true, thus contradiction.
It’s sufficient to allow an adversarial (and dishonest) speaker to force a contradiction, sure. But the theorem is completely subjective. It says that even from the agent’s perspective there is a problem. IE, even if we think the speaker to be completely honest, we can’t (computably) have (even minimally) consistent beliefs. So it’s more surprising than simply saying that if we believe a speaker to be honest then that speaker can create a contradiction by lying to us. (At least, more surprising to me!)
If we think the speaker to be completely honest (incapable of saying false things), and also we have nonzero belief that the speaker will enumerate axioms of Peano Arithmetic, then we already believe a contradiction (because assuming that a proof of X implies X creates all the contradictions).
Adding to the axioms of PA, the statement “a proof of X from the axioms of PA implies X”, does not create any contradictions. This is just the belief that PA is sound.
What would be contradictory would be for PA itself to believe that PA is sound. It is fine for an agent to have the belief that PA is sound.
There might be some technicality under which you’re not simply wrong. https://en.wikipedia.org/wiki/L%C3%B6b%27s_theorem