As an algebraist, I have some strong intuitions about what it means for an algebraic result to be true. In particular, my intuition would lead me to believe that if I cannot construct a counter-example to a claim, then the claim must be true. This is what motivated my previous question, where it became clear that my intuition needed some tweaking. (In other words, in some contexts it just doesn't hold true!) The answers given there are excellent, and I recommend people read them before continuing.

So here is a follow-up question to see just how far we can take the intuition.

Let $T_0$ be the theory ${\rm PA}$ over a countable language. Let $T_1$ be the extended theory obtained by adding as new axioms all $\Pi_1^0$ statement which are independent of ${\rm PA}$. This new theory is consistent and sound, assuming that the standard model of the natural numbers exists.

I would guess that this new theory $T_1$ is already not effectively computable. At any rate, suppose now that we let $T_2$ be the extended theory of $T_1$ obtained by adding as new axioms all $\Pi_2^0$ statement which are independent of $T_1$.

Is $T_2$ consistent? If so, is the theory $T_n$ (defined in the obvious recursive way) consistent? If so, is $\bigcup_{n\in \mathbb{N}}T_n$ the true theory of the standard model?

If $T_2$ is not consistent, is there some natural way to fix the problem?

Finally, what happens if we repeat these ideas for ${\rm ZFC}$ instead?

allindependent $\Pi^0_1$ sentences as new axioms. (The word "any" could be misunderstood as allowing you to add any such axioms you like.) $\endgroup$ – Joel David Hamkins Dec 28 '16 at 17:51