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](https://mathoverflow.net/questions/258132/platonic-truth-and-1st-order-predicate-logic), 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?



Platonic Truth and 1st Order Logic - Take 2
Yes, your theory is the same as the true arithmetic. In
particular, yes, it is consistent.


I claim that at stage $n$, your theory $T\_n$ consists of PA plus
the collection of true $\Pi^0\_n$ sentences (that is, true in the
standard model). This starts out true with $T\_0$. If $T\_n$ is like
that, then consider $T\_{n+1}$. If $\psi$ is a true $\Pi^0\_{n+1}$
statement, then in particular, it is consistent with $T\_n$, since
both are true in the standard model. So either it is provable from
$T\_n$, in which case it is already there, or else it is
independent, in which case you add it at stage $n+1$. Conversely,
if $\psi$ is a $\Pi^0\_{n+1}$ statement that is independent of
$T\_n$, then in particular, it asserts $\forall x\ \phi(x)$, where
$\phi(x)$ is $\Sigma\_n$. Since $\psi$ is true in a model of $T\_n$,
it follows that $\phi(m)$ must hold of every standard $m$, since if $\phi(m)$ failed then $\neg\phi(m)$ would be a true $\Pi^0\_n$ sentence and hence part of $T\_n$, preventing $\psi$ from being consistent with $T\_n$. So
$\psi$ is true.


Thus, by induction, you are adding all and only the true $\Pi^0\_n$
statements at each stage, and you've got the theory of true
arithmetic.


