Most true statements independent of PA that I know of is equivalent to some consistency statement. For example


* Con(PA), Con(PA + Con(PA)), Con(PA + Con(PA) + Con (PA + Con(PA)), $\dots$
* Goodstein's theorem is equivalent to Con(PA)
* Any conjunction or disjunction of the above.


Is every true statement independent of PA equivalent to some consistency statement?


By "equivalent to some consistency statement", I mean that $PA \vdash S \iff Con(T)$, for some theory $T$. Also, $T$ should be either finite, or specified by a Turing machine that outputs its axioms (and such that PA proves that the Turing machine never stops outputting statements), so that the description of $T$ doesn't throw PA off.


EDIT: In particular, are there are $\Pi^0\_1$ examples?



Is every true statement independent of $PA$ equivalent to some consistency statement?
The theory $PA + Con(PA)$ has the property you are asking for, this is the so called Friedman-Goldfarb-Harrington principle (see, e.g., [Fifty years of self-reference in arithmetic](https://projecteuclid.org/euclid.ndjfl/1093883515 "Fifty years of self-reference in arithmetic"), p. 366). Formally, for *every* $\Pi\_1$ sentence $\pi$, there is a $\Pi\_1$ sentence $\psi$ such that $PA + Con(PA) \vdash \pi \leftrightarrow Con(PA + \psi)$.


EDIT:


~~That being said, my hunch is that the answer must be "no" for plain $PA$ even though I can't produce a counterexample at the moment.~~


As observed by Will Sawin, we have that for every $\Pi\_1$ sentence $\pi$, there is a $\Pi\_1$ sentence $\psi$ such that $PA \vdash \pi \leftrightarrow Con(EA + \psi)$, which gives a positive answer to the OP's modified question.


My hunch is that $Con(EA + \psi)$ can not be replaced by $Con(PA + \psi)$ in the above.


