Let a formula $\phi$ of the language of first-order Peano arithmetic be total in a theory Th that extends PA iff, for any $k\_1, \dots, k\_n \in \omega$, Th $\vdash \phi(\bar k\_1, \dots, \bar k\_n)$ or Th $\vdash \neg \phi(\bar k\_1, \dots, \bar k\_n)$. Is it true that total formulae in Th are provably equivalent in Th to $\Delta\_0$ formulae?



Total formulae in a theory equivalent to $\Delta_0$ formulae in the theory?
No. 


Certainly the statement is false for arbitrary theories.


For computably axiomatizable theories, it's still false; what follows is informal. Consider a uniformly computably axiomatizable sequence of theories $T\_i$ which are *independent*: if I have a finite set $\{\varphi\_i: i<n\}$ of formulas with each $\varphi\_i$ either of the form $Con(T\_i)$ or $\neg Con(T\_i)$, then the conjunction of those formulas does not prove $Con(T\_{n})$ or $\neg Con(T\_{n})$.


Now let $Th$ be the theory gotten from PA by adjoining the axioms "$Con(T\_i)$" for each $i\in\omega$. Then $Th$ is consistent and computably axiomatizable, and the formula $Con(T\_x)$ is total in $Th$, but not provably $\Delta^0\_1$.


And this can even be extended to finitely axiomatizable theories: take $$Th=PA+\neg Con(PA)+\forall x(Con(T\_x)\text{ unless there is a $PA$-proof of 0=1 of length $<x$})$$ for appropriate choice of $T\_i$.


