$\newcommand\Tr{\text{Tr}}$My question is whether there can be a nonstandard model of PA having a unique inductive truth predicate.

**Background.** If $\mathcal{N}=\langle N,+,\cdot,0,1,<\rangle$ is a model of the
first-order PA axioms, then a *truth predicate* on $\mathcal{N}$, also commonly called a *satisfaction class*,
is a predicate $\Tr\subset N$ obeying the recursive Tarskian truth
conditions, that is:

- (atomic) The $\Tr$ predicate holds of the Gödel code of an atomic formula just in case that atomic formula is true. For example, $\Tr(\underline n+\underline k=\underline r)\iff \mathcal{N}\models n+k=r$, and so on, where $\underline n$ means the term $\underbrace{1+1+\cdots+1}_n$.
- (conjunction) $\Tr(\sigma\wedge\tau)$ holds if and only if $\Tr(\sigma)$ and $\Tr(\tau)$ both hold.
- (negation) $\Tr(\neg\sigma)$ holds if and only if $\Tr(\sigma)$ does not hold.
- (quantifiers) $\Tr(\exists x\varphi(x))$ holds if and only if there is some $n\in N$ such that $\Tr(\varphi(\underline n))$ holds.

Such a truth predicate is said to be *inductive*, if the model
$\langle N,+,\cdot,0,1,<,\Tr\rangle$ expanded by that predicate
satisfies induction in the expanded language.

For nonstandard models, the existence of an inductive truth
predicate implies that the model is computably saturated, and a
countable model is computably saturated if and only if there is an
inductive *partial* truth predicate, one which is defined only on
all sentences up to some nonstandard finite level of complexity.
Thus, some nonstandard models of PA, the non-computably saturated
ones, have no inductive truth predicate. Meanwhile, Krajewsky
proved that there can be nonstandard models of PA having more than
one distinct truth predicate.

I gave a talk yesterday for the NY Phil Logic Group in which such matters were an important part of the discussion, and Kit Fine asked a great question there:

**Question.** Can there be a nonstandard model of arithmetic
having a unique inductive truth predicate?

The standard model $\mathbb{N}$, of course, has a unique truth predicate, since one can prove by induction that every such predicate must agree with actual arithmetic truth. But for countable nonstandard models, the answer is definitely no, it does not happen. The reason is that if a nonstandard countable model of PA has any inductive truth predicate at all, then it is computably saturated, and so the usual back-and-forth constructions show that the model will have many automorphisms, and furthermore many automorphisms that take true sentences to false ones and vice versa. This is simply because the truth predicate cannot be definable in the base language of arithmetic, and so there must be two elements of the model having the same types in the model in the language of arithmetic, but one is the Gödel code of a true sentence and the other the code of a false sentence. By constructing a tree of such automorphisms, one can get continuum many distinct truth predicates this way. (See also Kossak and Schmerl, Minimal Satisfaction Classes with an Application to Rigid Models of Peano Arithmetic, NDJFL 32(3), 1991.) This back-and-forth argument provides an alternative proof of Krajewsky's result.

But can there be an uncountable affirmative instance of the question? We know that there can be rigid $\omega_1$-like computably saturated models, and that kind of situation is suggestive that an affirmative instance may be possible.