$\newcommand\Tr{\text{Tr}}$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](http://jdh.hamkins.org/definiteness-of-truth-definiteness-of-objects-nyu-november-2014/) for the NY Phil Logic Group in which such matters were an important part of the discussion, and [Kit Fine](http://philosophy.fas.nyu.edu/object/kitfine) 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](http://projecteuclid.org/download/pdf_1/euclid.ndjfl/1093635835#page=3).)
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.



Is there a nonstandard model of arithmetic having precisely one inductive truth predicate?

> 
> The answer to the question is in the positive.
> 
> 
> 


Let $(\cal{N}^\*,\textrm{Tr\*})$ be a *rather classless* elementary extension of $(\cal{N},\textrm{Tr})$, where $\cal{N}$ is a model of PA, and $\textrm{Tr}$ is a full truth predicate on $\cal{N}$, e.g., let $\cal{N}$ be the standard model of PA, and $\textrm{Tr}$ be the usual Tarskian truth predicate on $\cal{N}$.


In the above, "rather classless" means that $(\cal{N}^\*,\textrm{Tr\*})$ has the property that every subset $X$ of the universe $N^\*$ of $\cal{N}^\*$ that is piecewise coded in $\cal{N}^\*$ is first order definable in $(\cal{N}^\*,\textrm{Tr\*})$, where $X$ is piecewise coded means that the intersection of $X$ with every topped initial segment of $\cal{N^\*}$ is parametrically definable in $\cal{N^\*}$ (indeed, it is coded by a single element).
 See Theorem 2.2.14 (and the preceding two paragraphs) of the [Kossak-Schmerl monograph](http://sistemas.fciencias.unam.mx/~lokylog/images/stories/Alexandria/Oxford%20Logic%20Guides/%5BOLG%2050%5D%20The%20Structure%20of%20Models%20of%20Peano%20Arithmetic%20-%20Roman%20Kossak,%20James%20Schmerl%20%5BOxford%20Logic%20Guides%5D%20(2006)(T).pdf#page=53) on models of PA for the relevant existence theorem for rather classless models. Such models exist in every uncountable cardinality $\kappa$ of uncountable cofinality and indeed can be arranged to be $\kappa$-like.


On the other hand, it is easy to see that every inductive set is piecewise coded. Therefore, if $\textrm{T}$ is an inductive truth predicate on $N^\*$, then $\textrm{T}$ is piecewise coded and by rather classlessness it is definable in $(\cal{N}^\*,\textrm{Tr\*})$, and in particular the double-expansion $(\cal{N}^\*,\textrm{Tr\*}, \textrm{T})$ satisfies full induction.


Now, an easy induction carried out internally in $(\cal{N}^\*,\textrm{Tr\*}, \textrm{T})$ on the complexity of formulae shows that $\textrm{Tr\*} = \textrm{T}$. **QED**


