Recall that $I\Delta\_0$ is the theory in the language of arithmetic that consists of the axioms of $\mathsf{PA}$ with induction restricted to $\Delta\_0$ formulas (i.e., formulas where all quantifiers are bounded).


It is not too difficult to build a model $M$ of $I\Delta\_0$ with a proper cut $N \subset M$ such that $M \equiv N$. Moreover, we can ensure that $2^n$ exists for every $n \in N$ and that $2^x$ is not a total function on $N$ (and therefore not a total function on $M$ either). It is also a standard fact that the set $\{ x \in M : (\exists y \in N) x < 2^y\}$ is a cut that models $I\Delta\_0$ (this follows from the fact that it is downwards closed and closed under addition and multiplication). My question is about this last cut actually being all of $M$.


Let $C$ be a unary predicate symbol, and let $T$ be the theory in the language of arithmetic augmented with $C$ that contains $I\Delta\_0$ and says that


* $C$ is downwards closed,
* $C$ is not all of the structure,
* $2^x$ is not a total function,
* $2^x$ is total on $C$,
* the induced structure on $C$ has the same theory as the full structure (in particular, it is a model of $I\Delta\_0$), and
* for every $x$, there exists a $y \in C$ such that $x < 2^y$.


Obviously, the fifth bullet point must be stated as an axiom scheme.



> 
> **Question 1.** Is $T$ consistent?
> 
> 
> 



> 
> **Question 2.** Does $I\Delta\_0$ interpret $T$?
> 
> 
> 



> 
> **Question 3.** Does $T$ interpret $I\Delta\_0 + \mathrm{Exp}$ (where $\mathrm{Exp}$ is the statement that $2^x$ is a total function)?
> 
> 
> 


Note that since $I\Delta\_0$ does not interpret $I\Delta\_0 + \mathrm{Exp}$, these last two questions cannot both have a positive answer.



Consistency and consistency strength of certain special cuts in $I\Delta_0$
$T$ is inconsistent. The argument below is due to Robert Solovay (it is attributed to a letter from Solovay to Nelson in Visser’s [Peano Basso and Peano Corto](https://dspace.library.uu.nl/handle/1874/253993), see Lemma 3.7).


Let $2^x\_n$ denote the iterated exponential function $2^x\_0=x$, $2^x\_{n+1}=2^{2^x\_n}$. It is well known that the graph of $2^x\_n$ has a well-behaved $\Delta\_0$ definition in $I\Delta\_0$.



> 
> **Lemma (Solovay):** $I\Delta\_0+\neg\mathrm{Exp}$ proves that there exists a unique number $n$ such that $2^0\_n$ exists, but $2^0\_{n+1}$ does not.
> 
> 
> 


**Proof:** Take $x$ such that $2^x$ does not exist, and let $n$ be maximal such that $2^0\_n\le x$. Then $2^0\_{n+1}$ either does not exist, or satisfies $2^0\_{n+1}>x$, hence $2^0\_{n+2}$ does not exist. Thus, $n$ or $n+1$ satisfies the conclusion of the Lemma. QED


Let us call the $n$ from the Lemma as *Solovay’s number*.


Now, observe that the axioms of $T$ imply that $C=\{x:2^x\text{ exists}\}$. Thus, working in $T$ (which includes $I\Delta\_0+\neg\mathrm{Exp}$), if $n$ is Solovay’s number, then $n-1$ is Solovay’s number in $C$. In particular, the universe and $C$ disagree on the truth of the sentence “Solovay’s number is even”, contradicting the elementary equivalence axiom.


