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.

  • $\begingroup$ In the introduction, you required that $2^x$ exists in $M$ for every $x\in N$. This condition is missing in $T$ (and this will make a lot of difference). Is this omission intentional? $\endgroup$ Mar 31, 2021 at 7:22
  • $\begingroup$ No that was a mistake. $\endgroup$ Mar 31, 2021 at 8:01

1 Answer 1


$T$ 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, 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.