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Let $L_{E}$ be the language of discretely ordered rings together with an extra predicate symbol $E$. The system $A$ consists of the axioms of $I\Delta_{0}$ (basic arithmetic plus induction for bounded formulas) together with the statements:

(1) $E(0,1)$

(2) $\forall{x}\exists{y}E(x,y)$

(3) $\forall{x>0,y,z}E(x+1,y)\wedge E(x,z)\rightarrow y\geq 2z$

Is it then provable in $A$ that exponentiation (with base $2$) is total? (In any of the formulations as a bounded formula.)

This might be well-known, but I couldn't find a reference so far.

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Do you intend system $A$ to include only $I\Delta_0$, or actually $I\Delta_0(E)$? That is, do you allow induction for formulas involving $E$?

If you do allow it, then then you can prove easily $$\forall x,y\le w\,(E(x,y)\to x=y=0\lor\exists z\le y\,(z=2^x))$$ by induction on $w$, hence the theory proves exponentiation to be total.

If you only allow induction for $\Delta_0$ formulas without $E$, then the answer is negative, and in fact, $A$ is a conservative extension of $I\Delta_0$. This can be seen as follows: take any model $M\models I\Delta_0$, we’ll expand it to a model of $A$. For $n\in\mathbb N$, put $E(n,2^n)$. For nonstandard $x,y\in M$, let $x\sim y$ iff $x-y\in\mathbb Z$. For any equivalence class $C$ of $\sim$, choose its representative $a$ so that $C=a+\mathbb Z$, and choose an arbitrary nonstandard $b\in M$. Then for any $n\in\mathbb N$, put $E(a+n,b2^n)$ and $E(a-n,\lfloor b/2^n\rfloor)$.

One can do even better: for example, let $A^+$ be the extension of $A$ by the following axioms:

(4) $E(x,y)\land E(x,y')\to y=y'$

(5) $E(x,y)\land E(x',y')\to E(x+x',yy')$

(6) $E(x,y)\land E(x',y')\land x< y\to x'< y'$

(this would be more readable if we had a unary function symbol instead of $E$). Then $A^+$ is still a conservative extension of $I\Delta_0$. One way to prove this is as follows: take any countable recursively saturated model $M\models I\Delta_0$, and let $L=\{a\in M:M\models\exists y\,(y=2^a)\}$. The countable structures $(M,+,\le)$ and $(L,+,\le)$ are elementarily equivalent (as they are both models of Presburger arithmetic), and the joint model $(M,L,+,\le)$ is recursively saturated, hence there exists an isomorphism $f\colon(M,+,\le)\simeq(L,+,\le)$. Then putting $E(x,2^{f(x)})$ defines a model of $A^+$.

In fact, one does not need $M$ to be recursively saturated for this argument to work: it is enough if $M$ has no cofinal set of the form $\{a^n:n\in\mathbb N\}$. Then $(M,+,{\le})$ is recursively saturated by Cégielski, McAloon & Wilmers (improved to models of $IE_1$ by Wilmers), and then using the cofinality assumption, $(M,L,+,{\le})$ is recursively saturated using a variant of Theorem 5.3 in this paper of mine.

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  • $\begingroup$ Thanks for your answer. I was interested in the second variant, and after your first answer, I noticed that I should have added (5) in the original statement, so this extension is really interesting. $\endgroup$
    – M Carl
    Commented Apr 23, 2012 at 13:30
  • $\begingroup$ I realized the argument does not really give the axiom (7) $E(x,y)\to x< y$. While I am confident that the zig-zag construction of the isomorphism $f$ can be modified to yield $f(x)>\lfloor\log_2x\rfloor$ which guarantees (7), the proof gets messy, so I just deleted the axiom from the answer. $\endgroup$
    – Emil Jeřábek
    Commented Apr 24, 2012 at 9:30
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    $\begingroup$ The argument in Theorem 6.4 of arxiv.org/abs/2209.01197 does, in fact, give the axiom $E(x,y)\to x<y$. $\endgroup$
    – Emil Jeřábek
    Commented Sep 5, 2022 at 8:59

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