*I asked this over at [math.stackexchange](https://math.stackexchange.com/questions/873489), and though a number of people were interested enough to vote up the question, I didn't get an answer -- which makes me wonder whether it isn't quite so trivial/dumb as I originally feared it was. So let me try again in this more turbo-charged forum ...*


Take our old friend Robinson Arithmetic, and cut it down to a theory of successor and addition. 


To spell that out (just to ensure that we are singing from the same hymn sheet), take the first-order theory with $\mathsf{0}$ as the sole constant, and $\mathsf{S}$ and $+$ as the built-in function signs, with the five axioms 


1. $\mathsf{\forall x\ 0 \neq Sx}$
2. $\mathsf{\forall x\forall y\ Sx = Sy \to x = y}$
3. $\mathsf{\forall x(x \neq 0 \to \exists y\ x = Sy)}$
4. $\mathsf{\forall x\ (x + 0) = x}$
5. $\mathsf{\forall x\forall y\ (x + Sy) = S(x + y)}$


and whose deductive system is your favourite classical first-order logic with identity.


Since this cut-down theory doesn't represent the recursive functions, you can't use the usual proof of undecidability for an arithmetic. Since this cut-down theory doesn't even know that addition is commutative, you can't do the kind of manipulations inside the theory involved in a quantifier-elimination proof of decidability (cf. what happens when we add induction to this theory to get Presburger arithmetic, i.e. Peano Arithmetic minus multiplication). 


Ermmmm .... so .... Drat it, I *ought* to know how to prove that this cut-down theory is decidable or that it is undecidable. But I seem to have forgotten, assuming I ever knew, and searching around hasn't helped me out. OK folks, I'm more than likely to be having a senior moment here [*well, given the lack of answers on math.se maybe a forgivable senior moment?*] -- so be gentle! -- but how do we show the theory is (un)decidable? [*My bet is on undecidable, for what little that it is worth ...*]



The (un)decidability of Robinson-Arithmetic-without-Multiplication?
The theory $T'$ with axiom $Sx\ne x$ in place of $Sx\ne0$, as it was originally written, is undecidable, because the theory of groups with a distinguished nonidentity element $a$ is a conservative extension of the theory of nontrivial groups, shown undecidable by Tarski. Since this is a finite extension of $T'$ (modulo the translation of $Sx$ by $x+a$), the latter is also undecidable.


The theory with the proper axiom $Sx\ne0$ is undecidable as well. The reduction above doesn’t quite work as the offending axiom is incompatible with groups, nevertheless one can apply a minor modification of Tarski’s original argument.


Let $M$ be the set of affine functions $f\colon\mathbb Z\to\mathbb Z$ of the form $f(x)=ax+b$, where $a\in\mathbb N^{>0}$, $b\in\mathbb Z$, and if $a=1$, also $b\ge0$. It is easy to see that $M$ is closed under composition, and contains the identity and the function $s(x)=x+1$. The structure
$$\mathcal M=\langle M,\mathrm{id},\circ,S\rangle,$$
where $S(f)=f\circ s$, satisfies axioms 1–5: in particular, $f\circ s=g\circ s$ implies $f=f\circ s\circ s\_{-1}=g\circ s\circ s\_{-1}=g$ where $s\_n(x)=x+n$, which implies 2, and axioms 1 and 3 are consequence of the fact that $f\circ s\_{-1}\in M$ iff $f\ne\mathrm{id}$.


It thus suffices to prove that $\mathrm{Th}(\mathcal M)$ is hereditarily undecidable, and we can do this by interpreting $\langle\mathbb N,+,\cdot\rangle$ in $\mathcal M$. We embed $\mathbb N$ in $\mathcal M$ via $n\mapsto s\_n$. The range of the embedding is definable, as $f\in M$ is of the form $s\_n$ for some $n$ if and only if it commutes with $s$. Addition on $\mathbb N$ is definable in $\mathcal M$ as $s\_{n+m}=s\_n\circ s\_m$. Finally, notice that if $f(x)=ax+b$, we have $f\circ s\_n=s\_{an}\circ f$, hence
$$nm=k\iff\forall f\in M\,(f\circ s=s\_n\circ f\to f\circ s\_m=s\_k\circ f)$$
for $n>0$. This shows that multiplication on $\mathbb N$ is definable in $\mathcal M$.


