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Apr 22, 2013 at 11:25 comment added Emil Jeřábek Yes. This is provable by open (and therefore $\Sigma^b_0$) induction: if $x$ is neither $0$ nor a successor, we can prove $y\ne x$ for every $y$ by induction on $y$, and taking $y=x$ gives a contradiction.
Apr 16, 2013 at 21:11 comment added Colin McLarty $T^0_2$ does include Robinson's $Q$, right? Specifically, it includes $x=0\vee \exists y(x=Sy)$?
Apr 16, 2013 at 17:53 vote accept Colin McLarty
Apr 16, 2013 at 16:00 comment added Emil Jeřábek In light of the other answer, I should probably stress that $T^0_2$ is not a weak theory when it comes to Tennenbaum phenomena: it has no recursive nonstandard models, and in fact, every nonstandard countable model of $T^0_2$ has a nonstandard cut that is a model of PA. While the theory shows signs of pathological weakness (by Leszek’s results, it can’t prove that a power of $2$ is not divisible by $3$), the slightly stronger theory $T^0_2(\lfloor x/2^y\rfloor)$ is quite well-behaved, it can define all polynomial-time functions and prove induction for them (it is equivalent to $PV_1$).
Apr 16, 2013 at 13:46 history answered Emil Jeřábek CC BY-SA 3.0