Are all theorems of minimal arithmetic theorems of a given theory?

Question

I am working on some metamathematics revision and the following question came up. Let the theory $R_0$ be axiomatized by the following axiom schemata which hold for all $n,m \in \mathbb{N}$:

1. $\overline{m}+\overline{n} = \overline{m+n}$

2. $\overline{m}\cdot\overline{n} = \overline{m\cdot n}$

3. if $m<n$, then $\exists x \text{ } sx + \overline{m} = \overline{n}$

4. if $m\neq n$, then $\neg \overline{m} = \overline{n}$

($\overline{m}$ is a shorthand that abbreviates 0 prefixed my $m$ occurences of s)

Now take the theory $Q_0$ (which is a variant of minimal arithmetic -- note the absence of Ax. 3 of Robinson Arithmetic) axiomatized by:

1. $\forall x \neg 0 = sx$
2. $\forall x \forall y (sx=sy \rightarrow x=y)$
3. $\forall x + 0 = x$
4. $\forall x \forall y x+sy = s(x+y)$
5. $\forall x \cdot 0 = 0$
6. $\forall x \forall y x\cdot sy = (x\cdot y) +x$

Are all theorems of $Q_0$ theorems of $R_0$?

I'd be grateful for any pointers as to how to go about answering this kind of questions as well as full solutions. I am currently trying to derive the axioms of $Q_0$ as theorems of $R_0$ - is that the correct way forward?

Thanks for your help!

Best wishes,

Leon

Answer 1

Consider the interpretation with the domain $\mathbb N\cup \{*\}$ and the usual operations on naturals, extended with $$ s({*})=0 $$ $$ {*}+n = n+{*} = {*}+{*} = {*}\cdot n = n\cdot{*} = {*}\cdot{*} = {*} $$

This is a model of $R_0$ but not of $Q_0$ (it fails the first axiom). Therefore the axioms of $Q_0$ are not all theorems of $R_0$.