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In "Computability and logic" by Boolos et al. there is axiomatisation called axioms of minimal arithmetic Q: $$(Q1)\ 0\neq x'$$ $$(Q2)\ x'= y'\implies x=y$$ $$(Q3)\ x+0=y$$ $$(Q4)\ x+y'=(x+y)'$$ $$(Q5)\ x*0=0$$ $$(Q6)\ x*y'=(x*y)+x$$ $$(Q7)\ \neg x<0$$ $$(Q8)\ x<y'\iff(x<y\lor x=y)$$ $$(Q9)\ 0<y\iff y\neq 0$$ $$(Q10)\ x'<y \iff (x<y\land y\neq x')$$

It can be transformed to Robinson Arithmetic R by adding $Q0$ and replacing $Q7-Q10$ by $Q11$: $$(Q0)\ x=0\lor \exists y:x=y'$$ $$(Q11)\ x<y \iff \exists z:(z'+x=y)$$ It is easy to see along with authors, that ordinal arithmetic is indeed model of Q, and not R. What is beyond my comprehension is alleged interpretation of cardinal arithmetic as model of R, that further fails to satisfy $(Q10)$, thus is not model of Q.

Now, while historically the mere notion of cardinal varies, and it's order properties are further affected by the presence of particular axioms, like AC, or GCH, reasoning about $(Q10)$ without more concrete context is vague at best. It seems reasonable to assume AC and "standard" arithmetic interpretation of addition $+_C$ and cardinal successor $^+$ with domain being (proper class) $Card:=\{\alpha\in Ord|\forall \beta<\alpha:|\beta|<|\alpha|\}$. In my understanding, in this interpretation $(Q10)$ holds, but $(Q4)$, $(Q0)$, $(Q11)$ and possibly others fails. To wit: either of $(Q0)$, or $Q11$ implies, that there is no limit cardinal different than $0$. $(Q4)$ fails, because for $\alpha<\beta$, $\aleph_\beta+_C\aleph_\alpha^+=\aleph_\beta\neq (\aleph_\beta+_C\aleph_\alpha)^+=\aleph_{\beta+1}$.

That being said, I'm far from stating, that authors are incorrect, and there is no interpretation of cardinal arithmetics, that fits into R. But, because it is not the (more or less) standard cardinal arithmetics, and there is no further clues, how such model would look like, it is very confusing. And the question is: is there such model?

While it is understandable, that every book has to be constrained, and cannot expand topics indefinitely, in this particular example it is reminiscent of famous Fermat's problem of too small margin.

Przemek
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    "What is beyond my comprehension is alleged interpretation of cardinal arithmetic as model of R" Where is this alleged exactly? (Also, as an aside: does that book really use that notation for those theories? If so it's worth pointing out that $\mathsf{Q}$ and $\mathsf{R}$ are used differently in generally - see here for the standard meaning of $\mathsf{Q}$, for example.) – Noah Schweber Nov 22 '20 at 21:48
  • Excerpt from chapter 16.4: "There is also an extremely natural nonstandard model for R, called the system of cardinal numbers, in which (Q10) fails [...]".

    As for notion for Q, and R, it is nonstandard indeed, but the notion of R in this book is also referred in this wikipedia article: https://en.wikipedia.org/wiki/Robinson_arithmetic#Variant_axiomatizations

     – Przemek Nov 22 '20 at 22:00
  • In a more straightforward way: yes, this book uses exactly this notation, and it is explicitly stated, that usual symbol for Robinson Arithmetics is Q, contrary to the convention from the book. – Przemek Nov 22 '20 at 22:10
  • Wait, they explicitly state that they use nonstandard notation, but still use it? Gah ... I love that book (it's what I learned from), but ... why? – Noah Schweber Nov 22 '20 at 22:23
  • " the notion of R in this book is also referred in this wikipedia article" Actually, the wiki only says that Boolos et al call it that :(. Standardly, $\mathsf{R}$ is actually used to refer to the incredibly weak system consisting just of the true quantifier-free sentences of arithmetic together with, for each $k\in\mathbb{N}$, the sentence $$\forall x[(\bigwedge_{i\le k}x\not=\underline{i})\rightarrow \underline{k}<x].$$ (Here "$\underline{n}$" is the numeral corresponding to $n$.) – Noah Schweber Nov 22 '20 at 22:25
  • It looks, like authors are focused on Robinson Arithmetics with modified order axioms. Then, usual axiomatization is merely mentioned in the summary chapter, and some differences are given. – Przemek Nov 22 '20 at 23:02
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    See the answer to this post – Mauro ALLEGRANZA Nov 23 '20 at 07:02
  • That is an answer I've been looking for. Thank you. – Przemek Nov 23 '20 at 14:01

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