In the Robinson arithmetic, I wonder how the recursive definition of addition and multiplication can be well-defined.
Axiom 3 seems to prohibit other chains starting with an element which has no successor.
But it doesn't prohibit loops or chains which are infinite in both directions.
E.g. why should the following
$$ 0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow \cdots \\ \\ a {\rightarrow \atop \leftarrow} b$$
(that means Sa = b and Sb = a) not be compatible with Robinson arithmetic?
In the article,in the section Metamathematics, 4th paragraph, it states that the axioms cannot disprove $x=Sx,$ which I think should say "cannot disprove $\exists x\;(x=Sx).$"