According to Theorem 26H, if $\mbox{Th}\mathfrak{N}_{S}=\mbox{Cn}A_{S}$ were finitely axiomatizable, there would be a finite subset $A$ of $A_{S}$ such that $\mbox{Cn}A=\mbox{Cn}A_{S}=\mbox{Th}\mathfrak{N}_{S}$ . We show that this is not the case.

If a finite $A\subset A_{S}$ , then $\mbox{Cn}A\subseteq\mbox{Cn}A_{S}$ , and we need to show that $\mbox{Cn}A\subsetneq\mbox{Cn}A_{S}$ . That is, there is a sentence $\sigma$ true in $\mbox{Cn}A_{S}$ that is not true in $\mbox{Cn}A$ . At the same time, every model of $A_{S}$ is a model of $A$ , and every sentence true in $A_{S}$ will be true in such a model. Therefore, we need to find a model of $A$ that is not a model of $A_{S}$ , and a sentence $\sigma$ true in $\mbox{Cn}A_{S}$ that is not true in this particular model of $A$ .

Let us call each axioms S1-S4 as $\sigma_{1}$ , $\sigma_{2}$ , $\sigma_{3}$ and $\sigma_{4,n}$ for $n\ge1$ . Let $A_{n}=\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4,1},\ldots,\sigma_{4,n}\}$ , $n\in\mathbb{N}$ . Then, for a finite $A\subset A_{S}$ , there is some $n$ such that $A\subseteq A_{n}\subset A_{S}$ , and $\mbox{Cn}A\subseteq\mbox{Cn}A_{n}\subseteq\mbox{Cn}A_{S}$ . If we show that for every $n$ , $\mbox{Cn}A_{n}\subsetneq\mbox{Cn}A_{S}$ , then we will show the strict inclusion for every finite $A$ .

For $n\in\mathbb{N}$ , as the sentence that is true in $A_{S}$ but not in $A_{n}$ , consider $\sigma=\sigma_{4,n+1}$ . Then, obviously, $\sigma$ is true in $\mbox{Cn}A_{S}$ . We need to provide a model of $A_{n}$ such that $\sigma$ is not true in the model. The axioms of group 4 ensure that there is no loop of any finite length. In $A_{n}$ we excluded all such axioms for the lengths of the loop greater than $n$ . In particular, $\sigma$ says that there is no loop of length $n+1$ . So, our model of $A_{n}$ may have loops of any length greater than $n$ , and to ensure that $\sigma$ is false, it has to have a loop of length $n+1$ . At the same time, we still need to satisfy all axioms S1-S3: $0$ has no predecessor, $\mathbb{S}^{\mathfrak{A}}$ is one-to-one, and every element except $0$ has a predecessor. So, consider the set $N=\{1,\ldots,n+1\}$ , and let $S:\mathbb{N}\cup N\rightarrow\mathbb{N}\cup N$ be $S(k)=k+1\in\mathbb{N}$ for $k\in\mathbb{N}$ , and $S(k)=1+(k\mbox{ mod }(n+1))\in N$ for $k\in N$ . Then, let $\mathfrak{A}=(\mathbb{N}\cup N;0,S)$ . We check that all axioms S1-S4n hold for $\mathfrak{A}$ .

1. S1: $\forall x\mathbb{S}x\neq0$ . Indeed, $s(0)=0\in\mathbb{N}$ , and for every $d\in\mathbb{N}$ , $S(d)\ge1$ , while for every $d\in N$ , $S(d)\notin\mathbb{N}$ .
2. S2: $\forall x\forall y(\mathbb{S}x=\mathbb{S}y\rightarrow x=y)$ . Indeed, if for $d,d'\in\mathbb{N}\cup N$ , $S(d)=S(d')$ , then either $d,d'\in\mathbb{N}$ , and $d+1=d'+1$ implies $d=d'$ , or $d,d'\in N$ , and $1+d\mbox{ mod }(n+1)=1+d'\mbox{ mod }(n+1)$ implies $d\mbox{ mod }(n+1)=d'\mbox{ mod }(n+1)$ and $d=d'$ (all numbers in $N$ have different residuals).
3. S3: $\forall y(y\neq0\rightarrow\exists xy=\mathbb{S}x)$ . Indeed, if $d\in\mathbb{N}$ , $d>0$ , then $d-1\in\mathbb{N}$ , and $S(d-1)=d$ , and if $d\in N$ , then if $d=1$ , $d=S(n+1)$ , otherwise $d=S(d-1)$ .
4. S41-S4n: for every $k\in\mathbb{N}$ , $1\le k\le n$ , $\forall x\mathbb{S}^{k}x\neq x$ . Indeed, there are no loops in $\mathbb{N}$ , and $|N|=n+1$ , where each element except $n+1$ is mapped by $S$ to the next one, while $n+1$ is mapped to $1$ (since $N$ is finite, one can even check the finite number of axioms S41-S4n for each element in $N$ ).

Therefore, $\mathfrak{A}$ is indeed a model of $A_{n}$ . At the same time, $\sigma=\sigma_{n+1}$ is not true in $\mathfrak{A}$ , as there is a loop of length $n+1$ in $N$ (actually, for every $d\in N$ , $\vDash_{\mathfrak{A}}\mathbb{S}^{n+1}x=x[s(x|d)]$ , i.e. $\not\vDash_{\mathfrak{A}}\mathbb{S}^{n+1}x\neq x[s(x|d)]$ ).