A note on recognisability and definability

Definitions. An $n$-ary relation is associated with a set of $n$-tuples; an $n$-ary function $f$ is associated with a set of $(n+1)$-tuples $\{(a_1,...,a_n,f(a_1,...,a_n)\}$, which is called the graph of function $f.$ A sequence $s_1,s_2,...$ over $A$ is associated with a function $s:\mathbb N^n\rightarrow A$ such that $s(i)=s_i.$ Given an integer $n$, we use $(n)_p$ to denote its p-ary expansion; given a p-expansion $w\in\{0,...,p-1\}^*$, we use $[w]_p$ to denote its value $\sum_{i=0}^{|w|-1}p^{|w|-1-i}w[i].$ By convention, we set $(0)_p=\epsilon$ and $[\epsilon]_p=0.$

Recognisable Sets

Definitions. A set is called regular or recognisable if it can be recognised by a finite DFA. For $L\subseteq\Sigma^*$, $\sim_L$ is called a right congruence iff $u\sim_L v \iff uw\sim_L vw$ for all $w\in\Sigma^*.$ Note that a right congruence is an equivalence relation.

Myhill-Nerode Theorem. A set $L$ is regular iff $\sim_L$ has finite index over $\Sigma^*.$

Definitions. (p-recognisability) A set of integers is called p-recognisable iff the p-ary expansion of the integers is recognised by a finite automaton with alphabet $\{0,...,p-1\}.$ The automaton would read a word from left to right and ignores the leading zeros, so that if it accepts $w$ then it accepts all words in $0^*w.$
Fact. A sequence $s_1,s_2,...$ over $A$ is p-recognisable if $s^{-1}(a)$ is p-recognisable for each $a\in A.$ When $A$ is finite, it is easier to define a sequence in this way.

Definitions of recognisability generalise to n-ary relations naturally by replacing the alphabet $\{0,...,p-1\}$ to $\{0,...,p-1\}^n.$ All properties and facts listed above still hold in the general case.

Definitions. (p-automata) A p-automaton recognising a sequence $s_1,s_2,...$ over a finite domain is a complete deterministic Moore machine that reads $(n)_p$ and outputs $s_n$ for all $n\ge 1.$ In such case the sequence is called p-recognisable. Note that p-recognisability of sequences generalises p-recognisability of sets, since a set is p-recognisable iff there exists a p-automaton that outputs 1 on $(n)_p$ if $n$ belongs to the set and outputs 0 otherwise. Alternatively, we can describe a set of integers using a characteristic sequence. The characteristic sequence of a set $S$ refers to a sequence over $\{0,1\}$ where the nth bit is 1 iff $n\in S.$ A set is p-recognisable iff its characteristic sequence is p-recognisable.
A p-automaton can recognise an n-ary relation by recognising an n-dimensional sequence $s:\mathbb N^n\rightarrow\{0,1\}.$ The input language $((a_1)_p,...,(a_n)_p)$ is cleverly stipulated as a subset of $(\{0,...,p-1\}^n)^*$ instead of $(\{0,...,p-1\}^*)^n.$ Namely, components of a relation convoluted over alphabet $\{0,...,p-1\}^n$. In this way, theorems such as Myhill-Nerode still hold for $n\ge2.$

Definable Sets

Definitions. A set $L$ of $n$-tuples is called (first-order) definable by a formula $\phi$ in a first-order structure $\mathcal S$ iff $\phi$ has exactly $n$ free variables and $L=\{(a_1,...,a_n)\in D^n :$ $S\models \phi(a_1,...,a_n)\}$, where $D$ is the domain of $\mathcal S.$ A set is definable in a structure $\mathcal S$ if it is definable by a formula in $\mathcal S.$ When the structure is clear in the context we simply called a set definable. A sequence $s_1,s_2,...$ over $A$ is definable if the graph of the function $s:\mathbb N\rightarrow A$ associated with the sequence is definable.
Fact. A sequence $s_1,s_2,...$ over $A$ is definable iff $s^{-1}(a)$ is definable for each $a\in A.$ When $A$ is finite, it is easier to define a sequence in this way.

Definitions. (p-definability) Given $x\in\mathbb N$, define $V_p(x)\equiv$ the largest power of $p$ that divides $x.$ Let $V_p(0)\equiv 1.$ A subset of $\mathbb N$ is called $p$-definable if it is definable in $[\mathbb N,+,V_p].$ Given $x\in\mathbb N$, define $V_p(x)\equiv$ the largest power of $p$ that divides $x$ and $V_p(0)\equiv 1.$ Define a predicate $P_p$ that $P_p(x)$ holds iff $x$ is a power of $p.$ Define a binary relation $B_p$ such that $B_p(x,y)$ holds iff $y$ is a power of $p$ occurring in the p-ary expansion of $x.$ (Thus $B_p(x,y)$ iff $y<x$, $P_p(y)$, and there exits a $0\le z <y$ such that $y|(x-z).$)

Great efforts were made in the history to characterise 2-recognisable sets in an FO structure. Three candidate structures were considered for this purpose in order: $[\mathbb N, +, P_2]$, $[\mathbb N, +, B_2]$, and $[\mathbb N, +, V_2].$ We note two facts here about these structures:
Fact. $[\mathbb N, +, B_p]$ and $[\mathbb N, +, V_p]$ are (FO-)interpretable in each other.
Fact. $[\mathbb N, +, P_p]$ is strictly weaker than $[\mathbb N, +, V_p].$
To see the first, note that for $x>0$, $V_p(x)=y$ iff $B_2(x,y)\wedge \forall 0\le z<y.\ \neg B_p(x,z)$, and that $B_p(x,y)$ holds iff $\exists 0\le z <y.\ V_p(x-z, y).$ The second fact follows from an interesting decidability result. Let $p^x$ denote the exponential function. It is known that $[\mathbb N,+,V_p]$ and $[\mathbb N,+,p^x]$ are decidable, but $[\mathbb N,+,p^x,V_p]$ is undecidable. Clearly $P_p$ can be defined by $V_p$ and $p^x.$ If $[\mathbb N, +, P_p]$ is equal to $[\mathbb N, +, V_p]$, then $V_p$ will be definable by $p_x$, which implies $[\mathbb N,+,V_p]$ and $[\mathbb N,+,p^x,V_p]$ are mutually interpretable. This is impossible since the first is decidable and the second is not. The ultimate characterisation of 2-recognisable sets is given by the following theorem.
Theorem I. For any prime number p, a set is p-definable iff it is p-recognisable.
Proof. ($\Rightarrow$) It suffices to show that given any formula $\phi$ from a structure $\mathcal S$ equivalent to $[\mathbb N,+,V_p]$, we can construct a p-automaton $A_\phi$ recognising the set definable by $\phi$, i.e., $\mathcal S \models \phi(a_1,...,a_n)$ implies that $A_\phi$ accepts $([a_1]_p,...,[a_n]_p).$ To simply our proof, we choose $\mathcal S=[\mathbb N,R_1,R_2]$, where $R_1(x,y,z)$ holds iff $x+y=z$ and $R_2(x,y)$ holds iff $V_p(x)=y.$ Now we prove the claim by structural induction. First note that the three atom formulas $x=y$, $R_1(x,y,z)$, and $R_2(x,y)$ in $\mathcal S$ are p-recognisable (Exercise!). To complete the induction, it suffices to show that we can construct p-automata $A_{\neg\phi}$, $A_{\phi\vee\psi}$ and $A_{\exists x\phi}$ from p-automata $A_\phi$ and $A_\psi.$ Negation construction is immediate since p-automata is closed under complementation. For union, we modify the transition functions of the two p-automata so that they operate over the same set of input variables. For instance, suppose the free variables in $\phi$ are $x_1,...,x_m,y_1,...,y_n$ and the free variables in $\psi$ are $y_1,...,y_n,z_1,...,z_r.$ Then after the modification, the transition functions of the p-automata have the same domain $(x_1,...,x_m,$ $y_1,...,y_n,$ $z_1,...,z_r).$ The union of the two modified automata gives $A_{\phi\vee\psi}.$ For projection, to construct $\exists x.\phi(x,x_1,...,x_n)$ we just relabel the transitions of the p-automaton by replacing $(a,a_1,...,a_n)$ with $(a_1,...,a_n).$ Note that we may need to add new loops after the relabelling to recognise leading zeros caused by the projection. See [3] for details of this proof.

Corollary. The theory of $[\mathbb N,+,V_p]$ is decidable.
Proof. Given a sentence in $[\mathbb N,+,V_p]$, we can rewrite it to the form of either $\exists x.\phi(x)$ or $\neg\exists x.\phi(x).$ By Theorem I, $\phi(x)$ is p-recognisable. Hence we can construct an finite automaton $A$ that accepts a word $w$ iff $[\mathbb N,+,V_p]\models\phi([w]_p).$ Hence the truth of the sentence can be checked by checking emptiness of a recognisable language, which is decidable.

Corollary. The set of squares $\{n^2:n\in\mathbb N\}$ is not p-recognisable.
Proof. Suppose to the contrary that the set of squares is p-recognisable. By Theorem I, there exists a formula $\phi$ in $[\mathbb N,+,V_p]$ such that $\phi(x)$ holds iff $x$ is a square. Define $$\psi(c,d):=\phi(d)\wedge\forall d'.(d'>c\wedge\phi(d')\implies d'\ge d),$$ which asserts that $d$ is the smallest square larger than $c$. Observe that the set $\{(x,y)\in\mathbb N^2: y=x^2\}$ is p-definable by $$\lambda(x,y):=\phi(y)\wedge\exists a.\ \psi(y,a)\wedge a=y+2x+1.$$ Also, the set $\{(x,y,z)\in\mathbb N^3: z=xy\}$ is p-definable by formula
$\exists a,b,c.\ \lambda(a,x)\wedge\lambda(b,y)\wedge\phi(c)\wedge c=a+b+2z \wedge\exists d.\ \psi(c,d)\wedge d=c+2x+2y+1$
The validity of this formula exploits the fact that $c=(x+y)^2$ if $c$ is a square and the smallest square greater than $c$ is $c+2x+2y+1$ for some $x,y\in\mathbb N$. It then follows that multiplication is p-definable, which is impossible since $[\mathbb N,+,\cdot]$ is not decidable but $[\mathbb N,+,V_p]$ is.
(To be continued)

References and further reading

1. Recognizable sets of integers, Michel Rigo.
2. Languages, Automata, and Logic, Wolfgang Thomas.
3. Logic and p-recognizable sets of integers, V. Bruyère et al.