Fact 2.8 Let $\mathcal {A}$ be any NFA. There exists a deterministic finite automaton $\mathcal {A}'$ such that $\mathcal {A}$ and $\mathcal {A}'$ accept precisely the same words.

Proof Let $\mathcal {A} = (Q, I, \Sigma , \delta , F)$ . We define the DFA $\mathcal {A}'$ on $\Sigma $ as follows. First, add a new state $q_u$ to Q to obtain the set of states in $\mathcal {A}'$ ; $q_u$ will be the initial state. The final states of $\mathcal {A}'$ will be those in F; if (and only if) $u \in L$ , we also add $q_u$ as a final state. Finally, the transition function of $\mathcal {A}'$ agrees with that of $\mathcal {A}$ on its domain, and for each symbol $\sigma \in \Sigma $ , we add a transition from $q_u$ to $\delta (I, u\sigma )$ on $\sigma $ .

We claim that $\mathcal {A}'$ recognizes $\{v : uv \in L\}$ . Let $uv \in L$ . If v is empty, then $u \in L$ and hence the empty word is recognized by $\mathcal {A}'$ . Otherwise, let $v = \sigma w$ . Then $\delta (I, u\sigma w) \in F$ ; so $\delta (\delta (I, u\sigma ), w) \in F$ ; so $\delta (q_u, \sigma w) \in F$ , as needed. The proof when $uv \notin L$ is analogous.

Proof Let $\mathcal {A}_1 = (Q_1, I_1, \Sigma , \delta _1, F_1)$ and $\mathcal {A}_2 = (Q_2, I_2, \Sigma , \delta _2, F_2)$ . We will construct a new automaton $\mathcal {A}_3 = (Q_3, I_3, \Sigma , \delta _3, F_3)$ . Let $Q_3 = Q_1 \times Q_2$ , $I_3 = I_1 \times I_2$ , and let $\delta _3((q_1, q_2), \sigma ) = \delta _1(q_1, \sigma ) \times \delta _2(q_2, \sigma )$ . Note that this definition naturally extends to words; $\delta _3((q_1, q_2), w) = \delta _1(q_1, w) \times \delta _2(q_2, w)$ .

Let $F_3 = \{(q_1, q_2) \in Q_3 : q_1 \in F_1 \leftrightarrow q_2 \in F_2\}$ . Then we note that $\mathcal {A}_3$ recognizes w iff the single element of $\delta _3((q_1, q_2), w)$ is in $F_3$ . This is true if and only if the equivalence $\delta _1(q_1, w) \in F_1 \leftrightarrow \delta _2(q_2, w) \in F_2$ holds; but this is precisely when $w \in L_1 \leftrightarrow w \in L_2$ .

Fact 2.16 Let $V_k$ send every positive integer x to the largest power of k dividing x. A set $X \subseteq \mathbb {N}$ is k-automatic if and only if it is definable in the structure $(\mathbb {N}, +, V_k)$ .

Fact 2.17 [Reference Bès7, Theorem 3.1]

Let $X \subseteq \mathbb {N}$ be k-automatic but not definable in $(\mathbb {N}, +)$ (i.e., not eventually periodic). Then the structure $(\mathbb {N}, +, X)$ defines the set $k^{\mathbb {N}} = \{k^n : n \in \mathbb {N}\}$ .

Fact 2.19 [Reference Semenov18]

A one-dimensional set $X \subseteq \mathbb {N}$ is definable in $(\mathbb {N}, +, k^{\mathbb {N}})$ if and only if $X = [L]_k$ , where L is a finite union of k-Semenov-type languages. In particular, if U and V are, respectively, a sparse regular sublanguage of $\{0,\ldots ,k-1\}^*$ and a sublanguage of $\{0,\ldots ,k-1\}^*$ consisting of base-k expansions (without leading zeros) of a set definable in $(\mathbb {N},+,k^{\mathbb {N}})$ , then $[UV]_k$ is definable in $(\mathbb {N}, +, k^{\mathbb {N}})$ .

Proof Fix n. By the above remarks, this is equivalent to the claim that for all $R> M$ , $F_{M}(n) = F_{R}(n)$ .

We proceed by induction on R. Assume that $F_{M}(n) = F_{M+1}(n) = \dots = F_{R-1}(n)$ . We need to show that $F_{R-1}(n) = F_{R}(n)$ . Let $F_{R-1}(n) = k^i$ ; it suffices to show that for $r = R$ and for all words v with $|v| = j \leq r + i$ , the appropriate condition (either (3) or (4)) is satisfied.

We first consider the case where $|v| \leq r$ . In this case, we need to show that $k^r n + [a^{r-|v|}v]_k \in X \iff v \in L$ . Let $[u]_k = n$ ; then equivalently, we want $ua^{r-|v|}v \in L \iff v \in L$ .

Let $L' = \{a^{r'-|v'|}v' : ua^{r'-|v'|}v' \in L\}$ . By Lemma 2.9, the automaton for $L'$ has at most M states.

Consider the string $a^{r-|v|}v$ . Note that this string has length $r = R> M$ , so by the pumping lemma, we may write $a^{r-|v|}v = xyz,$ where y is nonempty and each $xy^mz \in L' \iff xyz \in L'$ . Then $xz = a^{r'-|v'|}v'$ for some $r' < r$ . We know that $ua^{r'-|v'|}v' \in L \iff v \in L$ by inductive assumption, so $ua^{r-|v|}v \in L \iff xyz \in L' \iff xz \in L' \iff ua^{r'-|v'|}v' \in L \iff v \in L$ .

Now we consider the case where $|v|> r$ . In this case, we need to show that $k^r n - [a^{j-r} 0^r]_k + [v]_k \in X \iff v \in L$ . Let $v = v_0 v_1$ , where $|v_1| = r$ ; then the condition $k^r n - [a^{j-r} 0^r]_k + [v]_k \in X$ may be rewritten as $k^r (n - [a^{j-r}]_k + [v_0]_k) + [v_1]_k$ . Let $w_{j,v_0}$ be a word such that $[w_{j,v_0}]_k = n - [a^{j-r}]_k + [v_0]_k$ ; then we want to show that $w_{j,v_0} v_1 \in L \iff v_0 v_1 \in L$ .

In this case, we set $L^{\prime }_{j,v_0} = \{v_1 : w_{j,v_0}v_1 \in L \leftrightarrow v_0 v_1 \in L\}$ . By Lemmas 2.9 and 2.10, the automaton for $L^{\prime }_{j,v_0}$ has at most M states. Because $|v_1| = r = R> m$ , we apply the pumping lemma and let $v_1 = xyz$ with y nonempty and $xy^mz \in L^{\prime }_{j,v_0} \iff xyz \in L^{\prime }_{j,v_0}$ for all m, in particular, $xz \in L^{\prime }_{j,v_0} \iff xyz \in L^{\prime }_{j,v_0}$ .

Let $v_1' = xz$ , and let $|v_1'| = r' < r$ . Let $j' = j + r' - r$ . Then:

$$ \begin{align*} [w_{j,v_0}v_1']_k &= k^{r'}(n - [a^{j-r}]_k + [v_0]_k) + [v_1']_k \\ &= k^{r'}n - k^{r'}[a^{j-r}]_k + k^{r'}[v_0]_k + [v_1']_k \\ &= k^{r'}n - k^{r'}[a^{j-r}]_k + [v_0v_1']_k \\ &= k^{r'}n - k^{r'}[a^{j'-r'}]_k + [v_0v_1']_k \\ &= k^{r'}n - [a^{j'-r'}0^{r'}]_k + [v_0v_1']_k. \\ \end{align*} $$

By inductive assumption, $k^{r'}n - [a^{j'-r'}0^{r'}]_k + [v_0v_1']_k \in X \iff v_0 v_1' \in L$ , that is, $w_{j,v_0}v_1' \in L \iff v_0 v_1' \in L$ . Therefore, $xz = v_1' \in L^{\prime }_{j,v_0}$ . It follows that $xyz = v_1 \in L^{\prime }_{j,v_0}$ , so $w_{j,v_0} v_1 \in L \iff v_0 v_1 \in L$ , as required.

Proof We will construct, step-by-step, several definable functions and relations that ultimately give us $F_R$ for a fixed R when put together. As $F = F_M$ by the previous lemma, we conclude that F itself is definable. Throughout this proof, we work in the expanded language $\{+,-,\equiv _n, (x \mapsto nx)_{n \in \mathbb {N}},<,X, k^{\mathbb {N}}\}$ . We do this without loss of generality because it is elementary to show that the relations $<,>, \leq , \geq , \neq $ , modular equivalence $\equiv _n$ modulo a constant n, multiplication $nx$ by a constant n, the graph of the partial function $-$ , and all natural number constants themselves are definable in Presburger arithmetic, and because Fact 2.17 guarantees that we can define $k^{\mathbb {N}}$ in $(\mathbb {N},+,X)$ .

• • Let $\ell (x)$ be the smallest $k^i \in k^{\mathbb {N}}$ such that $x < k^i$ . The graph of this function is defined by the formula $y \in k^{\mathbb {N}} \wedge \forall p_i \in k^{\mathbb {N}} \: (p_i> x \leftrightarrow p_i \geq y)$ . Note that if v starts with no leading zero, $\ell ([v]_k)$ is encoded by the word $10^{|v|}$ .

• • Let $L(x, y)$ be the binary relation that holds precisely when there is a word v with length j such that $[v]_k = x$ , $k^j = y$ , and $v \in L$ . The definition of this relation depends on whether we are in case (I) or (II). If case (I) holds, then for all v, we have $v \in L \iff 0v \in L$ . Therefore, in this case, $L(x, y)$ is defined by the formula $y \geq \ell (x) \wedge x \in X$ . If case (II) holds, then no word beginning with $0$ is in L. Therefore, in this case, $L(x, y)$ is defined by the formula $y = \ell (x) \wedge x \in X$ .

• • Let $a(x, y)$ be the partial function where, if $x = k^i$ and $y = k^j$ , then $a(x, y) = [a^{i-j}0^j]_k$ . For example, $a(7, 4) = [aaa0000]_k$ . Notice that $a(i, j) = [a]_k (k^j + \dots + k^{i-1}) = [a]_k \frac {k^i-k^j}{k-1}$ . Because $[a]_k$ and $(k-1)$ are constants, we therefore define the graph of a by the formula $x \in k^{\mathbb {N}} \wedge y \in k^{\mathbb {N}} \wedge (k-1) z = [a]_k (x - y)$ .

• • Fix an arbitrary constant r. Let the relation $E_{(3),r}(x, y, n)$ hold if there is a word v with length j such that $[v]_k = x$ , $k^j = y$ , and equivalence (3) holds. Note that $k^r$ is a constant here. Then $E_{(3),r}$ is defined by the formula $y \geq \ell (x) \wedge (k^r n + x - a(y, k^r) \in X \leftrightarrow L(x, y))$ .

• • Fix an arbitrary constant r. Let the relation $E_{(4),r}(x, y, n)$ hold if there is a word v with length j such that $[v]_k = x$ , $k^j = y$ , and equivalence (4) holds. Note that $k^r$ is a constant here. Then $E_{(4),r}$ is defined by the formula $y \geq \ell (x) \wedge (k^r n + a(k^r, y) + x \in X \leftrightarrow L(x, y))$ .

• • Fix an arbitrary constant r. Let the relation $E_r(x, y, n)$ hold if there is a word v with length j such that $[v]_k = x$ , $k^j = y$ , and equivalence (3) or (4) holds depending on whether $j> r$ . Then $E_r$ is defined by the formula $(y> k^r \wedge E_{(3),r}(x, y, n)) \vee (y \leq k^r \wedge E_{(4),r}(x, y, n))$ .

• • Let the relation $A_r(n, z)$ hold if $z = k^i$ , $k^{i-1} \leq n$ , and for all v with $|v| = j \leq r + i$ , we have $E_r(x, y, n)$ when $[v]_k = x$ and $k^j = y$ . Then $A_r$ is defined by the formula $z \in k^{\mathbb {N}} \wedge \ell (n) \geq z \wedge \forall x \: \forall y \: ((y \in k^{\mathbb {N}} \wedge \ell (x) \leq y \wedge y \leq k^r z) \to E_r(x, y, n))$ .

• • Let the relation $A_{\leq R}(n, z)$ hold if $A_{r}(n, z)$ holds for all $r \leq R$ . Then for a fixed R, $A_{\leq R}$ is defined by the formula $A_0(n, z) \wedge A_1(n, z) \wedge \dots \wedge A_R(n, z)$ .

• • The function $F_{R}(n)$ returns the largest $z = k^i \in k^{\mathbb {N}}$ such that $k^{i-1} < n$ (equivalently $k^i < kn$ ) and such that $A_{\leq R}(n, k^i)$ holds. Then the graph of $F_{R}$ is defined by the formula $z \in k^{\mathbb {N}} \wedge z < kn \wedge A_{\leq R}(n, z) \wedge \forall z' \: ((z' \in k^{\mathbb {N}} \wedge z' < kn \wedge A_{\leq R}(n, z')) \to z' \leq z)$ .

Proof Since there are only finitely many distinct vectors of sets $(\delta (q,a^r))_{q\in Q}$ for $r\in \mathbb {N}$ , by the pigeonhole principle, there exist natural numbers $r_1,r_2$ , with $r_1<r_2$ , such that $\delta (q,a^{r_1}) = \delta (q,a^{r_2})$ for all $q\in Q$ . We claim that we can take $\beta =r_2$ .

If $n=0,$ there is nothing to prove, so we may assume that $n\in X$ is positive. In addition, if the base-k expansion of n is of the form $a^m$ for some $m\ge 1$ , then it follows from the definition that $F(n)\ge V_{k,a}(n)/k=k^{m-1}$ . Thus, we may assume that n has at least one base-k digit not equal to a.

Let $wja^m$ denote a base-k of n that is in L, where $w\in \Sigma ^*$ is a possibly empty word and $j\in \Sigma _k\setminus \{ a\}$ . We claim that $F(n) \ge k^{m-r_2}$ . If $m\le r_2$ , the claim is clear, since $F(n)\ge 1$ for all $n\in X$ , so we assume that $m>r_2$ .

Since $\delta (p,wja^m)=\{p\}$ , by definition of $r_1$ and $r_2$ , there exists $m_0< r_2$ such that $\delta (p,wja^{m_0})=\{p\}$ . Moreover, since $\delta (p,a)=\{p\}$ , we in fact have $\delta (p,wja^{e})=\{p\}$ for all $e\ge m_0$ .

To show that $F(n)\ge V_{k,a}(n)/k^{r_2}= k^{m-r_2}$ , let $r\ge 0$ and let $v\in \Sigma _k^*$ be a word of length at most $m-r_2+r$

We must show that the conditions in Equations (3) and (4) hold.

First, if $|v|>r,$ then $k^r n - [a^{|v|-r}0^r]_k + [v]_k$ has base-k expansion of the form $wja^{e}v$ for some $e\ge m_0$ and since $wja^{e}$ is in L, we see that $wja^e v \in L$ if and only if $v\in L$ . Thus, since we are in either case (I) or (II), we see that

$$ \begin{align*}k^r n - [a^{|v|-r}0^r]_k + [v]_k\in X\iff v\in L\end{align*} $$

in this case.

Next, if $|v|\le r$ , then $k^r n + [a^{r-|v|} v]_k$ has base-k expansion $wja^e v$ with $e\ge m_0$ and so as in the previous case, we have that $k^r n + [a^{r-|v|} v]_k\in X\iff v\in L$ .

Thus, $F(n)\ge k^{m-r_2} = V_{k,a}(n)/k^{\beta }$ .

Proof Since $w\in L\implies wa\in L$ , it suffices to prove the result when $i=1$ and the general result will then follow by induction.

Let t be such that $k^t=F(n)$ , where $n=[w]_k$ with $w\in L$ . Then Equations (3) and (4) give that for all $r\ge 0$ and all words v of length at most $r+t$ with $|v|=j\le r+t$ , we have

$$ \begin{align*}k^r n - [a^{j-r}\cdot 0^r]_k + [v]_k \in X \iff v\in L\end{align*} $$

if $j>r$ ; and we have

$$ \begin{align*}k^r n + [a^{r-j} v]_k \in X \iff v\in L\end{align*} $$

if $j\le r$ .

Now let $m=kn+a$ , so $m=[wa]_k$ . Then for $r\ge 0$ and a word v of length at most $r+t+1$ , if $|v|>r,$ then

$$ \begin{align*}k^r m - [a^{|v|-r} 0^r]+[v]_k &= k^{r+1} n + ak^r - [a^{|v|-r} 0^r]_k+[v]_k\\& = k^{r+1}n - [a^{|v| - (r+1)}0^{r+1}]_k +[v]_k,\end{align*} $$

and so we see that this is in X if and only if $v\in L$ .

Similarly, if $|v|\le r,$ then

$$ \begin{align*}k^r m + [a^{r-j} v]_k = k^{r+1}n + [a^{r+1-j}v]_k,\end{align*} $$

and so this is X if and only if $v\in L.$ It follows that $F(kn+a) = F([wa]_k) \ge kF([w]_k)$ , and the general result now follows.

Proof We claim we can take $\alpha :=M+2$ , where M is the number of states in our minimal automaton $\mathcal {A}$ . By definition of $F(n),$ we have $n\ge k^{t-1}$ and so every base-k expansion of n has at least t digits. We pick a base-k expansion $wu$ of n in $L,$ where $w,u\in \Sigma _k^*$ and $|u|=t$ . When analyzing Equations (3) and (4), the main difficulty that arises is that carries can occur, either when performing additions or subtractions.

We first note that we may assume that the prefix w is non-empty: if not, $n=[u]_k <k^{t+1}$ and so if we take $n'=n,$ we have $n'< k^{\alpha } F(n)= k^{\alpha } F(n')$ , so the result is clear in this case.

Thus, we assume that w is non-empty and, in this case, it can be written uniquely in the form $w=w_0 c^{\ell }$ for some $\ell \ge 0$ , where $c\in \{0,k-1\}$ and $w_0$ is either empty or a word whose last letter is in $\{1,2,\ldots ,k-2\}$ .

Thus, we now write $n=[w_0 c^{\ell }u]_k$ . We consider the cases when $w_0$ is empty and when it is not empty separately. To ease arguments, we write $w_1\sim _L w_2$ to mean that either $w_1$ and $w_2$ are both in L or neither is in L.

Case A: $w_0$ is not the empty word.

In this case, we write $w_0 = w_0' s,$ where $s\in \{1,2,\ldots ,k-2\}$ . Since our minimal automaton $\mathcal {A}$ has at most M states, there is some word $w_0"$ of length $<M$ such that $\delta (q_0,w_0')=\delta (q_0,w_0")$ . We now take $n' = [w_0"s c^{\min (\ell ,2)} u]_k$ . Notice that $n' < k^{t+M+2}$ , so it suffices to show that $F(n')\ge k^t$ .

It now suffices to verify that Equations (3) and (4) hold for words v of length $t+r$ for $n'$ . We only consider Equation (3), since Equation (4) is verified in a similar manner.

Since $F(n)=k^t$ , by Equation (3), we have for $r\ge 0$ and $|v|\le t+r$ , if $|v|>r,$ then

$$ \begin{align*}k^r n - [a^{j-r}0^r]_k + [v]_k = k^r[c^{\ell} u]_k - [a^{j-r}0^r]_k + [v]_k\end{align*} $$

is in X if and only if $v\in L$ . Moreover, observe that the quantity $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k$ is greater than or equal to $-k^{t+r}+1$ and is at most $2\cdot k^{r+t}-1$ .

We now look at various subcases.

Subcase (i): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k<0$ , $c=k-1$ , and $\ell \ge 2$ . In this case,

$$ \begin{align*}k^r[w_0' s c^{\ell} u]_k - [a^{j-r}0^r]_k + [v]_k \end{align*} $$

either has base-k expansion of the form $w_0' s c^{\ell -1} (c-1) v'$ or of the form $w_0' s c^{\ell } v'$ for some $v'$ which depends on u and v but not on $\ell \ge 2$ . In particular, since $k-1$ is idempotent with respect to $\mathcal {A}' $ we see that $w_0' s c^{\ell -1}(c-1) v'\sim _L w_0" s c (c-1)v'$ and $w_0' s c^{\ell -1}v'\sim _L w_0" s ccv' $ . Thus, we see that $k^r n' - [a^{j-r}0^r]_k + [v]_k\in X \iff v\in L$ in this case.

Subcase (ii): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k<0$ , $c=k-1$ , and $\ell <2$ .

This case is handled similarly to Subcase (i), but where we now use $n'=[w_0" s c^{\ell } u]_k$ .

Subcase (iii): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k<0$ , $c=0$ , and $\ell \ge 2$ . In this case,

$$ \begin{align*}k^r[w_0' s c^{\ell} u]_k - [a^{j-r}0^r]_k + [v]_k \end{align*} $$

either has base-k expansion of the form $w_0' (s-1) (k-1)^{\ell } v'$ or of the form $w_0' s c^{\ell } v'$ for some $v'$ which depends on u and v but not on $\ell \ge 2$ . Then, using the fact that $c=0$ is idempotent with respect to $\mathcal {A}'$ we see as in Subcase (i) that if $n'=[w_0" s 0^{\ell }]_k$ , then $k^r n' - [a^{j-r}0^r]_k + [v]_k\in X \iff v\in L$ .

Subcase (iv): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k<0$ , $c=0$ , and $\ell <2$ .

This case is handled similarly to Subcase (ii), but where we now use $n'=[w_0" s c^{\ell } u]_k$ .

Subcase (v): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k\ge 0$ , $c=k-1$ , and $\ell \ge 2$ .

In this case,

$$ \begin{align*}k^r[w_0' s c^{\ell} u]_k - [a^{j-r}0^r]_k + [v]_k \end{align*} $$

either has base-k expansion of the form $w_0' (s+1) 0^{\ell } v'$ or of the form $w_0' s (k-1)^{\ell } v'$ for some $v'$ which depends on u and v but not on $\ell \ge 2$ . In this case, we argue as in Subcase (i) with $n'=[w_0" s c^{2}u]_k$ .

Subcase (vi): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k\ge 0$ , $c=k-1$ , and $\ell < 2$ .

Here, we argue similarly in Subcase (iv) with $n' = [w_0" s c^{\ell }u]_k$ .

Subcase (vii): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k\ge 0$ , $c=0$ , and $\ell \ge 2$ .

In this case,

$$ \begin{align*}k^r[w_0' s c^{\ell} u]_k - [a^{j-r}0^r]_k + [v]_k \end{align*} $$

either has base-k expansion of the form $w_0' s 0^{\ell -1} 1 v'$ or of the form $w_0' s 0^{\ell } v'$ for some $v'$ which depends on u and v but not on $\ell \ge 2$ . Then we argue as in the earlier subcases with $n'=[w_0" s 0^2 u]_k$ .

Subcase (viii): $k^r [u]_k - [a^{j-r}0^r]_k + [v]_k\ge 0$ , $c=0$ , and $\ell <2$ .

This is done as in the previous subcase with $n'=[w_0" s 0^{\ell } u]_k$ .

Case B: $w_0$ is the empty word. This case is done similarly to Case A, where we take $n' = [c^{\min (\ell , 2)} u]_k$ . Thus, we have found $n'\in X$ with $F(n')\ge k^t$ and $n'\le k^{M+2}k^t.$

Proof Let $h(t)\in \mathbb {Z}[[t]]$ be the characteristic function of T. Then h is k-Mahler and so it satisfies a nontrivial k-Mahler equation as given in Equation (6).

The condition that $j+p \in T\iff j\in T$ for $j+p\in [Cp,Np)$ says that $h(t) (1-t^p) = q(t) + O(t^{(N-1)\ell })$ for some polynomial q of degree $<(C+1)p$ . Then

$$ \begin{align*}\sum_{i=0}^r P_i(t) q(t^{k^i})/(1-t^{p k^i}) = O(t^{(N-1)\ell}).\end{align*} $$

Multiplying through by $1-t^{p k^r}$ , we then see

$$ \begin{align*}\sum_{i=0}^r P_i(t) q(t^{k^i})(1-t^{p k^r})/(1-t^{p k^i}) = O(t^{(N-1)p}).\end{align*} $$

Now let M denote the maximum of the degrees of $P_0,\ldots ,P_r$ . Then the left-hand side has degree at most $M+k^r p(C+1)$ . Thus, if $M+k^{r} p(C+1) < (N-1)p$ , then the left-hand side must be identically zero. In particular, taking $N\ge M+k^r(C+2)+1$ gives that the left-hand side vanishes. But now we claim that $h(t)$ must be equal to $q(t)/(1-t^{\ell })$ . To see this, we note that, if this is not the case, then we may write $h(t) = q(t)/(1-t^{\ell }) + t^e h_0(t)$ with $h_0(0)\neq 0$ and

$$ \begin{align*}e\ge (N-1)\ell \ge (M+k^L+1)\ell.\end{align*} $$

Then since

$$ \begin{align*}\sum_{i=0}^r P_i(t) h(t^{k^i})=0\end{align*} $$

and

$$ \begin{align*}\sum_{i=0}^r P_i(t) q(t^{k^i})/(1-t^{\ell k^i})=0,\end{align*} $$

we must have

$$ \begin{align*}\sum_{i=0}^r P_i(t) t^{ek^i} h_0(t^{k^i})=0.\end{align*} $$

But notice that all terms in the left-hand side other than $P_0(t) t^{e} h_0(t)$ vanish to order at least $ek$ at $t=0$ , while $P_0(t) t^{e} h_0(t)$ vanishes to order at most $t^e + \mathrm {deg}(P_0)\le e+M$ at $t=0$ . Since $e+M < ke$ , we get that $h(t)=q(t)/(1-t^p)$ , and so $h(t)$ is a rational function whose coefficients lie in $\{0,1\}$ and so the sequence of coefficients of $h(t)$ is eventually periodic. Thus, T is eventually periodic.

Proof Suppose that v and $v'$ are words with no leading zeros such that $[v']_k = [v]_k+P$ and $|v|,|v'| \le \log _k(\min (F(n_i)k^c, F(n_j)))$ . We now consider cases (I) and (II) separately.

When we are in Case (II): In this case, let $v,v'$ be words with no leading zeros such that $[v']_k = [v]_k+P$ and $[v']_k=m$ . If $i=|v|=|v'|,$ then by construction,

$$ \begin{align*}v\in L \iff n_j+P- [a^{|v|}]_k + [v]_k\in X \iff n_j - [a^p]_k + [v']_k \in X \iff v'\in L\end{align*} $$

and since we are in case (II), we see that $[v]_k\in X\iff [v']_k\in X$ . On the other hand, if v and $v'$ don’t have the same length, then we similarly have $0^{|v'|-|v|}v\in L \iff v'\in L$ and since we are in case (II), $0^{|v'|-|v|}v\not \in L$ and hence $v'\not \in L$ and so $[v']_k\not \in X$ in this case. Thus, we always have that if $m\in X$ and $m>P,$ then $m-P\in X$ . Now suppose that $m-P\in X$ but $m\not \in X$ . Now pick the smallest positive integer s such that $k^s> P$ . We claim that $X\cap [k^s, k^M)$ is empty. To see this, suppose that this is not the case and pick the smallest $m\in X\cap [k^s, k^M)$ . Since $m\in X\implies m-P\in X$ for $m>P$ , we see that $m\in [k^s, k^s+P)$ . Hence, $m-P<k^s$ and so if v and $v'$ are base-k expansions of $m-P$ and $m,$ respectively, and $v'$ has no leading zeros, then if $|v|=|v'|,$ then v necessarily has a leading zero. And so from the argument above since $v\not \in L,$ we then have $v'\not \in L$ and so $m\not \in X$ , a contradiction. Thus, $X\cap [k^s, k^M)$ is empty and since $k^s\le kP$ , $x\in X\iff x+P\in X$ for all $x\in [kP, k^M-P)$ . Thus, the result follows when we are in Case II. When we are in Case (I):

Let $m<k^{M}-P$ be a positive integer and let $v,v'$ be words with $[v]_k=m$ and $[v']_k=m+P$ . Since we are in case (I), we may pad v with leading zeros and assume that $|v|=|v'|$ . Then we see that

$$ \begin{align*}v\in L \iff n_j+P- [a^{|v|}]_k + [v]_k\in X \iff n_j - [a^p]_k + [v']_k \in X \iff v'\in L\end{align*} $$

and since we are in case (I), we see that $[v]_k\in X\iff [v']_k\in X$ . Hence, we have that $m\in X$ if and only $m+P\in X$ for all $m<k^M-P$ . So the result follows in this case.

Proof Suppose, toward a contradiction, that $F(n)/V_{k,a}(n)$ is not bounded above. Then there exists a sequence $n_1, n_2, n_3, \ldots $ in X with $F(n_i)/V_{k,a}(n_i)> k^{i}$ for all i. By Lemma 3.6, we can replace each $n_i$ by a corresponding $n_i'$ and assume additionally that $n_i < k^{\alpha } F(n_i)$ for some fixed $\alpha $ that does not depend on i and depends only on X.

In particular, by Lemma 3.5, we have

(7) $$ \begin{align} \frac{F\left(k^r n_i + \frac{k^r - 1}{k - 1}a\right)}{V_{k,a}\left(k^r n_i + \frac{k^r - 1}{k - 1}a\right)} \ge k^i \end{align} $$

for all $r\ge 0$ . Moreover, since $n_i < k^{\alpha } F(n_i)\in k^{\mathbb {N}}$ , we have

$$ \begin{align*}k^r n_i + \frac{k^r - 1}{k - 1}a < k^{r+\alpha}F(n_i).\end{align*} $$

Thus, by Equation (5), we see that

$$ \begin{align*}F\left(k^r n_i + \frac{k^r - 1}{k - 1}a\right) \le k^{r+\alpha+1}F(n_i)\end{align*} $$

and so

(8) $$ \begin{align} k^{r} F(n_i)\le F(k^r n_i+(k^r-1)a/(k-1)) \le k^{r+1+\alpha}F(n_i) \end{align} $$

for all $r\ge 0$ .

In particular, if $n_i < n_j$ and $n_i$ and $n_j$ are $(k,a)$ -equivalent, then we must have

$$ \begin{align*}n_j = n_i k^r+(k^r-1)a/(k-1)\end{align*} $$

for some $r\ge 0$ and so from the above, we have

$$ \begin{align*} F(n_j)/V_{k,a}(n_j) &= F(k^r n_i+(k^r-1)a/(k-1))/(k^r \cdot V_{k,a}(n_i))\\ & \le k^{\alpha+1+r} F(n_i)/(k^r\cdot V_{k,a}(n_i)) \\ & \le k^{\alpha+1} F(n_i)/V_{k,a}(n_i). \end{align*} $$

Thus, since $n_j\to \infty $ and $F(n_j)/V_{k,a}(n_j)\to \infty $ , we may refine our sequence $(n_i)$ if necessary that we may assume that they are pairwise $(k,a)$ -inequivalent. Now, for each positive integer n, we let $t(n)$ denote the unique natural number t such that $k^{t-1}\le n < k^{t}$ . Then the sequence $(n_i/k^{t(n_i)})$ lies in the compact set $[1/k,1]$ and so by the Bolzano–Weierstrass theorem, it has a convergent subsequence. In particular, since X is not eventually periodic, by Lemma 3.8, there is a natural number N such that whenever $p>0$ is such that $x\in X \iff x+p\in X$ for all $x\in [pk, M),$ we necessarily have $M<Np$ .

Choose $\epsilon \in (0,1/2k)$ so that

(9) $$ \begin{align} 2k(N+1)\epsilon < k^{-\alpha}/2 \end{align} $$

we see that we may find $(k,a)$ -inequivalent $n_i$ and $n_j$ in this subsequence (by assumption) such that

(10) $$ \begin{align} | n_i/k^{t(n_i)} - n_j/k^{t(n_j)}| < \epsilon, \end{align} $$

and we may assume in addition that $n_i<n_j$ and that $n_i> (a+1)(1+k\epsilon )/\epsilon $ . We let

$$ \begin{align*}c=t(n_j)-t(n_i).\end{align*} $$

Then, after multiplying both sides of Equation (10) by $k^{t(n_j)}$ , we have

$$ \begin{align*}|n_i k^c - n_j| < k^{t(n_j)} \epsilon \le kn_j \epsilon.\end{align*} $$

In particular,

(11) $$ \begin{align} k^c < (1+k\epsilon) n_j/n_i \end{align} $$

and

(12) $$ \begin{align} |n_i k^c + (k^c-1)a/(k-1) - n_j| < kn_j \epsilon + ak^c < kn_j\epsilon + a(1+k\epsilon) n_j/n_i. \end{align} $$

Now we let $P:=n_i k^c + (k^c-1)a/(k-1) - n_j$ , which is bounded above by $kn_j \epsilon + ak^c$ . We consider only the case when $P>0$ , with the case when $P<0$ being handled similarly. Then since $F(n_i)\ge k^i$ , we see that $F(k^c n_i+(k^c-1)a/(k-1))\ge k^{i+c}$ by Lemma 3.5. Thus, using Equations (3) and (4) with $r=0$ , we see that for all words v with $|v|=p \le \log _k(\min (F(n_i)k^c, F(n_j))),$ we have

$$ \begin{align*}(k^c n_i + (k^c-1)a/(k-1)) - [a^{p}]_k + [v]_k \in X \iff v \in L.\end{align*} $$

Thus, by our definition of $P,$ we have

$$ \begin{align*}(n_j + P) - [a^{p}]_k + [v]_k \in X \iff v\in L\end{align*} $$

for $|v| =p \le \log _k(\min (F(n_i)k^c, F(n_j)))$ . But by definition of $n_j$ , we also have

$$ \begin{align*}n_j- [a^{p}]_k + [v]_k \in X \iff v \in L\end{align*} $$

for $|v| =p \le \log _k(\min (F(n_i)k^c, F(n_j)))$ . It follows that if $v,v'$ are words of the same length $p\le \log _k(\min (F(n_i)k^c, F(n_j)))$ with $[v']_k=[v]_k+P$ then

$$ \begin{align*}v\in L \iff n_j+P- [a^{|v|}]_k + [v]_k\in X \iff n_j - [a^p]_k + [v']_k \in X \iff v'\in L.\end{align*} $$

It now follows from Lemma 3.9 that $x\in X\iff x+P\in X$ for all $x\in [kP, \min (F(n_i)k^c, F(n_j))-P)$ . Observe that, by Equation (12), we have

$$ \begin{align*}P\cdot (N+1) = (k^c n_i + (k^c-1)a/(k-1) - n_j)\cdot (N+1) < (kn_j \epsilon + a(1+k\epsilon) n_j/n_i) \cdot N.\end{align*} $$

Recall that we refined our sequence of $n_{\ell }$ ’s earlier and so we have $n_i> (a+1)(1+k\epsilon )/\epsilon $ and thus we have $P(N+1)< 2kn_j (N+1)\epsilon $ .

Now, by our choice of $n_{\ell }$ ’s, we have $F(n_j)> k^{-\alpha } n_j$ and, by our choice of $\epsilon ,$ we then have $2k (N+1)\epsilon n_j < F(n_j)$ . We similarly have

$$ \begin{align*}k^c F(n_i)> k^{c-\alpha} n_i > k^{\alpha} (n_j - k n_j\epsilon) < k^{-\alpha} n_j/2,\end{align*} $$

since $\epsilon <1/2k$ . Thus, we again have $2k(N+1)\epsilon n_j < k^c F(n_i)$ . Hence, $P(N+1) < \min (F(n_j), k^c F(n_i))$ . It follows that $x\in X\iff x+P\in X$ for $x\in [kP, PN)]$ , but by our choice of $N,$ this cannot hold since X is not eventually periodic.

Proof Let $n \in \tilde {X}$ and let t be the unique natural number such that $k^t \le n < k^{t+1}$ . We take $m := n + \frac {k^t - 1}{k - 1}a$ and note that m lies in X by definition of $\tilde {X}$ . By definition,

$$\begin{align*}\tilde{F}(n) = F(m). \end{align*}$$

Notice that if the base-k expansion of n ends with precisely s zeros, then the base-k expansion of m ends with exactly s copies of a. Hence,

$$\begin{align*}V_{k,a}(m) = V_k(n). \end{align*}$$

Since $F(m)/V_{k,a}(m)$ is bounded above and below for $m \in X$ by Lemmas 3.4 and 3.10, we deduce that $\tilde {F}(n)/V_k(n)$ is also bounded above and below. This proves part (i).

To prove part (ii), we note that, by Equation (15), we have a surjective map h from the positive elements of X to the positive elements of $\tilde {X}$ given by

$$ \begin{align*}m\mapsto m-\frac{k^s- 1}{k - 1} a,\end{align*} $$

where s is the unique integer satisfying

$$ \begin{align*}k^s + (k^s-1)a/(k-1)\le m < k^{s+1}+(k^s-1)a/(k-1).\end{align*} $$

As noted by Equations (14) and (15), the map h is injective and since $h(m)\le m$ , we see that the number of elements in $\tilde {X}$ up to a positive integer x is at least as large as the number of elements in X up to x, and so $\tilde {X}$ is not sparse since X is not sparse.

For part (iii), suppose $n \in \tilde {X}$ , so that $m = n + \frac {k^t - 1}{k - 1}a \in X$ . Then

$$\begin{align*}kn + \frac{k^{t+1} - 1}{k - 1}a = k\left(n + \frac{k^t - 1}{k - 1}a\right) + a = km + a. \end{align*}$$

Since $m \in X$ and X is closed under the map $n \mapsto kn + a$ , it follows that $km + a \in X$ , and hence $kn \in \tilde {X}$ . This proves (iii).

Finally, for part (iv), note from the above that

$$\begin{align*}\tilde{F}(n) = F(m), \quad \tilde{F}(kn) = F(km + a). \end{align*}$$

Since $F(kn + a) \ge kF(n)$ holds for all $n \in X$ and $m \in X$ , we conclude that

$$\begin{align*}\tilde{F}(kn) = F(km + a) \ge kF(m) = k\tilde{F}(n). \end{align*}$$

This completes the proof of part (iv).

Proof By Lemma 3.11, there exists a non-sparse set $\tilde {X}$ with a definable map $\tilde {F}: \tilde {X}\to k^{\mathbb {N}}$ such that $\tilde {F}(n)/V_k(n)$ is bounded above and satisfies properties (iii) and (iv) in the statement of the lemma. In particular, $\tilde {F}(n)/V_k(n)$ assumes finitely many values for $n\in X$ , say $k^{p_1},\ldots ,k^{p_e}$ with $p_1<p_2<\cdots < p_e$ . We let

$$ \begin{align*}T_j:=\{n\in \tilde{X}\colon \tilde{F}(n)/V_{k}(n) = k^{p_j}\}.\end{align*} $$

Then since $\tilde {X}$ is not sparse, at least one set $T_i$ is not sparse and we let j denote the largest index for which $T_j$ is not sparse. Then $T_i$ is sparse for $i>j$ and hence definable in $(\mathbb {N},+,X)$ , since X is not Presburger definable. We now take

$$ \begin{align*}T_i': = \{n\in \mathbb{N}\colon \exists m, k^m n \in T_i\}.\end{align*} $$

Then $T_i'$ is again sparse for $i>j$ and therefore definable in $(\mathbb {N},+,X)$ . Thus, we can define the set

$$ \begin{align*}X_0:=\tilde{X}\setminus \bigcup_{i>j} T_i'\end{align*} $$

in $(\mathbb {N},+,X)$ , and for $n\in X_0$ , we have $\tilde {F}(n)/V_k(n)$ is bounded above by $k^{p_j}$ and, moreover, this value is achieved on a non-sparse subset of $X_0$ .

We now claim that, if $n\in X_0$ and $s\ge 0,$ then $k^s n\in X_0$ . To see this, suppose that $n\in X_0$ and $k^s n\not \in X_0$ . Then $n\in X_0\subseteq \tilde {X}$ , so $k^s n\in \tilde {X}$ by Lemma 3.11(iii) and thus $k^s n\in \tilde {X}\setminus X_0$ . It follows that $k^s n\in T_i'$ for some $i>j$ and so there is some $\ell $ such that $k^{s+\ell } n\in T_i$ . But this gives $n\in T_i'$ by definition and hence $n\not \in X_0$ , a contradiction.

Since we can define $k^{\mathbb {N}}$ in $(\mathbb {N},+,X)$ by Fact 2.17, we see we can define the set

$$ \begin{align*}Y:=X_0\cup k^{\mathbb{N}}\end{align*} $$

and the map $H:Y\to k^{\mathbb {N}}$ given by

$$ \begin{align*}H(n)=\tilde{F}(n)\cdot k^{-p_j}\end{align*} $$

for $n\in X_0\setminus k^{\mathbb {N}}$ and $H(n)=n$ for $n\in k^{\mathbb {N}}$ . This completes the construction of the set Y and the map H, which satisfy properties (i)–(iv).

Proof of Theorem 3.1 in Cases I and II

We may assume without loss of generality that X is neither sparse nor eventually periodic. Hence, by Lemma 3.12, we can define a set Y such that properties (i)–(iv) from the statement of the lemma hold. So there is a set $Y_0$ (as described in Lemma 3.12) of $n\in Y$ for which $H(n)=V_k(n)$ contains $1$ and is not sparse. Note that this $Y_0$ is not a priori definable in $(\mathbb {N},+,X)$ since we do not assume that $V_k$ is definable in this structure, though $Y_0$ is clearly k-automatic. By [Reference Bell, Hare and Shallit5], there is a natural number P such that every natural number is a sum of at most P elements from $Y_0$ .

Now we define a map $G:\mathbb {N}\to k^{\mathbb {N}}$ by the rule

(17) $$ \begin{align} G(n):= \max_{r\le P} \max_{\{(i_1,\ldots ,i_r)\in Y^r\colon i_1+\cdots +i_r = n\}} \min(H(i_1),\ldots ,H(i_r)), \end{align} $$

for $n\ge 1$ and where we take $G(0)=1$ . By construction, this map G is definable in $(\mathbb {N},+,X)$ , since both Y and the map H are.

We claim that $G(n)=V_k(n)$ , which will complete the proof of the theorem, since we will then have defined $V_k$ and all k-automatic sets are definable from $V_k$ . Notice that if $V_k(n)=k^j$ and if $i_1+\cdots +i_r=n$ with $i_1,\ldots ,i_r\in Y$ , then we must have $V_k(i_s) \le k^j$ for some s. Since $H|_Y \le (V_k)|_Y$ , we then see that $H(i_s)\le k^j$ and so $\min (H(i_1),\ldots ,H(i_r))\le k^j$ . Thus, we see that $G(n)\le V_k(n)$ . We now show that $G(n)\ge V_k(n)$ . Write $n=k^s n_0$ with $s\ge 0$ , $n_0\ge 1$ , and $k\nmid n_0$ . Then we can write $n_0 = i_1+\cdots + i_r$ for some $r\le P$ and $i_1,\ldots ,i_r\in Y_0$ . Then $i_1 k^s, \ldots , i_r k^s \in Y_0$ and so by definition of the set $Y_0$ , we have $H(i_j k^s) = V_k(i_j k^s)\ge k^s$ for $j=1,\ldots ,r$ , and so $G(n)\ge \min (V_k(i_1 k^s), \dots , V_k(i_r k^s))\ge k^s$ . Thus, we obtain the desired lower bound. This completes the proof for the cases in question.

Proof Let E denote the language accepted by $\mathcal {A}$ . Fix $u\in \Sigma _k^*$ such that $\delta (q_0,u)=q$ . Since $\mathcal {A}$ is minimal, the mapping $q \mapsto \chi _q(w)$ , defined by

$$\begin{align*}\chi_q(w) := \begin{cases} 1 & \text{if } \delta(q, w) \in W, \\ 0 & \text{otherwise}, \end{cases} \end{align*}$$

separates the states. Therefore, there is a finite set of words $w_1, \dots , w_r \in \Sigma _k^*$ such that each state $q \in Q$ is uniquely determined by the vector $(\chi _q(w_1), \dots , \chi _q(w_r))$ .

Our goal, then, is as follows: given a word v, consider what happens when $\mathcal {A}$ is run on $uvw_i$ for each i. If v is in $L_{q \to q}$ , then the run of $uvw_i$ will begin at $q_0$ , proceed to q, perform some cycle on the word v that ends back at q, and then end at $\delta (q, w_i)$ . If v is not in $L_{q \to q}$ , then $uvw_i$ will instead end at $\delta (q', w_i)$ , where $q' = \delta (q, v)$ . By our choice of $w_i$ , this means that whether $v \in L_{q \to q}$ can be determined by whether each $uvw_i$ is accepted by $\mathcal {A}$ , or equivalently, whether each $[uvw_i]_k$ is in Z.

It suffices, therefore, to show that the relation between $[v]_k$ and the tuple $([uvw_i]_k)_i$ is definable in $(\mathbb {N}, +, Z)$ , as then, we can define whether $[v]_k \in Z_q$ by deciding whether for each i we have $[uvw_i]_k \in Z$ if and only if $\delta (q, w_i) \in W$ . Note that this relation is not a function, because $[\bullet ]_k$ is not one-to-one, due to the possibility of leading zeros. We will therefore instead define, for each i, the ternary relation of triples $([v]_k, [10^{|v|}]_k, [uvw_i]_k)$ , from which the previous relation is definable. This is given by the following formula in $(x, y, z)$ :

$$ \begin{align*}y \in k^{\mathbb{N}} \wedge y> x \wedge z = ([u]_k \cdot y) + (k^{|w_i|} \cdot x) + [w_i]_k.\end{align*} $$

Note that the multiplications by $[u]_k$ and $k^{|w_i|}$ are constant multiplications and hence Presburger-definable. The only remaining matter is how we mean to use $y \in k^{\mathbb {N}}$ in the above formula. But because Z is k-automatic and not Presburger-definable, $k^{\mathbb {N}}$ is definable in $(\mathbb {N},+,Z)$ by Fact 2.17.

Proof We will once again use Fact 2.17 to work in the expanded language $\{+,<,k^{\mathbb {N}}, \ldots \}$ . To prove (a), since $q'$ and q are in the same strongly connected component, there is some word $u\in L_{q'\to q}$ and so if w is a word with no leading zeros, then $[w]_k\in [L_{q\to q'}]_k$ if and only if $0^i w\in L_{q\to q'}$ for some i. By the pumping lemma, there is some pumping length p such that we can take $i<p$ . Then

$$ \begin{align*}[w]_k\in [L_{q\to q'}]_k \iff \exists i<p,~~ 0^i w\in L_{q\to q'} \iff \exists i<p,~~ u 0^i w \in L_{q'\to q'}.\end{align*} $$

Then $n=[w]_k\in [L_{q\to q'}]_k$ if and only if

$$ \begin{align*}\exists s\in k^{\mathbb{N}}, (n< s \le (k^{p} \cdot n)) \wedge (([u]_k \cdot s) + n\in [L_{q'\to q'} ]_k)\end{align*} $$

(here $k^{p}$ and $[u]_k$ are constants). Since $[L_{q'\to q'}]_k$ is definable in $(\mathbb {N},+,k^{\mathbb {N}})$ , we then see that $[L_{q\to q'}]_k$ is definable in $(\mathbb {N},+,k^{\mathbb {N}})$ .

Similarly, let $w'$ be a word such that $\delta (q",w')=q'$ , which is guaranteed to exist because $q', q"$ are in the same strongly connected component. Let v be a word without leading zeros that may or may not be in $L_{q\to q"}$ , and let $n = [v]_k$ . Then:

$$ \begin{align*}n\in [L_{q\to q"}]_k \iff vw' \in L_{q \to q'} \iff k^{|w'|} n + [w']_k\in [L_{q\to q'}]_k.\end{align*} $$

So $L_{q\to q"}$ is also definable in $(\mathbb {N},+,k^{\mathbb {N}})$ .

To prove (b), observe that by [Reference Bell and Moosa6, Proposition 7.1(3)], if $L_{q \to q'}$ is not sparse, then there exist words $u,v,a,b$ with $a,b$ distinct and of the same length such that $L_{q\to q'}\supseteq u\{a,b\}^*v$ . But now since $q,q'$ are in the same strongly connected component, there exists $w\in L_{q'\to q}$ and $w' \in L_{q'\to q"}$ and so $L_{q'\to q'}\supseteq wu\{a,b\}^*vw'$ , which, again using [Reference Bell and Moosa6, Proposition 7.1(3)], contradicts that $L_{q'\to q'}$ is sparse.

Proof of Theorem 3.1 in Case (III)

We let $p'\in Q$ be such that $\delta (p,0)=\{p'\}$ . Since $0$ is idempotent, we have $\delta (p',0)=\{p'\}$ and, by assumption, p and $p'$ are in the same strongly connected component. Since $\delta (p',0)=\{p'\}$ , we see that the cycle language $L_{p'\to p'}$ falls into Case I, and hence is either definable in $(\mathbb {N},+,k^{\mathbb {N}})$ , or it defines $V_k$ . If expanding $(\mathbb {N},+)$ by a predicate for $[L_{p'\to p'}]_k$ generates the same definable sets as $(\mathbb {N},+,k^{\mathbb {N}})$ , then by taking the subautomaton induced by this strongly connected component containing $p'$ (but where we take our initial state to be p and our final state to be p) we see by Lemma 3.13 that $[L_{p\to p}]_k$ defines $[L_{p'\to p'}]_k$ over $(\mathbb {N},+)$ , and hence also defines $V_k$ . Thus, we may assume that $[L_{p'\to p'}]$ is definable in $(\mathbb {N},+,k^{\mathbb {N}})$ , and hence, by Lemma 3.14, so is $[L_{p\to p}]_k$ . Thus, we have proven the dichotomy in this case.