Proof Note $\mathit {num}( \langle \mathit {bin}(n), x \rangle )< 2^{| \langle \mathit {bin}(n), x \rangle |+ 1}\leqslant 2^{O(\log n+ |x|)}$ . Choose a constant $d\in \mathbb {N}$ and a computable $h: \mathbb {N} \to \mathbb {N}$ such that $\mathit {num}( \langle \mathit {bin}(n), x \rangle )< n^d$ and $\mathit {num}(x)< n$ whenever $n\geqslant h(|x|)$ . It suffices to describe an interpretation of ${\mathcal S}(1^{\mathit {num}( \langle \mathit {bin}(n), x \rangle )})$ in ${\mathcal S}( \langle 1^n, x \rangle )$ whenever $n\geqslant h(|x|)$ . It will be clear that the interpretation does not depend on n, x.

Let $(n,\mathit {num}(x))$ be the expansion of the standard $L^{\text {r}}_{\text {ar}}$ -structure n that interprets a new constant by $\mathit {num}(x)\in [n]$ . This is interpretable in ${\mathcal S}( \langle 1^n, x \rangle )$ using $\mathit {BIT}$ . By Lemma 2.9, also $(n^d, \mathit {num}(x))$ is interpretable in ${\mathcal S}( \langle 1^n, x \rangle )$ . But this structure defines (n and) $\mathit {num}( \langle \mathit {bin}(n), x \rangle )\in [n^d]$ using $\mathit {BIT}$ . Thus, $\mathcal S(1^{\mathit {num}( \langle \mathit {bin}(n), x \rangle )})$ is interpretable in ${\mathcal S}( \langle 1^n, x \rangle )$ as claimed.

Finally, note there is a sentence $\varphi $ that is true exactly in structures of the desired form ${\mathcal S}( \langle 1^n,x \rangle )$ for $n\in \mathbb {N}$ and $x\in \{0,1\}^*$ . It is easy to modify the above interpretation to produce a structure isomorphic to ${\mathcal S}(0)$ given a structure that is not of the desired form.

Proof Let $p: \mathbb {N} \to \mathbb {N}$ be computable with $p(\mathit {num}(x))= \mathit {num}(P(x))$ for all ${x\in \{0,1\}^*}$ . We choose:

• • a quantifier-free $L^{\text {r}}_{\text {ar}}$ -formula $\varphi (x, y, \bar z)$ according to Remark 2.5,

• • a computable $f: \mathbb {N}\to \mathbb {N}$ so that for all $\ell \in \mathbb {N}$

  $$\begin{align*}\mathbb{N}\models \exists \bar z {<} f(\ell)\ \varphi(\ell,p(\ell),\bar z), \end{align*}$$

• • and a computable $h{\kern-1pt}:{\kern-1pt} \mathbb {N}\to \mathbb {N}$ such that $h(|x|){\kern-1pt}>{\kern-1pt} \mathit {num}(x), \mathit {num}(P(x)), f(\mathit {num}(x))$ for all $x\in \{0,1\}^*$ .

Assume $n\geqslant h(|x|)$ . Then ${\mathcal S}( \langle 1^n,x \rangle )$ interprets the expansion $(n,\ell )$ of the standard structure n by a constant c interpreting $\ell :=\mathit {num}(x)$ . In $(n,\ell )$ the formula $\exists \bar z\varphi (c,y,\bar z)$ defines $p(\ell )=\mathit {num}(P(x))$ . Using $\mathit {BIT}$ , thus ${\mathcal S}( \langle 1^n,x \rangle )$ interprets ${\mathcal S}( \langle 1^n,P(x) \rangle )$ .

Again it is easy to modify this interpretation to produce a structure isomorphic to ${\mathcal S}(0)$ given a structure that is not of the desired form.

Proof Choose ${\mathit I} , h$ for r and ${\mathit I} ', h'$ for $r'$ according to Definition 3.1. We can assume that $h'$ is nondecreasing. Choose $n_0, c\in \mathbb {N}$ such that $|r(x)|\geqslant |x|^{1/c}$ for all $x\in \{0,1\}^*$ with $|x|\geqslant n_0$ . Define $g: \mathbb {N}\to \mathbb {N}$ by

$$\begin{align*}g(k):= \text{max}\big\{(h'(f(k)))^c, h(k), n_0\big\}. \end{align*}$$

We claim that $I'\circ I$ and g witness that $r'\circ r$ is $\kappa $ -eventually definable. To verify this let $x\in \{0,1\}^*$ satisfy $|x|\geqslant g(k)$ where $k:= \kappa (x)$ . Then $|r(x)|\geqslant |x|^{1/c}\geqslant h'(f(k))\geqslant h'(\kappa '(r(x)))$ using that $h'$ is nondecreasing. Hence $\mathcal S(r(x))^{I'}\cong {\mathcal S}(r'(r(x)))$ . As $|x|\geqslant g(k)\geqslant h(k)$ , we conclude ${\mathcal S}(x)^I\cong {\mathcal S}(r(x))$ , which implies ${\mathcal S}(x)^{I'\circ I}\cong {\mathcal S}(r'(r(x)))$ .

Proof Let $P:\{0,1\}^*\to \{0,1\}^*$ map a nondeterministic Turing machine $\mathbb M$ to another $\mathbb M'$ that simulates $\mathbb M$ and, if $\mathbb M$ accepts, then $\mathbb M'$ nondeterministically makes any number of steps before it halts and accepts; strings x not encoding machines are mapped to themselves. This is clearly a parameterized reduction. By Example 3.3, $\langle 1^n,x\rangle \mapsto \langle 1^n,P(x)\rangle $ is eventually definable.

Recall that this function outputs $0$ on strings y not of the desired form $ \langle 1^n,x \rangle $ , i.e., the interpretation produces a structure isomorphic to ${\mathcal S}(0)$ . We change the interpretation to output ${\mathcal S}(y)$ in this case. This ensures honesty (we can assume $|\mathbb M'|\geqslant |\mathbb M|$ ) and thus gives a reduction as desired.

Proof (i) follows from Lemma 3.4. (ii) follows from (i) and Remark 3.7.

Proof (ii) follows from Example 3.6 and Lemma 3.8. Alternatively, let $(\mathsf C_{n,k})_{n,k}$ witness according to Proposition 2.2(b). Then $(\bigvee _{m\leqslant n}\mathsf C^n_{m,k})_{n,k}$ witnesses where $\mathsf C^n_{m,k}$ checks its input has the form $\langle 1^n,x\rangle $ for some $x\in \{0,1\}^k$ and then runs $\mathsf C_{m,k}$ on $\langle 1^m,x\rangle $ .

To prove (i), first assume $\mathsf {NE}\subseteq \mathsf {LINH}$ and let Q be the classical problem underlying but with input n encoded in binary:

Clearly, $Q\in \mathsf {NE}$ , so by assumption and Proposition 2.1(iii) we have $\mathit {un}(Q)\in \mathsf {AC}^0$ . Recall

$$ \begin{align*} \mathit{un}(Q)= \bigg\{1^{\mathit{num}(\langle \mathit{bin}(n), \mathbb M\rangle)} \;\Big|\; \begin{array}{l} \text{the nondeterministic Turing machine } {\mathbb M} \\ \text{accepts the empty input in exactly } n \text{ steps} \end{array}\!\!\! \bigg\}. \end{align*} $$

By Example 3.2 the map $ \langle 1^n, \mathbb M \rangle \mapsto 1^{\mathit {num}( \langle \mathit {bin}(n), \mathbb M \rangle )}$ is eventually definable with respect to the parameterization of . It is an honest parameterized reduction to $(\mathit {un}(Q), \kappa )$ where $\kappa $ maps $1^{\mathit {num}( \langle \mathit {bin}(n), \mathbb M \rangle )}$ to $|\mathbb {M}|$ . Since $(\mathit {un}(Q), \kappa )\in \mathsf {para}\text {-}\mathsf {AC}^0$ , Lemma 3.8 implies .

Conversely, assume . Let $Q\subseteq \{0,1\}^*$ be a problem in $\mathsf {NE}$ . To show that $Q\in \mathsf {LINH}$ , it suffices to prove $\mathit {un}(Q)\in \mathsf {AC}^0$ again by Proposition 2.1(iii).

As $Q \in \mathsf {NE}$ there is a constant $c\in \mathbb {N}$ and a nondeterministic Turing machine $\mathbb M$ accepting Q that on input x halts in time at most $\mathit {num}(x)^c-2|x|-2$ . Consider the nondeterministic Turing machine ${\mathbb M}^*$ that started with the empty input runs as follows:

Line 1 takes exactly $2|y|+2$ many steps by moving the head forth and back on some tape, so the dummy steps in line 4 are possible. Since $\mathit {num}$ is injective, we have

(2) $$ \begin{align} x \in Q & \iff {\mathbb M}^* \text{ accepts the empty input tape in exactly } \mathit{num}(x)^c+1 \text{ many steps}. \end{align} $$

Since $\mathbb M^*$ is a fixed machine, implies that the classical problem

$$\begin{align*}Q':=\Big\{ 1^n\;\Big|\; \mathbb M^* \text{ accepts the empty input tape in exactly } n+1 \text{ many steps}\Big\} \end{align*}$$

is in $\mathsf {AC}^0$ . Choose a first-order sentence $\varphi $ for $Q'$ according to Theorem 2.6. Lemma 2.9 gives an interpretation ${\mathit I} $ such that ${\mathcal S}(1^n)^{{\mathit I}}\cong \mathcal S(1^{n^c})$ for all $n>1$ . Then $1^{n^c}\in Q'$ is equivalent to $\mathcal S(1^n)\models \varphi ^{{\mathit I}} $ . Thus the r.h.s. in (2) is equivalent to ${\mathcal S}(1^{\mathit {num}(x)})\models \varphi ^{{\mathit I}}$ provided $\mathit {num}(x)>1$ , i.e., x is non-empty. The l.h.s. in (2) is equivalent to $1^{\mathit {num}(x)}\in \mathit {un}(Q)$ . Thus $\varphi ^{{\mathit I}}$ witnesses that $\mathit {un}(Q)\in \mathsf {AC}^0$ according to Theorem 2.6.

Proof Let $(Q, \kappa )\in \mathsf {para}\text {-}\mathsf {NP}$ be almost tally. The identity is a parameterized reduction from $(Q, \kappa )$ to its re-parameterization $(Q, \kappa ')$ where $\kappa '( \langle 1^n, x \rangle ):=|x|$ for all $n\in \mathbb {N}$ , $x\in \{0,1\}^*$ . Hence, the identity is an eventually definable reduction. We can therefore assume that $\kappa = \kappa '$ .

Let $\mathbb M$ be a nondeterministic Turing machine that accepts Q and on input $ \langle 1^n, x \rangle $ runs in time at most $f(k)\cdot n^{c}$ where $c\in \mathbb {N}$ , $f: \mathbb {N}\to \mathbb {N}$ is a computable function, and $k:=|x|$ .

Define $g:\mathbb {N}^2\to \mathbb {N}$ by

$$\begin{align*}g(m,k):= m^{c+1}+ 2m+ 2k+ 2. \end{align*}$$

For $x\in \{0,1\}^*$ with $k:=|x|$ , consider the nondeterministic Turing machine $\mathbb M_x$ that on the empty input runs as follows:

Step 1 can be implemented to take exactly $2+2m+2+2k$ many steps (recall (1)), so the dummy steps in line 4 are possible if $m>f(k)$ . Note that for each k, the function $m\mapsto g(m,k)$ is injective. Thus, if $n>f(k)$ , we have

Using Example 3.3, one easily constructs an eventually definable reduction that maps $ \langle 1^n,x \rangle $ to $ \langle 1^{g(n,k)+1}, \mathbb M_x \rangle $ .

Proof The l.h.s. is equivalent to by Theorem 1.2(i). And by Lemmas 3.8 and 3.12, is equivalent to the r.h.s.

Proof The proof of (i) is analogous to the proof of Corollary 3.13 using the subproblem of where the input machine $\mathbb M$ is deterministic. Similarly the proof of (iii) is analogous to the proof of (ii). We show how (ii) is proved by modifying the proof of Corollary 3.13.

Consider the following variant of :

Here, the space $\left \lfloor \log m\right \rfloor $ of a run bounds all work tapes together, that is, if $c_i$ is the maximal cell number visited on work tape i, then $\sum _ic_i\leqslant \left \lfloor \log m\right \rfloor $ .

It is clear that this problem is in $\mathsf {para}\text {-}\mathsf {NL}$ .

Proof of Claim 1

Assume $\mathsf {NLINSPACE}\subseteq \mathsf {LINH}$ and let Q be the classical problem underlying but with the inputs $n, m$ encoded in binary. Clearly, $Q\in \mathsf {NLINSPACE} \subseteq \mathsf {LINH}$ , so $\mathit {un}(Q)\in \mathsf {AC}^0$ by Proposition 2.1(iii). Similarly as Example 3.2 one sees that

$$\begin{align*}\langle 1^n, 1^m, \mathbb M\rangle\mapsto 1^{ \mathit{num}(\langle \mathit{bin}(n), \mathit{bin}(m), \mathbb M\rangle)} \end{align*}$$

is eventually definable. Then follows as in Theorem 1.2(i).

Conversely, assume and let $Q\in \mathsf {NLINSPACE}$ . Choose a nondeterministic Turing machine $\mathbb M$ accepting Q that on input $x\in \{0,1\}^*$ runs in time at most

$$\begin{align*}\frac{\mathit{num}(x)^{c}}{10c(|x|+2)} - 10c(|x|+ 2)- |x| \end{align*}$$

and uses space at most $c\cdot |x|$ ; here $c\in \mathbb {N}$ is a suitable constant. Define $\mathbb M^*$ as in the proof of Theorem 1.2 but with the following implementation details. For the simulation in line 2, first initialize a length $c(|y|+ 2)$ binary counter using exactly $10c(|y|+ 2)$ steps, and increase it using exactly $10c(|y|+ 2)$ many steps for each simulated step of $\mathbb M$ . In line 4 continue increasing the counter in this way until it reaches $\mathit {num}(y)^c/(10c(|y|+ 2))$ . For long enough y, the binary representation of this number can be computed in time at most $\mathit {num}(y)$ and space $O(|y|)$ (where the constant in the O-notation depends on c). This computation can be done in parallel to the simulation in lines 2 and 4. Hence, line 5 completes exactly $\mathit {num}(y)^c+1$ steps, and uses space at most $d\cdot |y|$ for a suitable $d\geqslant c$ .

Thus, we arrive at the following variant of (2). For long enough $x\in \{0,1\}^*$ :

$$ \begin{align*} x \in Q & \iff {\mathbb M}^* \text{ accepts the empty input in exactly } \mathit{num}(x)^c+1 \text{ many steps} \\ &\text{and space at most } \left\lfloor\log(\mathit{num}(x)^d)\right\rfloor. \end{align*} $$

Our assumption implies that the classical problem

$$ \begin{align*} Q':= \big\{\langle 1^n,1^m\rangle \;\big|\;\ & n\leqslant m \text{ and } \mathbb M^* \text{ accepts the empty input in exactly } n+1 \text{ many} \\ & \text{steps and space at most } \left\lfloor\log m\right\rfloor\big\} \end{align*} $$

is in $\mathsf {AC}^0$ . Now $\mathit {un}(Q)\in \mathsf {AC}^0$ (and hence $Q\in \mathsf {LINH}$ ) follows as in Theorem 1.2(i) using an interpretation ${\mathit I} $ such that ${\mathcal S}(1^n)^{{\mathit I}}\cong {\mathcal S}(\big \langle 1^{n^c}, 1^{n^d}\big \rangle )$ .

Proof of Claim 2

Let $(Q, \kappa )\in \mathsf {para}\text {-}\mathsf {NL}$ be almost tally and $\mathbb M$ be a nondeterministic Turing machine that accepts Q and that on input $ \langle 1^n, x \rangle $ runs in time at most $f(k)\cdot n^{c}$ and space at most $f(k)+ c\cdot \log n$ where $c\in \mathbb {N}$ , $f:\mathbb {N}\to \mathbb {N}$ is a computable function, and $k:= \kappa ( \langle 1^n, x \rangle )$ . We can assume $k= |x|$ (see the proof of Lemma 3.12).

For $x\in \{0, 1\}^*$ with $k:= |x|$ , define the nondeterministic Turing machine $\mathbb M_x$ as in the proof of Lemma 3.12 but with a different g (chosen below) and line 1 changed to nondeterministically write some $m\in \mathbb {N}$ in binary in exactly $2\left \lceil \log (m+ 1)\right \rceil + 2$ steps. The simulation in line 2 is done as in the previous claim maintaining a length $(c+1)\left \lceil \log (m+1)\right \rceil $ binary counter. It further maintains the position of $\mathbb M$ ’s head on the input tape which we can assume to be at most $| \langle 1^m,x \rangle |+1$ and uses it to compute the currently scanned bit. Counter and position are updated for each simulated step of $\mathbb M$ . If $k< m$ , then one step of $\mathbb M$ is simulated in exactly $10c\left \lceil \log (m+1)\right \rceil $ steps. In line 4 the binary counter is updated until it reaches $m^{c+1}$ . Hence line 4 is completed after exactly $g(m,k):=m^{c+1}\cdot 10c\left \lceil \log (m+ 1)\right \rceil + 2\left \lceil \log (m+1)\right \rceil + 2$ steps. The dummy steps in line 4 are possible if $m>f(k)$ . In this case the computation takes space at most $d\log m$ for suitable $d\in \mathbb {N}$ . Thus, if $n> f(k)$ , we have

Similarly as seen in the proof of Lemma 3.12, this implies the claim.

It now suffices to show that if and only if every almost tally problem in $\mathsf {para}\text {-}\mathsf {NL}$ is in $\mathsf {para}\text {-}\mathsf {AC}^0$ . The forward direction follows from Claim 2 and Lemma 3.8. And if , then we get an almost tally problem in $\mathsf {para}\text {-}\mathsf {NL} \setminus \mathsf {para}\text {-}\mathsf {AC}^0$ by rewriting inputs $ \langle 1^n, 1^m, \mathbb M \rangle $ of to $ \langle 1^{ \langle n, m \rangle }, \mathbb M \rangle $ where $ \langle n, m \rangle $ is a pairing function on $\mathbb {N}$ .

Proof For the second assertion define $\varphi (x)$ as $\psi ^{<x}$ with atoms rewritten in the functional language $L_{\text {ar}}$ . The first assertion is folklore, see [Reference Gaifman and Dimitracopoulos22, Proposition 2.2]. We give a brief sketch for completeness. It is routine to compute, given a $\Delta _0$ -formula $\varphi (\bar x)$ , a constant $c_\varphi>1$ and an $L^{\text {r}}_{\text {ar}}$ -formula $\psi _0(\bar x)$ such that

$$ \begin{align*} \mathbb{N} \models \varphi(\bar n) & \Longleftrightarrow \mathbb{N} \models \psi_0^{<m}(\bar n) \end{align*} $$

for all $\bar n,m\in \mathbb {N}$ with $m\geqslant \text {max}\{\bar n,2\}^{c_\varphi }$ . Hence, for $n>1$ , the truth of $\varphi (n)$ is equivalent to $n^{c_\varphi } \models \psi _0(n)$ . Since the number n is definable in the standard $L^{\text {r}}_{\text {ar}}$ -structure $n^{c_\varphi }$ (as the minimal element whose $c_\varphi $ th power does not exist), we can replace $\psi _0(n)$ by some sentence $\psi _1$ . Then set $\psi :=\psi _1^{{\mathit I} _{c_\varphi }}$ for the interpretation ${\mathit I} _{c_\varphi }$ from Lemma 2.9.

Proof Choose $\varphi _f(x,y,\bar z)$ according to Remark 2.5, and set $\chi _f(x,y):=\exists \bar z\varphi _f(x,y,\bar z)$ . In particular, $\varphi _f$ is quantifier-free, so $\chi _f^{<a}=\exists \bar z{<}a\ \varphi _f$ . Let $k\in \mathbb {N}$ and $ b\in M$ .

If $b=f(k)$ , then $\mathbb {N}\models \varphi _f(k,b,\bar m)$ for some $\bar m\in \mathbb {N}^{|\bar z|}$ since $\chi _f(x,y)$ defines f. Then $\bar m<^M a$ and $M\models \varphi _f(k,b,\bar m)$ , so $M\models \chi _f^{<a}(k,b)$ .

If $M \models \chi _f^{<a}(k,b)$ , then both $M\models \chi _f(k,b)$ and $M\models \chi _f(k,f(k))$ . But $\chi _f(x,y)$ defines a function in $\mathbb {N}$ and hence in M (by elementarity of the extension), so $b=f(k)$ .

Proof of Theorem 1.4

For contradiction, assume otherwise, so by Lemma 4.1. By Corollary 2.7, there is an increasing computable function $h: \mathbb {N}\to \mathbb {N}$ and a sentence $\textit {sat}$ such that for every $n\in \mathbb {N}$ and every $L^{\text {r}}_{\text {ar}}$ -sentence $\varphi $ with $n> h(\mathit {num}(\varphi ))$ we have

(5) $$ \begin{align} n \models \varphi & \iff {\mathcal S}\big(\langle 1^n, \varphi\rangle\big) \models \textit{sat}. \end{align} $$

Proof Since is an almost tally problem, (i) follows from Theorem 1.4 and Corollary 3.13. The other assertions follow using Lemma 3.14.

Proof Let r denote the reduction and choose f such that $\kappa '\circ r\leqslant f\circ \kappa $ . Choose an interpretation I and a function h witnessing that r is eventually definable. To show (i), assume $(Q',\kappa ')\in {\mathsf {XAC}^0}$ . We show that for every $k\in \mathbb {N}$ the kth slice of Q is in  $\mathsf {AC}^0$ .

Fix $k\in \mathbb {N}$ and let $x\in \{0,1\}^*$ with $\kappa (x)=k$ . Since $\kappa '$ is $\mathsf {AC}^0$ -computable, Theorem 2.6 implies that for every $k'\in \mathbb {N}$ there is a sentence $\chi _{k'}$ that is true in ${\mathcal S}(r(x))$ if and only if $\kappa '(r(x))=k'$ . For every $k'\in \mathbb {N}$ choose a sentence $\psi _{k'}$ that defines the $k'$ th slice of $Q'$ according to Theorem 2.6. If $|x|\geqslant h(k)$ , then

$$ \begin{align*}\textstyle \varphi:=\bigvee_{k'\leqslant f(k)} (\chi_{k'}\wedge\psi_{k'})^I \end{align*} $$

is true in ${\mathcal S}(x)$ if and only if $r(x)\in Q'$ , i.e., $x\in Q$ . Translating $\varphi $ gives an $\mathsf {AC}^0$ -family that decides the kth slice of Q on instances x with $|x|\geqslant h(k)$ . This can be extended to the whole slice by hardwiring instances of length $<h(k)$ .

For (ii) we assume there is $d\in \mathbb {N}$ such that every slice of $Q'$ is in $\mathsf {AC}^0_d$ . Now, in Theorem 2.6, the quantifier alternation rank of $\varphi $ depends only on the depth of the $\mathsf {AC}^0$ -family, and vice-versa; this follows from the proof of [Reference Barrington, Immerman and Straubing6, Theorem 8.1]. In particular, all $\psi _{k'}$ and $\chi _{k'}$ have quantifier alternation rank $\leqslant d'$ for some $d'$ that depends only on d. The depth of the $\mathsf {AC}^0$ -family translating the above $\varphi $ is $\leqslant d"$ for some $d"$ depending only on $d'$ . The hardwiring of instances of small length can be done by circuits of depth $2$ . Thus, $(Q,\kappa )\in {\mathsf {XAC}^0_{d"}}$ .

Proof It suffices to show that for every $\Delta _0$ -formula $\varphi (x)$ the problem $\{1^n\mid \mathbb {N}\models \varphi (n)\}$ belongs to $\mathsf {AC}^0$ . But this problem is $\mathit {un}(Q)$ for $Q:= \{x\in \{0,1\}^*\mid \mathbb {N}\models \varphi (\mathit {num}(x))\}$ . Clearly $Q\in \mathsf {LINH}$ , so $\mathit {un}(Q)\in \mathsf {AC}^0$ follows from Proposition 2.1.

Proof (i) For fixed $k\in \mathbb {N}$ , let $\mathbb M_{k,0}, \ldots , \mathbb M_{k,\ell _k-1}$ list all nondeterministic Turing machines of size k and let $n_{k,i}$ be the minimal n such that $\mathbb M_{k,i}$ accepts the empty input in n steps; if there is no such n, let $n_{k,i}:=\infty $ . Then, on instances $(1^n,\mathbb M)$ with parameter $|\mathbb M|=k$ , is decided by the following family of simple Boolean functions:

$$ \begin{align*} F_{n,k}(& x_0\ldots x_{n-1}, y_0\ldots y_{k-1}) = \bigvee_{\substack{i<\ell_k \text{ such} \\ \text{that } n_{k,i}\leqslant n}}\big(x_0\ldots x_{n-1}= 1^n \wedge y_0\ldots y_{k-1}= \mathbb M_{k,i}\big). \end{align*} $$

Observe that $F_{n,k}$ can be understood as a circuit of depth $2$ and size $O(k\cdot \ell _k\cdot n)$ .

(ii) By Remark 3.10, $\mathsf {NE}\subseteq \mathsf {LINH}$ is equivalent to both and . By (14) it is equivalent to .

To see (iii), assume $\mathsf {LINH}$ collapses. Paris and Dimitracopolous [Reference Paris and Dimitracopoulos34, proof of Proposition 4] showed that this implies the following. There is an $L^{\text {r}}_{\text {ar}}$ -formula $\lambda (x,y)$ such that for every $\Delta _0$ -formula $\varphi (x)$ there are $c_\varphi ,d_\varphi ,e_\varphi \in \mathbb {N}$ such that for all $n\geqslant c_\varphi $

$$ \begin{align*} \mathbb{N}\models \varphi(n) & \iff n^{d_\varphi}\models\lambda(n,e_\varphi). \end{align*} $$

For each fixed $\varphi $ there is an $\mathsf {AC}^0$ -family that given $1^n$ decides whether n satisfies the r.h.s. The size of this family is bounded by $n^{f_\varphi }$ for some $f_\varphi \in \mathbb {N}$ depending on $\varphi $ , but the depth of this family is determined by the quantifier alternation rank of $\lambda $ and, in particular, does not depend on $\varphi $ . This implies for some $d\in \mathbb {N}$ .

Conversely, assume and let $Q\in \mathsf {LINH}$ . It is well known (see, e.g., [Reference Hájek and Pudlák25, Chapter V, Lemma 2.13]) that there is a $\Delta _0$ -formula that is satisfied by $\mathit {num}(x)$ if and only if $x\in Q$ . Fixing this formula in the input to , the assumption implies that there is a dlogtime uniform circuit family $(C_{n})_n$ of polynomial size and depth d such that for all $x\in \{0,1\}^*$ :

$$ \begin{align*} x\in Q & \iff C_{\mathit{num}(x)}(1^{\mathit{num}(x)})=1. \end{align*} $$

It suffices to show that, given x, the r.h.s. can be checked by an alternating machine in linear time with d alternations. This is straightforward by guessing a path through $C_{\mathit {num}(x)}$ . For example, if the output gate is a $\vee $ -gate, the machine existentially guesses an input gate $g_1$ to it, and if it is a $\wedge $ -gate it universally guesses $g_1$ . Depending on the type of $g_1$ it either existentially or universally guesses an input gate $g_2$ to $g_1$ , and so on. When reaching (with $g_{d-1}$ or earlier) an input gate or a negation thereof, the machine accepts or rejects, respectively. Each guess requires $O(|x|)$ bits. Checking that, e.g., $g_2$ is an input to $g_1$ can be done in time logarithmic in the size of $C_{\mathit {num}(x)}$ , that is, in time $O(|x|)$ . We omit further details.

Proof By Theorem 7.4(i), . If is reducible to , then by Lemma 7.2(ii). This implies $\mathsf {NE}\subseteq \mathsf {LINH}$ by Theorem 7.4(ii).

Proof (iii) follows from (i) and (ii). For (i), assume is reducible to . Then and $\mathsf {NE}\not \subseteq \mathsf {LINH}$ follows by Theorem 1.3.

For (ii), assume is reducible to . Then by Proposition 7.3 and Lemma 7.2(i). This implies $\mathsf {NE}\subseteq \mathsf {LINH}$ by Remark 3.10.

Proof It is straightforward to compute from a nondeterministic Turing machine $\mathbb M$ a first-order sentence $\varphi _{\mathbb M}$ that has a model of size n if and only if $\mathbb M$ accepts the empty input in exactly n steps.

Concerning , by Lemma 4.1, it suffices to show that is reducible to : map an instance $(1^n,\varphi )$ of to $(1^n,\varphi \wedge \psi )$ where $\psi $ is an $L^{\text {r}}_{\text {ar}}$ -sentence whose finite models are exactly those isomorphic to some standard finite $L^{\text {r}}_{\text {ar}}$ -structure.