Zero terminal weights do not help: natural and nonnegative real matrix interpretations of the open Collatz rewriting systems at exponential growth

1. Result

Yolcu, Aaronson and Heule [1] encoded the shortcut Collatz map, and several restrictions of it, as string rewriting systems. Their Conjecture 4.14 is equivalent to the termination of the system for the partial map $$H(n)=3n/4\ (n\equiv0\bmod4),\qquad H(n)=(9n+1)/8\ (n\equiv7\bmod8),$$ which is the problem collatz-T-5or7mod8 of the Termination Problem Database [1, Proposition 4.15]. Nashida [2] proved that natural-number matrix interpretations in the form of Endrullis, Waldmann and Zantema [9] with all top-left entries at least $1$ make no progress on collatz-T, and that arctic interpretations make no progress on collatz-T and on $H$. He lists natural-number interpretations of $H$ and of its 12-rule form $R_H$, and rational and real interpretations, as not covered [2, §1, Q5].

For max-plus (arctic) interpretations, lyapunov_weaver [3] showed that a normal positive-drift witness in the sense of [4] excludes every finite reader, with no completion condition on the terminal weights. lyapunov_weaver's collatz_45 [5] treats natural readers read from the least significant end whose terminal weights are at least $1$ on reachable states, and it records that natural evaluation from the most significant end, whose terminal weights may vanish, is open [5, §12].

We treat sum-product readers: nonnegative real (in particular natural) digit matrices $N_0,N_1$ of any dimension $D$, any nonnegative terminal vectors (zero entries allowed), read from either end, with no monotonicity condition. Write $\Gamma_{\rm lo}$ for the following minimum over reachable recurrent supports, after the normalization of Section 2 (Section 4). For each support, take the largest growth rate, in the sense of expected $\log$, among its classes that exit through the fixed suffix of the witness; this is $-\infty$ when none of them exits. $\Gamma_{\rm lo}>0$ thus requires every reachable recurrent support to exit, and to do so at a positive rate.

Theorem 1. Let the system be the kept system of one of the four open classes of [1], or $H$ in the string form of [2, §2.3], or $R_H$. Let $V$ be a sum-product reader with $\Gamma_{\rm lo}>0$. Then $V$ does not weakly orient all rules used by the witness derivations of Section 5. In particular, a rule-removal proof whose stages use only such interpretations fails at the first stage that would remove a used rule.

At $\Gamma_{\rm lo}\le0$ weak orientation alone cannot be excluded. Remark 1 gives an example at $\Gamma_{\rm lo}=0$, and a natural interpretation of $H$ in dimension $3$ with $\Gamma_{\rm lo}=-\infty$ that weakly orients all eleven rules although its values grow exponentially. For $H$ and $R_H$ in dimension $2$ after homogenization, Section 7 treats this case using strictness.

Theorem 7 (Section 7). For $H$ and $R_H$, no natural interpretation whose binary digits have homogenized dimension $2$ weakly orients every used rule and lowers the value at every use of some used rule along the witness derivations.

Dimension $2$ covers linear $2\times2$ and one-dimensional affine interpretations of the binary digits. In dimension $3$ and above, Corollary 11 gives the same conclusion when the support graph of the binary digit matrices is strongly connected. The reducible case at $\Gamma_{\rm lo}=0$, which includes every affine interpretation of dimension at least $2$, is not covered by this paper (Section 8). For natural-number interpretations of $H$ and $R_H$ whose strict rule lowers the value at every use in the canonical derivations, which includes every monotone interpretation [2, Definition 9.1], joint work of lyapunov_weaver and the author [13] treats it in every dimension. Theorem 7 and Corollary 11 need strictness only along the witness derivations of Section 5.

What is new.

• Theorem 3, the sum-product form of Nashida's minimal-rate theorem [2, Theorem B.3, Corollary B.4]: on a strongly connected component that no digit word kills, every nonzero entry of a typical product grows at the common rate.
• Lemma 4, the sum-product form of the window-frequency upper bound [2, Lemma 6.5], by a decomposition over component itineraries, and Lemma 6, the matching lower bound.
• For $H$: the conjugacy $H(n)=(f(6n+1)-1)/6$ to the compressed map $f$ of the class avoiding $\{5,7\}\bmod8$, and the transfer of normal witnesses through it (Lemma 5).
• The observation that at $\Gamma_{\rm lo}>0$ the contradiction of [3] already excludes weak orientation.
• Theorem 7, from two inputs: along every lift of the witness every rule of $H$ is used linearly often (Lemma 8), and in dimension $2$ the source value is bounded with positive probability when $\Gamma_{\rm lo}=0$ (Lemma 9).
• Lemma 10: for natural matrices with strongly connected support that no word kills, the Lyapunov exponent is $0$ if and only if every product is a $0/1$ matrix. It gives Corollary 11 in every dimension. The assembly of Theorem 1 is that of [3, §§4–7], with these inputs in place of the max-plus ones.

Status. Nothing here is formalized. Theorem 3 was checked line by line against the written proof of [2, Theorem B.3]. The numerical checks for $H$ are in the supplement. Theorem 1 has been posted for audit [6]; no second reader has checked it. Section 7 has had two readings by the author only.

2. Setting

Fix $D$ and nonnegative real $D\times D$ matrices $N_0,N_1$. For a binary word $w=w_1\cdots w_L$ put $N_w=N_{w_1}\cdots N_{w_L}$. A reader is $V(n)=u^{\top}N_{\mathrm{bin}'(n)}c$ with nonnegative vectors $u,c$, where $\mathrm{bin}'(n)$ is the binary word of $n$ below its leading one, read from the most significant end. Reading from the least significant end is the same with $N_s^{\top}$ and the roles of $u,c$ exchanged. Affine interpretations are homogenized as in [3, §7], and fixed end markers are absorbed into $u,c$.

The support graph has an edge $i\to j$ when $(N_0)_{ij}+(N_1)_{ij}>0$. For a strongly connected component $C$ let $N^C_w$ be the restriction of $N_w$ to $C\times C$. A word $w$ kills $C$ if $N^C_w=0$. Since all entries are nonnegative, there is no cancellation: the support of $N_{xy}$ is the Boolean product of the supports of $N_x$ and $N_y$. Hence the type monoid, the minimal idempotent $E$, its classes, and the permutation group $\Pi$ of [2, Lemma 6.2 and Step 3 of Theorem B.3] are the same as for any max-plus reader with the same support.

Put $f_{\max}(w)=\log\max_{ij}(N^C_w)_{ij}$ and $f_{\min}(w)=\log\min\{(N^C_w)_{ij}:(N^C_w)_{ij}>0\}$. Then $$f_{\min}(xy)\ge f_{\min}(x)+f_{\min}(y),\qquad f_{\max}(xy)\le f_{\max}(x)+f_{\max}(y)+\log D.\tag{F1$'$}$$ The first holds because a nonzero entry of $N^C_{xy}$ is a sum of nonnegative terms, at least one of which is a nonzero product $(N^C_x)_{ik}(N^C_y)_{kj}$. The second is the bound of a sum of $D$ terms by $D$ times the largest. By Fekete's lemma applied to $a_{\max}(L)+\log D$, where $a_{\max}(L)=2^{-L}\sum_{|w|=L}f_{\max}(w)$, the limit $$\Lambda_C=\lim_{L\to\infty}a_{\max}(L)/L$$ exists. Normalization. Fix $\kappa\ge1$ such that every nonzero entry of $\kappa N_0$ and $\kappa N_1$ is at least $1$ ($\kappa=1$ for natural matrices), and replace $N_s$ by $\kappa N_s$. Then $f_{\min}\ge0$ and $f_{\max}\ge0$ on words that do not kill $C$. For a word that kills $C$ put $f_{\max}(w)=f_{\min}(w)=-\infty$. Then $a_{\max}(L)=-\infty$ for all large $L$, so $\Lambda_C=-\infty$ whenever some word kills $C$; this is the convention used in Sections 7 and 8. The second inequality of (F1$'$) holds for all $x,y$. The first is claimed only when $N^C_{xy}\ne0$, and then $N^C_x$ and $N^C_y$ are nonzero as well. Assigning the value $0$ to zero products would break it: for $N_0=\begin{pmatrix}0&2\\0&0\end{pmatrix}$ and $N_1=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ the support is strongly connected, $f_{\min}(0)=\log2$, and $N_{00}=0$. Theorem 3 concerns components that no word kills, where every product is nonzero, so there (F1$'$) holds for all words. The value of $n$ becomes $\tilde V(n)=\kappa^{|\mathrm{bin}'(n)|}V(n)$, and Section 6 accounts for this factor. All rates below are those of the scaled matrices.

3. Minimal rates for sum-product products

Theorem 3. Let $C$ be strongly connected, and let no digit word kill $C$. Then for every $\varepsilon>0$ there is $L_0$ such that for all $L\ge L_0$, at least $(1-\varepsilon)2^L$ words $w$ of length $L$ have every nonzero entry of $N^C_w$ at least $e^{(\Lambda_C-\varepsilon)L}$. The same holds for row blocks $\sum_{i\in S}(N^C_w)_{i\cdot}$ with $S\subseteq C$ nonempty and such that no digit word kills $C$ from $S$, as in [2, Corollary B.4].

Proof. We follow the proof of [2, Theorem B.3, Theorem B.1 and Corollary B.4] and record only the changes. Every step there uses (F1), Boolean support facts, or the second-moment bound (F3), which is pure probability. (F3) and the support facts are unchanged. (F1) is replaced by (F1$'$), which costs $\log D$ per concatenation on the $f_{\max}$ side only.

• Proposition B.2 and Step 1: $a_{\max}-a_{\min}$ is subadditive up to $\log D$, so the spread rate $s_C=\lim(a_{\max}(L)-a_{\min}(L))/L$ exists, and $s_C=0$ gives MinRate as before.
• Step 2 (concentration of the spread): choose the block length $K$ with $(\log D)/K<\varepsilon'/4$. The extra $\log D$ per block is then absorbed.
• Step 4 (lineage values, now $\mathrm{lin}_i(x)=\log\max\{(N^C_{xu^*})_{ct}:c\in\beta_i\}$, with $e^*=M'|u^*|$ and $M'$ as in Step 6): (L1) is Boolean. In (L2) and (L4), each place where [2] splits a finite entry into two finite parts now splits a nonzero entry into a sum of at most $D$ nonnegative products. The largest product is at least $1/D$ of the entry. So (L2) and (L4) hold with $e^*$ replaced by $e^*+\log D$. (L3) becomes the two-sided inequality $$\mathrm{lin}_i(x)+\mathrm{lin}_{\pi_x(i)}(y)-e^*-\log D\ \le\ \mathrm{lin}_i(xu^*y)\ \le\ \mathrm{lin}_i(x)+\mathrm{lin}_{\pi_x(i)}(y)+\log D.\tag{L3$'$}$$ For the upper side, $(N^C_{xu^*yu^*})_{ct}$ is a sum of at most $D$ products $(N^C_{xu^*})_{ck}(N^C_{yu^*})_{kt}$. By (L1), every nonzero one has $k\in\beta_{\pi_x(i)}$, so each is at most $e^{\mathrm{lin}_i(x)+\mathrm{lin}_{\pi_x(i)}(y)}$. The upper side needs its own $\log D$: for $D=2$, $N_0=N_1$ the all-ones matrix, $u^*=0$ and empty $x,y$, it reads $\log2\le\log2$, while the max-plus form would assert $\log2\le0$. For the lower side, keep the single product through a $t$ that attains $\mathrm{lin}_{\pi_x(i)}(y)$, with $c$ given by (L2). (L5) holds after scaling.
• Step 5 is algebra. In Step 6, each junction applies (L3$'$) twice. It therefore costs at most $\mathrm{lin}(z_k)+2\log D$ from above and $2e^*+2\log D$ from below, and (J1) holds with $J'_{\rm jn}=M'(2h+N_{\rm jn}+3|u^*|)+2\log D$ in place of $J_{\rm jn}$. (J2) and (J3) are counting. (J4) uses $f_{\min}(w_lmw_r)\ge f_{\min}(m)$, which holds by (F1$'$) and $f_{\min}\ge0$, and $f_{\max}(w_lmw_r)\le f_{\max}(m)+2M'h+2\log D$ with $M'=\log(D\max_{s,ij}(N_s)_{ij})$.
• Steps 7 and 8: the main inequality is unchanged with $M_{\max}$ replaced by $M'$ and $e^*$ by $e^*+\log D$. The largeness conditions on $L$ gain terms of order $\log D$. The choices are made in the same order ($\eta,n,\delta,n_1,L_0,L,h$), and the final contradiction $s^2(n+1)\le1024rM'^2$ is the same.
• Theorem B.1 needs $f_{\min}$ exactly superadditive and $f_{\min}\ge0$, and both hold. Corollary B.4 follows as in [2]. $\square$

4. The upper bound, and the rate $\Gamma_{\rm lo}$

We use the notation of [3, §§3–4] and [2, §6]: a fixed word $u^*$ of idempotent type, its occurrences, the occupied set $I_\theta$ of classes at an occurrence ending at time $\theta$, and the value vector $v_\theta=u^{\top}N_{\omega[0,\theta)}$. For orientation, here is what these objects are. $E$ is an idempotent of the minimal ideal of the Boolean monoid generated by the supports of $N_0,N_1$. Its classes $Q_1,\dots,Q_r$ have rows $\beta_1,\dots,\beta_r$, ordered by $i\le j$ if and only if $\beta_j\subseteq\beta_i$, and $u^*$ is a fixed word whose support type is $E$. If $\omega[\theta-|u^*|,\theta)=u^*$, the occupied set $I_\theta$ is an upper set of classes, and the support of $v_\theta$ is $\bigcup_{i\in I_\theta}\beta_i$. From one occurrence to the next, the classes are carried by order automorphisms that keep each class in its component $C(j)$ [2, Lemma 6.3]. The window hypothesis of [2, Lemma 6.5] says that the windows of a fixed length $J$ of $\omega$ on the relevant intervals have an empirical distribution within total variation $\delta$ of the uniform distribution on $\{0,1\}^J$. A class $p$ exits by $z$ if a path reading $z$ joins an index of $\beta_p$ to $\mathrm{supp}(c)$. For an occupied set $I$ put $\alpha_z(I)=\max\{\Lambda_{C(j)}:j\le p,\ j,p\in I,\ p\text{ exits by }z\}$, with $\max\varnothing=-\infty$. Let $\alpha_{\rm lo}$ be the minimum of $\alpha_z(I(X))$ over reachable recurrent supports $X$, with $z$ chosen as in [3, §4]. These are the definitions of [3, (F1)] with $\Lambda_C$ from Section 2, for the scaled matrices. Put $$\Gamma_{\rm lo}=\alpha_{\rm lo}-\log\kappa.$$ For natural matrices $\Gamma_{\rm lo}=\alpha_{\rm lo}$. Theorem 1 holds for every admissible $\kappa$ for which $\Gamma_{\rm lo}>0$.

Lemma 4. There is $C_1$, depending only on the reader, such that for every $\varepsilon>0$ there are $J,\delta>0$ and $C_u(\varepsilon)$ with the property of [2, Lemma 6.5], with sum-product values: under the window hypothesis on a word $\omega$ of length $N$, every index $b$ with $v_\theta(b)>0$ satisfies $\log v_\theta(b)\le(\Lambda_{I_\theta(b)}+C_1\varepsilon)\theta+C_u(\varepsilon)$.

Proof. The max-plus proof follows one path of maximal weight. A sum-product entry is a sum over up to $D^\theta$ paths, so that argument cannot be applied term by term. Instead, group the paths by itinerary. A path stays in a component or moves down the component order, so it visits components $C_1,\dots,C_m$ with $m\le D$ and switch times $t_1<\dots<t_{m-1}$. There are at most $2^D(\theta+1)^D$ itineraries. The sum over the paths of one itinerary is at most $(D^2M)^D$, where $M$ is the largest entry of $N_0,N_1$, times the product over $k$ of $\max(N^{C_k}_{\omega[t_{k-1},t_k)})$. Hence $$\log v_\theta(b)\le D\log(\theta+1)+O(1)+\max_{\text{nonzero itineraries}}\sum_kf^{C_k}_{\max}(\omega[t_{k-1},t_k)).$$ Each segment is cut along the intervals of the partition, which adds at most $\varepsilon^{-2}+D$ cuts in total, each costing $\log D$ by (F1$'$). On a non-exceptional piece of length at least $L_0$ the block bound of [2, Lemma 6.5] holds with $J_C$ chosen so that also $(\log D)/J_C<\varepsilon/2$. Exceptional and short pieces contribute at most $M'(\varepsilon N+\varepsilon^{-2}L_0)$. The at most $D$ pieces containing a switch time contribute at most $DM'\varepsilon N$; this is the role of the potential $\xi\psi$ in [2]. For a nonzero itinerary some path realizes it, and the rate-carrying step of [2] ("long stays carry the rate", Lemma 6.3) is a statement about supports. So $\Lambda_{C_k}\le\Lambda_{I_\theta(b)}$ for every component with a long non-exceptional piece. Collecting terms gives the claim with $C_1=2+2(D+2)M'$; the term $D\log(\theta+1)$ is absorbed into $\varepsilon\theta$ for large $\theta$. $\square$

The lower bound is [2, Lemma 6.6] with Theorem 3 in place of [2, Corollary B.4].

Lemma 6. (F3) of [3] holds for sum-product readers in this form: under its hypotheses, for at least $(1-\delta_1-\delta_2-\eta)|\mathcal X|$ elements, $u^{\top}N_{\omega(\xi)}c\ge\exp\bigl((\alpha_z(I)-\eta)|\omega(\xi)|-C_{\rm lo}\bigr)$.

Proof. The proof of [2, Lemma 6.6] combines Corollary B.4 on the uniform blocks with the stage propagation through $u^*$ and the fixed words, which only uses supports and the bounded fixed lengths. In the sum-product form each concatenation is bounded below by one nonzero product term, as in (F1$'$), and the fixed words contribute $O(1)$ to $\log$. Theorem 3 supplies Corollary B.4. If $\alpha_z(I)>-\infty$, some occupied class exits by $z$, so the value is nonzero. $\square$

In [3] the reader is assumed finite on every positive integer. A sum-product value is always finite, and Lemma 6 shows that it is nonzero on the endpoint family, so no such hypothesis is needed here.

5. The witnesses

The four classes. For the kept systems of the four open classes of [1] we use the binary-normal positive-drift witnesses of [4, Corollary 2], the simulation property (R1) of [4, §5.1], and the return times of [4, §5.4], exactly as in [3, §8, proof of Corollary 2]. These give, for every $g\ge g_A$, unbounded times $S$ with (2) of [3] and the lift identity (P1) of [3].

The map $H$. Let $f(x)=(3x+1)/4$ for $x\equiv1\bmod8$ and $f(x)=(9x+5)/8$ for $x\equiv11\bmod16$, with $f$ undefined elsewhere. This is the shortcut map of the class avoiding $\{5,7\}\bmod8$ compressed to valuation blocks $2$ and $12$. For $x\equiv1\bmod8$, $v_2(3x+1)=2$. For $x\equiv11\bmod16$, $(3x+1)/2$ is odd and $\equiv1\bmod4$, so one more odd step and two halvings give $(9x+5)/8$.

Lemma 5. (i) $H(n)$ is defined iff $f(6n+1)$ is, and then $6H(n)+1=f(6n+1)$. The same holds on $\mathbb Z_2$. (ii) If $x\in1+2\mathbb Z_2$ has binary-normal digits, so does $n=(x-1)/6$. (iii) Let $n\in\mathbb Z_2$ have $s$ forward $H$-steps, $a_s$ of type $3n/4$ and $b_s$ of type $(9n+1)/8$. Put $Q=3^{a_s+2b_s}$ and $\beta=2a_s+3b_s$. For $g\ge0$, $r=n\bmod2^{\beta+g}$ and every integer $y\ge1$, $$H^s(r+2^{\beta+g}y)=e+2^gQy,\qquad 0\le e<Q2^g+Q.$$ (iv) For every $n\ge1$, $n\ne7$, with $H(n)$ defined, the string system of $H$ [2, §2.3, Table 3] rewrites the binary string of $n$ to that of $H(n)$. One dynamic rule acts at the low end ($\mathtt{ff.}\to\mathtt{0.}$ or $\mathtt{ttt.}\to\mathtt{22.}$). Carry rules then move each new ternary digit up to the leading one, and one left-end rule ($/0\to/\mathtt t$, $/1\to/\mathtt{ff}$, $/2\to/\mathtt{ft}$) converts it.

Proof. (i) If $x=6n+1$, then $n\equiv0\bmod4$ iff $x\equiv1\bmod8$, and in that case $(3x+1)/4=6\cdot(3n/4)+1$. Also $n\equiv7\bmod8$ iff $x\equiv11\bmod16$, and then $(9x+5)/8=6\cdot(9n+1)/8+1$. Both $H(n)$ and $f(6n+1)$ depend only on $n\bmod8$, so these residues cover every case. The map $n\mapsto6n+1$ is a bijection from $\mathbb Z_2$ onto $1+2\mathbb Z_2$.

(ii) Here $(x-1)/2$ changes one digit of $x$ and then shifts, so it is normal. Division by $3$ on $\mathbb Z_2$, read from the least significant digit, is a Mealy machine. Write $x=3y=y+2y$. Then $y_k\equiv x_k-y_{k-1}-c_k\pmod 2$, the state is $(y_{k-1},c_k)$, and $c_{k+1}=\lfloor(y_k+y_{k-1}+c_k)/2\rfloor$. Four states are reachable. In each state the input digit determines the output digit bijectively, so for every state and every $J$ the map from input $J$-blocks to output $J$-blocks is a bijection. By Schnorr and Stimm [7], when a finite automaton reads a normal sequence, the joint frequency of the state at time $k$ and the input block $x_{[k,k+J)}$ is $\pi_s2^{-J}$ for a probability vector $\pi$. Summing over the states, every output $J$-block has frequency $2^{-J}$.

(iii) By induction on $t\le s$, for all $z\equiv n\bmod2^{\beta_t}$ we have $H^t(z)=(Q_tz+c_t)/2^{\beta_t}$ with $0\le c_t<Q_t2^{\beta_t}$. An $A$-step sends $(Q,c,\beta)$ to $(3Q,3c,\beta+2)$. A $B$-step sends it to $(9Q,9c+2^\beta,\beta+3)$, and $9c+2^\beta<9Q2^{\beta+3}$. The $t$-th iterate of $z=r+2^{\beta+g}y$ is congruent to $H^tn$ modulo $2^{\beta-\beta_t+g}$. While a step remains, this modulus is at least $4$ before an $A$-step and at least $8$ before a $B$-step, which is enough to fix the step type. So $z$ follows the itinerary of $n$. Put $e=(Qr+c_s)/2^\beta=H^s(r)$. Then $e<Q2^g+Q$. Every iterate is at least $(3/4)^s2^{\beta}\ge3^s$, so the orbit stays above $1$.

(iv) Read from the most significant end, a binary digit $b$ acts as $v\mapsto2v+b$ and a ternary digit $d$ as $v\mapsto3v+d$. The dynamic rules are exact: $4m\mapsto3m+0$ and $8m+7\mapsto9m+8=3(3m+2)+2$. A binary digit $b$ followed by $d$ acts as $v\mapsto6v+3b+d$. Writing $3b+d=2d'+b'$ with $d'\in\{0,1,2\}$ and $b'\in\{0,1\}$ gives the carry rule $bd\to d'b'$. At the leading one, $3+d\in\{3,4,5\}$ is $11,100,101$ in binary, which gives the left-end rules. For $n=7$ the string $/\mathtt{tt.}$ has no suffix $\mathtt{ttt.}$; the lifts used below are large. This is [2, Lemma 2.3(ii)], which is kernel-checked there; [2] takes $\mathrm{Dom}(H)=\{n\ge8\}$ and so excludes $4$ and $7$. For $R_H$, whose left marker $L$ carries no digit, the canonical string is $L\,\mathrm{bin}(n).$ and the simulating derivations are those of [2, Lemma 2.3(iii)]. Its reader is $u'^{\top}N_{\mathrm{bin}'(n)}c$ with $u'^{\top}=u^{\top}N_{\mathtt t}$, again a sum-product reader. $\square$

Witnesses for $H$. Let $x$ be a witness of [4, Corollary 2] for the class avoiding $\{5,7\}\bmod8$. Its digits are normal, and its valuation blocks $2$ and $12$ are independent with $P(12)=p$. By Lemma 5(i), $n=(x-1)/6$ has an infinite $H$-orbit, with $A$-steps at the blocks $2$ and $B$-steps at the blocks $12$, and by (ii) it is normal. Since $3$ is a unit in $\mathbb Z_2$, $H^sn\equiv n\bmod2^g$ iff $f^sx\equiv x\bmod2^{g+1}$, so the return times of [4, §5.4] carry over. The drift is $$\log_2Q-\beta=a_s(\log_2 3-2)+b_s(2\log_2 3-3),$$ which is at least $d\beta$ for some fixed $d>0$ at large times exactly when $p>(2-\log_2 3)/(\log_23-1)=0.7095$. This is the threshold of [4] for these blocks. At a return time, $e=H^sr\equiv H^sn\equiv n\pmod{2^g}$. So with $s_0=n\bmod2^g$ we can write $e=s_0+2^ge'$ with an integer $e'$, and Lemma 5(iii) gives $0\le e'<2Q$, which is (P1) of [3] with $S$ replaced by $\beta$.

6. Proof of Theorem 1

An interpretation assigns nonnegative matrices to all letters, including the ternary letters of $H$ and the markers. The reader $V$ is its restriction to binary strings between the markers. Suppose the interpretation weakly orients every rule used by the witness derivations: $[\ell]\ge[r]$ entrywise for the homogenized matrices and end vectors. Nonnegative contexts preserve entrywise inequalities, so $V$ does not increase along these derivations. By (R1) for the four classes, by Lemma 5(iv) for $H$, and by [2, Lemma 2.3(iii)] for $R_H$, for $z_0=r+2^{S+g}y$ and $z_1=C^Sz_0$ (with $C=H$ and $S$ replaced by $\beta$ for $H$ and $R_H$) we get $$V(z_0)\ge V(z_1).\tag{6.1}$$ We follow [3, §§4–7] and keep their notation.

• (F1), (F4) and (F5) concern supports, digits and counting, and are used unchanged. So are the choice of $z$ and of $X_{\rm lo}$ in [3, §4], the integer family (P2)–(P6) in [3, §5], and the event (E1) in [3, §6].
• For $H$, (P1) is the identity at the end of Section 5. The rest of [3, §5] sees the map only through (P1), the word lengths and $Q$. Each $H$-step multiplies $Q$ by at most $3^{2/3}$ per consumed bit, so $\log_2Q\le\tfrac23\beta\log_23<1.06\beta$. Hence the bounds $N_0\le4\beta$ and $N_1\le5\beta$ hold for large times.
• The upper bound (P7) becomes $\log V(z_0)\le(\alpha_{\rm lo}+C_1\varepsilon)N_0+O_\varepsilon(1)$, by Lemma 4. Only classes that exit through $z$ contribute to $V(z_0)$, which is why $\alpha_{\rm lo}$ bounds the rate.
• The lower bound (P8) becomes $\log V(z_1)\ge(\alpha_{\rm lo}-\varepsilon)N_1-O(1)$, by Lemma 6 and $\alpha_z(I_1)\ge\alpha_{\rm lo}$ [3, §6]. In particular $V(z_1)>0$.

Here $V$ stands for the scaled value $\tilde V$, so (6.1) reads $\log\tilde V(z_0)-\log\tilde V(z_1)\ge-(N_1-N_0)\log\kappa$. For a $\upsilon$ in the intersection of these events this gives $$-(N_1-N_0)\log\kappa\le-\alpha_{\rm lo}(N_1-N_0)+C_1\varepsilon N_0+\varepsilon N_1+O_\varepsilon(1),$$ that is, $\Gamma_{\rm lo}(N_1-N_0)\le C_1\varepsilon N_0+\varepsilon N_1+O_\varepsilon(1)$. By (P6), $N_1-N_0\ge dS-1$, so $\Gamma_{\rm lo}dS\le(4C_1+5)\varepsilon S+O_\varepsilon(1)$. Choose $\varepsilon<\Gamma_{\rm lo}d/(8C_1+10)$. Then the left side exceeds the right side by at least $\Gamma_{\rm lo}dS/2-O_\varepsilon(1)$, which is positive for large $S$, a contradiction. Here $C_1$ depends only on the reader and $d$ only on the witness, so $\varepsilon$ can be fixed first, as in [3].

For rule removal: at the first stage that removes a rule used by the witness, all used rules are present, and that stage's interpretation must weakly orient them all. $\square$

Strictness was not used. This differs from [3, §7], where $\alpha_{\rm lo}=0$ is allowed and the additive term $fS$ from strict uses carries the contradiction. In the sum-product setting $V$ is not on a log scale, and $\log(V+fS)-\log V\to0$ when $V$ grows exponentially. So the strict term gives nothing at $\Gamma_{\rm lo}>0$, and the argument must come from the drift alone.

7. The polynomial core in dimension 2

Here $D$ is the dimension after homogenization (Section 2). So $D=2$ covers linear natural $2\times2$ interpretations of the binary digits, and one-dimensional affine natural interpretations $x\mapsto m_sx+v_s$, whose homogenized matrices are $\begin{pmatrix}m_s&v_s\\0&1\end{pmatrix}$. Two-dimensional affine matrix interpretations in the sense of [9] homogenize to $D=3$ and are not covered.

Call a rule strict along the witnesses if every application of it in the canonical derivations from $\mathrm{can}(z_0)$ to $\mathrm{can}(z_1)$ of Section 6 lowers the value. For natural interpretations each such drop is at least $1$. This includes every notion of strictness under which a strictly oriented rule lowers the value of each string it is applied to, in particular that of [9] after homogenization: $[\ell]\ge[r]$ entrywise, a strict top-right entry, and diagonal corner entries at least $1$.

Theorem 7. Let the system be $H$ or $R_H$, and take a natural interpretation with $D=2$ for the binary digits; the ternary digits and the markers may have any natural matrices and vectors of compatible shape. If it weakly orients every rule of $H$ (respectively every used rule of $R_H$), then none of these rules is strict along the witnesses. In particular, the first stage of a rule-removal proof for $H$ or $R_H$ that removes a used rule is not a natural interpretation with $D=2$.

The proof combines a count of rule uses (Lemma 8) with a bound on the source value (Lemma 9).

Lemma 8. For almost every witness $n$ of Section 5 there are $f>0$ and $s_0$ such that the following holds for all $s\ge s_0$, $g\ge0$ and integers $y\ge1$. Put $z=r+2^{\beta+g}y$ as in Lemma 5(iii). The canonical derivation from $\mathrm{can}(z)$ to $\mathrm{can}(H^sz)$, that is, the derivations of [2, Lemma 2.3(ii)] for $z,Hz,\dots,H^{s-1}z$ in turn, uses each of the eleven rules of $H$ at least $fs$ times. The same holds for the ten used rules of $R_H$ and the derivations of [2, Lemma 2.3(iii)]. One can take $f=p(1-p)^3/2$.

Proof. By Lemma 5(iii), $z$ follows the itinerary of $n$ for $s$ steps. So $H^tz$ has the step type of $H^tn$, and if the itinerary of $n$ from time $t$ begins with a word $w$ of total valuation $\beta(w)\le\beta-\beta_t$, then $H^tz$ lies in the residue class modulo $2^{\beta(w)}$ of the points that follow $w$. The steps of the witness are independent, with $P(B)=p$, so for almost every witness each finite word $w$ occurs as such a prefix at a fraction $P(w)+o(1)$ of the times $t\le s-|w|$.

Dynamic rules. $\mathtt{ff.}\to\mathtt{0.}$ is used once per $A$-step and $\mathtt{ttt.}\to\mathtt{22.}$ once per $B$-step, so $a_s=(1-p+o(1))s$ and $b_s=(p+o(1))s$ times.

Carry rules. A sweep moves a ternary digit $d$ across the binary digits of a number $m$, from the least significant end. At each digit $b$ it applies $bd\to d'b'$ with $3b+d=2d'+b'$, and $d'$ is the carry into the next digit. Take $w=BAAA$. Its class is $455\bmod2^9$: indeed $455\equiv7\bmod8$, and $(9\cdot455+1)/8=512$, then $384$, $288$, $216$. If $H^tz\equiv455\bmod2^9$, the first sweep of step $t$ moves $d=2$ across $m=(H^tz-7)/8\equiv56=111000_2\bmod2^6$. Its six lowest digits, read upward, are $0,0,0,1,1,1$, and the carries into them are $2,1,0,0,1,2$. So this sweep applies $\mathtt{f2}\to\mathtt{1f}$, $\mathtt{f1}\to\mathtt{0t}$, $\mathtt{f0}\to\mathtt{0f}$, $\mathtt{t0}\to\mathtt{1t}$, $\mathtt{t1}\to\mathtt{2f}$ and $\mathtt{t2}\to\mathtt{2t}$, which are all six carry rules. Hence each carry rule is used at least $(p(1-p)^3+o(1))s$ times.

Left-end rules. Number the sweeps $k=0,\dots,N-1$, where $N=a_s+2b_s\ge s$, and let sweep $k$ move $d_k$ across $m_k$, with $\ell_k=|\mathrm{bin}'(m_k)|$. It ends with the left-end rule $/e_k\to/\mathrm{bin}'(3+e_k)$, where $3m_k+d_k=(3+e_k)2^{\ell_k}+(\text{lower digits})$. With $\psi_k=\log_2m_k-\ell_k\in[0,1)$ this reads $e_k=\lfloor3\cdot2^{\psi_k}+d_k2^{-\ell_k}\rfloor-3$. So $e_k=0,1,2$ according as $\psi_k$ lies in $[0,\log_2\tfrac43)$, $[\log_2\tfrac43,\log_2\tfrac53)$ or $[\log_2\tfrac53,1)$, up to an error of $O(2^{-\ell_k})$ at the endpoints. Consecutive sweeps satisfy $m_{k+1}=(3m_k+d_k-c_k)/2^{j_k}$ with $c_k\in\{0,7\}$ and $j_k\in\{0,2,3\}$: $j_k=0$ between the two sweeps of a $B$-step, and otherwise the next dynamic rule removes $2$ or $3$ low digits. Every iterate is at least $3^s$ (Lemma 5(iii)), so $m_k\ge3^s/8-1$ and $$\psi_k\equiv\psi_0+k\log_23+\epsilon_k\pmod1,\qquad|\epsilon_k|\le Cs3^{-s}.$$ Since $\log_23$ is irrational, Weyl's theorem gives that the number of $k<N$ with $\psi_0+k\log_23$ in an arc $I$ is $N|I|+o(N)$, uniformly in $\psi_0$ and $I$. The arcs have lengths $0.415$, $0.322$, $0.263$, and the perturbations move them by $O(s3^{-s})$. So each left-end rule is used at least $0.26N$ times for large $s$, whatever $y$ is.

For $R_H$, [2, Lemma 2.3(iii)] replaces each left-end step $/e\to/\mathrm{bin}'(3+e)$ by the carry step $\mathtt te\to e'b'$ with $3+e=2e'+b'$, followed by $\mathtt Le'\to\mathtt L\,\mathrm{bin}(e')$. So $\mathtt{L1}\to\mathtt{Lt}$ is used at the sweeps with $e_k=0$, and $\mathtt{L2}\to\mathtt{Ltf}$ at those with $e_k\in\{1,2\}$. The other rules are used as for $H$. Since $p(1-p)^3$ is smaller than $p$, $1-p$ and $0.26$, $f=p(1-p)^3/2$ works for every rule. $\square$

Lemma 9. Let $D=2$ and let the matrices be natural. Suppose $\alpha_{\rm lo}\le0$ and that (6.1) holds for the witness lifts. Then there are $B$ and $c_0>0$ such that $P_\upsilon(V(z_0)\le B)\ge c_0$ for all large selected times.

Proof. We use (P2)–(P6) of [3] with $S$ replaced by $\beta$. Thus $\mathrm{bin}'(z_0)=w_{\rm lo}\,\mathrm{bits}_F(\upsilon)\,v\,q$ with $q=S_1w^*u^*z$, where $w_{\rm lo}$ takes $\mathrm{supp}(u)$ to $X_{\rm lo}$, $\upsilon$ is uniform, and $v$ and $q$ are digits of $n$. The choice of $g$ in [3, §4] allows any further finite requirement; we take $g$ so large that $S_1$ contains both binary letters and the word $\rho$ of case (A) below. Let $G$ be the least $t$ such that the first $t$ digits of $\mathrm{bits}_F(\upsilon)$ contain both letters, and also $\rho$ and a killing word of $\{1,2\}$ when these exist. $G$ has a geometric tail, so for some $t_0$ the event $\mathcal A=(\mathrm{E1})\cap\{G\le t_0\}$ has probability at least $c_0=p_*/2$. We bound $V(z_0)$ on $\mathcal A$.

Visible rates. Write $\upsilon_0$ for the first $t_0$ digits of $\mathrm{bits}_F(\upsilon)$, and $\theta$ for the end of the occurrence of $u^*$ just before $z$. Let $\pi$ be a path from $\mathrm{supp}(u)$ to $\mathrm{supp}(c)$ that reads $\mathrm{bin}'(z_0)$ with nonzero weight; if there is none, $V(z_0)=0$. On (E1) the value vector at $\theta$ has support $\bigcup_{j\in I}\beta_j$ with $I=I(X_{\rm lo})$, an upper set (F1 of [3]). The path $\pi$ is at $\theta$ at an index $\iota$ of this support, and from $\iota$ it reaches $\mathrm{supp}(c)$ by reading $z$. So every $j\in I$ with $\iota\in\beta_j$ exits by $z$, and then $\Lambda_{C(j')}\le\alpha_{\rm lo}\le0$ for every $j'\in I$ with $j'\le j$. These rates are also nonnegative. A component with a killing word would have rate $-\infty$, but $C(j')$ contains a class of $E$, and for every word $x$ the product $E\bar N_xE$ permutes the core classes of $C(j')$ inside the same component [2, Lemma 6.3(ii)]. A path cannot leave a component and return to it, so every word has a nonzero restricted product in $C(j')$, and a nonzero natural matrix has an entry at least $1$. So these rates are $0$. We use this in the following form.

(V) If $\pi$ stays in a component $C'$ during all of $u^*$, or during all of $\upsilon_0$, then $\Lambda_{C'}=0$.

If $C'$ is the whole support graph, every class lies in $C'$ and this is immediate. Otherwise $C'=\{i\}$ is a one-state component. If $\pi$ stays at $i$ during $u^*$, then $(N_{u^*})_{ii}>0$, so $E_{ii}=1$ and $\{i\}$ is a class. Then $\pi$ is at $i$ at $\theta$, and $i\in\beta_j$ for some $j\in I$, so $\{i\}\ge j$ and $\{i\}\in I$ because $I$ is an upper set; and $\{i\}$ exits by $z$ because $i\in\beta_{\{i\}}$. If $\pi$ stays at $i$ during $\upsilon_0$, which contains both letters, then both letters have nonzero weight at $i$, so again $E_{ii}=1$. Since edges join the two states in one direction only, either $\pi$ stays at $i$ until $\theta$, which is the first case, or it started at $i$ and moved to the other index $i'$ before $\theta$. In the second case $i$ keeps positive value up to $\theta$, so $\{i\}\in I$, and $\iota=i'$. It suffices to show $E_{ii'}=1$, since then $i'\in\beta_{\{i\}}$ and $\{i\}$ exits by $z$. If $i'\in\beta_{\{i\}}$ this holds. Otherwise $i'\in\beta_{\{i'\}}$, so $E_{i'i'}=1$. Take a word $x$ with $(N_x)_{ii'}>0$; one exists because $\pi$ moved from $i$ to $i'$. The minimal ideal of a finite monoid is completely simple, so $X=EN_xE$ lies in the maximal subgroup at $E$, and $X^m=E$ for some $m\ge1$. As Boolean entries $X_{ii}=1$, because every letter has nonzero weight at $i$, and $X_{ii'}=1$. So $E_{ii'}=(X^m)_{ii'}\ge X_{ii}^{m-1}X_{ii'}=1$.

Since $D=2$, there are three cases.

(A) The support graph is strongly connected. If a word kills $\{1,2\}$, it occurs in $\upsilon_0$ and $V(z_0)=0$. Otherwise the rate is $0$ by (V). If some product $N_h$ has full support, then $N_{hh}\ge J^2=2J$, where $J$ is the all-ones matrix, and $JN_xJ=(\textstyle\sum N_x)J\ge J$ for every word $x$. A typical word contains linearly many disjoint copies of $hh$, and $m$ copies give an entry at least $2^m$, so the rate is positive, which is excluded. A support with three ones that is not full has a full-support product. The two supports with a zero diagonal entry square to full support. For the upper unitriangular support and any $N_s$ with a nonzero lower-left entry, which exists by strong connectivity, the triple product has full support; the lower case is symmetric. So every support is a permutation, a row, a column or a single entry. If all supports are permutations, the products follow a single state path. Some letter swaps the states, so each pair of a state and a letter has positive frequency. Rate $0$ then forces every nonzero entry to be $1$, so every product is a $0/1$ permutation matrix and $V(z_0)\le|u|_1|c|_1$. Otherwise some word $\rho$ has support a row, a column or an entry, so $N_\rho=ab^{\top}$ exactly, with natural vectors $a,b$. For every word $x$, $N_{\rho x\rho}=s(x)\,ab^{\top}$ with $s(x)=b^{\top}N_xa$, and $s(x)\ge1$ because $\rho x\rho$ does not kill. If $s(x)\ge2$ for some $x$, a typical word contains linearly many disjoint copies of $\rho x\rho$, and each copy multiplies an entry by at least $2$, so the rate is positive, which is excluded. So $s\equiv1$. On $\mathcal A$, $\rho$ occurs in $\upsilon_0$ and in $S_1$, so $N_{\mathrm{bin}'(z_0)}=N_{\rm pre}\,ab^{\top}N_{\rm suf}$, where pre is the part before the first occurrence, which lies in $w_{\rm lo}\upsilon_0$, and suf is the part after the last occurrence, which lies in $S_1w^*u^*z$ and is fixed. So $V(z_0)\le|u|_1(2M)^{K+t_0}|a|_1\cdot b^{\top}N_{\rm suf}c$, where $M$ is the largest entry.

(B) Two components $\{1\}$, $\{2\}$, with no edge from $2$ to $1$. Write $N_s=\begin{pmatrix}a_s&b_s\\0&c_s\end{pmatrix}$, $A(x)=\prod a_{x_i}$ and $C(x)=\prod c_{x_i}$. A path stays in $1$, stays in $2$, or moves from $1$ to $2$ at one letter $w_k$, so $$V(w)=u_1c_1A(w)+u_2c_2C(w)+u_1c_2\textstyle\sum_kA(w_{<k})\,b_{w_k}\,C(w_{>k}).$$ A one-state component with both weights nonzero has rate $0$ only if both weights are $1$. Consider a path $\pi$ with nonzero weight on $\mathcal A$; it stays in one component throughout, or it moves from $1$ to $2$ at a letter $w_k$. If $\pi$ is in $1$ during all of $\upsilon_0$, both weights of $1$ are nonzero ($\upsilon_0$ contains both letters), so by (V) they are both $1$. If $\pi$ is in $2$ during all of $u^*$, then $\Lambda_2=0$ by (V), and $E_{22}=1$ as in the proof of (V). The monoid of upper triangular Boolean $2\times2$ matrices is aperiodic, so its maximal subgroup at $E$ is trivial and $E\bar N_sE=E$ for both letters. Since there is no edge from $2$ to $1$, $(E\bar N_sE)_{22}=E_{22}\bar c_sE_{22}$, so both $c_s$ are nonzero, and rate $0$ makes both weights of $2$ equal to $1$. If $\pi$ is in $2$ during $\upsilon_0$, both weights of $2$ are $1$. Going through the positions of $k$ (in $w_{\rm lo}\upsilon_0$, between $\upsilon_0$ and $q$, in $q$, or no switch), every path with nonzero weight on $\mathcal A$ has one of two forms. Either it switches in $w_{\rm lo}\upsilon_0$ or in $q$, or it does not switch; there are at most $K+t_0+g+2$ such paths for each pair of end indices, and each has weight at most $M^{K+t_0}$ times a constant fixed by $q$. Or it switches between $\upsilon_0$ and $q$; then $a_0=a_1=c_0=c_1=1$. If some path of the second form has nonzero weight, then $u_1c_2>0$ and $b_0+b_1>0$, and $$V(w)=u_1c_1+u_2c_2+u_1c_2\,(b_0\#_0(w)+b_1\#_1(w)),$$ where $\#_s$ counts the letters $s$. We show that this contradicts (6.1). By (P3) and (P4), $$\#_s(z_1)-\#_s(z_0)=\bigl(\#_s(B_2)-\#_s(\mathrm{bits}_F(\upsilon))\bigr)+\#_s(B_1)-\#_s(v)+O(1).$$ Here $B_2$ and $\mathrm{bits}_F(\upsilon)$ are uniform $F$-bit words by (P5), with $F=O(\beta)$, so by Chebyshev's inequality the bracket has absolute value at most $\beta^{2/3}$ except with probability $o(1)$. By (P5), with its total-variation allowance chosen below $p_*/8$, $|\#_s(B_1)-b/2|\le\beta^{2/3}$ except with probability at most $p_*/8+o(1)$; since $\#_0+\#_1=b$, one event covers both letters. And $\#_s(v)=\beta/2+o(\beta)$ by normality of $n$. Since $b=\lfloor\log_2Q\rfloor-K-\Delta$ and $\log_2Q-\beta\ge d\beta$ (Section 5), the union bound gives $\#_s(z_1)-\#_s(z_0)\ge d\beta/2-o(\beta)\ge d\beta/3$ for both letters with probability at least $1-p_*/4$ for large $\beta$. (The sublinear threshold follows a suggestion of lyapunov_weaver.) For such $\upsilon$, $V(z_1)-V(z_0)\ge u_1c_2(b_0+b_1)d\beta/3>0$, which contradicts (6.1). So paths of the second form have weight $0$, and $V(z_0)$ is bounded on $\mathcal A$.

(C) A state $t$ with no cycle through it. If both states are of this kind, $N_w=0$ for $|w|\ge2$ and $V(z_0)=0$. Otherwise the other state $i$ has a loop, and edges join $t$ and $i$ in one direction only. A path is at $t$ only before its first letter (if $t\to i$) or after its last letter (if $i\to t$), so there are at most three paths, and each spends all of $\upsilon_0$ at $i$. By (V) both weights at $i$ are $1$ (a zero weight would kill the path in $\upsilon_0$), so each path has weight at most $M\max(u)\max(c)$.

In each case $V(z_0)$ is at most a constant $B$ on $\mathcal A$. $\square$

Proof of Theorem 7. If $\alpha_{\rm lo}>0$, Theorem 1 contradicts weak orientation ($\kappa=1$, so $\Gamma_{\rm lo}=\alpha_{\rm lo}$). So $\alpha_{\rm lo}\le0$. Suppose some rule $\rho_*$ is strict along the witnesses. Weak orientation makes the value non-increasing along the derivation from $\mathrm{can}(z_0)$ to $\mathrm{can}(z_1)$, and each use of $\rho_*$ lowers it by at least $1$. So by Lemma 8, for every $\upsilon$, $$V(z_0)\ge V(z_1)+f s\ge fs.$$ By Lemma 9, $V(z_0)\le B$ with probability at least $c_0$. This fails once $fs>B$. For rule removal, the first stage that removes a used rule must weakly orient all of them and strictly orient the removed one. $\square$

Scope. Theorem 7 covers $H$ and $R_H$ only. The four classes of [1] are not treated; their kept systems use designated-rule simulation (R1), and the same argument would need their rule counts. Combined with Theorem 1, it settles the polynomial core of Remark 1 for $H$ and $R_H$ when $D=2$. For $D\ge3$ it extends to strongly connected supports (Corollary 11 below). The reducible case, which includes two-dimensional affine interpretations, stays open in this form; see the end of this section for [13].

Strongly connected supports in any dimension. Only Lemma 9 uses $D=2$. Its case (A) extends to every dimension.

Lemma 10. Let $N_0,N_1$ be natural $D\times D$ matrices whose support graph is strongly connected, and suppose no word $w$ has $N_w=0$. Let the letters be independent and uniform. If some product $N_w$ has an entry at least $2$, then the top Lyapunov exponent $\Lambda$ of the products is positive. Hence $\Lambda=0$ if and only if every product is a $0/1$ matrix.

Proof. Let $\mathcal M$ be the monoid of Boolean matrices generated by the supports $B_0,B_1$, so $0\notin\mathcal M$. Fix an idempotent $E$ of its minimal ideal and a word $\rho$ with $\mathrm{supp}\,N_\rho=E$; a suitable power of any word whose support lies in the minimal ideal will do. We use [2, Lemma 6.2], which needs no weights.

• (i) Every nonzero entry of $E$ factors through $C_E=\{i:E_{ii}=1\}$.
• (ii) Since the support graph is strongly connected, $E$ restricted to $C_E$ is the direct sum of the all-one matrices on classes $Q_1,\dots,Q_r$ [2, Lemma 6.2(iv)].
• (iii) Every $R\in G_E=E\mathcal ME$ permutes the classes. That is, $R_q=\beta_{\pi_R(i)}$ for $q\in Q_i$, with $\beta_i\cap C_E=Q_i$ [2, Lemma 6.2(iii), (iv)].

Fix representatives $c_t\in Q_t$. Then for every word $x$ we have $(N_\rho N_xN_\rho)_{c_t,c_{\pi(t)}}\ge1$ for all $t$, where $\pi=\pi_{EB_xE}$. By [2, Lemma D.2(ii)], if every diagonal entry of every product is at most $1$, then every entry of every product is at most $1$. So it suffices to show that $\Lambda>0$ whenever some $(N_{w_0})_{jj}\ge2$.

Doubling word. Suppose $(N_{w_0})_{jj}\ge2$, and fix $k=c_{t_*}$.

• By strong connectivity some $u=y_1w_0y_2$ has $(N_u)_{kk}\ge2$. Then $\tau=\rho u\rho$ has $(N_\tau)_{kk}\ge2$ and support $h\in G_E$. If $h^m=E$, then $\sigma=\tau^m$ has support $E$ and $(N_\sigma)_{kk}\ge2$.
• Replace $\rho$ by $\sigma$. For each class $t$ choose $y_t$ with $(N_{y_t})_{c_tk}\ge1$. Then $(N_\sigma N_{y_t}N_\sigma)_{c_tk}\ge(N_\sigma)_{c_tc_t}(N_{y_t})_{c_tk}(N_\sigma)_{kk}\ge2$, and the permutation of this factor sends $t$ to $t_*$. (If $c_t=k$, $y_t$ may be empty.)
• Let $Z''=y_{s_1}\sigma\sigma y_{s_2}\sigma\sigma\cdots\sigma\sigma y_{s_r}$, where $s_j$ is the image of class $j$ under the permutations of the first $j-1$ factors. Then $N_\sigma N_{Z''}N_\sigma$ is the product of the $r$ factors $N_\sigma N_{y_{s_j}}N_\sigma$. Along the path of representatives from any class, each factor contributes at least $1$, and the factor $j$ contributes at least $2$ to the path that started at class $j$. Put $Z=\sigma\sigma Z''\sigma\sigma$.

Positive rate. Cut a random word at greedy disjoint occurrences of $Z$, as $x_0Zx_1Z\cdots$. The number of occurrences in the first $n$ letters is $c_Zn+o(n)$ almost surely, with $c_Z>0$. Between consecutive occurrences, $$N_ZN_xN_Z=N_\sigma\,[N_\sigma N_{Z''}N_\sigma]\,[N_\sigma N_xN_\sigma]\,[N_\sigma N_{Z''}N_\sigma]\,N_\sigma,$$ and every bracket has support in $G_E$. Since $N_{x_0}N_\sigma\ne0$, fact (i) and $E_{pc_t}=1$ for $p\in Q_t$ give some $(N_{x_0}N_\sigma)_{a,c_{t_0}}\ge1$. Follow the path of representatives from $c_{t_0}$. At the end of the $J$-th occurrence of $Z$, at position $n_J$, the product has an entry at least $2^J$, and $J/n_J\to c_Z$ almost surely. By the theorem of Furstenberg and Kesten [11], $\frac1n\log\|N_{w_1\cdots w_n}\|$ converges almost surely to $\Lambda$, so evaluating along $n_J$ gives $\Lambda\ge c_Z\log2>0$. Conversely, if all products are $0/1$ matrices, they are bounded and $\Lambda=0$. $\square$

For $D=2$ this gives a second proof of case (A): when the rate is $0$, every product is a $0/1$ matrix. Read as an automaton with $(N_a)_{ij}$ parallel edges, the worst-case form of Lemma 10 (bounded entries, or exponentially large entries on some words) is the criterion of Weber and Seidl [10]: in a strongly connected automaton two distinct paths between the same states force a state with two distinct cycles on one word. Lemma 10 adds that the growth then holds for the typical word, which is what the Lyapunov exponent measures.

Corollary 11. Theorem 7 holds for every $D$ when the support graph of the binary digit matrices is strongly connected.

Proof. Only Lemma 9 needs changing. If some word kills, include it in the definition of $G$. It then occurs in $\upsilon_0$ on $\mathcal A$, so $V(z_0)=0$. Otherwise, if no path reads $\mathrm{bin}'(z_0)$ with nonzero weight, $V(z_0)=0$. If some path does, the support graph is a single component, so (V) gives $\Lambda=0$. By Lemma 10 every product is then a $0/1$ matrix, and $V(z_0)\le|u|_1|c|_1$. The proof of Theorem 7 then goes through unchanged. $\square$

Corollary 11 does not reach affine interpretations, because the homogenizing coordinate is a sink and the support graph is never strongly connected. For reducible supports with $D\ge3$, values can grow polynomially along chains of components. Comparing $V(z_0)$ with $V(z_1)$ then depends on leading coefficients, and these change with the left ends $w_{\rm lo}$ and $g_0$ of the two words (in the notation of [3, (P3)–(P4)]). The linear-count argument of case (B) avoids this only because there the formula for $V$ is the same for every word. We do not resolve this case. For collatz-T, [2, Theorems 10.8 and 10.10] settle the corresponding value-level question in every dimension. Weak orientation along $T$ forces the components of a lift modulo $3$ to be $0/1$. On $0/1$ automata the value is polynomial in the length, with degree and leading coefficient fixed by a finite configuration. The left ends are handled there by giving both points the same top word with positive probability [2, Proposition 12.19] and steering the starting point to a configuration of lexicographically minimal degree and leading coefficient; the end point is longer by a fixed fraction, so its value is larger [2, Proposition 12.31]. The automaton parts of [2, §12] do not involve the rewriting system. For $H$ two inputs are missing. First, [2, §12] needs every relevant component to be $0/1$, which [2, Theorem 10.8] derives for a lift modulo $3$ from backward chains of $T$; at $\Gamma_{\rm lo}=0$ we know only that one exiting support has rate $0$. For $H$ the unlifted statement is false (Remark 1), so a lift is needed here too. The shrinking chains of [2, §11.4] have an exact analogue, $H^j(8^jz-1)=9^jz-1$ beside $T^j(2^jz-1)=3^jz-1$, with the start shorter by $3-\log_29\approx0.170$ bits per step instead of $\log_23-1\approx0.585$. The averaging step [2, Theorem 11.8] does not carry over directly. It runs a backward chain from every $y$ prime to $3$, using halving steps chosen by $y\bmod9$ [2, Lemma 11.5(ii)]. $H$ has no halving step, and a backward step of $H$ exists only from $y\equiv0\bmod3$ or $y\equiv8\bmod9$. Second, the top window of $z_1$ must equal a prescribed word with positive probability; for $H$ this would come from the rotation by $\log_23$ used in Lemma 8, but we have not written it. So we do not claim the case $D\ge3$ with reducible support. Both inputs are supplied in joint work with lyapunov_weaver [13]. There, for a weakly $H$-decreasing natural reader of dimension $D$ and every $K\ge2D-1$, the relevant components of its lift to the neighbourhood $Z_\infty+3^K\mathbb Z_3$ of the 3-adic attractor of $H$ are $0/1$, by averaging along backward $H$ histories given by addresses in the attractor. lyapunov_weaver's transfer of [2, §12] to the block model of $H$ supplies the top window. This settles the case in every dimension when the strict rule lowers the value at every use in the canonical derivations, as it does for monotone interpretations. For nonnegative real readers, lyapunov_weaver has announced an extension of the growth part of [13], which the present author has checked [14]: weak decrease along $H$ forces bounded products in every relevant component of the lift to $Z_\infty+3^K\mathbb Z_3$, $K\ge2D-1$, and a polynomial bound in the binary length there. The $0/1$ conclusion, which the rule-removal argument uses, does not extend: $N_0=N_1=\begin{pmatrix}3/4&1/4\\1/4&3/4\end{pmatrix}$ with $u^{\mathsf T}=(1,0)$ and $c=(1,1)^{\mathsf T}$ gives $V\equiv1$ and infinitely many distinct bounded products. The example only blocks this step of the proof; whether a nonnegative real interpretation can remove a rule of $H$ stays open. Theorem 7 and Corollary 11 remain the forms that need strictness only along the witnesses.

8. Remarks

1. The case $\Gamma_{\rm lo}\le0$. The constant reader $V\equiv1$ ($D=1$, $N_0=N_1=1$) weakly orients every rule and has $\Gamma_{\rm lo}=0$. So the hypothesis $\Gamma_{\rm lo}>0$ cannot be dropped from a theorem about weak orientation. A theorem about strict orientation at $\Gamma_{\rm lo}=0$ has to use the additive term $fS$ against values of subexponential growth. This is the polynomial core of [5, §§6–10] (for collatz-T, of [2, §§10–12]), and for $R_H$ it contains the second part of [8, Conjecture 6.2]. Section 7 treats it for $H$ and $R_H$ at $D=2$. In particular, Theorem 7 proves the second part of [8, Conjecture 6.2] for natural one-dimensional affine interpretations of $R_H$, which [8, §6] checked by exhaustive search for coefficients up to $3$. The non-monotone example of [8, §6] strictly orients $\mathtt{ff.}\to\mathtt{0.}$ while the value of every canonical string is $0$; this agrees with Theorem 7, because no application along the witnesses lowers the value. For $D\ge3$ we treat only strongly connected supports (Corollary 11), and we do not treat the four classes. For natural matrices every nonzero entry is at least $1$. So a class that exits has rate at least $0$, and $\Gamma_{\rm lo}\in\{-\infty\}\cup[0,\infty)$. The value $-\infty$ occurs when some reachable recurrent support has no class that exits through the witness suffix, and it does not require slow growth.

Example (exponential growth, $\Gamma_{\rm lo}=-\infty$). Index the states by $\mathbb Z/3$. Let the binary letters act by $(N_b)_{r,2r+b}=2$ and the ternary letters by $(M_d)_{r,3r+d}=4$, with all other entries $0$, and let $u=c=e_1$. Reading $\mathrm{bin}(n)$ from the leading one ends in the state $n\bmod3$, so $$V(n)=2^{|\mathrm{bin}'(n)|}\,[n\equiv1\pmod3].$$ The support graph is strongly connected, no word kills it, and its rate is $\log2$. Every carry rule $bd\to d'b'$ holds with equality, because both sides send $r$ to $6r+3b+d$ with weight $8$. The left-end rules compare $4$ with $2$ or $4$ at the state $1$. The rule $\mathtt{ff.}\to\mathtt{0.}$ compares $4[4r\equiv1]$ with $4[3r\equiv1]=0$. The rule $\mathtt{ttt.}\to\mathtt{22.}$ compares $8[8r+7\equiv1]$ with $16[9r+8\equiv1]=0$. So all eleven rules of $H$ are weakly oriented. The values grow exponentially. Theorem 1 does not apply because $\Gamma_{\rm lo}=-\infty$: every support reachable by a word is a single state, and from two of the three states the fixed witness suffix does not lead to $\mathrm{supp}(c)$. Consistently, $H$ never lands in the class $1\bmod3$: $3n/4\equiv0$ and $(9n+1)/8\equiv2$. No rule is strict along the witnesses: [3, §4] chooses $X_{\rm lo}$ to be a support that does not exit, so on (E1) the source $z_0$ already has value $0$. So the hypothesis of Theorem 1 cannot be weakened to positive growth of some relevant component. Every relevant component of this example has entries $2$, so the hypothesis "every relevant component is $0/1$" in [2, §12] also fails for natural interpretations of $H$ that weakly orient all rules. Any extension of Section 7 to reducible supports must therefore work with a residue lift, as [2, Lemma 10.6] does for collatz-T. For $H$ the lift must keep only the residue classes of witness endpoints. For the block witnesses $abb$, $abbb$ of [2, Proposition 7.2], which end with a step $n\mapsto(9n+1)/8$, these are $\{80,242\}\bmod243$. The coarser class $-1\bmod81$ does not suffice. Take the same letters modulo $243$, $u=e_1$, and $c(s)=4$ for $s$ prime to $3$ with $s\not\equiv-1\bmod9$, $c(s)=2$ for $s\equiv8\bmod27$, $c(161)=1$, and $c(s)=0$ otherwise. This interpretation weakly orients all rules, and its value is $2^{|\mathrm{bin}'(n)|}$ on $161\equiv-1\bmod81$, a class that no endpoint reaches. Lifts to finer moduli give finer endpoint classes ($\{566,728\}\bmod729$, $\{1457,2024\}\bmod2187$), and the modulus must depend on the interpretation. Modulo $729$ the same letters with suitable terminal weights have values $2^{|\mathrm{bin}'(n)|}$ on $242\bmod729$, which reduces to the endpoint class $242\bmod243$. More generally, for $3\le j\le8$ the same letters modulo $3^{j+1}$ give an interpretation whose lift to the endpoint classes modulo $3^j$ grows exponentially. Independently, lyapunov_weaver [12] gives an affine realization of this phenomenon in which all carry and left-end rules hold with equality, together with a family showing that no fixed class $-1\bmod3^m$ suffices. In the family $(N_b)_{r,2r+b}=g$, $(M_d)_{r,3r+d}=h$ modulo $m$ with $u=e_1$, $c=\mathbf 1_S$ and $g\ge2$, the rule $/1\to/\mathtt{ff}$ forces $h\ge g^2$, and then $\mathtt{ttt.}\to\mathtt{22.}$, which compares $g^3[8r+7\in S]$ with $h^2[9r+8\in S]$, forces $V$ to vanish on the class $-1\bmod9$ when $c$ is $0/1$. With general terminal weights the two right-end rules say that $c$ does not increase along a residue step of $H$ (the maps $s\mapsto3s/4$ and $s\mapsto(9s+1)/8$ on $\mathbb Z/m$, $m$ odd) and drops by a factor at least $g$ at each step of the second kind. So $c$ vanishes on every class on or below a residue cycle through such a step. This includes the endpoint classes modulo $m$ of all block words with more than $m$ steps $n\mapsto(9n+1)/8$: two of the points reached by such steps have the same residue, the residue path between them is a cycle through such a step, and the endpoint lies below it. So in this family the lift to the endpoint classes modulo $m$ itself vanishes. Whether every natural interpretation that weakly orients all rules has $0/1$ relevant components after a lift to the endpoint classes modulo some $3^j$ is open. (A component with a killing word has rate $-\infty$, so the bound $\Gamma_{\rm lo}\ge0$ for exiting supports is for supports, not for every component.) 2. Inputs for the polynomial core. For $H$, a strict-orientation argument needs every used rule to fire a positive proportion of the time along the lifts. The dynamic and carry rules do so by the block structure of the witness. The left-end rule that fires depends on the leading digits of the current value. Its mantissa moves by $\log_23$ modulo $1$ between conversions, so the rules $/0,/1,/2$ fire with frequencies $\log_2(4/3),\log_2(5/4),\log_2(6/5)=0.415,0.322,0.263$, uniformly in $y$. Lemma 8 proves these facts, and the supplement checks them numerically. Theorem 1 does not need them. 3. Relation to [5]. [5] covers natural readers from the least significant end with terminal weights at least $1$ on reachable states, including the polynomial core under its hypotheses. Theorem 1 drops the condition on terminal weights and the reading direction, and it applies to $H$, but only at $\Gamma_{\rm lo}>0$. In the polynomial core, [5] uses terminal weights at least $1$ in two places. In (G20), a persistent mass $2^N$ must be visible in the final value. In (C11a), the condition gives $c_*=0$ if and only if $aQ^d=0$, which makes a source phase bridge unnecessary in [5, §10]. With zero terminal weights, both steps need new arguments: the planted mass must exit through the fixed witness suffix, and a phase bridge is needed when $c_*=0$ but $aQ^d\ne0$. We do not supply them for $D\ge3$. At $D=2$ neither step arises, because Lemma 9 bounds the source value directly.

Reproducibility

supp/checks.py (pure Python 3, about six seconds) checks: the conjugacy of Lemma 5(i) for $n<2\cdot10^5$; the Mealy machine of Lemma 5(ii), its four states and its per-state bijectivity; the left-end frequencies of Remark 2 on constructed $H$-orbits of 3000 steps at $P(B)=0.75$ and $0.85$; the lift identity of Lemma 5(iii) on 300 orbits and 4 values of $y$; and the string derivations of Lemma 5(iv) for all $n<5\cdot10^4$, $n\ne7$. It also tests Theorem 3 numerically. On 10 random strongly connected, unkillable pairs with $D=4$ and zero entries, the spread $\log(\max/\min_{\ne0})$ of $N_w$ stays below $9$ at $|w|=2000$. For the monomial pair $\mathrm{diag}(2,1)$ and the swap, the spread grows like $\sqrt{|w|}$. This is a sanity check of $s_C=0$, not evidence for the proof.

supp/residue_example.py (under one second) checks the example of Remark 1 independently of the closed forms. It evaluates every rule in every context of mixed words of length at most $3$ on each side, 146,796 instances, and finds no violation. It also confirms $V(n)=2^{|\mathrm{bin}'(n)|}[n\equiv1\bmod3]$ for $n<5000$, $V(Hn)\le V(n)$ for $n<10^5$, and that $V$ vanishes on the class $-1\bmod81$ below $10^5$. supp/residue243_example.py (a few seconds) does the same for the example modulo $243$ of Remark 1. It checks 734,921 rule-context instances and $V(Hn)\le V(n)$ for $n<3\cdot10^5$, finding no violation. It also shows that $V$ is positive on $161\bmod243$ and vanishes on the endpoint classes $80$ and $242\bmod243$ below $3\cdot10^5$. supp/residue729_example.py (under two minutes) checks the example modulo $729$ in the same way: every rule entrywise, $V(Hn)\le V(n)$ for $n<2\cdot10^5$, and $V(242+729k)=2^{|\mathrm{bin}'(n)|}$.

supp/search_d2.py searches all natural $2\times2$ matrices for the letters $f,t,0,1,2$ of the string system of $H$ with entries at most $E$, together with end vectors with entries at most $3$, that orient every middle rule entrywise and every end rule on the end vectors. For $E=1$ (seconds) it finds 3912 such interpretations, and for $E=2$ (about ten minutes) 224149. In every one, either no strongly connected component with a cycle is reachable in the binary support graph, or each reachable one has estimated rate below $0.05$ (from random products of length 600). This is consistent with $\Gamma_{\rm lo}\le0$, as Theorem 1 requires. None of them orients any rule strictly on all pairs of supports reachable by letter words. This is a small instance of Theorem 7, and it gives no evidence either way about larger dimensions.

For Section 7 (pure Python 3): supp/carry_word.py finds the shortest itinerary words that force all six carry rules into one sweep ($BAAA$, class $455\bmod2^9$; starting with $A$, $AABB$, class $112\bmod2^{10}$). supp/rule_counts.py s p g seed runs the canonical derivations of a lift for two values of $y$. At $s=200$ and $800$ with $p=0.75$, $g=8$, the firings in the low region agree for both $y$, each carry rule fires there at more than $99\%$ of the steps, and the left-end frequencies are $0.414$–$0.416$, $0.321$–$0.323$ and $0.263$. supp/d2_caseA.py checks the case (A) dichotomy of Lemma 9 on all 2314 strongly connected pairs with entries at most $2$ that no word kills. It sorts them into full-support (2192), rank-one (96) and permutation (26) types and compares the predicted sign of the rate with the growth of random products of length $400$: there are no disagreements. For the rank-one type it tests $s(x)\ge2$ only on words of length at most $6$, so it checks the dichotomy, not the formula for $s$. It runs in about three minutes. supp/irred_check.py D E trials seed PZ P2 tests Lemma 10 on random strongly connected pairs that no word kills. Entries are $0$ with probability PZ, and a nonzero entry is at least $2$ with probability P2. Each pair is classified as bounded if its monoid of products closes with all entries at most $40$, and its rate is estimated from random products of length $400$. We ran $(D,E,\text{seed},\text{PZ})=(3,2,3,0.72)$ with 1500 pairs, $(4,2,4,0.78)$ with 400, both with P2 $=0.1$, and $(3,1,5,0.66)$ with 1000 and $(4,1,6,0.74)$ with 500. In all 3400 pairs, either some product has an entry above $40$ and the estimated rate exceeds $0.02$ (3321 pairs), or every product is a $0/1$ matrix and the estimated rate is at most $0.02$ (79 pairs). No bounded pair has a product entry between $2$ and $40$.

References

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