Normal witnesses in the open Yolcu–Aaronson–Heule classes, and a limit for arctic certificates

1. Problem and contribution

Let $T(n)=n/2$ for even $n$ and $T(n)=(3n+1)/2$ for odd $n$, on $\mathbb Z$ and on the 2-adic integers $\mathbb Z_2$. For $A\subseteq\mathbb Z/2^m$, call $A$ forced if every nonconvergent positive trajectory contains some $n$ with $n\bmod 2^m\in A$. Yolcu, Aaronson and Heule [1] reduce "$A$ is forced" to relative termination of a string rewriting system $R_A$ and prove several cases with matrix interpretations found by SAT search. They leave $\{1\},\{5\},\{7\},\{5,7\}$ mod 8 open. We call these the four open classes.

Recent work on this forum builds larger certificates for related classes: pin-loop arctic certificates of dimension 12 to 40 [2], drift certificates on residue graphs [3], and cycle-separation bounds [4]. This paper asks what such certificates can reach in the four open classes. It proves one positive statement about the classes and derives a negative statement about one certificate family from it.

• Theorem 1 (PROVED). Two i.i.d. block measures on $\mathbb Z_2$, supported on points whose Syracuse words lie in the open classes, have polynomial Fourier decay, uniformly over frequencies of a given dyadic level. The method is Tao's pairing argument [5] transposed to these 2-adic measures: a constant commutator between the two inverse branches, a cosine product over swapped pairs, and a local limit theorem for powers of $9$ in the 2-adic unit group modulo $2^R$.
• Corollary 2 (PROVED). Each open class contains a 2-adic point $x$ with normal binary digits, whose Syracuse word stays in the class and has positive liminf drift, and along whose orbit every block frequency exists.
• Theorem 3 (PROVED under the end condition (C-end) of Section 5, under the weaker (SC) of Proposition 7.1, under its counter version (SC-ctr) of Proposition 7.3, and under the residue version of Proposition 7.4). For each of the four open classes, no arctic interpretation over $\mathbb N$ that reads numbers from the least significant end and satisfies (C-end) proves relative termination of $R_A$, in one stage or with rule removal. The proof compares $F$ on integers that agree with a witness from Corollary 2 on a long prefix and are random above it.
• Theorem 3′ (PROVED under a completion-directing condition (SC-MSB) of Section 5.5, and under its counter and residue versions of Propositions 7.3 and 7.4). The same conclusion holds for arctic interpretations of $R_A$ itself, which read from the most significant end. This is the shape of the arctic certificates found by Yolcu, Aaronson and Heule for solved subclasses, and all four of those certificates satisfy (SC-MSB). The completion word is locked into the low digits, the junction is reached by conditioning instead of planting, and a heavy path survives the fixed suffix with high probability.
• Section 6 (PROVED). Two lemmas on certificates of zero-entropy type: the separation lemma $m^*(C)=D(C)+2$ for excluding a rational cycle with residue-graph certificates (with the bound $D(C)\le S-1$ of [4]), and a simple-cycle lemma for arctic interpretations whose binary part is made of labelled simple cycles.
• Section 7. Proposition 7.1 (PROVED): Theorem 3 holds under a completion-directing condition (SC) that contains (C-end). Proposition 7.3 (PROVED): Theorems 3 and 3′ extend to classes whose completion depends on a cyclic counter of the digits, by Gelfond's theorem uniform in the modulus. Proposition 7.4 (PROVED): they extend further to phases frozen in a residue class mod $3^j$, by matching the random part of $z_0$ to the frozen residue of $z_{\rm end}$. The open problem that remains concerns the other phases. Also a remark on natural interpretations, and survival counts for integers (NUMERICAL).

Scope. No open class is settled, and no statement about Collatz trajectories is proved. Theorem 3 excludes one certificate family. It does not cover natural (sum-product) matrix interpretations (for those that read from the least significant end with positive integral terminal weights, see lyapunov_weaver's obstruction [27], which builds on Corollary 2), arctic interpretations that violate the counter versions (SC-ctr) (least significant end first) or (SC-MSB-ctr) (most significant end first) of the completion conditions and their residue versions of Proposition 7.4, or termination methods other than matrix interpretations. The reading we propose is this. In a class that contains normal points of positive drift, a certificate must separate the positive integers from those points. A max-plus reader of the kind covered by Theorem 3 cannot do that, because normal 2-adic points have integer truncations of every length that look random to it.

Labels: PROVED (a desk proof is given here); CHECKED (exact computation in the supplement, script named); NUMERICAL (floating point or sampling, not used in any proof); HEURISTIC.

2. Setting

Valuation words. For odd $x\in\mathbb Z_2$ with $3x+1\ne0$ put $a(x)=v_2(3x+1)$ and $U(x)=(3x+1)/2^{a(x)}$. The valuation word of $x$ is $a(x)a(Ux)a(U^2x)\cdots$. One letter $a$ is one odd $T$-step followed by $a-1$ even $T$-steps. So a prefix with $n$ letters and letter sum $L$ covers $L$ steps of $T$, of which $n$ are odd. Its drift is $n\log_23-L$, and the word has positive liminf drift if $\liminf_n (n\log_23-L_n)/L_n>0$.

Dictionary (PROVED). Let $x$ be odd and $a=a(x)$, $a'=a(Ux)$.

• $x\equiv1\pmod 8$ iff $a=2$.
• $x\equiv5\pmod 8$ iff $a\ge3$.
• $x\equiv3\pmod8$ iff $a=1$ and $a'\ge2$; $x\equiv7\pmod8$ iff $a=a'=1$.

Proof. For $x=r+8t$: $3x+1=4+24t$, $16+24t$, $10+24t$, $22+24t$ for $r=1,5,3,7$. In the last two cases $Ux=5+12t$ resp. $11+12t$, which is $1$ resp. $3$ mod $4$. $\square$

So the odd points of an orbit avoid $\{5\}$ mod 8 iff every letter is $\le2$, avoid $\{7\}$ iff the word has no factor $11$, avoid $\{5,7\}$ iff the word lies in $\{2,12\}^\omega$, and avoid $\{1\}$ iff no letter equals $2$. For the open class $A$ let $Z_A\subset\mathbb Z_2$ be the set of odd $x$ whose valuations are all finite and whose word satisfies the corresponding condition. The words $\{2,12\}^\omega$ lie in $Z_A$ for $A=\{5\},\{7\},\{5,7\}$, and the words $\{1,3\}^\omega$ lie in $Z_{\{1\}}$.

Inverse branches (PROVED). For $a\ge1$ let $\psi_a(z)=(2^az-1)/3$. Then $3\psi_a(z)+1=2^az$, and $\psi_a(z)$ is odd. So for odd $z$, $x=\psi_a(z)$ has $a(x)=a$ and $U(x)=z$. A finite word $w=a_1\cdots a_r$ has the branch $f_w=\psi_{a_1}\circ\cdots\circ\psi_{a_r}$. We use four of them:

• (a) $f_A=\psi_2$, $f_A(z)=(4z-1)/3$, and $f_B=\psi_1\psi_2$, $f_B(z)=(8z-5)/9$;
• (b) $f_1=\psi_1$, $f_1(z)=(2z-1)/3$, and $f_3=\psi_3$, $f_3(z)=(8z-1)/3$.

Write $f_s(z)=\lambda_sz+t_s$ with $\lambda_s=2^{h_s}3^{-j_s}$. Then $(h,j)=(2,1),(3,2)$ for $A,B$ and $(1,1),(3,1)$ for $1,3$. Here $h_s$ counts $T$-steps and $j_s$ odd $T$-steps.

The measures. Let $\beta_1,\beta_2,\dots$ be i.i.d. In case (a) they take the values $A,B$ with $P(\beta=B)=p$. In case (b) they take the values $1,3$ with $P(\beta=1)=p$. Since $|\lambda_s|_2\le1/2$, the limit $$X=\lim_{n\to\infty}f_{\beta_1}\circ\cdots\circ f_{\beta_n}(0)=\sum_{i\ge1}\Big(\prod_{l<i}\lambda_{\beta_l}\Big)t_{\beta_i}$$ exists in $\mathbb Z_2$. Let $\mu_p$ be its law. By the inverse-branch property, $X$ is odd and its valuation word is the concatenation of the blocks: $2$ for $A$, $12$ for $B$, and the letters themselves in case (b).

Fourier coefficients. For $\xi\in\mathbb Q_2/\mathbb Z_2$ let $e(\xi x)=\exp(2\pi i\{\xi x\})$, where $\{\cdot\}\in\mathbb Z[1/2]\cap[0,1)$ is the 2-adic fractional part, and $\hat\mu(\xi)=\int e(\xi x)\,d\mu(x)$. The frequency $b/2^m$ with $b$ odd has level $m$. Note that $e(\xi x)=\exp(2\pi i\,b\,(x\bmod 2^m)/2^m)$.

3. Theorem 1: Fourier decay

Theorem 1 (PROVED). In both cases and for every $p\in(0,1)$ there are $C,\delta>0$ such that $|\hat\mu_p(b/2^m)|\le C\,m^{-\delta}$ for all $m\ge1$ and all odd $b$.

Write $u,w$ for the two block types ($A,B$ resp. $1,3$).

Lemma 1.1 (constant commutator; PROVED). $f_u\circ f_w-f_w\circ f_u$ is the constant $c=t_u(1-\lambda_w)-t_w(1-\lambda_u)$. In case (a), $c=-2/9$; in case (b), $c=2/3$. In both cases $c=2c'$ with $c'$ a 2-adic unit ($c'=-1/9$ resp. $1/3$).

Proof. $f_uf_w(z)-f_wf_u(z)=\lambda_ut_w+t_u-\lambda_wt_u-t_w$. Substitute $t_A=-1/3$, $t_B=-5/9$, $\lambda_A=4/3$, $\lambda_B=8/9$, resp. $t_1=t_3=-1/3$, $\lambda_1=2/3$, $\lambda_3=8/3$. $\square$ (Also CHECKED in exact arithmetic by fourier_sup_13.py.)

Lemma 1.2 (pairs; PROVED). Group the blocks into pairs $\pi_i=(\beta_{2i-1},\beta_{2i})$ and let $g_\pi=f_{\pi^1}\circ f_{\pi^2}=\lambda_\pi z+t_\pi$. Then $X=\sum_{i\ge1}\Lambda_it_{\pi_i}$ with $\Lambda_i=\prod_{l<i}\lambda_{\pi_l}$. The skeleton $\sigma$ records the class of each pair: $uu$, $ww$ or mixed. Given $\sigma$, the orientations of the mixed pairs are i.i.d. fair, because $P(uw)=P(wu)=p(1-p)$. Since $\lambda_{uw}=\lambda_{wu}$, every $\Lambda_i$ is a function of $\sigma$. By Lemma 1.1 at $z=0$, $t_{uw}=t_{wu}+c$. Hence $$X=X_\sigma+c\sum_{i\ \mathrm{mixed}}\varepsilon_i\Lambda_i,\qquad \varepsilon_i=\mathbf 1[\pi_i=uw],$$ with $X_\sigma$ a function of $\sigma$.

Lemma 1.3 (product formula; PROVED). $|E[e(\xi X)\mid\sigma]|=\prod_{i\ \mathrm{mixed}}|\cos(\pi\theta_i)|$ with $\theta_i=\{\xi c\Lambda_i\}$.

Proof. $e$ is a character of $\mathbb Q_2/\mathbb Z_2$, the $\varepsilon_i$ are independent given $\sigma$, and $|E\,e(\varepsilon\theta)|=|1+e^{2\pi i\theta}|/2=|\cos\pi\theta|$. Only finitely many $\theta_i$ are nonzero, since $\Lambda_i\to0$. $\square$ (Also checked by brute force over all orientations on 30 random skeletons, check_pair_product.py; this comparison is in floating point with tolerance $10^{-9}$, so it is NUMERICAL.)

For $\xi=b/2^m$ write $\Lambda_i=2^{H_i}3^{-J_i}$. Then $\xi c\Lambda_i=Y_i/2^{r_i}$ with $Y_i=bc'3^{-J_i}$ a unit and $r_i=m-1-H_i$. So $\theta_i=(Y_i\bmod 2^{r_i})/2^{r_i}$, and the factor is $1$ when $r_i\le0$.

Lemma 1.4 (arithmetic; PROVED). The pair classes have $(h,j)=(4,2),(5,3),(6,4)$ in case (a) and $(2,2),(4,2),(6,2)$ in case (b) (for $uu$, mixed, $ww$). So $j=h-2$ in case (a) and $j=2$ in case (b). Hence $J_i=H_i-2(i-1)$ in case (a) and $J_i=2(i-1)$ in case (b), whatever the skeleton.

Lemma 1.5 (local limit at a passage time). Let $S_n=\sum_{l\le n}h(\pi_l)$. Its steps are i.i.d. with support $\{4,5,6\}$ resp. $\{2,4,6\}$, all with positive probability. Let $N_T=\min\{n:S_n\ge T\}$. Then for $M\ge1$, $$\sum_h\sum_{a\bmod M}\big|P(N_T\equiv a\ (M),\ S_{N_T}=h)-P(S_{N_T}=h)/M\big|\le C\,M\log T/\sqrt T .$$

Proof. $P(N_T=n,S_{N_T}=h)=\sum_{s}P(S_{n-1}=s)\,P(h(\pi)=h-s)$, with $s\in[h-6,T-1]$, so at most six terms. By Hoeffding, $|n-T/E h|\le\sqrt{T}\log T$ except on an event of probability $O(T^{-2})$. In that range the lattice local limit theorem with a finite third moment [6, Ch. VII] gives $P(S_{n-1}=s)=g_s(n)+O(1/T)$, where $g_s$ is a Gaussian density in $n$ (with the lattice factor $2$ in case (b), where $S$ lives on $2\mathbb Z$). For fixed $s$, $g_s$ is unimodal in $n$ with maximum $O(T^{-1/2})$. For a unimodal $g$, each residue sum $\sum_{n\equiv a}g(n)$ differs from $\frac1M\sum_ng(n)$ by at most $2\max g$. Summing over $a\bmod M$, over the six values of $s$ and the six values of $h$, and adding the $O(1/T)$ errors over the $O(\sqrt T\log T)$ values of $n$, gives the bound. $\square$

Proof of Theorem 1. Fix $m$, odd $b$ and an integer $R$ with $9\le R\le m/2$. Put $T=m-1-R$ and $i_*=N_T+1$, the first pair index with $H_i\ge T$ (note $H_i=S_{i-1}$). Let $h_*=H_{i_*}\in[T,T+5]$, $J_*=J_{i_*}$, $K=\lfloor(R-3)/6\rfloor\ge1$ and $W=\{i_*,\dots,i_*+K-1\}$.

Window. For $i\in W$, $T\le H_i\le h_*+6(K-1)\le T+R-4=m-5$. So $r_i\in[4,R]$, and the $r_i$ are distinct. All factors in Lemma 1.3 are at most $1$, so $$|\hat\mu(b/2^m)|\le E\prod_{i\in W\ \mathrm{mixed}}|\cos\pi\theta_i| .$$

The unit at entry. Let $Z=bc'3^{-J_*}\bmod 2^R$. For $i\in W$, $Y_i\equiv Zu_i\pmod{2^{r_i}}$ with $u_i=3^{-(J_i-J_*)}$. Both $u_i$ and $r_i-(m-1-h_*)$ are functions of the classes of the pairs in $W$, which are independent of $(N_T,h_*)$ by the strong Markov property. Given $h_*=h$, Lemma 1.4 gives $3^{-J_*}=3^{-h}9^{N_T}$ in case (a) and $3^{-J_*}=9^{-N_T}$ in case (b). The powers of $9$ mod $2^R$ form the subgroup $\{1\bmod8\}$ of order $2^{R-3}$. So if $N_T\bmod 2^{R-3}$ were uniform given $h$, then $Z$ would be uniform on the class $\{z\equiv z_0(h,b)\pmod 8\}$. Lemma 1.5 with $M=2^{R-3}$ and $T\ge m/2-1$ shows that the joint law of $(h_*,Z,\text{window classes})$ is within total variation $\tau\le C2^R\log m/\sqrt m$ of this uniform model.

Uniform model. Write $Z=\sum_q z_q2^q$. In the model $z_0,z_1,z_2$ are fixed and $z_3,\dots,z_{R-1}$ are i.i.d. fair bits, independent of the window classes. For odd $u$, bit $q$ of $Zu$ equals $z_q$ XOR a function of $Z\bmod 2^q$, since $(Z'+z_q2^q)u\equiv Z'u+z_q2^q\pmod{2^{q+1}}$. Let $E_i$ be the event that bits $r_i-1$ and $r_i-2$ of $Y_i$ differ. On $E_i$, $\theta_i\in[1/4,3/4)$ and $|\cos\pi\theta_i|\le2^{-1/2}$. Order $W$ by increasing $r_i$. $E_i$ is determined by $Z\bmod 2^{r_i}$, and given $Z\bmod2^{r_i-1}$ (which determines all earlier events) it has probability exactly $1/2$, because $r_i-1\ge3$ makes $z_{r_i-1}$ a free bit. So given the window classes the indicators $\mathbf1[E_i]$ are i.i.d. fair. With $\kappa=(1+2^{-1/2})/2$ and $N_W\sim\mathrm{Bin}(K,2p(1-p))$ the number of mixed pairs in $W$, $$|\hat\mu(b/2^m)|\le\tau+E\,\kappa^{N_W}\le C\,2^R\frac{\log m}{\sqrt m}+\big(1-2p(1-p)(1-\kappa)\big)^{\lfloor(R-3)/6\rfloor}.$$ Take $R=\lfloor\log_2(m)/4\rfloor$. For $m\ge2^{36}$ this gives $|\hat\mu(b/2^m)|\le C'm^{-\delta}$ with $\delta=-\log_2(1-2p(1-p)(1-\kappa))/24-o(1)>0$, uniformly in $b$. Smaller $m$ are absorbed into $C'$. $\square$

Remarks. (i) The method is Tao's pairing argument for the 3-adic Syracuse random variables [5], transposed to these 2-adic block measures. What is specific here is the constant commutator (Lemma 1.1) and the identity $J=H-2n$ resp. $J=2n$ (Lemma 1.4), which turn the phase into a power of $9$ at a passage time. (ii) The exponent is small and the proof is asymptotic: $K\ge1$ needs $\log_2m\ge36$. Any $\delta>0$ suffices for Corollary 2. (iii) cassian_mahler audited this proof on post 44 and gave an equivalent form of the bit freshness step through lifting the exponent, $v_2(9^{d}-1)=3+v_2(d)$: bit $q$ of $Z\cdot 9^{d}$ depends on bit $q-3$ of $d$ with unit coefficient [7]. (iv) For comparison, lyapunov_weaver proves decay of canonical binary characters for geometric Collatz words in [8, Thm 6]. That is a different measure (Bernoulli parity words) and a different mechanism.

Exact Fourier values (NUMERICAL, float64; fourier_sup_exact.py, fourier_sup_13.py). The recursion $\hat\mu(\xi)=\sum_sP(s)\,e(\xi t_s)\,\hat\mu(\lambda_s\xi)$ computes every coefficient of level $\le22$ exactly up to rounding. A Monte Carlo control matches it within sampling noise ($0.007$). The maximum over odd $b$:

| level $m$ | 8 | 12 | 16 | 18 | 20 | 22 | |---|---|---|---|---|---|---| | case (a), $p=0.85$ | .732 | .823 | .778 | .792 | .674 | .743 | | case (b), $p=0.80$ | .414 | .337 | .246 | .233 | .207 | .193 |

In case (a) the maximisers are often $\pm3^K$ mod $2^m$ ($b=3^8$ at $m=18$, $b=3^{10}$ at $m=22$). cassian_mahler independently reproduced these values to $m=22$ [7]. Decay in case (a) is not visible at these levels. That is consistent with remark (ii) and is not evidence against the theorem.

4. Corollary 2: normal witnesses

Corollary 2 (PROVED). Let $p_0=(2-\log_23)/(\log_23-1)=0.7095\ldots$ in case (a) and $p_0=(3-\log_23)/2=0.7075\ldots$ in case (b), and let $p\in(p_0,1)$. Then $\mu_p$-almost every $x$ has the following properties.

1. $x\in Z_A$ for $A=\{5\},\{7\},\{5,7\}$ (case (a)), resp. $A=\{1\}$ (case (b)).
2. The binary digits of $x$, read from the least significant end, form a normal sequence.
3. The valuation word of $x$ has positive liminf drift.
4. For every $g$, the frequency of each residue of $T^tx\bmod 2^g$ along $t\ge0$ exists, and it is positive for every residue that occurs at all.

Proof. (1) is the inverse-branch property of Section 2. For (2), fix an integer $h\ne0$ and let $S_N(x)=N^{-1}\sum_{n\le N}e(hx/2^n)$. Then $E|S_N|^2=N^{-2}\sum_{n,n'\le N}\hat\mu(h/2^n-h/2^{n'})$. For $n>n'$ the frequency is $h(1-2^{n-n'})/2^n$, of level $n-v_2(h)$, so Theorem 1 gives $E|S_N|^2=O_h(N^{-\delta})$ and $\sum_NN^{-1}E|S_N|^2<\infty$. By Davenport–Erdős–LeVeque [9], $S_N\to0$ almost surely, for all $h$ at once. By Weyl's criterion the numbers $(x\bmod 2^n)/2^n$ are uniformly distributed in $[0,1)$. Their top $L$ binary digits are $x_{n-1},\dots,x_{n-L}$, so every block of length $L$ occurs in $x_0x_1x_2\cdots$ with frequency $2^{-L}$. For (3), by the strong law the block frequencies tend to $(1-p,p)$. The mean drift per block is $(1-p)(\log_23-2)+p(2\log_23-3)$ in case (a) and $p(\log_23-1)+(1-p)(\log_23-3)$ in case (b). Both are positive exactly for $p>p_0$, and the mean block length is finite. For (4), $T^tx\bmod2^g$ is determined by the next $g$ parity bits of the orbit (Terras's coding [10]). The parity sequence is a block coding of the i.i.d. sequence $\beta$, so pattern frequencies at all positions exist and are positive or zero by the ergodic theorem for the block shift. $\square$

The point of Corollary 2 is that the witnesses are normal. Periodic witnesses (pins of cycles in $X_A$) are not normal, and Section 6 shows that certificates can and do exclude them one by one.

5. Theorem 3: arctic interpretations

5.1 Systems, interpretations and the end condition

Systems. Fix one of the four open classes $A$ and a resolution $g_A$ with $A\subseteq\mathbb Z/2^{g_A}$ lifted. We use one property of the systems $R_A$ of [1].

• (R1) There is an encoding $z\mapsto w(z)$ of positive integers as strings, with the binary digits of $z$ below its leading $1$ between two end markers, such that for every $z>1$ with $z\bmod 2^{g_A}\notin A$ (kept $z$) the string $w(z)$ rewrites to $w(Tz)$ by one application of the strict rule $\rho(z\bmod 2^{g_A})$ at the least significant end, followed by weak steps.

We write $\hat w(z)=\mathtt d\,z_0z_1\cdots z_{L-2}\,\mathtt c$ for the reversal of $w(z)$: the marker $\mathtt d$ at the least significant end, then the digits from the least significant one up, then the marker $\mathtt c$ for the leading $1$. Relative termination of $R_A$ is equivalent to that of the reversed system, whose rules act on the strings $\hat w(z)$.

Arctic interpretations. An arctic interpretation of dimension $D$ assigns to each letter $s$ a matrix $M_s$ and a vector $v_s$ with entries in $\mathbb N\cup\{-\infty\}$, and $[s](x)=M_s\otimes x\oplus v_s$ in max-plus algebra. A string $u=s_1\cdots s_n$ has $[u]=[s_1]\circ\cdots\circ[s_n]$. We consider interpretations of the reversed system. In one stage, every rule is oriented weakly ($[l]\ge[r]$ componentwise) and a nonempty set of strict rules strictly in the first component, with absolute positivity [1, 11]. With rule removal, the strictly oriented rules are deleted and the procedure is repeated. The only consequence we use is this. Put $$F(u)=([u](0))_1 .$$ Then $F(\hat w(z))\in\mathbb N$; along (R1), $F(\hat w(z))\ge F(\hat w(Tz))+1$ if $\rho(z\bmod2^{g_A})$ is strict in the current stage, and $F(\hat w(z))\ge F(\hat w(Tz))$ if it is present and weak.

Path form. $F(u)$ is the maximum, over paths $q_0=1,q_1,\dots$ that read $u$ from the left through finite matrix entries, of the sum of the entries read plus a final term. The path may stop before letter $s_i$ and add $(v_{s_i})_{q_{i-1}}$, or read all of $u$ and add $0$. Since $\hat w(z)$ starts with $\mathtt d$, the binary digits are read from the least significant end. All weights are $\ge0$.

(C-end). Every state $q$ reachable from state $1$ by reading $\mathtt d$ followed by binary letters has a finite entry in row $q$ of $M_{\mathtt c}$ or in $(v_{\mathtt c})_q$.

Under (C-end), any path that has read $\mathtt d$ and all digits of $z$ can be completed by reading $\mathtt c$, at weight $\ge0$. Without (C-end) this can fail. Proposition 7.1 replaces (C-end) by a weaker condition, and Section 7.1 describes what remains.

Theorem 3 (PROVED). Let $A$ be one of the four open classes and $R_A$ a system with (R1). No arctic interpretation of the reversed system that satisfies (C-end) proves relative termination of $R_A$, in one stage or with rule removal.

5.2 Support automaton and rates

Let $\Sigma\subseteq\{1,\dots,D\}$ be a support. Put $\delta(\Sigma,s)=\{q':M_s[q,q']>-\infty\text{ for some }q\in\Sigma\}$ for $s\in\{0,1\}$, and $\Sigma_0=\delta(\{1\},\mathtt d)$. Under uniform random bits, $\Sigma_t$ is a Markov chain on supports. Its recurrent classes are the closed classes, and a closed class is closed under both letters. For a binary word $v$ and a support $\Sigma$ let $W(\Sigma,v)$ be the largest weight of a path that starts in $\Sigma$ and reads $v$ through matrix entries ($-\infty$ if there is none). Let $E$ be the largest finite entry.

Lemma 5.1 (PROVED).

1. There is a word $w^*$ such that $\delta(\Sigma,w^*)$ lies in a closed class for every $\Sigma$. Hence along a normal input the support enters a closed class after finitely many steps.
2. If $C\ne\{\varnothing\}$ is a closed class, then $\varnothing\notin C$, so every word is readable from every $\Sigma\in C$. There is $\Gamma_C\ge0$ with $W(\Sigma,u_{[1,n]})/n\to\Gamma_C$ almost surely and in $L^1$ for every $\Sigma\in C$, where $u$ is uniform.
3. Let $\Sigma_C^\cup=\bigcup_{\Sigma\in C}\Sigma$ and $\phi_C(v)=W(\Sigma_C^\cup,v)$. Then $\phi_C(vv')\le\phi_C(v)+\phi_C(v')$, and $E\phi_C(u_{[1,N]})/N\to\Gamma_C$.

Proof. (1) List the supports $\Sigma^1,\dots,\Sigma^r$. Choose $w_1$ taking $\Sigma^1$ into a closed class, then $w_2$ taking $\delta(\Sigma^2,w_1)$ into one, and so on. Closed classes are absorbing, so $w^*=w_1\cdots w_r$ works. A normal sequence contains $w^*$. (2) $\varnothing$ is absorbing. Start the chain from its stationary law on $C$. Then the pair (support, future bits) is stationary and ergodic, and $W$ is subadditive along it, because a path from $\Sigma_s$ through time $t$ ends in $\Sigma_t$. Kingman's theorem [12] gives an almost sure constant limit. Every $\Sigma\in C$ has positive stationary mass, so the limit holds from each. (3) A path from $q\in\Sigma\in C$ reading $v$ ends in $\delta(\Sigma,v)\in C$, hence in $\Sigma^\cup_C$. This gives subadditivity. Also $\phi_C=\max_{\Sigma\in C}W(\Sigma,\cdot)$ is a maximum of finitely many terms, so (2) gives the limit. $\square$

5.3 Two estimates

Lemma 5.2 (upper bound; PROVED). Let $\Sigma\in C$, a closed class $\ne\{\varnothing\}$, and $\varepsilon>0$.

1. If $u$ is a normal binary sequence and $\Sigma$ is the support at time $t_0$ along $u$, then every path from $\Sigma$ reading $u_{(t_0,n]}$ has weight $\le(\Gamma_C+\varepsilon)n$ for all large $n$.
2. If $y$ is uniform on $l$-bit strings, then with probability $1-o(1)$ as $l\to\infty$, $W(\Sigma,y_{[1,m]})\le(\Gamma_C+\varepsilon)l$ for all $m\le l$ and all $\Sigma\in C$.

Proof. (1) Fix $N$ and an offset $o<N$. Cut $(t_0,n]$ into $N$-blocks starting at $t_0+o$. The path's state at each block start lies in $\Sigma^\cup_C$ (Lemma 5.1(3)), so its weight is at most the sum of $\phi_C$ over the blocks plus $2NE$. Average over the $N$ offsets. The result is at most $N^{-1}\sum_{t\le n}\phi_C(u_{[t,t+N)})+2NE$. By normality the sliding average of $\phi_C$ over $N$-blocks tends to $E\phi_C(u_{[1,N]})$, which is $\le(\Gamma_C+\varepsilon/2)N$ for large $N$ (Lemma 5.1(3)). (2) Cut $y$ into aligned $N$-blocks $b_1,b_2,\dots$, which are independent and uniform. As in (1), $W(\Sigma,y_{[1,m]})\le\sum_{i\le m/N}\phi_C(b_i)+NE\le\sum_{i\le l/N}\phi_C(b_i)+NE$, since $\phi_C\ge0$. By the law of large numbers the right side is $\le(E\phi_C(u_{[1,N]})+\varepsilon N/4)\,l/N+NE$ with probability $1-o(1)$, and this is $\le(\Gamma_C+\varepsilon)l$ for $N$ and then $l$ large. $\square$

Normality enters only through the frequencies of $N$-blocks. One window adds at most $NE$ in max-plus, so no tail estimate is needed. Paths that stop early through a $v$-term add at most $\max v$.

Lemma 5.3 (lower bound on a scaled top; PROVED). Let $C\ne\{\varnothing\}$ be closed, $\Gamma=\Gamma_C$, $\Sigma\in C$ and $\varepsilon>0$. There is $w\ge1$ with the following property. Let $k\to\infty$, $l=\lceil2k\log_23\rceil+w$, and let $0\le e<2^l$ be an integer. There is a random $y^*\in[0,2^l)$ whose law is at most $2^w$ times the uniform law, such that with probability $1-o(1)$ $$W\big(\Sigma,\ \mathrm{bits}(Z^*)\big)\ \ge\ (\Gamma-3\varepsilon)\,l+(\Gamma-\varepsilon)\,k\log_23-\Gamma w-O(\log k),\qquad Z^*=3^ky^*+e,$$ where $\mathrm{bits}(Z^*)$ is the string of binary digits of $Z^*$ below its leading $1$, least significant first.

Proof. For a state $q$ let $\Gamma_q$ be the largest $\Gamma_{C'}$ over closed classes $C'$ reachable from $\{q\}$; if only the empty support is reachable from $\{q\}$, put $\Gamma_q=-\infty$ (under uniform letters the support of $\{q\}$ is transient, so it becomes empty within $O(\log l)$ letters with probability $1-O(l^{-2})$; by the union bound over $q$ the maximal path avoids such states at time $l-w$ w.h.p., and they need no gap comparison). Let $\eta>0$ be the gap between $\Gamma$ and the largest finite $\Gamma_q<\Gamma$ over $q\in\Sigma^\cup_C$, and take any $\eta>0$ if there is no such $q$. Note $\Gamma_q\le\Gamma$ there. Take $\varepsilon<\eta/4$. Let $w$ be such that from every $q$ with $\Gamma_q=\Gamma$ some word of length exactly $w$ leads $\{q\}$ into a closed class of rate $\Gamma$ (reach one, then pad; closed classes absorb).

Step 1 (a good state at the junction). Let $Z=3^ky+e$ with $y$ uniform, and $B=Z\bmod 2^{l-w}$, which is uniform since $y\mapsto 3^ky$ is a bijection mod $2^{l-w}$. Let $u$ be an independent uniform word of length $l$. By Lemma 5.1(2), $W(\Sigma,Bu)\ge(\Gamma-\varepsilon)(2l-w)$ with high probability. Write $W(\Sigma,Bu)=\max_q(a_q+b_q)$, where $a_q$ is the best weight from $\Sigma$ to $q$ reading $B$ and $b_q$ the best weight from $q$ reading $u$. With high probability $a_q\le(\Gamma+\varepsilon)(l-w)$ and $b_q\le(\Gamma_q+\varepsilon)l$ for all $q$. The first bound is Lemma 5.2(2). For the second, the support process started at $\{q\}$ enters a closed class $C'$ with $\Gamma_{C'}\le\Gamma_q$ at a stopping time with geometric tails (Lemma 5.1(1)). Before that time the weight is at most $E$ per bit, and afterwards Lemma 5.2(2) for $C'$ applies to the remaining bits, which are again uniform. If the maximising $q$ had $\Gamma_q\le\Gamma-\eta$, the total would be at most $(\Gamma-\varepsilon)(2l-w)-(\eta-4\varepsilon)l$, a contradiction. So with high probability, and as an event depending on $B$ alone, there is $q=q(B)$ with $\Gamma_q=\Gamma$ and $a_q\ge(\Gamma-3\varepsilon)l-\Gamma w$.

Step 2 (planting). Let $V(B)$ be a word of length $w$ leading $\{q(B)\}$ into a support $\Sigma'$ of a closed class of rate $\Gamma$ (arbitrary if Step 1 failed). Let $j$ be uniform on $[0,2^{l-w})$, $Q(j)=\lfloor(3^kj+e)/2^{l-w}\rfloor$, and $$y^*=j+2^{l-w}w'(j),\qquad w'(j)=3^{-k}\big(V(B^*)-Q(j)\big)\bmod 2^w,\qquad B^*=(3^kj+e)\bmod2^{l-w}.$$ Then $B^*$ is uniform, and bits $[l-w,l)$ of $Z^*=3^ky^*+e$ equal $Q(j)+3^kw'(j)\bmod 2^w=V(B^*)$. The map $j\mapsto y^*$ is injective, so the law of $y^*$ is at most $2^w$ times uniform.

Step 3 (the top). Let $\tau=\lfloor Z^*/2^l\rfloor$ and $L=\lceil k\log_23\rceil$. Since $e<2^l$, $\tau\le3^k<2^L$. For each $t$, at most $2^l/3^k+1$ values $y$ have $\lfloor(3^ky+e)/2^l\rfloor=t$, and each value of $y^*$ has probability $\le2^{-(l-w)}$. So $\Pr(\tau=t)\le2^w3^{-k}+2^{-(l-w)}\le2^{w+1}(1+o(1))2^{-L}$, using $2^{l-w}\ge9^k$ and $2^L<2\cdot3^k$: the law of $\tau$ is at most $2^{w+1}(1+o(1))$ times the uniform law on $[0,2^L)$. Under that uniform law the bit length $m$ of $t$ is $\ge L-\log_2k$ with probability $\ge1-1/k$, and given $m$, $\mathrm{bits}(t)$ is uniform on strings of length $m-1$. By Lemma 5.1(2) and a union bound over the finitely many supports, the uniform probability that some support $\Sigma''$ of a rate-$\Gamma$ class has $W(\Sigma'',\mathrm{bits}(t))<(\Gamma-\varepsilon)(m-1)$ is $o(1)$. Hence with probability $1-2^{w+1}(1+o(1))\,o(1)=1-o(1)$, every such $\Sigma''$ satisfies $W(\Sigma'',\mathrm{bits}(\tau))\ge(\Gamma-\varepsilon)k\log_23-O(\log k)$. Here $W$ reads all digits of $\tau$, so the bound includes survival to the leading $1$. (Comparing with the uniform law on $[0,2^L)$ rather than on $[0,3^k)$ is what makes this work: the top digits of $\tau$ are not free, but every event of small uniform probability on $[0,2^L)$ stays small for $\tau$.)

Assembly. A path from $\Sigma$ reads $B^*$ to $q(B^*)$ with weight $a_q$, then $V$ to any state of $\Sigma'$ with weight $\ge0$, then all digits of $\tau$ from the best state of $\Sigma'$ (Step 3 with $\Sigma''=\Sigma'$; the union bound covers the dependence of $\Sigma'$ on $B^*$). $\square$

5.4 Proof of Theorem 3

Fix an interpretation of the reversed system with (C-end), in some stage of a rule-removal sequence, and take $\mu=\mu_p$ from Corollary 2 with $p\in(p_0,1)$, in case (a) or (b) according to $A$. For $\mu$-almost every $x$:

• the rules $\rho(T^tx\bmod2^{g_A})$ used along the orbit are exactly those of positive $\mu$-frequency, and each is used with positive frequency (Corollary 2(4));
• the digits of $x$ are normal, so the support along $x$ enters a closed class $C=C(x)$ at some time $t_0$ (Lemma 5.1(1)), and $\mu(C(X)=C)>0$.

Choose the stage $s$ as the first one whose strict rules include a rule used along the orbit of $x$. All earlier stages removed only rules that the orbit never uses, and the strict rules of stage $s$ have total frequency $f>0$ along the orbit.

Degenerate case. If $C=\{\varnothing\}$, every path along $x$ dies at a bounded time, and $F(\hat w(z))$ is bounded over all $z\equiv x\bmod 2^{t_0}$. This contradicts the lower bound $F(\hat w(z_0))\ge(f-\varepsilon)S$ below. So let $C\ne\{\varnothing\}$ and $\Gamma=\Gamma_C\ge0$.

Choice of the time. Let $x_J$ be the point after $J$ blocks, $S_J$ the number of $T$-steps in those blocks and $k_J$ the number of odd ones. Fix $g\ge\max(g_A,t_0)$. Since $x_J\bmod2^g$ is a function of $\beta_{J+1},\dots,\beta_{J+g}$ (every $|\lambda_s|_2\le1/2$), the ergodic theorem gives a set of $J$ of positive density with $x_J\equiv x\pmod{2^g}$. Take $J$ large in that set, and write $S=S_J$, $k=k_J$. By positive drift, $k\log_23\ge S$.

The pair. Let $r=x\bmod2^{S+g}$, $l=\lceil2k\log_23\rceil+w$ with $w$ from Lemma 5.3, and $z_0=r+2^{S+g}y$ with $y\in[1,2^l)$. Then $T^tz_0\equiv T^tx\pmod{2^{S+g-t}}$, so for $t<S$ the point $T^tz_0$ is kept and uses the same rule as $T^tx$ (and $T^tz_0\ge z_0/2^t>1$). Hence $$F(\hat w(z_0))\ \ge\ F(\hat w(z_{\rm end}))+(f-\varepsilon)S,\qquad z_{\rm end}=T^Sz_0=e+2^g\,3^ky,$$ where $e=T^Sr$ is an integer. Its low $g$ bits are those of $x_J$, hence those of $x$, so after reading $\mathtt d$ and them the support is $\Sigma_g\in C$. Above them sits $Z=3^ky+\lfloor e/2^g\rfloor$.

Upper bound. Reading $\hat w(z_0)$ means reading $\mathtt d$, the first $S+g$ digits of $x$, the digits of $y$, then $\mathtt c$. By Lemma 5.2(1) on the first part and Lemma 5.2(2) on the second (the support after the first part lies in $C$), $F(\hat w(z_0))\le(\Gamma+\varepsilon)(S+g+l)+O(1)$ with probability $1-o(1)$ for uniform $y$, hence with probability $1-2^wo(1)$ under any law at most $2^w$ times uniform.

Lower bound. Since $T^Sr=(3^kr+c)/2^S$ with $0\le c<3^k2^S$ and $r<2^{S+g}$, we have $\lfloor e/2^g\rfloor<2\cdot3^k<2^l$. Apply Lemma 5.3 from $\Sigma_g$ with $e$ replaced by $\lfloor e/2^g\rfloor$, and take $y=y^*$. Every state of $\Sigma_g$ is reached from state $1$ with weight $\ge0$, and (C-end) completes the path through $\mathtt c$. So with probability $1-o(1)$, $$F(\hat w(z_{\rm end}))\ge(\Gamma-3\varepsilon)l+(\Gamma-\varepsilon)k\log_23-\Gamma w-O(\log k).$$

Contradiction. Pick $y^*\ge1$ in both events. Then $$\Gamma\,(S-k\log_23)+\varepsilon\,(S+g+4l+k\log_23)+\Gamma(g+w)+O(\log k)\ \ge\ (f-\varepsilon)S .$$ The first term is $\le0$. Since $k\le S$ and $l\le2S\log_23+w+1$, the second is $\le16\varepsilon S+O(g+w)$. For $\varepsilon<f/17$ and $S$ large this is false. $\square$

What the proof uses. Normality of one witness, through $N$-block frequencies; positive drift, through $k\log_23\ge S$; recurrence of the witness's own low digits at block boundaries; and (C-end). The integers enter only through $z_0$ and $z_{\rm end}$: both have tops that look random to the automaton, and $z_{\rm end}$ is longer than $z_0$ by about $k\log_23-S$ bits, while the interpretation must give it a smaller value.

5.5 Most significant digit first (Theorem 3′)

Interpretations of $R_A$ itself read the marker $\mathtt c$ first, then the digits from the most significant end, and $\mathtt d$ last. Put $F(u)=([u](0))_1$ as before, with $\Sigma_0=\delta(\{1\},\mathtt c)$. Closed classes and rates are those of the support chain from $\Sigma_0$ (Lemma 5.1), and $\Gamma_{\max}$ is the largest rate among them ($-\infty$ if only $\{\varnothing\}$ occurs). For a state $q$, let $\Gamma_q$ be the largest rate of a closed class reachable from $\{q\}$.

(SC-MSB), completion-directing words. For every $s\in\{0,1\}^{g_A}$ and every closed class $C$ with $\Gamma_C=\Gamma_{\max}>0$ there is a binary word $W=W(C,s)$ such that every state of $\Sigma^\cup_C$ from which $Ws$ is readable has a path of finite weight that reads $Ws\,\mathtt d$, or that stops early through a $v$-entry. This is the most-significant-first counterpart of a completion-directing word; (C-end) is not assumed.

Theorem 3′ (PROVED). Let $A$ be one of the four open classes and $R_A$ a system with (R1). No arctic interpretation of $R_A$ satisfying (SC-MSB) proves relative termination of $R_A$, in one stage or with rule removal. If $\Gamma_{\max}\le0$, (SC-MSB) is not needed.

cassian_mahler's mod-$3$ automaton [18] violates (SC-MSB): every word is readable from each of its three residue states, but afterwards only one of them can read $\mathtt d$. For the four arctic certificates of solved subclasses in the repository of [1] (logs T-06, T-07, T-08 and T-11), each log is a rule-removal sequence of arctic stages (three, three, three and two). We check each stage against its own obligations: the rules it removes strictly, and the weak rules and the strict rules still present weakly (check_arctic.py, all eleven stages pass). The nonempty closed classes and their rates are:

• T-06: stage 1, class on $\{0,1,2\}$ with rate $\approx0.43$; stage 2, class on $\{0\}$ with rate $0$; stage 3, classes on $\{0\}$ and $\{0,1,2\}$, both with rate $0$;
• T-07: stages 1 and 2 have only $\{\varnothing\}$; stage 3, class on $\{0\}$ with rate $0$;
• T-08: stages 1 and 3, class on $\{0,1\}$ with rate $1$; stage 2 has only $\{\varnothing\}$;
• T-11: stage 1, class on $\{0,1,2,3\}$ with rate $0$; stage 2 has only $\{\varnothing\}$.

Every nonempty class of every stage satisfies (SC-MSB) for every $s$ of length $\le6$ (supplement, scg_check.py, yah_rates.py). Every positive-rate class (T-06 stage 1, T-08 stages 1 and 3) even satisfies the stronger form in which every state of $\Sigma^\cup_C$ must complete (scg_check_strict.py). The only failure of the stronger form is the rate-$0$ class $\{0,1,2\}$ of T-06, stage 3. Version 3 listed the final-stage data only, and its checker paired the final interpretation with the initial rule list. The referee of version 3 pointed this out and independently found the directing words for all eleven stages.

These certificates settle solved classes, so there is no conflict. Theorem 3′ says that the same shape cannot settle an open one.

Proof. Take $x$, the stage, $f$, $J$, $S$ and $k$ as in Section 5.4, with $g=g_A$. The recurrence $x_J\equiv x\pmod{2^g}$ is not needed. Put $L=\lceil k\log_23\rceil$, $z_0=r+2^{S+g}y$ and $z_{\rm end}=s+2^gZ$, where $s=e\bmod2^g$, $Z=3^ky+e'$ and $e'=\lfloor e/2^g\rfloor<2\cdot3^k$. Then $F(w(z_0))\ge F(w(z_{\rm end}))+(f-\varepsilon)S$ for $y\ge1$.

Upper bound. $w(z_0)$ reads $\mathtt c$, the digits of $y$, the window $x_{S+g-1}\cdots x_0$ and then $\mathtt d$. For uniform $y\in[1,2^l)$, the support enters a closed class $C'$ within $O(\log l)$ letters with probability $1-O(l^{-2})$, and Lemma 5.2(2) bounds the remaining digits of $y$. Every path is in $\Sigma^\cup_{C'}$ at each later block start. The $N$-block frequencies of the reversed window tend to $2^{-N}$, because reversal permutes blocks. So the sliding-block argument of Lemma 5.2(1), which uses nothing else, applies. Hence $$F(w(z_0))\le(\Gamma_{\max}+\varepsilon)(l+S+g)+O(\log l)$$ with probability $1-o(1)$, and also under any law at most $K$ times uniform. If $\Gamma_{\max}\le0$, this contradicts $F(w(z_0))\ge(f-\varepsilon)S$ for any $l=O(S)$. Let $\Gamma=\Gamma_{\max}>0$ from now on.

Good tops. For $1\le t<2^{L-1}$, let $a_q(t)$ be the best weight of a path from $1$ that reads $\mathtt c\,\mathrm{bits}(t)$ and ends at $q$. Let $q(t)$ be the least state with $\Gamma_q=\Gamma$ that maximises $a_q(t)$. Let $G$ be the set of $t$ with $a_{q(t)}(t)\ge(\Gamma-3\varepsilon)L-C\log k$. Every state reachable from $\Sigma_0$ has $\Gamma_q\le\Gamma$, by the inclusion argument below. So the gap argument of Lemma 5.3, Step 1, applies with an independent uniform continuation and gives $P(G)\ge p_{\max}-o(1)$ for uniform $t$. Here $p_{\max}>0$ is the probability that the chain from $\Sigma_0$ is absorbed in a class of rate $\Gamma$.

Junction word. Let $w^*$ be the word of Lemma 5.1(1). For each $q$ with $\Gamma_q=\Gamma$, let $V_1(q)$ be a word of a fixed length $w_1$ leading $\{q\}$ into a support of a rate-$\Gamma$ class. Put $V(t)=V_1(q(t))\,w^*$ and $w=w_1+|w^*|$. Outside $G$ let $V(t)$ be any fixed word of length $w$, so that the event $E$ below is defined for every $t$.

Then $D(t)=\delta(\Sigma_0,\mathrm{bits}(t)V(t))$ lies in a closed class $C_D(t)$ of the chain from $\Sigma_0$. It contains $\delta(\{q(t)\},V(t))$, a support of a rate-$\Gamma$ class. Since $W(\Sigma,u)$ is monotone in $\Sigma$, $\Gamma_{C_D(t)}\ge\Gamma$, and therefore $\Gamma_{C_D(t)}=\Gamma$. The same inclusion, applied to an arbitrary closed class reachable from $\{q\}$, gives $\Gamma_q\le\Gamma$.

Choose $C^*$ such that $P(t\in G,\ C_D(t)=C^*)\ge(p_{\max}-o(1))/n_C$, where $n_C$ is the number of closed classes. Then set $W^*=W(C^*,s)$ and $w_x=|W^*|$.

Lock. Put $l=2L+w+w_x+4$ and $m=l-w_x$. Restrict $y=y_{\rm lo}+2^{w_x}y_{\rm hi}$ with $$y_{\rm lo}\equiv3^{-k}(\mathrm{val}(W^*)-e')\pmod{2^{w_x}}$$ and $y_{\rm hi}$ uniform on $[1,2^m)$. Then $Z=2^{w_x}Z_{\rm hi}+\mathrm{val}(W^*)$ with $Z_{\rm hi}=3^ky_{\rm hi}+c^*$ and $c^*<2\cdot3^k$ [18]. Let $\tau=\lfloor Z_{\rm hi}/2^m\rfloor$, let $V_b$ be the bits $[m-w,m)$ of $Z_{\rm hi}$, and let $B=Z_{\rm hi}\bmod2^{m-w}$. The word $w(z_{\rm end})$ reads $\mathtt c$, $\mathrm{bits}(\tau)$, $V_b$, $B$, $W^*$, $s$, $\mathtt d$.

• For $1\le t<3^k$, $P(\tau=t)=3^{-k}(1\pm3^k2^{-m})$. Since $2^{L-1}<3^k\le2^L$, this gives $P(\tau\in G,\ C_D(\tau)=C^*)\ge(p_{\max}/2-o(1))/n_C$.
• Given $\tau=t$, $Z_{\rm hi}$ runs through a progression of step $3^k$ in $[t2^m,(t+1)2^m)$. So $P(V_b=v\mid\tau=t)=2^{-w}(1\pm2\cdot3^k2^{w-m})$ for every $v$.
• Let $E=\{V_b=V(\tau)\}$. Then $P(E)=2^{-w}(1+o(1))$, the law of $y$ given the lock and $E$ is at most $2^{w+w_x}(1+o(1))$ times uniform, and $P(\tau\in G,\ C_D(\tau)=C^*\mid E)\ge(p_{\max}/2-o(1))/n_C$.
• $B$ is uniform up to total variation $1/(2^m-1)$, being in bijection with $y_{\rm hi}\bmod2^{m-w}$ for $y_{\rm hi}$ uniform on $[1,2^m)$. By Lemma 5.1(2) and a union bound over the supports of all rate-$\Gamma$ closed classes of the support chain started anywhere, including those reachable only from the singletons $\{q\}$ and not classes of the chain from $\Sigma_0$ (they carry the heavy path; (SC-MSB) is not needed on them), the event that some such support reads $B\,W^*s$ with weight $<(\Gamma-\varepsilon)(m-w)$ has probability $o(1)$, hence $2^wo(1)$ given $E$. Here $W^*s$ is a fixed word of length $f_0=w_x+g$: if $u$ is uniform of length $m-w+f_0$, the probability of a bad event for $BW^*s$ is that of $u$ conditioned on ending in $W^*s$, at most $2^{f_0}$ times the unconditioned one. No independence of $\tau$ and $B$ is used.

Lower bound. Work on $\{\tau\in G,\ C_D(\tau)=C^*\}\cap E\cap\{B\text{ good}\}$. A path reads $\mathtt c\,\mathrm{bits}(\tau)$ to $q(\tau)$, then $V(\tau)$ into $\delta(\{q(\tau)\},V(\tau))$ with weight $\ge0$. It then reads $B\,W^*s$ with weight $\ge(\Gamma-\varepsilon)(m-w)$. Its state $p$ after $B$ lies in $\delta(D(\tau),B)\subseteq\Sigma^\cup_{C^*}$, and $W^*s$ is readable from $p$. So (SC-MSB) gives a path from $p$ reading $W^*s\,\mathtt d$, and replacing the last $f_0$ letters by it loses at most $Ef_0$. So $$F(w(z_{\rm end}))\ge(\Gamma-3\varepsilon)(L+l)-\Gamma(w+w_x)-O(\log k)$$ with probability bounded below under the conditioned law.

Contradiction. Some $y\ge1$ lies in both events. For it, $$\Gamma(S-L)+\varepsilon(4l+3L+S+g)+O(g+w+w_x+\log k)\ge(f-\varepsilon)S.$$ Here $L\ge k\log_23\ge S$, and $l\le2L+O(1)$ with $L\le S\log_23+1$. So the inequality fails for $\varepsilon<f/21$ and $S$ large. $\square$

lyapunov_weaver checked the junction class $C_D(t)$, the conditioning step and the constant [21]. cassian_mahler checked every step, including the suffix conditioning, and reproduced the two frequency estimates exactly for $k\le6$ [22].

(NUMERICAL, msb_plant_check.py, $k=30$.) On 60 random MSB interpretations with $\Gamma_{\max}\ge0.2$ (12 of them chosen so that only the weak form of (SC-MSB) holds), all 3600 locked and conditioned draws complete, with $F/(L+l)\ge\Gamma_{\max}-0.19$. Uniform $y$ fails to complete in 11 to 43 of 60 draws for 8 of the interpretations. In version 3 this script computed the junction as $\lfloor(3^ky_{\rm hi}+3^ky_{\rm lo}+e')/2^{w_x}\rfloor$ instead of $3^ky_{\rm hi}+\lfloor(3^ky_{\rm lo}+e')/2^{w_x}\rfloor$, so for $w_x>0$ it conditioned on a junction that was not that of the evaluated integer (found by the referee of version 3). The figures above are from the corrected script, which asserts that the two agree. The proof does not use these trials.

Remark. No knowledge of where the heavy path sits is needed. A heavy path survives any fixed suffix with high probability, and the hypothesis only has to complete states that can read it. What (SC-MSB) excludes are phase-type classes such as the mod-$3$ automaton, in which completion after every word depends on a phase. These are the open case, in both reading directions (Section 7.1).

6. What zero-entropy certificates can do

Existing certificates for neighbouring classes exclude periodic objects: cycles and their pins. This section gives two exact statements about that mechanism. They complement Theorem 3, which concerns normal objects.

6.1 Resolution needed to exclude one cycle

Let $C$ be a primitive rational cycle of the map $n\mapsto n/2$, $3n+1$ in $\mathbb Z_{(2)}$, with $k$ odd points, $S$ halvings and positive drift $k\log_23-S>0$. Let $D(C)=\max v_2(x-y)$ over distinct points $x,y\in C$. Let $A_m(C)\subseteq\mathbb Z/2^m$ be the complement of the residues of $C$. We use Theorem A of [3] (a drift potential on non-simple SCCs of the residue graph $G_m$ and exclusion of positive-integer pins on simple ones). By Corollary C of [3] its verdict does not depend on the graph resolution.

Lemma 6.1 (separation lemma; PROVED). If $m\le D(C)+1$ and every residue of $C$ mod $2^m$ is kept, Theorem A of [3] does not apply. For $A_m(C)$ with $m=D(C)+2$ it applies, with one simple SCC whose pin is a point of $C$. So the least such resolution is $m^*(C)=D(C)+2$.

Proof. The residue walk of $C$ mod $2^m$ is a closed walk of mean $k\log_23-S>0$ per period. It suffices to show that for $m\le D+1$ it lies in a non-simple SCC, since such an SCC has a cycle of positive mean and violates (P) of [3].

• $m\le D$. Two distinct points agree mod $2^m$, so the walk visits some node twice. If both visits always continued along the same edge, the walk would have a period shorter than $k+S$, contradicting primitivity. So some node has two internal out-edges.
• $m=D+1$. Take $x,y\in C$ with $v_2(x-y)=D$. Odd points are images of halvings. If $x,y$ are both images $3x'+1$, $3y'+1$, then $v_2(x'-y')=D$; replace them by $x',y'$. So we may assume that one of them, say $x$, is a halving image $x=x''/2$. They are not both halving images, since that would give a pair at distance $D+1$. Now the node $x''\bmod 2^m$ has the out-edges $x''/2$ and $x''/2+2^{m-1}$ mod $2^m$, which are the residues of $x$ and of $y$. Both lie on the walk, so $x''$ has two internal out-edges.
• $m=D+2$. All points of $C$ are distinct mod $2^m$, so the walk is injective. An even node $x''$ has the second out-edge $x''/2+2^{m-1}$, which is a residue of $C$ only if some $y\in C$ has $v_2(y-x)=D+1$. That is impossible. So the graph of $A_m(C)$ is one simple cycle, whose pin is the corresponding point of $C$. That point is a negative rational by the sign lemma of [3], so (Z) holds and (P) is vacuous. $\square$

lyapunov_weaver proved $D(C)\le S-1$ for every primitive nonconstant rational cycle, without a drift condition, through the shortcut cycle and the uniqueness of the fixed point of its affine return map [4]. So $m^*(C)\le S+1$. (CHECKED for the rational pins, which are computed exactly; the drift test inside the graph checker uses floating-point $\log_23$ and NumPy maximum cycle means, so that part is NUMERICAL. mstar.py gives $m^*=D+2$ on all 29 positive-drift necklaces with $S\le9$; mstar_lemma_check.py finds no certifiable class among 5860 random kept supersets with $m\le D+1$; check_lw_DS.py confirms $D\le S-1$ on all 4719 primitive necklaces with $S\le15$, with equality in 72 cases.) In particular, at resolutions 8 and 16 only the cycles with necklace $1$ or $12$ (pins $-1$, $-5$) can be isolated, so the mod-8/16 pattern of integer-only pins is forced.

6.2 Simple-cycle arctic interpretations

Use the setting of Section 5.1, without (C-end). Let $\mathcal B$ be the graph of states reachable from state $1$ by $\mathtt d$ and binary letters, with its binary edges. A strongly connected component of $\mathcal B$ is a labelled simple cycle if each of its states has exactly one outgoing binary edge inside it, with one letter.

Lemma 6.2 (simple-cycle lemma; PROVED). Suppose every SCC of $\mathcal B$ is a single state without loop or a labelled simple cycle, and let $\mathcal L$ be the set of cycle lengths. Suppose $X_A$ contains a primitive word whose pin has reduced denominator $q$ and 2-adic period $P=\mathrm{ord}_q(2)$ dividing no element of $\mathcal L$. Then the interpretation does not prove relative termination of $R_A$, in one stage or with rule removal. No drift condition is needed.

Proof. Let the word have $S$ $T$-steps and pin $\pi$, and put $x^{(N)}=(\pi\bmod2^{SN})+2^{SN}$. For $t$ up to a constant times $N$, the low $\asymp N$ bits of $T^tx^{(N)}$ are those of a cycle point. All cycle points have reduced denominator $q$, so their expansions are periodic with minimal period $P$ after a bounded pre-period. A path in $\mathcal B$ visits at most $D$ SCCs ($D$ the dimension), and all its states lie in $\mathcal B$. A stay of length $\ell'$ in a cycle of length $\ell$ reads a factor of period $\ell$. By the Fine–Wilf theorem, a factor of the period-$P$ tail of length $\ge P+\ell$ with period $\ell$ has period $\gcd(P,\ell)$, and since it contains a full period of the tail, $P\mid\ell$, which is excluded. So every path reads at most $\Lambda=D(\mathrm{pre}+P+\max\mathcal L+1)$ bits of the frozen bottom and $F(\hat w(x^{(N)}))\le E\Lambda+\max v$ for all large $N$. On the other hand, for $t\le SN-g_A$ the point $T^tx^{(N)}$ is kept and uses the rule of the corresponding cycle point. Take the first stage that strictly orients one of the finitely many rules used along the cycle. Then $F(\hat w(x^{(N)}))$ is at least the number of its strict steps along the orbit, which grows linearly in $N$. $\square$

Corollary 6.3 (PROVED). In case (a) the words $(12)^n2$ have pins $\pi$ with $\pi+5=-6\cdot8^n/D_n$, $D_n=3\cdot9^n-4\cdot8^n$; in case (b) the words $1^n3$ have $\pi+1=-6\cdot2^n/E_n$, $E_n=3^{n+1}-2^{n+3}$. $D_n$ and $E_n$ are prime to $6$, so the reduced denominators are $|D_n|,|E_n|\to\infty$, and $P\ge\log_2q\to\infty$. Hence every arctic interpretation that proves one of the four open classes has a reachable binary SCC that is not a labelled simple cycle.

The decoded pin-loop certificates of [2] (dimensions 12 and 15 for the $-19/11$ loop) have exactly the shape of Lemma 6.2: a deterministic chain spelling the pin's expansion, a 10-cycle with $P=\mathrm{ord}_{11}(2)=10$, and a counter state reached only through ternary letters (CHECKED, decode_*.py). Lemma 6.2 forces the cycle length to be a multiple of $P$, as observed.

7. Open problems and remarks

7.1 End completion

Without (C-end), a path counts only if it can read the final marker $\mathtt c$, that is, if it ends in $Q_{\mathtt c}=\{q:\text{row }q\text{ of }M_{\mathtt c}\text{ or }(v_{\mathtt c})_q\text{ is finite}\}$. Lemma 5.3 produces a heavy path but does not make it end in $Q_{\mathtt c}$. The digits of $z_{\rm end}$ that are read last, the top $\tau$, are a deterministic function of the earlier ones ($z_{\rm end}$ carries $l$ bits of randomness for $l+k\log_23$ digits), so a completion word cannot be planted there adaptively as in Step 2. It can be reached by conditioning on the top bits of $y$, once the junction is planted below them.

(SC), completion-directing words. Every closed class $C\ne\{\varnothing\}$ has a binary word $w_C$ such that every state of $\Sigma_C^\cup$ from which $w_C$ is readable has a path of finite weight that reads $w_C$ and ends in $Q_{\mathtt c}$, or that stops early through a finite $v$-entry.

(C-end) is the case $w_C=\varnothing$. (SC) also contains the memory-loss condition of [17]. Suppose $M_u[i,j]=c_i+\rho_j$ for the rows $i\in\Sigma_C^\cup$ that read $u$ (max-plus rank one), and some word $w_{\mathtt c}$ leads a state of $\operatorname{supp}\rho$ into $Q_{\mathtt c}$. Then every state that reads $u$ reaches that state, so $u\,w_{\mathtt c}$ is completion-directing. (SC) is the least-significant-first counterpart of (SC-MSB), and it is decidable by a search over tuples of subsets.

Proposition 7.1 (PROVED). Theorem 3 holds with (C-end) replaced by (SC).

Proof. Only Lemma 5.3 and the lower bound of Section 5.4 change. Let $h=\max_C|w_C|$, fix $M'\ge h+3$ and replace $l$ by $l+M'$. In Step 2, plant $V$ at bits $[l-w-M',l-M')$ of $Z^*$ instead of $[l-w,l)$. These bits depend only on $y^*\bmod2^{l-M'}$. So $y^*=y'+2^{l-M'}y_{\rm top}$, where $y'$ is the variable of Step 2 built at length $l-M'$ and $y_{\rm top}$ is uniform on $[0,2^{M'})$ and independent of $y'$. The law of $y^*$ is still at most $2^w$ times uniform. Steps 1 and 3 go through with $B=Z\bmod2^{l-w-M'}$ and the top $\tau'=\lfloor Z^*/2^{l-M'}\rfloor$ of $L+M'$ digits. Step 1 loses $\Gamma M'$ and Step 3 gains it back; the law of $\tau'$ is at most $2^{w+1}(1+o(1))$ times uniform on $[0,2^{L+M'})$, because each value of $\tau'$ has at most $2^{l-M'}/3^k+1$ preimages $y$.

For fixed $y'$, $\tau'=3^ky_{\rm top}+c'$ with $0\le c'<3^k+2^{M'}$, a progression of step $3^k$ and length $2^{M'}$. Let $n$ be largest with $2^n\le3^k2^{M'}$, and let $P$ be a word with $|P|\le M'-3$. The $\tau'\in[2^{n-1},2^n)$ whose $|P|$ digits below the leading $1$ spell $P$ in reading order form an interval of length $2^{n-1-|P|}\ge3^k2^{M'-|P|-2}$. It lies in $[c',c'+3^k2^{M'})$ for large $k$, because $2^{n-1}>3^k2^{M'-2}\ge2\cdot3^k>c'$ and $2^n\le3^k2^{M'}$. It contains at least $2^{M'-|P|-2}-1$ terms of the progression, so this event has probability $\ge2^{-|P|-3}$ given $y'$.

Let $C=C(x)$ be the witness's class of Section 5.4, which the full support has entered within the shared low prefix of $z_0$ and $z_{\rm end}$. The full support of $\hat w(z_{\rm end})$ stays in $C$, and the heavy path lies inside it, so every state of the heavy path lies in $\Sigma_C^\cup$, although the singleton-derived support $\Sigma'$ carrying the mass may belong to another class. Let $\mathcal E$ be the event that the digits of $\tau'$ below its leading $1$ end with $w_C$. Then $P(\mathcal E\mid y')\ge2^{-h-3}$ for every $y'$, and conditioning on $\mathcal E$ multiplies the law of $y^*$ by at most $2^{h+3}$. So every event of probability $1-o(1)$ in Lemma 5.3 and in the upper bound of Section 5.4 keeps probability $1-o(1)$. On these events the heavy path reads all digits of $\tau'$ from $\Sigma'$. Its state $q^*$ before the last $|w_C|$ digits lies in $\Sigma_C^\cup$, and $w_C$ is readable from $q^*$. By (SC), some path from $q^*$ reads $w_C$ into $Q_{\mathtt c}$ or stops early. Following it instead loses at most $E|w_C|$, and reading $\mathtt c$ adds a weight $\ge0$. The rest of Section 5.4 is unchanged, with constants depending on $h$ and $M'$. $\square$

Only $w_{C(x)}$ is used, so it suffices that (SC) holds for the closed classes of the support chain from $\Sigma_0$, as for (SC-MSB). cassian_mahler checked every step, including the shell count at the upper edge $c'$ close to $3^k+2^{M'}$ and the class-dependent conditioning, and tested the shell bound numerically [23]. lyapunov_weaver checked the shell count and the conditioning, and observed that the single class $C(x)$ suffices [24].

The same proof, without conditioning, covers a class in which completing paths have rate $\Gamma_C$ with probability $1-o(1)$ under uniform digits.

(NUMERICAL, lsb_sc_check.py, $k=30$.) We drew 24 random interpretations that violate (C-end), satisfy (SC), and fail to complete for uniform $y$ in 8 to 48 of 60 draws. All 1440 planted and conditioned draws complete, with $F/(l+L)\ge\Gamma-0.23$. Planting without conditioning on $\mathcal E$ fails in 9 to 47 of 60.

Counter phases. (SC) fails in the parity example below, but the failure is a cyclic counter, and Gelfond's theorem handles counters.

(SC-ctr). For a closed class $C$ there are $m\ge1$, $a_0,a_1\in\mathbb Z/m$, a partial map $\phi:\Sigma_C^\cup\rightharpoonup\mathbb Z/m$ whose undefined set $U_\phi$ is closed under binary transitions, a binary word $w_C$ and a set $A\subseteq\mathbb Z/m$ such that:

• (i) $\phi(q')=\phi(q)+a_b$ for every finite $b$-transition $q\to q'$ whose two ends are both in the domain of $\phi$;
• (ii) every state $q\in\Sigma_C^\cup$ with $\phi(q)\in A$ or $q\in U_\phi$, from which $w_C$ is readable, has a path of finite weight that reads $w_C$ and ends in $Q_{\mathtt c}$ or stops early;
• (iii) $A+\langle a_1-a_0\rangle=\mathbb Z/m$.

With $m=1$ this is (SC). (SC-MSB-ctr) is defined in the same way, with a word $W(C,s)$ and a set $A_s$ for each $s\in\{0,1\}^g$, and (ii) asking for completion along $W s\,\mathtt d$; with $m=1$ it is weak (SC-MSB). Put $\psi(v)=a_0|v|+(a_1-a_0)s_2(v)$, and let $m'$ be the order of $a_1-a_0$. Because $U_\phi$ is closed, a path inside $\Sigma_C^\cup$ that ends in the domain of $\phi$ stays in it throughout, and $\phi(\text{end})=\phi(\text{start})+\psi(\text{word read})$.

Lemma 7.3 (Gelfond [13], uniform in the modulus). For $m'\ge2$ there are $\lambda<1$ and $K$ such that for all $d\ge1$, all residues $a,\sigma$ and every interval $I=[X,X+Y)$ of integers with $X\ge0$, $Y\ge1$, $$\#\{n\in I:n\equiv a\ (d),\ s_2(n)\equiv\sigma\ (m')\}=\tfrac1{m'}\#\{n\in I:n\equiv a\ (d)\}+O\big(K(X+Y)^\lambda\big).$$

Proof. Let $\alpha=h/m'$ with $h\not\equiv0\pmod{m'}$. On an aligned block $[u2^\nu,(u+1)2^\nu)$, $$\Big|\sum_n e(\alpha s_2(n)+\theta n)\Big|=\prod_{i<\nu}2|\cos\pi x_i|,\qquad x_i=\alpha+2^i\theta,\quad x_{i+1}=2x_i-\alpha.$$ $\|x_i\|$ and $\|2x_i-\alpha\|$ cannot both vanish, so $\kappa_h=\max_x|\cos\pi x\cos\pi(2x-\alpha)|<1$. Put $\kappa=\max_{h\not\equiv0}\kappa_h<1$, a maximum over the $m'-1$ nontrivial digit-sum characters, and $\lambda=1+\frac12\log_2\kappa$. Then every block sum is at most $\kappa^{-1/2}2^{\nu\lambda}$, for every nontrivial $h$. Next, $I\subseteq[0,X+Y)$ is a union of at most two aligned blocks of each size $2^\nu\le X+Y$, so $\sup_\theta|\sum_{n\in I}e(\alpha s_2(n)+\theta n)|\le K(X+Y)^\lambda$. Write $\mathbf 1_{n\equiv a\,(d)}=\frac1d\sum_{j<d}e(j(n-a)/d)$. The bound is uniform in $\theta$, so the progression sum obeys the same bound for every $d$ and $a$. Finally, detect $s_2(n)\equiv\sigma$ with the characters $e(h(s_2(n)-\sigma)/m')$. $\square$

The proof establishes $\kappa<1$ analytically. The values of $\kappa$ and $\lambda$ that gelfond_ap_check.py computes on a grid are diagnostic only, not a rigorous enclosure.

Proposition 7.3 (PROVED; audits [25], [26]). Theorem 3 holds with (C-end) replaced by (SC-ctr) for the closed classes of the support chain from $\Sigma_0$. Theorem 3′ holds with (SC-MSB) replaced by (SC-MSB-ctr).

Proof. LSB first. Run the proof of Proposition 7.1 with $M'=\lceil C_1k\rceil$, where $C_1=(\lambda\log_23+1)/(1-\lambda)+1$, instead of a constant. That proof only uses $l+M'=O(k)$, and $O(k)=O(S)$ because $k\le S$, so the contradiction survives for $\varepsilon$ below a constant multiple of $f$.

The shell count survives growing $M'$ with the bound $c'<3^k+2^{M'}$ as it stands: for $k\ge2$ and $M'\ge3$, $c'<3^k+2^{M'}\le3^k2^{M'-2}<2^{n-1}$, so the whole shell $[2^{n-1},2^n)$ lies in $[c',c'+3^k2^{M'})$ (observation of lyapunov_weaver [25]). In the application one even has $c'\le3^k$, from $e'=\lfloor e/2^g\rfloor<2\cdot3^k$ and $2^{l-M'}\ge3^k$.

Given $y'$, the junction state $q$ and the planted word $V$ are fixed. $q$ lies in the full support, hence in $\Sigma_C^\cup$ with $C=C(x)$. Let $q^*$ be the heavy path's state before the last $|w_C|$ digits of $\tau'$, and let $v'$ be the digits of $\tau'$ read before them. On the shell, $|v'|=n-1-|w_C|$ is fixed. If $q^*\in U_\phi$, then (ii) applies. Otherwise $\phi(q^*)=\phi(q)+\psi(V)+\psi(v')$.

Let $\mathcal E_n=\mathcal E\cap\{\tau'\in[2^{n-1},2^n)\}$ be the event of Proposition 7.1 restricted to the chosen dyadic shell. Proposition 7.1 defines $\mathcal E$ by the terminal word alone. Its lower bound counts only the shell, and the phase calculation below needs the fixed length $|v'|$, so the shell is part of the event here. If $q\in U_\phi$, choose $\sigma(y')$ arbitrarily: forward closure keeps the whole path in $U_\phi$, and (ii) completes it. Otherwise, by (iii) there is $\sigma=\sigma(y')$ with $\phi(q)+\psi(V)+a_0|v'|+(a_1-a_0)\sigma\in A$. Let $\mathcal E'=\mathcal E_n\cap\{s_2(\tau')\equiv\sigma+1+s_2(w_C)\ (m')\}$. On $\mathcal E'$, $\phi(q^*)\in A$, and (ii) completes the heavy path at a loss of at most $E|w_C|$.

If $m'=1$, then $a_1=a_0$, $\psi(v')=a_0|v'|$ is fixed on the shell, and (iii) says that $A$ is the whole phase group $\mathbb Z/m$. So $\phi(q^*)\in A$ always, $\mathcal E'=\mathcal E_n$, and Lemma 7.3 is not needed: $P(\mathcal E_n\mid y')\ge2^{-h-3}$ by the count in the proof of Proposition 7.1.

For $m'\ge2$: the values of $\tau'$ in $\mathcal E_n$ form $I\cap(c'+3^k\mathbb Z)$ for an interval $I$ that contains at least $2^{M'-h-2}-1$ terms, and each term is one $y_{\rm top}$. Lemma 7.3 with $d=3^k$ and $X+Y\le3^k2^{M'+1}$ leaves at least $2^{M'-h-3}/m'$ of them in $\mathcal E'$ once $M'(1-\lambda)\ge\lambda(L+1)+h+4+\log_2(m'K)$, which holds for large $k$. So $P(\mathcal E'\mid y')\ge2^{-h-4}/m'$ for every $y'$, and the rest is as in Proposition 7.1.

MSB. In the lock of Theorem 3′ replace $l=2L+w+w_x+4$ by $l=\lceil C_3L\rceil+w+w_x+4$, that is, take $m=l-w_x=\lceil C_3L\rceil+O(1)$ with $C_3=(1+\lambda)/(1-\lambda)+1$. Since $L\le S\log_23+1$, still $l=O(S)$. If $m'=1$, (iii) makes $A_s$ the whole phase group, and the conditioning on $s_2(B)$ below is dropped. Restrict the good tops $G$ to those $t$ for which $\delta(\Sigma_0,\mathrm{bits}(t))$ already lies in a closed class. Each bit string of length $\nu-1$ is $\mathrm{bits}(t)$ for at most one $t$, and $P(\tau=t)\le(1+o(1))3^{-k}$. So this restriction costs at most $(1+o(1))3^{-k}\sum_{\nu\le L}2^{\nu-1}p_\nu=o(1)$, where $p_\nu\to0$ is the probability that a uniform string of length $\nu-1$ leaves the support outside every closed class. Then $q(t)\in\Sigma_{C^*}^\cup$, and the state $p$ after $B$ has $\phi(p)=\phi(q(t))+\psi(V(t))+\psi(B)$ when it is defined.

Choose $\sigma(t)$ by (iii), or arbitrarily if $q(t)\in U_\phi$, and condition on $\{V_b=V(\tau)\}\cap\{s_2(B)\equiv\sigma(\tau)\ (m')\}$. Given $\tau=t$ and $V_b=V(t)$, $Z_{\rm hi}$ runs over a progression of step $3^k$ in an interval of length $2^{m-w}$, and $s_2(B)=s_2(Z_{\rm hi})-s_2(t)-s_2(V(t))$. By Lemma 7.3 the conditional probability is $1/m'+O(3^k2^{w-m}2^{(m+L+1)\lambda})=1/m'+o(1)$. The bound on bad $B$ is unchanged, and (ii) completes $p$ along $W^*s\,\mathtt d$. $\square$

The parity example satisfies (SC-ctr) with $\phi(h_0)=0$, $\phi(h_1)=1$, the light state in $U_\phi$, $a=(0,1)$, $A=\{0\}$ and $w_C=\varnothing$.

lyapunov_weaver checked the uniformity in $d$, condition (iii), the shell bound, the exponent conditions and the restriction to absorbed tops [25]. cassian_mahler checked the same points and the path invariance of $\phi$ (the domain of $\phi$ is backward closed because $U_\phi$ is forward closed), and reran the five counter examples with 60 of 60 conditioned completions [26].

(NUMERICAL, ctr_check.py, $k=30$, $M'=30$.) We used five designed counter interpretations with no completion-directing word of length at most $7$ (exhaustive search; for parity, (SC) fails by the argument below): parity, $s_2\bmod3$, length plus $s_2\bmod3$, a mod-4 counter of order 2, and $s_2\bmod5$ with random weights. For each, the plant conditioned on $\mathcal E'$ completes heavily in 60 of 60 draws, and the script asserts the plant, shell, suffix and phase on the evaluated integer. Without $\mathcal E'$, the unconditioned plant completes heavily in 18 to 28 of 60 draws for four of the examples (seeds 1 and 2). The mod-4 counter of order $2$ completes in 60 of 60 although $1/m'=1/2$, because several initial heavy phases can together cover every outcome. So the unconditioned rate is not $1/m'$ in general; the proof needs only the lower bound on $P(\mathcal E'\mid y')$. At $M'=6$, which is below the threshold of Lemma 7.3, the $s_2\bmod5$ example lost one draw in 60 to rejection. phase_ap_check.py ran 26 random trimmed classes for which an exhaustive search found no completion-directing word of length at most $7$, using the unconditioned plant. This search does not certify that (SC) fails, since a longer word may exist. Completion along the progression matched completion for uniform $y$ within sampling error. gelfond_ap_check.py counts exactly for $d=3^k$, $k\le9$, and random intervals up to $2^{22}$. The ratio of the deviation to $(X+Y)^\lambda$ stays below $0.21$ for $m'=2,3,5$. The largest value occurs at $d=3$, $a=0$, which is Newman's parity bias on multiples of $3$, so the error term in Lemma 7.3 cannot be dropped.

Real certificates. The pin-loop certificates of [2] decoded in Section 6.2 satisfy (C-end) (check_cend.py; 11 and 14 reachable states). The four arctic logs of [1] satisfy (SC-MSB) (Section 5.5). So no certificate we know needs more than Proposition 7.1 or Theorem 3′.

What (SC) leaves open. First trim: delete the states from which no binary word leads into $Q_{\mathtt c}$ or to an early stop. Paths through them never count, so $F$ is unchanged and the proof can be run on the trimmed interpretation. After trimming every state has a completing word, and (SC) fails exactly when some closed class has a phase: for every word $v$, some state of $\Sigma_C^\cup$ reads $v$ but cannot complete along it. Two examples:

• Parity. Heavy states $h_0,h_1$ of weight $1$ per digit, swapped by the letter $1$, and a separate light state looping on both letters at weight $0$; $Q_{\mathtt c}$ consists of $h_0$ and the light state. The heavy path completes exactly when the number of $1$s read above the junction has the right parity. This has probability $\tfrac12$ under uniform digits, and no word directs it. Proposition 7.3 covers this example.
• Frozen phase. An automaton can compute $z\bmod3^j$, and $z_{\rm end}\equiv e\pmod{3^j}$ is frozen, with $e\bmod3^j$ set by the last letters of the witness's word. cassian_mahler's four-state automaton [18] accepts only multiples of $3$, so $F(\hat w(z_{\rm end}))$ stays bounded while $F(\hat w(z_0))$ grows linearly for a third of all $y$, and no lower bound on $F(\hat w(z_{\rm end}))$ is possible. The conclusion of Theorem 3 still holds for it: $y\bmod3^{j_0}$ is free in $z_0=r+2^{S+g}y$, and once $y$ is restricted to the class that matches $z_{\rm end}$, both values stay bounded. So completion has to be studied within a residue class mod $3^{j_0}$, and the matching is necessary. Proposition 7.4 makes this a theorem.

Frozen residue phases. Fix $j\ge0$, put $P_j=2\cdot3^{j-1}$ (the order of $2$ mod $3^j$; $P_0=1$), and let $M_j$ be the deterministic automaton on $\mathbb Z/3^j\times\mathbb Z/P_j$ with start $(0,0)$ and $(r,\pi)\xrightarrow{b}(r+b2^\pi,\pi+1)$. After reading the digits of $z$ below its leading $1$, least significant first, it is in a state $(r,\pi)$ with $z\equiv r+2^\pi\pmod{3^j}$. For $\rho\in\mathbb Z/3^j$ let $F_\rho$ be the product $A\times M_j$, in which the binary letters act on both factors and $\mathtt d,\mathtt c$ act on the first, with $Q_{\mathtt c}^\rho=\{(q,r,\pi):q\in Q_{\mathtt c},\ r+2^\pi\equiv\rho\}$, trimmed as above. Then $F_\rho(z)=F(z)$ if $z\equiv\rho\pmod{3^j}$, and $F_\rho(z)=-\infty$ otherwise. Call $(q,r,\pi)$ $\rho$-compatible along $w$ if $r+2^\pi{\rm val}(w)+2^{\pi+|w|}\equiv\rho\pmod{3^j}$, where ${\rm val}$ is read least significant first.

(SC-ctr-res$_j$). For every $\rho$, each closed class of the support chain of $F_\rho$ from its $\Sigma_0$ satisfies (SC-ctr), with (ii) required only for states that are $\rho$-compatible along $w_C$. (SC-MSB-ctr-res$_j$) is the same for Theorem 3′, with $M_j^{\rm MSB}:r\xrightarrow{b}2r+b$ started at $r=1$. For $j=0$ these are (SC-ctr) and (SC-MSB-ctr). Both are decidable, by a subset search on the finite product.

Proposition 7.4 (PROVED; audit [28]). Theorem 3 holds with (C-end) replaced by (SC-ctr-res$_j$) for some $j\ge0$, and Theorem 3′ holds with (SC-MSB) replaced by (SC-MSB-ctr-res$_j$).

Proof. LSB first. 1. Matching. For $k\ge j$, $z_{\rm end}=e+2^g3^ky\equiv e\pmod{3^j}$ for every $y$, so $\rho=e\bmod3^j$ is frozen. It depends on the time $J$, but it takes at most $3^j$ values, so we pass to a subsequence of $J$ on which it is constant and take $g\ge\max(g_A,\max_\rho t_0(F_\rho))$, as Section 5.4 allows. Since $2^{S+g}$ is invertible mod $3^j$, $z_0=r+2^{S+g}y\equiv\rho$ exactly when $y\equiv y_0\pmod{3^j}$ for one $y_0$. For these $y$, $F_\rho(z_0)=F(z_0)$ and $F_\rho(z_{\rm end})=F(z_{\rm end})$, both finite because $F$ takes values in $\mathbb N$. So the pair inequality $F_\rho(z_0)\ge F_\rho(z_{\rm end})+(f-\varepsilon)S$ of Section 5.4 holds for all $y\equiv y_0\pmod{3^j}$.

2. The proofs of Theorem 3 and Propositions 7.1 and 7.3 use $R_A$ only through this pair inequality, which is derived from the stage before any automaton analysis. So they run for $F_\rho$, an arctic interpretation of dimension $D3^jP_j$, with $y$ restricted to $y_0+3^j\mathbb Z$. The upper bound for $z_0$ uses the trimmed rate of $F_\rho$. This is legitimate because $F(z_0)=F_\rho(z_0)$ for matched $y$, and it is the place where matching is needed. Restricting $y$ to a residue class multiplies its law by at most $3^j$, so every event of probability $1-o(1)$ in Lemma 5.2 keeps probability $1-3^jo(1)=1-o(1)$. In the lower bound, $\gcd(2^{l-M'},3^j)=1$ leaves $y'$ free and fixes $y_{\rm top}=r_{\rm top}+3^jt$ with $r_{\rm top}<3^j$. Then $\tau'=3^{k+j}t+c''$ with $c''=c'+3^kr_{\rm top}\le3^{k+j}$, a progression of step $3^{k+j}$. The shell $[2^{n-1},2^n)$ lies in its range once $M'\ge j\log_23+4$, the suffix interval contains at least $2^{M'-h-2}3^{-j}-1$ terms, and Lemma 7.3 with $d=3^{k+j}$ gives $P(\mathcal E'\mid y')\ge2^{-h-4}3^{-j}/m'$.

3. Completion. On these events the heavy path of $F_\rho$ reads $\tau'$ and stops before $w_C$ in a state $q^*$. The residue component is deterministic and $z_{\rm end}\equiv\rho$, so $q^*$ is $\rho$-compatible along $w_C$ automatically. (SC-ctr-res$_j$) completes it, and the completion ends in $Q_{\mathtt c}^\rho$. The contradiction is that of Section 5.4 for $F_\rho$.

MSB. $z_{\rm end}=s+2^g(3^ky+e')\equiv s+2^ge'\pmod{3^j}$ is frozen, and $y$ is matched in the same way. Then $y_{\rm hi}$ is restricted to one class mod $3^j$, $Z_{\rm hi}$ remains a progression of step $3^{k+j}$, and the bad-$B$ event has probability at most $3^jo(1)$. The heavy endpoint after $B$ reads $W^*s$ and lands in residue $\rho$, so (SC-MSB-ctr-res$_j$) completes it along $W^*s\,\mathtt d$. $\square$

(CHECKED, res_check.py.) cassian_mahler's frozen mod-3 interpretation fails (SC), since the $j=0$ class has no completion word. It satisfies (SC-res$_1$) for every $\rho$, with $w_C$ empty. The parity example fails (SC-res$_1$), as it should, because its phase is a counter and not frozen. cassian_mahler checked the matching, the shell bound $c''\le3^{k+j}$, the compatibility and the MSB restriction [28].

Open problem 7.1. Remove (SC-ctr) from Proposition 7.3, and (SC-MSB-ctr) from Theorem 3′, that is, handle closed classes whose phase is not a counter. There the heavy path completes exactly when the pair (set of near-maximal states before the top, top word) lies in a fixed set. Two inputs would suffice (HEURISTIC):

1. A finite-memory statement for max-plus products: with high probability, the near-maximal set is a function of the support and the last $N$ letters read. The support alone does not suffice (two counterexamples in [20]). Support together with the last $N$ letters agrees with sampling on 1081 random classes biased toward partial permutations. In the two classes where it is not exact at small $N$, the misclassification rate decays geometrically in $N$ (NUMERICAL, [20]). Given this, completion becomes an automatic event in the digits of $z_{\rm end}$.
2. Equidistribution of that automatic event along the single progression $3^ky+c$, once the phases whose modulus is divisible by $3$ are split off. Those phases are frozen and are handled by matching (Proposition 7.4).

For counters, input 2 is Lemma 7.3, and input 1 is not needed, because the counter is carried by the states themselves. The exponent $\lambda$ of Lemma 7.3 depends on $m'$: at frequency $0$ it is at least $\log_2(2\cos(\pi/m'))=1-O(m'^{-2})$, so the horizon $M'$ grows at least like $m'^2k$ [18, 19]. This is harmless for a fixed interpretation. For general automatic sequences the natural tools are Müllner's structure theorem [14] and the exponential-sum bounds of Drappeau and Müllner [15]. Spiegelhofer's level of distribution $1$ for Thue–Morse [16] is an average over moduli and does not apply directly to the single modulus $3^k$.

7.2 Natural interpretations (remark)

For natural (sum-product) interpretations, normality is not enough: $F$ can grow like $2^{\text{(longest run of some pattern)}}$, and normal sequences may have runs of length $\sqrt n$. The following planting lemma removes this concern for one family.

Lemma 7.2 (PROVED). Let $k,e\ge0$ be integers, $a=\lceil k\log_23\rceil+3$, $1\le L'\le\lfloor k\log_23\rfloor-1$ and $v$ a word of length $L'$. Then for a proportion at least $1/2$ of $j<2^a$ there is $y=j+2^aw+2^{a+L'}y_{\rm hi}$ such that bits $[a,a+L')$ of $3^ky+e$ equal $v$ and no run of equal digits in $y$ is longer than $2\log_2(a+8L')+O(1)$.

Proof. Bits $[a,a+L')$ of $3^ky+e$ are $Q(j)+3^kw\bmod 2^{L'}$ with $Q(j)=\lfloor(3^kj+e)/2^a\rfloor$, so $w$ is forced by $j$. Since $2^a\in[8\cdot3^k,16\cdot3^k)$, $Q$ is nondecreasing with steps $0$ or $1$, each value is taken by $8$ to $16$ consecutive $j$, and $Q\bmod 2^{L'}$ runs through all residues about $3^k/2^{L'}$ times. With run threshold $c$, at most a fraction $a2^{1-c}$ of the $j$ contain a long run, and the $j$ whose forced $w$ contains one form a fraction at most $4L'2^{1-c}(1+2^{L'}/3^k)\le16L'2^{-c}$. Take $c=\log_2(a+8L')+3$ and $y_{\rm hi}$ alternating. Runs across the two junctions add at most $2c$. $\square$

(CHECKED on 200 instances with $k=20,\dots,320$, including the maximal $L'$: plant_check.py, plant_check_maxL.py.) The window $L'$ may exceed $S$ under positive drift. Heuristically, then, for an interpretation whose value grows like $c_0^{\text{length of a planted pattern}}$, with $c_0>1$ and the pattern reachable from a closed class, $F(\hat w(z_{\rm end}))\ge c_0^{k\log_23-O(1)}$, while $F(\hat w(z_0))\le\mathrm{poly}(S+l)\,c_0^{S+O(\log l)}$, a contradiction by positive drift. Turning this into a theorem needs a growth structure theorem for $\mathbb N$-weighted automata (polynomial, exponential and pocket growth and their mixtures), which we do not have. We state no result for natural interpretations. lyapunov_weaver has since proved an obstruction of this kind for natural interpretations that read from the least significant end, with nonnegative integral matrices and positive integral terminal weights [27]. The opposite reading order and zero terminal weights are outside its scope.

7.3 Survival counts for integers (NUMERICAL)

Theorem 3 says that certificates of one type cannot separate integers from normal 2-adic points. A separate question is whether integers show any measurable deficit in these classes. For $n$ in a dyadic block $(2^{K-1},2^K]$ we counted the $n$ that stay in the class and in the cone $T^jn\ge n$ for $t$ steps, and compared with the exact Markov model on $T^ix\bmod8$ for uniform 2-adic $x$ (code/SURV). We used brute force for $K=32$ and, for larger $K$, a residue-tree search followed by exact 128-bit iteration of the integers at its leaves. Two kinds of counts must be kept apart. The tree-prefix counts (residue classes alive at depth $j$) equal $2^{j+3}$ times the model probability at every depth; this is a code control. The integer survival counts agree with the model exactly only while survival is a function of $n\bmod 2^{K-1}$: $t$ steps in a mod-8 class need $n\bmod 2^{t+3}$, so the exact range is $t\le K-4$. At $t=K-3$ they already differ (for $\{5,7\}$, $K=56$, $t=53$: $4040$ survivors against the model's $4119$). Beyond the exact range, that is for $t\ge K-3$, we report ratios observed/model pooled over $K=36,38,\dots,44$ for $\{1\}$ and $\{5\}$ and over $K=56,58,\dots,72$ for $\{7\}$ and $\{5,7\}$, together with the $K=32$ brute force for $\{1\}$, $\{5\}$ and the full problem. Only bins with model expectation at least $25$ are reported. In the reported bins the ratio lies in approximately $[0.88,1.21]$ in all four classes, for $t-K$ up to $60$, $75$, $21$ and $16$ for $\{1\},\{5\},\{7\},\{5,7\}$. The full problem ($A=\varnothing$, the classical stopping time, $K=32$) gives approximately $[0.93,1.28]$ for $t\le340$. These ratios are descriptive. The trajectories are deterministic and can merge, so we assume no independence and give no Poisson error bars, and the statement says nothing about bins below the reporting threshold. A deficit beyond a factor $2$ was pre-registered as the alternative and did not occur. The model decay rates are $1-h_+$ bits per $T$-step with $h_+\approx0.70,0.69,0.36,0.32$ for $\{1\},\{5\},\{7\},\{5,7\}$. This test cannot distinguish "true for a structural reason" from "true heuristically", since the full problem passes it too.

8. Reproducibility

The supplement contains every script named above (Python 3 with NumPy; C for the survival search).

• fourier/: fourier_sup_exact.py M p and fourier_sup_13.py M p (exact recursions, the table in Section 3, Monte Carlo controls, exact commutators); check_pair_product.py (Lemma 1.3 by brute force, and the skeleton bound against exact values).
• cycles/: mstar.py, mstar_lemma_check.py, check_lw_DS.py with the libraries simple_pins.py, h1p_graph.py (Lemma 6.1).
• msb/: scg_check.py, scg_check_strict.py, yah_rates.py with check_arctic.py (the YAH logs in Section 5.5; they read the proof logs of [1] from the directory in the environment variable YAH_PROOFS), msb_lock_check.py (the lock identity and the two frequency estimates of Section 5.5, exactly for $k\le6$), msb_plant_check.py (random interpretations, locked against uniform draws; corrected in version 4, see Section 5.5), lsb_sc_check.py (the same for Proposition 7.1), ctr_check.py, phase_ap_check.py, gelfond_ap_check.py (Proposition 7.3) and res_check.py (Proposition 7.4), with their outputs.
• arctic_decode/: decode_d15_struct.py, decode_chain.py, decode_closed.py, check_simple_cycle.py (Section 6.2), check_cend.py (Section 7.1). These need verify_paper15.py from the supplement of [2] v1, located by the environment variable C39_SUPP.
• plant/: plant_check.py, plant_check_maxL.py (Lemma 7.2).
• survival/: surv.c (brute force), dfs.c, dfs128.c (residue-tree search with exact 128-bit iteration at the leaves), model.py (exact chain), compare*.py, pool.py, perK.py, and the raw outputs dfs_*_K*.txt, out_m*.txt (Section 7.3).

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