Proof. Choose a smooth cutoff \(\chi_i\) which is one on the \(\eta\) enlargement of \(K_i(t)\) and supported in its \(2\eta\) enlargement, using products of the effective switches. Put \(y=x-c_i\) and \(L_i=\dot D_iD_i^{-1}\); then \(\operatorname{tr}L_i=0\). Take the curl of \[ \chi_i\{\tfrac12\dot c_i\times y+\tfrac13(L_i y)\times y\}. \tag{6}\] The identity \(\nabla\times(G\times y)=2G-y\operatorname{div}G+(y\cdot\nabla)G\) shows that this curl equals \(\dot c_i+L_i y\) on the plateau. Disjoint supports permit summation over \(i\). The path \(c_i(t)+D_i(t)D_i(t_0)^{-1}(x_0-c_i(t_0))\) has exactly that derivative and remains in its prescribed box. ODE uniqueness proves the formula. On the compact time interval the normalized matrix norm has a finite bound \(M\), computable under the effective-data assumptions; an initial perturbation less than \(\eta/(2M)\) remains on the plateau. This proves the neighborhood claim, also for zero-width initial boxes. The potential is zero near time endpoints because the centers and scales are constant there. Its support is bounded. Within a torus chart extend the potentials by zero; periodic curls have zero integral. ◻

Proof. The residual identity and Lemma 2 give existence, uniqueness and the material flow. In the whole-space pressure refinement use case (ii): the reference field has compact support and smooth bounded mixed derivatives, hence belongs to \(CH^2\cap C^1L^2\) with bounded velocity and gradient; the competing velocity and pointwise \(H^1\) pressure are exactly that case’s hypotheses. No pressure continuity in time is added. Integrating the residual on a torus uses \(\int U=0\), \(\int\Delta U=0\) and \(\int(U\cdot\nabla)U=\int\operatorname{div}(U\otimes U)=0\).

Each derivative of \(e^{-1/s}\) for \(s>0\) is a polynomial in \(1/s\) times \(e^{-1/s}\). The estimate \(v^m e^{-v}\le(m+k)!v^{-k}\) supplies effective endpoint error bounds, and the denominator defining \(\sigma\) is at least \(e^{-2}\). Under those effective-data assumptions all positive scales have effective positive lower bounds. Finite compositions, differentiation, and the logarithms of positive computable scales are therefore effective. At an approximate time, a coarse bound gives a finite superset of possible translates of the zero-extended unit template. This evaluates the periodic field without deciding whether time is an integer. None of these operations iterates the machine execution: the program contains the entire finite instruction list. ◻

Proof. In the first construction choose padding smaller than a quarter of the within-family horizontal gaps and choose height differences larger than twice the padded vertical half-width. During lift and descent, horizontal separation suffices. During transfer, vertical separation suffices. The two choices of \(g_i\) are positive and monotone between \(1\) and \(\lambda_i\); their reciprocals are monotone as well. Thus the width claim holds. The affine path is \(c_i(s)+\operatorname{diag}(g_i,g_i^{-1},1)(X_0-c_i(0))\). Its first coordinate gives (8) exactly for the linear choice. Lemma 5 now applies.

For the second construction choose padding smaller than the gaps in each union of source projections with reserved parking rectangles, and of target projections with those rectangles. In the evacuation pass, vertical motion is separated from uncollected sources and parked boxes. During delivery it is separated from already filled targets and boxes still parked. At travel height, the moving box is above every resting box. A shape change stays within its own reserved projection. These are the occupied sets at each actual time; source–target intersection causes no interference because all sources have left before delivery starts. Lemma 5 applies to each motion with every other box fixed. Finite concatenation gives the required neighborhood map. ◻

Proof. For an empty family take the zero field. Otherwise let \(R=\max(1,h_{ij},k_{ij})\). Choose a rational \(\Lambda>10\) so that centers in each of the lists \((\Lambda a_i)\) and \((\Lambda b_i)\) are separated by more than \(20R\) in maximum norm. This is possible because distinct disjoint boxes have distinct centers. Let \[K=\max(4\Lambda,|\Lambda a_{i1}|,|\Lambda b_{i1}|), \qquad d_i=(K+30Ri,0,0).\] The \(d_i\) are parking centers, separated from each other and both scaled families by at least \(30R\) in their first coordinate.

First multiply all source centers by a common factor increasing from \(1\) to \(\Lambda\), while keeping shapes fixed. Next change all shapes at those separated centers by factors \(s_{ij}^{\theta}\), \(0\le\theta\le1\). Each half-width is the geometric interpolation \(h_{ij}^{1-\theta}k_{ij}^{\theta}\le R\). Move the boxes one at a time from \(\Lambda a_i\) to \(d_i\), then, after all have been parked, from \(d_i\) to \(\Lambda b_i\). Finally shrink all target centers by a common factor from \(\Lambda\) to \(1\), with target shapes fixed.

Here is a finite construction of every single-box path. Around each stationary center place the closed cube of maximum-norm radius \(4R\). These cubes are disjoint since stationary centers are separated by more than \(20R\), and neither endpoint is in a cube. Start with the straight segment between the endpoints. Its intersections with the cubes are rational intervals, found by linear inequalities. Replace each portion inside a cube by a polygonal path on its boundary: join the entry point to a vertex of a containing face, use cube edges to a face containing the exit point, and join to that exit point. The cube is convex face by face, so these segments remain on its boundary; no other cube meets them. A tangency needs no replacement. The resulting rational polygonal path has distance at least \(4R\) from every stationary center. Traverse its segments with flat switches, stopping to all orders at vertices. Its geometric image is unchanged, so smoothing the time parameter does not reduce clearance.

For a uniform localization margin, let \(g_a\) be the minimum over \(i<j\) of \[\max_\ell(|a_{i\ell}-a_{j\ell}|-h_{i\ell}-h_{j\ell}),\] and define \(g_b\) similarly. These are positive; for a singleton list set both to \(R\). During outward expansion a separating coordinate gap can only increase; during final contraction the target gap is its minimum. During deformation centers are more than \(20R\) apart. During a single-box path the center distance is at least \(4R\), leaving a coordinate gap at least \(2R\). Hence padding smaller than \(\tfrac18\min(R,g_a,g_b)\) satisfies Lemma 5 throughout. Put all stages, and all straight segments, in consecutive subintervals of \([1/4,3/4]\). Their flat switches give a smooth finite motion, with positive diagonal determinant-one linear parts. That lemma provides the velocity and the neighborhood assertion. All choices use rational inequalities and effective elementary functions, which also supply bounds for every derivative. ◻

Proof. The identity \(E(aA)=(d(a)+E(A))/B\) gives all three formulas and their action on the independent tails. At the first differing digit, incompatible prefix intervals have a positive gap: the digit difference is at least two while the child interval has length one in that scale. Thus the source state, read letter, and any left split separate sources. At one target state the displacement is fixed. For a right or stationary move the written letter separates the relevant target coordinate; for a left move the pair \((c,b)\) separates the two-letter prefix. The incoming inverse condition rules out duplicates. These arguments concern the complete intervals, including their endpoints. For odd digits the infinite tail lies between \(1/(B-1)\) and \((2m-1)/(B-1)<1\), proving interiority. The longest target prefix has length two, giving the asserted gap. Uniformly scaling state squares by \(s\) scales gaps by \(s\) and leaves the linear parts unchanged. ◻

Proof. Use letters \((c,\ell,j)\) with work letter \(c\), history \(\ell\in\{e,F\}\sqcup\{r\}\) and mark \(j\in\{0,1\}\). The controls are \(C_q,J_r,S_r,P_p,T_p\). The complete partial table is \[\begin{array}{lll} C_q:(a,\ell,0)&\mapsto J_r:(b_r,\ell,0),d_r&\ell\ne F,\\ J_r:(c,\ell,0)&\mapsto S_r:(c,\ell,1),1&\ell\ne F,\\ S_r:(c,\ell,0)&\mapsto S_r:(c,\ell,0),1&\ell\ne F,\\ S_r:(c,F,0)&\mapsto P_{p_r}:(c,r,0),1,&\\ P_p:(c,e,0)&\mapsto T_p:(c,F,0),-1,&\\ T_p:(c,\ell,0)&\mapsto T_p:(c,\ell,0),-1&\ell\ne F,\\ T_p:(c,\ell,1)&\mapsto C_p:(c,\ell,0),0&\ell\ne F. \end{array}\] Initialize the frontier at two, all other history empty, and all marks zero. At checkpoint \(n\) the head satisfies \(|h_n|\le n\). The first row updates the work tape and moves to \(h'\le n+1\), strictly left of the frontier \(f=n+2\). The second row marks \(h'\). The forward scan writes \(r\) at \(f\), puts a new frontier at \(f+1\), and the backward scan stops exactly at the mark, which it clears. Its length is \(2(f-h')+4\), so it is finite and positive. This proves the invariant inductively, including the exact work update and halt equivalence.

For the full-domain inverse, \(r\) recovers the old work pair into \(J_r\); mark one distinguishes entry into \(S_r\) from its mark-zero loop; a written record determines entry into \(P_p\); a written \(F\) distinguishes entry into \(T_p\) from its loop; and entry into \(C_p\) uniquely restores mark one. Every destination has the single incoming displacement displayed in the table. No invariant on remote cells was used. ◻

Proof. Use main controls \(M(q,e)\), forward controls \(G(r)\), and \(A(p,d),B(p,d)\) for every work state \(p\) and \(d\in D\), including pairs not produced by a work instruction. Copy the \(O\) component in each row. In the first three rows use every \(r=(q,e,a)\in\mathcal R\) with \(\delta(q,a)=(p,c,d)\). The final three rows apply independently to every \((p,d)\in Q\times D\), including pairs that are not outputs of any work instruction. The variables \(v\in\Gamma\) and \(h\in\{\bot,F\}\sqcup\mathcal R\) range over all letters subject to the displayed restrictions. The whole table is \[\begin{array}{lll} M(q,e):(a,h,0)&\mapsto G(r):(c,h,1),1&h\ne F,\\ G(r):(v,h,0)&\mapsto G(r):(v,h,0),1&h\ne F,\\ G(r):(v,F,0)&\mapsto A(p,d):(v,r,0),1,&\\ A(p,d):(v,\bot,0)&\mapsto B(p,d):(v,F,0),-1,&\\ B(p,d):(v,h,0)&\mapsto B(p,d):(v,h,0),-1&h\ne F,\\ B(p,d):(v,h,1)&\mapsto M(p,d):(v,h,0),d&h\ne F. \end{array}\] Starting at \(M(q_0,0)\), the first row writes and marks the old head. The scan reaches frontier \(j=n+1\), stores its record, creates the next frontier, returns to the unique mark and makes the work move. The count is \(2(j-i)+3=5+2(n-i)\). Since \(i\le n<j\), every test is satisfied and both scans are finite.

Into \(G(r)\) the written mark distinguishes entry and loop; \(r\) recovers the old state and work symbol. Into \(A(p,d)\) the record recovers the old \(G(r)\). Into \(B(p,d)\) a written frontier distinguishes arrival and loop. Into \(M(p,d)\) only clearing a mark is possible. All remaining components were copied. This proves the inverse on arbitrary tapes and the fixed incoming displacement. Deleting the passive \(O\) coordinate is an exact specialization of every rule, not merely an equivalence of initialized runs. When \(O=\{0,1\}\), a unique initial one at physical cell zero records the absolute origin. ◻

Proof. For the stopping version let \(J=(Q\setminus\{h\})\times\Gamma\). For the continuing version take \(J=Q\times\Gamma\) and add \(\delta(h,a)=(h,a,0)\). Use letters \((c,\eta,j)\in\Gamma\times(\{\bot,F\}\sqcup J)\times\{0,1\}\), controls \(C_q,D_q,T_q,P_r,S_r\), and precisely \[ \begin{aligned} C_q:(a,\eta,0)&\mapsto P_r:(b_r,\eta,0),d_r,\\ P_r:(c,\eta,0)&\mapsto S_r:(c,\eta,1),-1&&\eta\ne F,\\ S_r:(c,\eta,0)&\mapsto S_r:(c,\eta,0),-1&&\eta\ne F,\\ S_r:(c,F,0)&\mapsto D_{p_r}:(c,r,0),-1,\\ D_q:(c,\bot,0)&\mapsto T_q:(c,F,0),1,\\ T_q:(c,\eta,0)&\mapsto T_q:(c,\eta,0),1&&\eta\ne F,\\ T_q:(c,\eta,1)&\mapsto C_q:(c,\eta,0),0&&\eta\ne F. \end{aligned} \tag{20}\] In the first row only, the stopping version requires \(\eta\ne F\); the continuing version allows every \(\eta\). These domains remain distinct even though initialized runs never use that difference.

Initialize \(F\) at \(-2\), empty history elsewhere and zero marks. At checkpoint \(n\), \(i\ge-n=f_n+2\). After the work update, \(i'\ge f_n+1\). Marking \(i'\) and scanning left therefore reaches \(f_n\), records \(r\), puts the next frontier at \(f_n-1\), and returns right to the mark without crossing the new frontier. Clearing it completes the next work checkpoint. The two finite traversals and four endpoint actions give the stated count. The continuing version has a next rule even after every halt checkpoint.

Into \(P_r\) the record index recovers the work pair. Into \(S_r\) mark one distinguishes entry from loop; into \(D_q\) the written record recovers the scan; into \(T_q\) a written \(F\) distinguishes entry from loop; into \(C_q\) there is only stationary unmarking. This proves the incoming conditions for both full partial domains. It also gives the configuration inverse: reverse the displacement, inspect the written cell, and invert its uniquely determined rule. ◻

Proof. The first rule writes and marks \(h_n\). Since \(p_n>h_n\), the forward scan reaches the frontier, moves it one cell right, and the return scan reaches the mark. Its neighboring-cell guard holds because the new frontier is \(p_n+1\). The last rule clears the mark and performs the work displacement. Counting the two traversals and endpoint actions gives \(2(p_n-h_n)+1\); induction gives \(|h_n|\le n\).

For injectivity on all relative tapes, into Go the old cursor is output offset \(-1\): a mark there distinguishes entry from the loop, and \(r\) restores the overwritten work letter and old Ready control. Into Back the old cursor is offset \(+1\). An output pointer at \(+2\) identifies a log insertion, with record at \(+1\) recovering \(r\) and old pair \((P,E)\). A nonpointer at \(+2\) identifies the guarded loop. Into Ready the corresponding Back control uniquely restores the cleared mark. These are all possibilities, proving the full inverse. ◻

Proof. After the writes the new left prefix consists of old indices \(e-1,e-2,\ldots,-a\), and the right prefix of \(e,e+1,\ldots,b-1\). The two remaining tails are free and unchanged, so the image is the whole output cylinder. Distinct source cylinders have disjoint images by injectivity. If both pairs of output prefixes were compatible, a common extension would belong to both images; hence at least one coordinate has incompatible prefixes. The gapped interval argument then separates the complete closed rectangles. The prefix lengths give the two reciprocal scales directly. ◻

Proof. Specialize Lemma 12 to a singleton passive track \(O\) and delete that coordinate. Rename its controls \((M(q,e),G(\tau),A(q,d),B(q,d))\) as \((M_{q,e},R_\tau,F_{q,d},L_{q,d})\), and its frontier \(F\) as \(P\). The full alphabet is \(A\times(\{\bot,P\}\sqcup\mathcal T)\times\{0,1\}\), where \(\mathcal T=(Q\setminus H)\times\{-1,0,1\}\times A\). Every row and guard is the same under this bijection, so both incoming properties hold on the entire partial domain, including unused controls. Only \(M_{q,e}\) with \(q\in H\) are halting controls.

Initialize \(M_{q_0,0}\) with all marks zero, frontier at \(r_0\), and empty history elsewhere. The case \(r_0=1\) is the initialization already proved in the shared lemma. For either allowed \(r_0\), checkpoint \(n\) has frontier \(g=r_0+n>h_n\). Mark the old head, scan right to \(g\), write the record and install the new frontier at \(g+1\), then return to the mark. All intervening history is nonfrontier, the new frontier cell is empty, and there is only one mark. Hence the same rows apply when \(r_0=2\). There are \(g-h_n-1\) continuing right moves, \(g-h_n\) continuing left moves, and four other instructions, giving \(2(g-h_n)+3\). The last row clears the mark and makes the work movement; the invariant is restored with frontier \(g+1\). This proves finite completion, indefinite legality for a nonhalting run, and the stated halt equivalence, including initial halting. ◻

Proof. Use Lemma 10 with digits \(d(a)=e(a)\). Its digit hypotheses hold for every \(B\ge2m+1\). In the notation \[C(a_1a_2\cdots)=\sum_{j\ge1}e(a_j)B^{-j},\quad h_a(v)=\frac{e(a)+v}{B},\quad I_{a_1\cdots a_l}=h_{a_1}\cdots h_{a_l}([0,1]),\] the left stream starts immediately left of the head and the right stream starts at the head. Place state \(s\) in \(k_s+a[0,1]^2\) with a common rational \(a>0\). The minimal branch maps are exactly (19); scaling both charts by \(a\) leaves \(\operatorname{diag}(B^{-d},B^d)\) unchanged.

For the additional full subdivision, restrict the right- and stay-move sources to \(x\in I_\gamma\), for every full letter \(\gamma\). Their targets become \(I_{\beta\gamma}\times[0,1]\) and \(I_\gamma\times I_\beta\), respectively, for written letter \(\beta\). The already split left-move targets remain \([0,1]\times I_{\gamma\beta}\). At a fixed target state the incoming direction is fixed, and the pair \((\beta,\gamma)\) distinguishes the branches in every case. Thus the added subdivision preserves separate target separation, while source separation follows by restriction of the shared source rectangles. All these statements concern the filled closed rectangles and arbitrary symbolic tails.

For completeness the quantitative gaps remain valid for the entire range \(B\ge2m+1\): distinct first-digit intervals have gap at least \(B^{-1}\) and distinct second-digit intervals at least \(B^{-2}\). The source and target families therefore have within-state gaps at least \(aB^{-2}\); different state squares have their own chosen gaps. Tail values lie between \(1/(B-1)\) and \((2m-1)/(B-1)<1\). An initial constant tail after \(l\) digits contributes \(B^{-l}e(a_{\rm tail})/(B-1)\), so initialized codes are rational. At checkpoints the history records, or the known absolute origin of the record block, also recover the absolute work-head position. ◻

Proof. Each pulse is a finite expression in rational data, effective smooth switches, their derivatives, positive diagonal scales and logarithms. Finite positive separation margins bound every cutoff derivative. A finite maximum bounds a period and the loading interval, and repetitions preserve that bound. To evaluate near a join, a coarse enclosure of time gives a finite superset of possibly active pulses. Summing them avoids an exact comparison of a computable time with an integer. The flat collars give smoothness. Compact support gives every required spatial Sobolev bound. Lemma [lem:p2-realization] gives the asserted solution and comparison; on a torus the integral of each curl or Hamiltonian derivative vanishes, and the convection term is a divergence. Its flow satisfies \(\det D\Phi_t=1\), because differentiating this determinant gives \((\mathop{\mathrm{div}}U)(t,\Phi_t)\det D\Phi_t=0\) and its initial value is one. The velocity and acceleration identities in that lemma consequently apply to these volume-preserving diffeomorphisms. ◻

Proof. For \(N=0\) take the zero field. For \(N\ge1\) apply Lemma 7(i) with vertical half-width one. Use the shared phase switches with \(\eta=w_2\) and set \[C_i(s)=\big((1-\eta)c_i^S+\eta c_i^T,\,6i(w_1-w_3)\big), \qquad D_i(s)=\operatorname{diag}(r_i,r_i^{-1},1).\] The affine path is \(G_i(s,y)=C_i(s)+D_i(s)(y-(c_i^S,0))\). Choose \(\epsilon\) to be one quarter of the minimum of 1 and all pairwise max-norm distances in the two planar lists, omitting empty pair lists. The \(\epsilon\)-padded moving boxes are disjoint: source and target gaps are at least \(4\epsilon\) during the vertical phases, and the middle-phase height spacing is \(6>2(1+\epsilon)\).

For later loading formulas, record the centered cutoff explicitly: \[\chi(C,L,\epsilon;x)= \prod_{j=1}^3\sigma\!\left(\frac{L_j+\epsilon+x_j-C_j}{\epsilon/2}\right) \sigma\!\left(\frac{L_j+\epsilon-x_j+C_j}{\epsilon/2}\right).\] Here \(L\) is the vector of half-widths; the plateau has padding \(\epsilon/2\) and the support has padding \(\epsilon\). If \(L_i(s)\) are the moving half-widths, \(Y=x-C_i(s)\) and \(A_i=D_i'D_i^{-1}\), the field supplied by (6) is \[ V(s,x)=\sum_i\nabla\times\left\{ \chi(C_i,L_i,\epsilon;x) \left[\tfrac12 C_i'\times Y+\tfrac13(A_iY)\times Y\right] \right\}. \tag{23}\] Indeed \(\det D_i=1\), and the preceding separation verifies Lemma 5 with localization parameter \(\epsilon/2\). Its neighborhood conclusion applies with \(L_* =\max_i\max(1,\lambda_i,\lambda_i^{-1})\) because \(\|D_i(s)\|_\infty\le L_*\) and \(D_i(0)=I\). It therefore gives the stated affine motion on the full boxes and on their initial max-norm neighborhoods of radius \(\epsilon/(4L_*)\). The paths are stationary for \(s\le1/12\) and \(s\ge11/12\). Their finite center ranges and endpoint width bounds give common compact support. The effective positive lower bound \(r_i\ge\min(1,\lambda_i)\) verifies the scale requirement. For the linear scale, the first-coordinate identity is \[(G_i(s,y))_1=(1-\eta)y_1+\eta(F_i(y_1,y_2))_1.\] The second coordinate need not interpolate its own endpoint values. Instead, if \(b_{i2}\) is the source’s second half-width, convexity of the reciprocal function gives \[ \frac{b_{i2}}{(1-\eta)+\eta\lambda_i} \le (1-\eta)b_{i2}+\eta\frac{b_{i2}}{\lambda_i}. \tag{24}\] The first half-width is exactly its endpoint interpolation. For the exponential scale, convexity of the exponential gives the corresponding upper bounds in both coordinates, since \(\lambda_i^{\pm\eta}\le(1-\eta)+\eta\lambda_i^{\pm1}\). In either case the center is interpolated linearly and each half-width is at most the interpolation of its endpoint half-widths. Therefore each lower edge is at least the interpolation of the endpoint lower edges, and each upper edge is at most the corresponding interpolation of upper edges. This proves the claimed bounding rectangle for the entire sweep. The individual endpoint-width bounds also follow directly from monotonicity of the positive scales and their reciprocals. ◻

Proof. Use the height phases in Section 7.1. For source center \(p_i\), end its center path at its actual image \(F_i(p_i)\); in the onto case this is the named target center. Choose separated heights \(H_i\) as above and set \[c_i=\bigl((1-w_2)p_i+w_2F_i(p_i),\,H_i(w_1-w_3)\bigr),\] \[\lambda_{ij}=1-w_2+w_2\lambda_{ij}^*,\qquad D_i=\operatorname{diag}\bigl( \lambda_{i1},\lambda_{i2},(\lambda_{i1}\lambda_{i2})^{-1}\bigr).\] The volume-preserving path has logarithmic derivative \[T_i=\dot D_iD_i^{-1}=\operatorname{diag}\left( \frac{\dot\lambda_{i1}}{\lambda_{i1}}, \frac{\dot\lambda_{i2}}{\lambda_{i2}}, -\frac{\dot\lambda_{i1}}{\lambda_{i1}} -\frac{\dot\lambda_{i2}}{\lambda_{i2}}\right).\] This matrix has trace zero, so (6) localized around the moving plate gives the desired field. Its planar motion is \(c_i+(\lambda_{i1}(y_1-p_{i1}),\lambda_{i2}(y_2-p_{i2}),0)\). Normal displacement changes by the factor \((\lambda_{i1}\lambda_{i2})^{-1}\). The factors have positive lower bounds; choose the initial thickness less than their product’s minimum times the normal plateau width. A small enlargement of the planar rectangle is allowed by the positive collars as well. Thus a whole thin neighborhood follows the volume-preserving affine motion. The normal factor returns to one at the endpoint. The final plate image is contained in \(Q_i\times\{0\}\). Source and target-container gaps protect the vertical phases and distinct heights protect the middle phase. Linear interpolation of each planar factor gives both convex-coordinate identities on the plate. This proves smooth existence for arbitrary real data. Under the effective input assumptions of Lemma 5, the positive scale bounds and finite gap margins can be computed and the construction is effective. ◻

Proof. Fixed compact support and bounded mixed derivatives imply all reference velocity hypotheses of Lemma 2. Case (i) gives exactly the stated whole-space comparison class, without a competing gradient bound; its torus clause gives the stated classical comparison class. The same lemma provides the pressure normalization and global material flow. The product rule applied to (25) proves the force’s support and derivative bounds. A chart curl has zero mean, and integration of the residual uses \(\int\Delta W=0\) and \(\int(W\cdot\nabla)W=\int\operatorname{div}(W\otimes W)=0\). For these effective-data instances, gate derivatives and seam evaluation are those of (5) and Proposition 6. The explicit affine scales and every reciprocal denominator have positive computable bounds. Thus differentiating the finite formulas, using a finite superset of possible period indices, evaluates every derivative without a machine execution or a real equality test at a seam. ◻

Proof. Use Lemma 11, let \(\Sigma\) be its full cell alphabet, and enumerate its \(N\) controls by \(i=1,\ldots,N\). Set \(s=1/[16(N+1)]\) and give the \(i\)th control a square of side \(s\) centered at \[m_i=(o_i+i/[4(N+1)],1/2),\qquad o_i=\begin{cases}5/8,&C_q\text{ with }q\text{ halting},\\1/8,&\text{otherwise}. \end{cases}\] Use odd digits and \(B=2|\Sigma|+2\). The code at height \(z_0=1/4\) is \((m_i+s(E(L)-1/2,E(R)-1/2),z_0)\). It is rational initially. Lemma 10 gives \(J\) branches with separately separated rectangles, all side lengths at most \(s\) and endpoints on the grid of mesh \([32(N+1)B^2]^{-1}\). For branch \(j\) choose height \(z_j=1/2+j/[4(J+1)]\). Use the three-phase lift of Lemma 7, with exponential middle scale \(\lambda_j^\theta\) and center interpolation. A padding parameter \[\epsilon=\min\{1/128,[320(N+1)B^2]^{-1},[40(J+1)]^{-1}\}\] separates the enlarged sheets in the source, height, and target phases and keeps them inside the unit cube. Their thickness may then be chosen positive by Lemma 5. For \(J=0\) the repeated field is zero.

During \([0,1]\) a localized curl translation takes the fixed label along the segment to the initial code. Both segment and a small padding lie in the chart. Repeat the branch pulse thereafter. At each subsequent integer time the particle is the next table code. If both endpoint controls are nonhalting, the interpolated center and half-width bounds give \(x_1<3/8+s<1/2\) throughout the motion. The loader also stays below \(1/2\) in a nonhalting run. A halting checkpoint lies in the upper half, reached in finitely many positive local steps; an initial halt is reached at the end of loading. This proves the event for every real time. ◻

Proof. Choose \(r_0=2\) in Lemma 14: initially the frontier is at cell two, all other history cells are empty, and all marks are zero. Thus \(p_n=n+2\), with records in cells \(2,\ldots,n+1\). Split the full three-cell windows \([-1,1]\) and apply Lemma 15 with \((a,b)=(1,2)\). Let \(\mathcal A\) be the full cell alphabet of this window table and use odd digits in base \(B=2|\mathcal A|+1\). For \(S\) controls let \(\epsilon=1/[16(S+1)]\) and place their squares at \(o_s+\epsilon[0,1]^2\), where \[o_s=(\xi_s,1/4+2\kappa(s)\epsilon),\quad \kappa(s)\in\{1,\ldots,S\},\quad \xi_s=\begin{cases}5/8,&s\text{ halting Ready},\\1/4,&\text{otherwise}. \end{cases}\] Write \(J\) for the number of branches and \(e_j\) for branch \(j\)’s cursor displacement. The full source and target rectangle gaps are at least \(\epsilon B^{-3}\). Put their common height at \(1/4\), assign branch \(j\) height \(1/2+j/[4(J+1)]\), and use the exponential middle scale \(\operatorname{diag}(B^{-e_j\theta},B^{e_j\theta},1)\). Take one tenth of \(\min(1/32,\epsilon B^{-3},1/[4(J+1)])\) as padding. Horizontal source and target gaps and the private heights separate the three phases, while all widths remain at most \(\epsilon\). The strain can equivalently be generated by \((0,0,-e_j\dot\theta\log B\,(x_1-c_{j1})(x_2-c_{j2}))\), retaining the explicit diagonal-strain implementation.

A first unit interval loads the label along the segment to the rational initial code. Thereafter the exact rectangle action and the compiler lemma give checkpoint times \(t_0=1\) and \(t_{n+1}=t_n+2(p_n-h_n)+1\). A nonhalting center stays in \([1/4,1/4+\epsilon]\), with half-width at most \(\epsilon/2\); thus every intermediate first coordinate remains below \(1/2\). The loader has the same exclusion. Halt codes are in \([5/8,5/8+\epsilon]\), inside the observer, including an initial halt. The history block starts at the known absolute cell two, so the head-relative code also recovers absolute indexing at checkpoints. ◻

Proof. For the first case introduce a fresh terminal state \(\dagger\) and replace every missing instruction and every originally terminal state–symbol pair by a write-preserving stationary transition into it. The original initial control is therefore nonterminal in this normalized table, even when the original machine is initially halted. Apply Lemma 12 with its passive origin track, and let \(\mathcal A\) be the resulting full cell alphabet. Use odd digits in base \(B=2|\mathcal A|+1\) and initial fractions \(x_*,y_*\). Index the initial main state by zero, the main \(\dagger\) states by distinct negative integers, and the rest by distinct positive integers. State \(s\) has placement \[(x,y)\longmapsto(10k(s)+x-x_*,y-y_*).\] The initialized point is zero; nonhalt codes have first coordinate \(>-1\), and halt codes \(<-5\). Thicken the rectangles of Lemma 10 by \([-1,1]\), give branch \(i\) height \(10i\), and use the linear scale \(g_i=1+\theta(\lambda_i-1)\) in the lift schedule. A padding smaller than one tenth of the minimum of one and all within-family planar gaps suffices. The middle strain has potential \((0,0,(\dot g_i/g_i)(x_1-c_{i1})(x_2-c_{i2}))\). Equation (8) excludes every nonhalting visit to \(x_1<-5\). The table and routing are fixed for the machine; only \(x_*,y_*\) change with the input. The passive origin bit remains at physical cell zero, so no absolute tape information is discarded.

For the second case first merge all halting work states into a single state \(q_h\), redirecting incoming transitions and retaining initial halting. Delete the passive track from the direction recorder, and now let \(\mathcal A\) denote that full alphabet. Use odd digits with \(B=2|\mathcal A|+2\) and \(\lambda=1/[16(J+1)]\), where \(J\) counts the controls. Number them from zero and set \[S_s(x,y)=(A_s+\lambda x,1/4+\lambda y),\qquad A_s=2\lambda\operatorname{index}(s)+ \begin{cases}5/8,&s=M(q_h,e),\\1/4,&\text{otherwise}. \end{cases}\] The nonhalt squares lie in \([1/4,3/8]\times[1/4,1/4+\lambda]\); halt squares lie in \([5/8,3/4]\times[1/4,1/4+\lambda]\). Write \(m\) for the number of branches. Use coding height \(1/4\), branch heights \(1/2+i/[4(m+1)]\), and the same linear first scale. One hundredth of the minimum of \(1/16\), \(1/[4(m+1)]\), and the source and target planar gaps is valid padding; absent pair gaps are omitted. All supports lie inside the cube, and all horizontal widths are at most \(\lambda\). A localized translation with padding \(1/1000\) loads the rational initial point from the fixed label in one unit interval.

In a nonhalting period the center has first coordinate at most \(3/8\) and half-width at most \(\lambda/2\), hence is below \(1/2\); loading has the same bound. Halting codes lie in the observer. The exact sampling induction gives one table step per period, with checkpoint cycle \(2(k+1-h_k)+3\). The record block beginning at cell one recovers absolute indexing at checkpoints. An initial halt is observed at time one. ◻

Proof. Use the single-halt move-first recorder of Lemma 11, without an entry state. Let \(\Gamma\) be its full alphabet and \(N\) its number of controls. Put \(B=2|\Gamma|+1\), \(a=1/(16N)\), the halt square at \((1/8,1/4)+a[0,1]^2\), and the other squares at \((1/2+2ja,1/4)+a[0,1]^2\), \(0\le j\le N-2\). Use the minimal subdivision, so right and stay moves are not split by the left letter. Its gaps are at least \(a/B^2\). For \(M\) branches choose \[h=\frac1{64(M+1)},\quad Z_i=\frac12+\frac{i}{4(M+1)},\quad \varepsilon=\min\{1/128,h/4,a/(4B^2)\}.\] Apply linear-first-scale full-box routing from height \(z_0=1/4\). Middle heights differ by \(16h\); source and target gaps handle the vertical phases. Centers lie in \([1/8,5/8]\times[1/4,5/16]\), widths are at most \(a\), and all padded boxes lie in \((1/16,15/16)^3\). This is a construction on the specified unit torus, with no change of physical viscosity.

In the first unit interval translate a box of side \(1/64\) from the fixed label to the rational initial code, using Lemma 5. The segment and a small padding lie inside the same chart. Repeat only the instruction cycle afterward. Code induction proves the checkpoint identities. A halting code has first coordinate in \([1/8,3/16]\), including an initial halt reached by loading. For a nonhalting run both centers of every used branch have first coordinate at least \(1/2\), with half-width at most \(a/2\); loading has the same lower center bound. Hence the particle always stays to the right of \(1/4\). Residual realization proves every analytic assertion. The separate loading field accounts for the qualification \(t\ge1\) in the periodicity statement. ◻

Proof. Normalize to one halt state and prepend a new nonhalting state \(q_I\) whose rule preserves the scanned letter, stays, and enters the original initial state. No machine rule targets \(q_I\). Apply the move-first recorder of Lemma 11 and omit the unused controls \(P_{q_I},T_{q_I}\). Then the entry control \(C_{q_I}\) has no incoming local instruction. Write \(\Gamma\) for this recorder’s full alphabet. Put \(B=2|\Gamma|+1\), let \(N\) be the number of controls, and set \(a=1/(16(N+1))\). Use the halt square \((3/4,1/4)+a[0,1]^2\) and nonhalting squares \((1/4+2al,1/4)+a[0,1]^2\), \(0\le l\le N-2\). The full subdivision of Lemma 17 gives source half-widths \(a/(2B)\) in both coordinates.

Add a square with those half-widths centered at \((1/8,1/8)\) and a translation to the square centered at the initial entry code. Each initial tail is at distance at least \(1/(B-1)>1/(2B)\) from the endpoints of \([0,1]\), so the new target lies strictly inside the entry square. There is no other target in that square. Both enlarged branch lists therefore remain separately separated, with gaps at least \(a/B^2\). The extra source is separated from all state squares.

Number the branches, including loading, by \(1\le i\le M\). Set \[z_0=\tfrac14,\qquad h=\frac1{16(M+1)},\qquad Z_i=\tfrac12+\frac{i}{4(M+1)},\qquad \varepsilon=\frac1{10}\min(h,a/B^2).\] Thicken each planar source by \([z_0-h,z_0+h]\). Lift the resulting full box to center height \(Z_i\), move it horizontally, and lower it to \(z_0\). For an instruction with displacement \(d_i\), use \(D_i=\operatorname{diag}(B^{-d_iw_2},B^{d_iw_2},1)\); loading has \(d_i=0\). Its center is the planar endpoint interpolation plus \((Z_i-z_0)(w_1-w_3)e_3\). Adjacent middle heights are \(4h\) apart, while the padded half-thickness is \(h+\varepsilon<2h\). Source and target gaps handle the other phases. Horizontal half-widths never exceed \(a/2\), horizontal centers lie between \(1/8\) and \(3/4+a\), and vertical centers lie between \(1/4\) and \(3/4\). The padded motions are thus strictly inside the unit cube. Lemma 5 gives a chart field, extended periodically in space and time. Its zero time collars give zero initial velocity.

The first cycle loads the fixed particle. Each subsequent cycle maps its exact code to the next local code until the first halt, or forever in a nonhalting run. Lemma 11 identifies the finite checkpoint times with machine steps. The added entry state handles an original initial halt. A halt code is in the target slab. For a nonhalting execution every used source and target center, including the loader, has first coordinate at most \(3/8\). Exponential interpolation keeps the center below that bound and the half-width below \(a/2\); the particle cannot reach \(5/8\) at any intermediate time. The final transition into a halt is not subject to this nonhalting-endpoint bound. Lemma 21 gives the force, its mean, its effective derivatives, and uniqueness. Repeating the loading field is harmless to the initialized trajectory and makes the force periodic from zero. ◻

Proof. Normalize to one halting work state \(q_h\), redirecting incoming transitions to it and replacing missing nonhalting rules by unchanged-letter stay moves into it. This preserves initial halting. Write \(\Gamma\) for the work alphabet, \(b\) for its blank, \(q_0\) for the initial control, and \(\omega\) for the finite input word. For top-first stacks \(L,R\), a rule \((s,\alpha,\beta)\rightsquigarrow(r,\gamma,\eta)\) replaces the two displayed prefixes and leaves their independent infinite tails unchanged. Introduce fresh state \(s_0\), boundary letter \(S\) and mark letter \(M\). The preliminary rules are \[(s_0,\varepsilon,\varepsilon)\rightsquigarrow(q_0,S,\omega)\] and, for each work transition \((q,a)\mapsto(r,c,d)\), \[\begin{array}{c|l} d&\text{preliminary rules}\\\hline 1&(q,\varepsilon,a)\rightsquigarrow(r,c,\varepsilon)\\ 0&(q,\varepsilon,a)\rightsquigarrow(r,\varepsilon,c)\\ -1&(q,v,a)\rightsquigarrow(r,\varepsilon,vc)\quad(v\in\Gamma),\\ &(q,S,a)\rightsquigarrow(r,S,bc). \end{array}\] The final row exposes the implicit blank beyond the left boundary. Index each preliminary occurrence by \(j\), with target \(r(j)\), and introduce a private control \(g_j\) and record letter \(\ell_j\). Replace that occurrence by \((s,\alpha,\beta)\to(g_j,\gamma,M\eta)\) and add \[ \begin{aligned} (g_j,v,\varepsilon)&\to(g_j,\varepsilon,v)&&v\in\Gamma,\\ (g_j,S,\varepsilon)&\to(k_{r(j)},S\ell_j,\varepsilon),\\ (k_q,\varepsilon,v)&\to(k_q,v,\varepsilon)&&v\in\Gamma,\\ (k_q,\varepsilon,M)&\to(q,\varepsilon,\varepsilon). \end{aligned} \tag{28}\] Sources are separated by control and tested top symbols. Into \(g_j\), entry has right top \(M\), whereas a loop has a work letter there. Into \(k_q\), entry has left prefix \(S\ell_j\), whereas a loop has a work letter at left top; distinct entries have distinct \(\ell_j\). Into a work state only the last rule enters. Thus both source and target cylinders are disjoint for arbitrary tails.

After a preliminary replacement the stacks have form \(wSB,MY\), with \(w\in\Gamma^*\). The forward loop transfers \(w\), leaving \(SB,w^{\rm rev}MY\). The boundary rule inserts \(\ell_j\) behind \(S\), the reverse loop restores \(w\), and the final rule removes \(M\). After \(2|w|+2\) further instructions the stacks are \(wS\ell_jB,Y\) at control \(r(j)\). At a work control, \(w\) lists represented cells left of the head, nearest first; \(R\) lists the current cell and right tail, and cells beyond \(S\) on the left are implicit blanks. The preliminary rules perform exactly the work write and move. Starting at \((s_0,b^\infty,b^\infty)\), the first work visit is at rule count three, and consecutive work visits differ by \(3+2|w_{\rm new}|\). This proves finite continuation and the halting equivalence, including an initially halted input. The retained records also recover all head displacements.

Enumerate the enlarged alphabet, blank first, by even digits \(0,2,\ldots,2m-2\) in base \(K=2m\). Use prefix intervals \(I(v)\) and codes \(E\) as in Lemma 10. Set offsets \(o_{q_h}=-3\), \(o_{s_0}=0\), and \(3,6,9,\ldots\) for the other controls. Encode \((s,L,R)\) by \((o_s+E(L),E(R),0)\); its initial value is zero. For a rule \((s,\alpha,\beta)\to(s',\gamma,\eta)\) put \[a_i=K^{|\alpha|-|\gamma|},\quad b_i=K^{|\beta|-|\eta|},\quad c_i=(a_ib_i)^{-1}.\] The full source box is \((o_s+I(\alpha))\times I(\beta)\times[-1,1]\); the target is \((o_{s'}+I(\gamma))\times I(\eta)\times[-c_i,c_i]\). The center-to-center map has matrix \(\operatorname{diag}(a_i,b_i,c_i)\) and determinant one. It implements the prefixes on all codes. Disjoint cylinders give separated full planar projections in each family, including blank-tail endpoints.

Park box \(i\) at \((4i,4,0)\), reserving the horizontal rectangle \([4i-1/2,4i+1/2]\times[7/2,9/2]\). Let \(R=\max(1,c_1,\ldots,c_N)\) and travel at height \(Z=2R+3\). Evacuate all sources to their parking sites, reshape there by \(\operatorname{diag}(a_i^s,b_i^s,c_i^s)\), then deliver all targets. The two horizontal widths remain at most one, and vertical half-widths at most \(R\). Let \(g\) be the minimum of one and the within-family gaps in the source-plus-parking and target-plus-parking projections. Padding smaller than \(g/8\) works during vertical motion and strain; at travel height the moving bottom is at least \(Z-R>R+2\). Lemma 7 gives the full \(7N\)-motion schedule.

Repeat this unit pulse from zero. Nonhalting endpoint boxes and all parking boxes have nonnegative first coordinate. Translations preserve that bound, and strain takes place in positive parking envelopes. Therefore a nonhalting path never enters the negative observation box. A halting code lies in \([-3,-2]\times[0,1]\times\{0\}\) inside it. The residual proposition finishes the proof. Loading was a repeated instruction, so no exceptional first-time force is required. ◻

Proof. Since \(s'=\rho\) and \(\rho'=-\rho^2\), \[ \partial_t\{\rho^jY(s(t),x)\} =\rho^{j+1}(\partial_s-j)Y(s(t),x). \tag{29}\] For \(P_{k,j}(D)=\prod_{r=0}^{k-1}(D-j-r)\), with \(P_{0,j}=1\), induction yields \[ \partial_t^k\{\rho^jY(s(t),x)\} =\rho^{j+k}P_{k,j}(\partial_s)Y(s(t),x). \tag{30}\] Spatial differentiation commutes with this identity. Taking \(j=1\) proves the velocity bound at every order. The residual is exactly \[ F=\rho^2\{W_s-W+(W\cdot\nabla)W\}(s(t),x) -\nu\rho\Delta W(s(t),x). \tag{31}\] Every mixed derivative of the braces is bounded by the finite product rule and the hypotheses. Apply (30) with \(j=2\) and \(j=1\) to obtain orders \(\rho^{2+k}\) and \(\rho^{1+k}\), respectively. This gives the force bound, including every spatial multi-index. In particular no temporal derivative of the convection term has been omitted.

Supports do not increase under differentiation or multiplication. For either field \(Z\), the squared integral of an indicated derivative is at most \[|K|C_{k,\alpha}^2\int_0^\infty(1+t)^{-2-2k}\,dt =\frac{|K|C_{k,\alpha}^2}{1+2k}.\] The zero datum remains zero, so residual realization applies. The chain rule shows that \(\Xi(s(t),a)\) solves the \(U\) trajectory equation with initial label \(a\); uniqueness identifies it with the actual trajectory. The clock is increasing onto \([0,\infty)\) with inverse \(e^s-1\), so it loses no finite computational time. Finally logarithm on positive arguments, the rational powers of \(1+t\), and the finite-index evaluation of the template and its derivatives are effective. This proves all assertions. ◻

Proof. Write \(\Gamma\) for the full alphabet of the recorder being used. For the six-rule table use initial frontier 2 and base \(B=2|\Gamma|+2\). Give nonhalting controls lanes \(3i+[0,1]\) and halting controls lanes \(-3i+[0,1]\), enumerating each group from 1; the second coordinate is \([0,1]\). The minimal subdivision has gap \(B^{-2}\). Use full thickness \([-1,1]\), heights \(10i\), and the transported-cutoff construction in Section 7.2, with initial padding \(1/(10B^3)\). Target horizontal padding is at most \(1/(10B^2)\), below half the gap; vertical middle intervals are also separated.

For the move-first table use a single halt lane \(-3+[0,1]\), other lanes \(2+2j+[0,1]\), base \(B=2|\Gamma|+1\), the minimal subdivision, thickness \([-1,1]\), and heights \(4i\). Use current-box padding \(1/(10B^2)\). The middle-phase separation is \(2(1+1/(10B^2))<4\); the planar gaps handle the other phases. This localization has a fixed current padding, as distinct from the transported initial padding in the six-rule version.

The rational initial code \(P_*\) is reached from zero along \(\theta(t)P_*\) by a localized translation during \([0,1]\). Repeating the chosen cycle produces a template \(W\) with one compact support and bounded mixed derivatives. The code identity at times \(1+n\) follows by induction through the local rules; the respective recorder invariant supplies every next checkpoint in finite positive time. A halt lies below \(-1\) in the first coordinate. In a nonhalting run all branch endpoints have positive first coordinate, and (8) keeps it positive throughout every cycle. Loading is nonnegative. Lemma 21 completes the PDE argument. Uniform support and boundedness give a uniform energy bound. The separate square-integrable choice follows from Lemma 30, without claiming periodicity in physical time for that choice. ◻

Proof. Let \(\Sigma\) be the full work–history–mark alphabet of the chosen left-frontier table, including its record symbols and any halting idle records in the continuing version. Use odd digits in base \(B=2|\Sigma|+2\), offset \(-2\) for \(C_h\) and \(2,4,6,\ldots\) for other controls. The two head-relative fractions and height zero give a rational initial code \(Z_*\). The rightmost nonempty history cell is the original cell \(-2\), an absolute landmark.

For the stopping table, apply the stream-function construction (9) at heights \(z_i=3i\). Choose source and target column paddings no greater than \(1/2\) and one third of their respective family gaps. A middle layer padding \(1/2\) keeps distinct layers disjoint. The explicit thickening bound following that construction proves the neighborhood action. Write \(Z_*=(x_*,y_*,0)\). With a flat switch \(\theta\) from zero to one, load the origin along \(\theta(t)Z_*\) using the velocity \(\nabla\times(0,0,H)\), where \(H(t,x,y,z)=\theta'(t)\chi(x,y,z)(x_*y-y_*x)\). Choose \(\chi\) compactly supported in \(\mathbb R^3\) and equal to one near the entire segment. The gate \(\theta'\) vanishes on the time collars. All nonhalt source and target lower-left first coordinates are at least two. Inequality (10) therefore excludes the observer during each period; loading has nonnegative first coordinate. Halt codes lie in \((-2,-1)\).

For the continuing table retain its unrestricted first-row history symbol and its idle work rules. Thicken every rectangle by \([-1,1]\). Choose \(L\) ten larger than the maximum of zero and every source or target right endpoint. Reserve parking projections \((L+4i,0)+[-1,1]^2\) and let \(\delta\) be the minimum of one and the gaps in each union of one rectangle family with these reservations. Use padding \(\delta/10\), travel height five, and \(7N\) slots: evacuate all sources by three translations each, reshape all parked boxes by \(\operatorname{diag}(1+\theta(\lambda_i-1),[1+\theta(\lambda_i-1)]^{-1},1)\), and deliver by three translations each. Horizontal widths stay at most one, vertical half-width is one, and the reservations miss both original families. Lemma 7 proves exact action on the whole boxes and their neighborhoods even with source–target overlap. A first localized translation loads \(Z_*\) from the origin. The sampling identity is now valid for every \(k\ge0\), not just until halting: \(\Phi_{1+k}(0)=Z(C_k)\) for the continued table.

For a nonhalting run all used endpoint centers have first coordinate at least two, parking centers are farther right, and half-widths are at most \(1/2\). Translation, parking strain, and waiting therefore keep first coordinate positive. A halt checkpoint lies in \((-2,-1)\). Proposition 6 proves the analytic assertions for both constructions. Their compact support also gives uniformly bounded kinetic energy. ◻

Proof. A full symbol is \((a,e,s)\in\Gamma\times\{0,1\}\times\mathcal H\), where \(\mathcal H=\{E,P\}\sqcup\{r_\gamma\}\) records empty history, a pointer, or an instruction. All displayed control families are disjoint. The following is the complete table; free cell indices range over their finite alphabets: \[\begin{array}{c|c|c|c|r} \text{control}&\text{read and guard}&\text{target}&\text{write}&d\\\hline A_q&(a,1,s),\ s\ne P&B_\gamma&(a^+_\gamma,0,s)&d_\gamma\\ B_\gamma&(c,0,s),\ s\ne P&R_\gamma&(c,1,s)&1\\ R_\gamma&(c,0,s),\ s\ne P&R_\gamma&(c,0,s)&1\\ R_\gamma&(c,0,P)&C_{q^+_\gamma}&(c,0,r_\gamma)&1\\ C_q&(c,0,E)&L_q&(c,0,P)&-1\\ L_q&(c,0,s),\ s\ne P&L_q&(c,0,s)&-1\\ L_q&(c,1,s),\ s\ne P&A_q&(c,1,s)&0 \end{array}\] The first row has \(q\ne h\) and \(\gamma=(q,a)\); there is no rule from \(A_h\). Initially the actual head and the unique mark are at zero, the work tape is the input, and history is \(E\) except for \(P\) at cell 2.

At checkpoint \(n\), \(|w_n|\le n\), so after the first row the head is at \(w'\le n+1<p_n\). That row clears the old mark, and the second installs the new one at \(w'\). The right scan reaches the pointer without meeting another mark, writes \(r_\gamma\), and visits the empty cell \(p_n+1\). The next row installs the new pointer there. The left scan returns through the newly written record to the unique mark and enters \(A_{q^+_\gamma}\) without clearing it. Hence every invoked rule is defined, the work instruction is correct, and the checkpoint invariant is restored. The two continuing scans take \(p_n-w'-1\) and \(p_n-w'\) steps; the other five steps give the stated total. This also handles \(d_\gamma=0\). Induction gives a defined execution until and including the first halting checkpoint, and indefinitely in the nonhalting case.

We check the inverse independently of this initialization. Targets \(B_\gamma,R_\gamma,C_q,L_q,A_q\) have incoming displacements \(d_\gamma,1,1,-1,0\), respectively. Undo that displacement to locate the written cell. At \(B_\gamma\), the index \(\gamma\) recovers the old control and overwritten work symbol, and the old mark was one. At \(R_\gamma\), the written mark is one for entry and zero for the scan, so it identifies the source and the old mark. At \(C_q\), the written record identifies \(\gamma\) and its former value \(P\). At \(L_q\), a written pointer identifies entry, whereas every loop has history different from \(P\). At \(A_q\), the unique source control is \(L_q\) and the full symbol is unchanged. All unspecified tracks are retained. Thus every possible output has at most one preimage on the full domain. ◻

Proof. Prepend a nonhalting state \(q_{\rm in}\) which, on every work symbol, preserves it, stays put, and enters the original initial state. This includes original initial halting while keeping the new initial state nonhalting. Apply Lemma 33. For its full alphabet of size \(m\) use odd digits \(e(\sigma)\in\{1,3,\ldots,2m-1\}\) in base \(B=2m+2\), and write \[c(\omega)=\sum_{j\ge0}e(\omega_j)B^{-j-1},\qquad (\xi,\eta)= \big(c(\sigma_{-1}\sigma_{-2}\cdots), c(\sigma_0\sigma_1\cdots)\big).\] The initial values \(\xi_0,\eta_0\) are rational, because the initialized streams have finite prefixes and known constant tails. Set \[O(A_{q_{\rm in}})=(2-\xi_0,-\eta_0),\qquad O(A_h)=(-2,0),\qquad O(P)=(4+2j,0)\] for the remaining controls in a finite ordering. All squares \(O(P)+[0,1]^2\) are separated. Every nonhalting square lies strictly to the right of \(x_1=1\), and the halting square is \([-2,-1]\times[0,1]\). The initial code at height zero is exactly \((2,0,0)\).

Apply Lemma 10 with digits \(d(\sigma)=e(\sigma)\), base \(B=2m+2\), and the state translations just specified. Its incoming hypotheses are exactly those of Lemma 33. With \(I_\sigma=(e(\sigma)+[0,1])/B\), use the minimal left-split branches of (19), including every left letter for a left move. The state translations conjugate these maps without changing their reciprocal diagonal linear parts. The shared lemma therefore gives the exact updates and separate source and target separation on full closed rectangles. If an absolute head position is wanted at a checkpoint, the number of contiguous records gives \(n\), and the pointer’s relative position locates the head relative to its known absolute position \(n+2\). The gapped digits permit each finite prefix needed for this recovery to be read with increasing precision.

Apply the exponential interpolation of Lemma 19, with heights \(10i\) and the following separated phase intervals: lifting on \([1/8,2/8]\), horizontal motion on \([3/8,5/8]\), and descent on \([6/8,7/8]\). On each interval use a rescaled flat step \(\sigma\) from (5). Let \(\delta\) be one quarter of the minimum of 1 and the positive planar source and target gaps. Use plateau padding \(\delta/3\) and support padding \(2\delta/3\). The same gap and height estimates in the lemma make these collars disjoint. Here the local potential can be written \[\tfrac12 C_i'\times(x-C_i) +(0,0,\eta'\log\lambda_i\,(x_1-C_{i1})(x_2-C_{i2})).\] Its curl is the required translation plus \(\eta'\log\lambda_i\,(x_1-C_{i1},-(x_2-C_{i2}),0)\). It realizes the whole cuboid and a neighborhood by the plateau argument.

Repeat this unit-time field as \(U\), and put \(f=U_t+(U\cdot\nabla)U-\nu\Delta U\) at the prescribed physical viscosity. The zero collars give \(U(0)=0\) and smooth period-one gluing. Lemma 21 gives the solution, uniqueness, support and effective mixed-derivative bounds. Its hypotheses hold because this is a finite curl formula on a compact spatial set and one compact time period. The formula includes every branch and never requires the run on the selected input.

At integer times, induction gives the exact local configuration through the first halt, or at every integer time for a nonhalting run. A halt therefore reaches the negative square. During a nonhalting run, both endpoint squares of every used branch are nonhalting. Every moving center has first coordinate greater than one and every first half-width is at most \(1/2\), because the endpoint widths are at most one and the exponential scale is monotone between them. The actual particle stays positive during all phases. This proves the equivalence for all real times, including the added entry step.

For the separate decaying choice apply Lemma 30: \(\widehat U(t,x)=(1+t)^{-1}U(\log(1+t),x)\), with its own residual at the same \(\nu\). That lemma gives the stronger temporal-order-\(k\) bound \(O((1+t)^{-1-k})\) for every spatial derivative, hence the stated bound and space-time \(L^2\) conclusion on the common compact support. The clock is onto \([0,\infty)\), and the chain rule identifies the new trajectory with \(X(\log(1+t);(2,0,0))\). The fixed event is unchanged. ◻

Proof. Use symbols \(c_i=(a_i,\ell_i,\mu_i)\) in \(C=A\times(\{\star,\perp\}\sqcup\mathcal R)\times\{0,1\}\) and controls \(N_q,B_q,J_r,S_r\). Only \(N_q\) with \(q\in H\) halt. After writing, a move \(d\) reindexes a relative tape by \(c'_j=\bar c_{j+d}\). The following rules give the complete domain:

1. At \(N_q\), \(q\notin H\), require \(\mu_0=0\) and \(\ell_0\ne\perp\). With \(r=(q,a_0)\) write \(b(r)\), move \(d(r)\), and enter \(J_r\).

2. At \(J_r\), require \(\mu_0=0\) and \(\ell_0\ne\perp\). Mark cell zero, move right, and enter \(S_r\).

3. At \(S_r\), require \(\mu_0=0\). If \(\ell_0\ne\perp\), scan right. If \(\ell_0=\ell_1=\perp\), write record \(r\) at zero, move left, and enter \(B_{q^+(r)}\).

4. At \(B_q\), require \(\ell_0,\ell_1\ne\perp\). Scan left if \(\mu_0=0\); if \(\mu_0=1\), clear the mark, stay, and enter \(N_q\).

Initialize history as \(\star\) below absolute cell 2 and \(\perp\) from 2 onward, all marks zero, the prescribed work tape, and control \(N_{q_0}\). At checkpoint \(n\) the first fresh cell is \(F=n+2\), records occupy \(2,\ldots,F-1\), and the work head satisfies \(h\le F-2\). The work move reaches \(k\le F-1\). Mark \(k\), scan to \(F\), write the record, and return from \(F-1\) to \(k\). The return’s two-history guard holds even at its first cell because cell \(F\) has just been filled. Clearing the mark restores the invariant with frontier \(F+1\). There are \(F-k-1\) instructions in each continuing scan and four others, so the duration is \(2(F-k)+2\). This proves finite completion and the checkpoint halt equivalence.

For the full-domain inverse, targets \(J_r,S_r,B_q,N_q\) determine moves \(d(r),1,-1,0\), respectively. Undo the indicated shift. At \(J_r\), \(r\) recovers the overwritten letter and old control. At \(S_r\), the written mark distinguishes entry from the zero-mark scan. At \(B_q\), target offset \(+2\) is the former offset \(+1\); its history is fresh for a recording entry and nonfresh for a return scan. In the entry case target offset \(+1\) is the former scanned cell; its record identifies \(r\). At \(N_q\), restore the mark to one. All other tracks are preserved. The neighbor guard is what makes this inverse valid on arbitrary allowed tapes.

Use odd digits in base \(B=2|C|+1\) and state offsets \(e_s\) whose unit intervals have gaps at least two. A permitted tuple \((s,c_{-1},c_0,c_1)\) has source \((e_s+I_{c_{-1}})\times I_{c_0c_1}\). The full inverse just proved checks the injectivity hypothesis of Lemma 15 with \((a,b)=(1,2)\) and \(d\in\{-1,0,1\}\). Only cell zero is written, which is within that window. The lemma therefore gives full target cylinders with prefix lengths \((1+d,2-d)\), exact affine factors \((B^{-d},B^d)\), and separate closed-rectangle images. For one target state the incoming \(d\) is fixed, so its prefix lengths agree across incoming branches. An incompatible digit among at most three positions yields the additional quantitative target gap \(B^{-3}\). This is separation of the filled rectangles, including their boundaries. ◻

Proof. Use the full alphabet \(C\) of Lemma 35 and \(B=2|C|+1\). Place halting controls at \(e_s=-3j_s\) and all others at \(3j_s\), with distinct positive integer labels. Use the three-cell rectangles as plates at height zero. Branch \(i\) has middle height \(4i\) and padding \(B^{-3}/10\). Interpolate both planar factors independently and apply Lemma 20, including its reciprocal normal strain. Planar gaps separate the vertical phases; middle heights separate the padded plates, whose collar half-thickness is less than one. Thus the specified motion holds on full neighborhoods. It retains independent planar interpolation throughout the cycle.

The initial code \(P_*\) is rational. A point-centered curl translation, with center \(\theta(t)P_*\) and padding one, loads zero in one unit. Repeat the plate cycle thereafter. The finite-step invariant and code maps identify the trajectory at all times \(1+n\) until halt and for all \(n\) otherwise. A halt code has first coordinate below \(-1\). For a nonhalting run, loading is nonnegative and each later pair of coded endpoints is positive; the two-coordinate convex interpolation of Lemma 20 proves the all-time exclusion. Use residual realization for the periodic choice and Lemma 30 for the separate decay choice. ◻

Proof. Full symbols are \((a,h,m)\), with work letter \(a\), mark \(m\in\{0,1\}\), and history \(h\in\{\#,E\}\sqcup\mathcal R\), where \(\mathcal R=(Q\setminus H)\times\Gamma\). History is called occupied when \(h\ne E\). A move \(d\) reindexes the post-write relative tape as \(\xi^{\rm new}_k=\widetilde\xi_{k+d}\). The complete table is \[\begin{array}{c|l|l|c|r} \text{control}&\text{guard}&\text{write at }0&\text{target}&d\\\hline \mathsf{Run}_q&m_0=0,\ h_0\ne E&a_0\leftarrow w_\tau& \mathsf{Mark}_\tau&d_\tau\\ \mathsf{Mark}_\tau&m_0=0,\ h_0\ne E&m_0\leftarrow1& \mathsf{Seek}_\tau&1\\ \mathsf{Seek}_\tau&m_0=0,\ h_0\ne E&\text{unchanged}& \mathsf{Seek}_\tau&1\\ \mathsf{Seek}_\tau&m_0=0,\ h_0=h_1=E&h_0\leftarrow\tau& \mathsf{Back}_{q^+_\tau}&-1\\ \mathsf{Back}_q&m_0=0,\ h_0\ne E,\ h_1\ne E&\text{unchanged}& \mathsf{Back}_q&-1\\ \mathsf{Back}_q&m_0=1,\ h_0\ne E&m_0\leftarrow0& \mathsf{Run}_q&0. \end{array}\] The first row requires \(q\notin H\) and \(\tau=(q,a_0)\); all unmentioned tracks are unchanged. Initialize the original work tape with head zero, all marks zero, history \(\#\) at every index below 2, and \(E\) elsewhere.

At checkpoint \(n\) the original head has \(|j_n|\le n\). The work step ends at \(j\le n+1<F_n\), on occupied history. Mark that cell and scan right across occupied unmarked cells to \(F_n\). Both \(F_n\) and \(F_n+1\) are unused, so the recording row applies and moves left. Every unmarked return-loop cell has occupied history at itself and its right neighbor, including the new record at \(F_n\). The return reaches \(j\), clears the mark, and enters the new work state. The continuing scans each take \(F_n-1-j\) steps, with four other steps. This proves finite completion, the duration, and the new occupied boundary \(F_n+1\). Induction gives the stated computation and halt equivalence.

For the full-domain inverse, target controls determine displacement. The moves are \(d_\tau\) into \(\mathsf{Mark}_\tau\), one into \(\mathsf{Seek}_\tau\), minus one into \(\mathsf{Back}_q\), and zero into \(\mathsf{Run}_q\). At \(\mathsf{Mark}_\tau\), undoing the move and restoring the read work symbol encoded by \(\tau\) recovers the predecessor. At \(\mathsf{Seek}_\tau\), target offset \(-1\) is the written cell; its mark distinguishes the marking entry from the scan. At \(\mathsf{Back}_q\), target offset 2 is the former offset 1. Its history is \(E\) for a recording entry and is occupied for a return loop. In the recording case, target offset 1 contains \(\tau\), identifying the old control and restoring its history to \(E\). Finally, the only entry into \(\mathsf{Run}_q\) clears a mark without moving, so the mark is restored to one. These inverses use only the table’s guards and retained symbols, not the special shape of initialized history. ◻

Proof. Apply Lemma 15 with window parameters \((a,b)=(1,2)\) to the complete table of Lemma 37. That table reads only offsets \(-1,0,1\), writes only offset zero, and has displacements \(d\in\{-1,0,1\}\subset[-1,2]\); its full-domain injectivity was proved above. Thus every hypothesis of the window lemma holds.

To fix the actual rectangles, choose odd digits \(\gamma(a)\in\{1,3,\ldots,2|\mathcal A|-1\}\) and \(B=2|\mathcal A|+1\). Write \[e(P_1\cdots P_k)=\sum_{j=1}^k\gamma(P_j)B^{-j},\qquad I(P)=e(P)+[0,B^{-k}],\qquad I(\varnothing)=[0,1],\] using the infinite series for an infinite word. Choose separated state intervals \(o_s+[0,1]\) and use the code \[r(s,\xi)= (o_s+e(\xi_{-1}\xi_{-2}\cdots),\ e(\xi_0\xi_1\cdots),0).\] For each permitted triple \((l,a,b)\) at offsets \(-1,0,1\), let \(a'\) be its written full symbol. The source prefix pair is \(((l),(a,b))\), and the target pair is \[\begin{cases} ((a',l),(b)),&d=1,\\ ((l),(a',b)),&d=0,\\ (\varnothing,(l,a',b)),&d=-1. \end{cases}\] Take the corresponding products of prefix intervals with the source and target state offsets. Both arbitrary tails are unchanged, so the window lemma gives the exact center-to-center affine map with factors \(B^{-d},B^d\), onto the whole specified target cylinder and filled rectangle. Its separation conclusion applies to these complete closed rectangles, including their boundaries. All endpoints and positive within-family gaps are effectively rational. The odd digits also put all infinite-word codes strictly in \((0,1)\) in each normalized coordinate. ◻

Proof. Use the occupied-prefix recorder. Assign offsets \(2j\), \(j\ge1\), to halting run controls and \(-2j\) to all other controls, with distinct indices within each group. The initial code \(r_{\rm in}\) is rational: the left stream is constant and the right stream is eventually constant. If \(N\) is the number of cylinders, set \[O=1+\max_s|o_s|,\qquad K=O+6\max(1,N)+4,\qquad \sigma=(8K)^{-1},\qquad b_*=(-K/2,0,0).\] Use the linear version of Lemma 19 for the branch boxes of third interval \([-1,1]\), obtaining \(V_1\). Use the same lemma with scale one on the single pair of cubes of half-width one centered at \(b_*\) and \(r_{\rm in}\), obtaining a loader \(V_0\). Extend both fields by zero outside their unit time intervals. Their supports lie in \((-K,K)^3\). For \(V_1\), first-coordinate centers have magnitude at most \(O\), second-coordinate centers lie in \([0,1]\), horizontal half-widths are at most \(1/2\), and heights are at most \(6N\). For \(V_0\), the horizontal center is bounded by \(\max(K/2,O)\), half-widths are one, and height is at most six. The padding is at most \(1/4\) in either application. Each bound is strictly less than \(K\).

Periodize the rescaled fields in space and the step field in time: \[U(t,[x])=\sigma\sum_{k\in\mathbb Z^3} \left[V_0(t,(x-k)/\sigma)+ \sum_{n\ge1}V_1(t-n,(x-k)/\sigma)\right].\] Spatial support lies in chart boxes of radius \(1/8\), so distinct spatial copies are disjoint. The time collars make the gluing smooth, initially zero and periodic after time one. Every mixed derivative is bounded. The rescaled blocks remain curls of compactly supported potentials, so \(U\) has zero spatial mean. Define the force at the physical viscosity by \(f=U_t+(U\cdot\nabla)U-\nu\Delta U\). It has zero mean because the convection term is \(\operatorname{div}(U\otimes U)\) and the Laplacian integrates to zero. Lemma 21 now gives the unique solution \((U,0)\). Recomputing the residual after the chart scaling is essential: it does not identify two different viscosities.

The prescription is effective without deciding a space or time seam. Given approximations \(\widehat x,\widehat t\) with error at most \(1/4\), all possibly active spatial indices satisfy \(\|\widehat x-k\|_\infty\le1\), and repeated time indices satisfy \(|\widehat t-n|\le2\). Evaluate this finite superset using the flat cutoffs and their derivative bounds. The finite table, not an executed machine history, determines every coefficient.

Since \(\sigma b_*=(-1/16,0,0)\), loading sends the fixed particle to \([\sigma r_{\rm in}]\). At time \(1+j\) it is the code after \(j\) local steps, through the first halt or forever in the nonhalting case, by the exact whole-box maps. Halting, including initial halting, gives a positive first-coordinate representative below \(1/8\), inside the target half-circle. For a nonhalting input both loading endpoints and both endpoints of every subsequent used branch have negative first coordinate. The linear-scale identity keeps the first coordinate negative at every intermediate time. All representatives stay in \((-1/8,0)\), disjoint from the target modulo one. This proves the all-time equivalence. ◻

Proof. Add a fresh nonhalting initial state \(*\), without incoming transitions, which writes back the scanned symbol and stays while entering the original initial state. This handles an originally halted input. Use work, flag and history tracks \(\Sigma=\Gamma\times\{0,1\}\times(\{\perp,E\}\sqcup\mathcal R)\), and controls \(R_q,A_r,B_r,C_q,D_q\). Its complete table is \[ \begin{array}{lll} R_q:(a,0,g)&\mapsto A_r:(b_r,0,g),d_r,\\ A_r:(a,0,g)&\mapsto B_r:(a,1,g),1,\\ B_r:(a,0,g)&\mapsto B_r:(a,0,g),1&g\ne E,\\ B_r:(a,0,E)&\mapsto C_{p_r}:(a,0,r),1,\\ C_q:(a,0,\perp)&\mapsto D_q:(a,0,E),-1,\\ D_q:(a,0,g)&\mapsto D_q:(a,0,g),-1&g\ne E,\\ D_q:(a,1,g)&\mapsto R_q:(a,0,g),0&g\ne E. \end{array} \tag{33}\] In particular its first two rows do not exclude the frontier symbol; this is a wider domain than the move-first table. Into \(A_r\) the record identifies the work instruction; into \(B_r\) the output flag distinguishes entry and loop; into \(C_q\) the record determines the incoming scan; into \(D_q\) a written \(E\) distinguishes entry and loop; into \(R_q\) only clearing a flag is possible. The destination fixes the incoming displacement. These checks include every copied track and every permitted frontier case.

Initialize at \(R_*\), with all flags zero and only \(E\) at cell two. At ready step \(n\) the frontier is \(e_n=n+2\), the records are at \(2,\ldots,n+1\), and the head is \(h_n\), \(|h_n|\le n\). The work move reaches \(h'=h_{n+1}<e_n\); the scan marks that cell, records at \(e_n\), creates the next frontier, returns to the mark and clears it. Its count is \(2(e_n-h')+4\) and all its tests hold. Thus it has exactly the required initialized behavior despite its wider full domain.

Use even digits in base \(b=2|\Sigma|\) and initial fractions \(\ell_*,r_*\). Let \(S\) be the control set and \(m\) count the branches in Lemma 10, and put \(L=100(1+|S|+m)\). Give \(R_*\) offset \(-\ell_*\), other nonhalt states offsets \(-4i\), and ready halt states offsets \(L+4i\). The code is \((o_s+\ell,r-r_*,0)\), so the initial point is zero. The incoming conditions give separately disjoint full rectangle families. Thicken each by \([-1,1]\), lift to height \(10i\), and use the linear reciprocal middle scale. Choose padding at most \(1/10\) small enough for both family gaps. Vertical phases can use fixed column potentials \((0,\dot Z x\chi,0)\), with columns covering \([-3,10i+3]\); middle phases use horizontal Hamiltonians in disjoint height bands. All paths and supports are contained in \((-2L,2L)^3\): their coordinate bounds are \([-4|S|-3,L+4|S|+4]\), \([-4,4]\), and \([-3,10m+3]\). This verifies the hypotheses of the lift schedule on neighborhoods of the whole boxes.

Scale by \(\gamma=(8L)^{-1}\) into the torus patch \((-1/2,1/2)^3\): \(U(t,x)=\gamma W(t,x/\gamma)\). The support lies inside \((-1/4,1/4)^3\), the potentials scale by \(\gamma^2\), and the residual is recomputed at viscosity \(\nu\) by Lemma 3. Periodicity starts at zero because the input is already encoded there and the pulse has idle endpoint collars. At integer times the particle is the correct code. Halt first coordinates before scaling lie between \(L+4\) and \(2L\). For a nonhalt branch its endpoint first coordinates are at most one; the linear-scale identity keeps them at most one, and the weaker bound \(3/2\) also follows from center and width estimates. After scaling every nonhalt path lies in \((-1/4,1/16)\), disjoint modulo one from the event. The initialized dummy step makes the implication valid even for an originally halted machine.

For the optional projection solve \(\Delta q=\operatorname{div}g\), \(\int q=0\), where \(g=\mathcal F_\nu[U]\). Proposition 4 gives the effective force \(f=g-\nabla q\) and pressure \(p=-q\), with all mixed derivative bounds and periodicity retained. The pressure change is part of this alternative; the direct residual had pressure zero. ◻

Proof. Normalize to one halt state and use work, permanent work-head marker, and history tracks \((a,m,g)\in\mathcal A:=\Gamma\times\{0,1\}\times (\{\bot,\#\}\sqcup\mathcal R)\). Let \(S\) be the set of controls \(M_q,L_q,R_r\). With the cursor at zero, make exactly the following tests and writes, then reindex by \(T'_i=\bar T_{i+e}\) for the displayed shift \(e\).

1. In \(M_q\), \(q\) nonhalting, require \(m_0=1\) and, if \(d_r\ne0\), \(m_{d_r}=0\). Write \(b_r\) at zero, move the marker to \(d_r\) if necessary, and enter \(R_r\) with shift \(e=d_r+1\).

2. In \(R_r\), a cell with \(m_0=0,g_0\ne\#\) gives shift \(1\) in the same control. If \(m_0=0,g_0=\#,g_1=\bot\), write \(g_0=r,g_1=\#\) and enter \(L_{p_r}\) with shift zero.

3. In \(L_q\), if \(m_0=0,g_0\ne\#\), shift \(-1\) in that control. If \(m_0=1\), enter \(M_q\) with shift zero.

All unmentioned entries are retained. Into \(R_r\), a main entry has output \(m_{-1}=1\), whereas a scan has \(m_{-1}=0\). The record recovers the overwritten work letter, displacement and marker changes. Into \(L_q\), a log write has output \(g_1=\#\) and record \(r\) at zero, whereas a scan has \(g_1\ne\#\). Its tests recover the overwritten pair. Into \(M_q\) only the matching control change is possible. This is a full-domain inverse; uniqueness of a marker was not assumed on arbitrary tapes.

Initially the sole marker is at zero, the sole pointer at two, and all other history is blank. At work step \(n\), head and marker are at \(h\), pointer at \(n+2\), and records at \(2,\ldots,n+1\). The work update puts the marker at \(h'=h+d_r\) and the cursor at \(h'+1\le n+2\). The forward scan reaches the pointer, logs, and returns to \(h'\). All tests hold since \(h'<n+2\). The cycle length is \[1+[n+2-(h'+1)]+1+[(n+2)-h']+1.\] This proves finite continuation, exact work simulation and halting.

Use even digits, blank zero, in base \(B=2|\mathcal A|+1\). Enumerate every full word on \([-2,2]\) with an applicable rule, giving \(J\) branches. Lemma 15 gives prefix lengths \((2,3)\) and \((2+e,3-e)\), with \(e\in\{-1,0,1,2\}\). Each output length is at most four, so gaps after chart scaling by \(l=[100(|S|+1)]^{-1}\) are at least \(lB^{-4}\). Place the halt square at \((3/4,1/4)+l[0,1]^2\) and other squares at \((1/4+2il,1/4)+l[0,1]^2\); the latter lie below \(x_1=1/3\). At base height \(1/4\), lift branch \(j\) to \(1/2+j/[4(J+1)]\), use exponential scales \(B^{-e_j\theta},B^{e_j\theta}\), and lower. Padding \(\min(1/200,1/[32(J+1)],lB^{-4}/3)\) separates supports. The middle Hamiltonian is \[\dot c_x(y-c_y)-\dot c_y(x-c_x) -e_j\dot\theta\log B\,(x-c_x)(y-c_y).\] It gives the intended derivative on the full moving rectangle. The plateau margins and finite scale bounds extend this to an open neighborhood, as in Lemma 5.

Load the fixed label along its segment to the rational initial code in time one, then repeat the unit field. Integer samples give the full local execution. In a nonhalt period the center is below \(1/3\) and the width at most \(l\), so the whole path remains below \(1/2\); loading also avoids the event. Halt codes lie in \([3/4,3/4+l]\). Thus every real time is accounted for. The pointer site \(n+2\), record count \(n\), and relative pointer offset recover the absolute head position at main checkpoints. The residual construction proves the remaining assertions. ◻

Proof. Use exactly the direction table of Lemma 12 without its passive track, renaming the controls by \(M\mapsto S\), \(G\mapsto R\), \(A\mapsto I\), and \(B\mapsto J\), and the frontier by \(F\mapsto\#\). This is a bijection of its entire partial table, including the nonfrontier tests. Write \(\Sigma\) for the resulting three-track alphabet. Initialize frontier at one and zero flags. At work step \(j\) the frontier is \(j+1\) and the cycle length is \[2(j-h_j)+5.\]

Use odd digits in base \(B=2|\Sigma|+1\), offsets \(3,6,\ldots\) for nonhalt controls and \(-3,-6,\ldots\) for halt main controls. In this realization split every one-cell rule by every left symbol \(c\), including the stationary and right-moving rules. Its source is \(I(c)\times I(a)\), and the three target rectangles, for moves \(1,-1,0\), are respectively \[\varphi_b\varphi_c([0,1])\times[0,1],\quad [0,1]\times\varphi_c\varphi_b([0,1]),\quad I(c)\times I(b), \qquad \varphi_a(t)=(d(a)+t)/B.\] The maps themselves are the restrictions of (19). Source separation follows from \((c,a)\), target separation from the fixed incoming move and \((c,b)\). Their gaps are at least \(B^{-2}\), and their linear parts are \(\operatorname{diag}(B^{-d},B^d)\).

Thicken by \([-1,1]\), lift branch \(i\) to \(4i\), use exponential middle scale and center interpolation, then descend. Padding smaller than \((4B^2)^{-1}\) is valid, by source gaps, height gaps, and target gaps in the three phases. A first localized translation takes \((1,0,0)\) to the rational initial code. Exact full-box transport then gives the local execution at every subsequent integer time. Nonhalt centers have first coordinate at least three and horizontal width at most one, so every intermediate point stays positive; the nonhalt loading segment does too. A halt sample has negative first coordinate, including initial halting after loading. Proposition 6 gives the force, pressure zero, and uniqueness. The decaying alternative follows from Proposition 73 below, applied to this same loaded template. ◻

Proof. Add a fresh initial state which writes back and stays, and replace all originally terminal state-symbol pairs by stay transitions into one fresh terminal state \(q_h\). Normalize the other missing instructions similarly. Let \(r=(q,a)\mapsto(p_r,b_r,d_r)\) denote the resulting nonterminal instructions, and let \(\mathcal A\) be the work alphabet. The full cell alphabet is \(\Gamma=\mathcal A\times\{0,1\}\times(\{e,p\}\sqcup\mathcal R)\). Use its work, flag and log tracks \(W,F,L\), and controls \(N_q,A_r,R_r,B_q\). The complete rules are:

1. At \(N_q\), \(q\ne q_h\), write \(W_0=b_r\), enter \(A_r\), and shift \(d_r\), with no test on the other tracks.

2. At \(A_r\), require \(F_0=0\), set it to one, enter \(R_r\) and shift one.

3. At \(R_r\), if \(F_0=0,L_0\ne p\), shift one in that control. If \(F_0=0,L_0=p,L_1=e\), write \(L_0=r,L_1=p\), enter \(B_{p_r}\) and shift minus one.

4. At \(B_q\), if \(F_0=0,L_1\ne p\), shift minus one in that control. If \(F_0=1\), clear it and enter \(N_q\) without moving.

The first output control records the overwritten work pair and shift. Into \(R_r\), output flag one at \(-1\) distinguishes entry from its loop. Into \(B_q\), a log insertion has pointer at output index two and record at one, whereas a return loop has a nonpointer at two; the record and tests recover both overwritten log entries. Into \(N_q\) only unmarking is possible. Thus this map is injective on its whole domain, including the unrestricted first-row tracks.

Initialize zero flags and a sole pointer at cell three, all other log cells empty. At work step \(n\), records occupy \(3,\ldots,n+2\) and pointer \(g=n+3\). The work move reaches \(h'\) with \(g-h'\ge2\). The finite scan flags \(h'\), logs at \(g\), advances the pointer, and returns to the flag. The return guard holds because the new pointer is at \(g+1\). The cycle has \(2(g-h'-1)+4\) steps. This verifies the work invariant and eventual terminal entry.

Use odd digits in radix \(D=2|\Gamma|+1\) and write \(\phi(\xi)=\sum_{j\ge1}d(\xi_j)D^{-j}\) for the radix sum. Let \(L_*,R_*\) be the initialized left and right streams, ordered as in Lemma 10. Give the terminal state chart offset \((-2,0)\), the initial chart offset \((2-\phi(L_*),-\phi(R_*))\), and all other charts \((4+2j,0)\). The initial code is \((2,0,0)\), every nonterminal interval lies in \(x_1>1\), and terminal codes lie strictly in \((-2,-1)\times(0,1)\). The infinite tails are in \([\delta,1-\delta]\), \(\delta=1/(D-1)\), so finite-prefix decoding of length \(b\) is possible from error less than \((\delta/2)D^{-b}\) without a boundary ambiguity.

Split full triples \([-1,1]\). By Lemma 15, source prefix lengths \((1,2)\) become \((1+d,2-d)\), target cylinders are complete, and both rectangle families have gaps at least \(D^{-3}\). Thicken by \([-1/4,1/4]\), lift to height \(4i\), use linear middle scale \(\mu_i=1+(D^{-d_i}-1)\theta\), and descend. Padding smaller than \(D^{-3}/4\) separates all phases. Repeating the pulse, zero on \([0,1/12]\) and \([11/12,1]\), gives periodicity from zero without loading. At a nonhalting branch’s lift and descent the first coordinate is greater than one; during the middle transfer the particle’s third coordinate is exactly \(4i\), outside \((-1,1)\). Thus even a horizontal route passing over the observation box cannot cause a false visit. The integer samples and the compiler invariant prove halting equivalence. The record count and pointer site \(n+3\) recover absolute indexing at checkpoints.

For uniqueness apply Lemma 2(iii). The explicit compactly supported smooth field lies in \(CH^4\cap C^1H^2\) on every finite interval and has pressure zero. The competing class in the theorem is precisely case (iii), including its distributional \(CL^2\) gradient. Lemma 1 cancels that gradient against the solenoidal velocity difference without introducing a pressure \(L^2\) norm. Thus the velocity and pressure gradient are unique, while pressure itself remains free up to a time-dependent spatial constant. Proposition 73 gives the stated decaying alternative. ◻

Proof. Merge terminal states into \(q_\dagger\), send missing instructions there by a stay step, and add a fresh initial state \(q_*\), with no incoming instruction, which writes back and stays while entering the old initial state. For each normalized nonterminal rule \(r=(q,a)\mapsto(p_r,b_r,d_r)\) use cell letters \[(w,h,j)\in\Gamma:=\mathcal A\times (\{\circ,\star\}\sqcup\{[r]:r\in\mathcal R\})\times\{0,1\},\] where \(\mathcal A\) is the work alphabet and \(\mathcal R\) is the finite set of normalized instruction labels. Thus \(\Gamma\) includes all three tracks. The complete table, with \(h\ne\star\) wherever \(h\) occurs, is \[\begin{array}{lll} M_q:(a,h,1)&\mapsto A_r:(b_r,h,0),d_r,\\ A_r:(w,h,0)&\mapsto R_r:(w,h,1),1,\\ R_r:(w,h,0)&\mapsto R_r:(w,h,0),1,\\ R_r:(w,\star,0)&\mapsto B_{p_r}:(w,[r],0),1,\\ B_q:(w,\circ,0)&\mapsto L_q:(w,\star,0),-1,\\ L_q:(w,h,0)&\mapsto L_q:(w,h,0),-1,\\ L_q:(w,h,1)&\mapsto M_q:(w,h,1),0. \end{array}\] The destination fixes the incoming displacement. The record index recovers the first row’s old work pair; a flag distinguishes the entries into \(R_r\); a record distinguishes the entries into \(B_q\); a written frontier distinguishes entry and loop into \(L_q\); and entry into \(M_q\) is the unique cell-preserving last row. This proves the full one-cell inverse conditions.

Initialize the sole flag at zero and the sole frontier at three, with empty history elsewhere. At ready step \(k\), only the work head \(i\) is flagged, records occupy \(3,\ldots,k+2\), and the frontier is \(e_k=k+3\). The work rule clears the flag and moves to \(j=i+d_r\le e_k-2\), which is unflagged even for \(d_r=0\). It is flagged anew; the scan logs at \(e_k\), advances the pointer, returns, and preserves the ready flag. The two loop lengths are \(e_k-j-1\) and \(e_k-j\), with five other rules, for a total \(2(e_k-j)+4\). Thus all scans terminate and ready halting is exact.

Use odd digits in base \(K=2|\Gamma|+2\), initial fractions \(x_*,y_*\), and state offsets \(a_{M_{q_*}}=0\), \(a_{M_{q_\dagger}}=-5\), and \(4,8,12,\ldots\) for the others. The code \((a_s-x_*+x,y-y_*,0)\) starts at zero. Lemma 10 gives \(N\) full rectangle maps with scales \(\lambda_i\in\{K^{-1},1,K\}\); thicken by \([-1/4,1/4]\). The entire terminal block is inside the observation box because \(0<x_*,y_*<1\). Every nonterminal block is in \(x_1>-1\). Eventually constant tails are rational even for an arbitrary finitely specified initial tape, and the pointer and record count recover its absolute head position at ready checkpoints.

For the required two-pass geometry, let \(g\) be the minimum of one and all positive separating coordinate gaps in either planar family, and set \(\epsilon=g/20\). Choose \[C=20+\max(\{0\}\cup\{\text{right endpoints of source and target boxes}\}), \qquad S_i=(C+4i,0,0).\] In a first pass give each source four consecutive slots: lift by ten, translate above \(S_i\), descend, and reshape there by the linear reciprocal scale. In a second pass give each parked box three slots: lift by ten, translate above its target, and descend. The total is \(7N\) slots, and every source is gone before a target is filled. Widths remain at most one. Source gaps protect the first lift, parking spacing protects strain and parking moves, target gaps protect delivery, and travel height separates every horizontal transfer from resting boxes. These inequalities remain strict under \(2\epsilon\) padding. Applying Lemma 5 in chronological order fixes every stationary box and gives the exact full-box map to the moving one. Thus overlapping source and target families cause no interference.

Repeat the flat-ended template from zero and take its residual. The integer sampling induction gives the table execution. In a nonhalting run every endpoint box lies in \(x_1>-1\); translations interpolate corresponding endpoint points, strain occurs far to the right, and waiting leaves the point fixed. The entire two-pass path therefore remains in \(x_1>-1\). A terminal sample lies inside the observer. This proves the event for all times.

For the exact pressure class use Lemma 2(iv). The prescribed compactly supported smooth velocity satisfies the reference \(CH^2\cap C^1L^2\) and bounded-gradient conditions. The competing velocity has precisely these bounds and a \(CL^2\) pressure representative, as that case requires. Its cutoff argument uses the pressure itself, with error \(CR^{-1}\|p\|_2\|u-U\|_2\), and assumes no extra integrability of its derivative. The lemma gives velocity uniqueness; the \(L^2\) pressure representative is then zero. Compact support and bounded derivatives of the prescribed velocity give its uniform energy bound and its global material flow. ◻

Proof. Let \(a\) be the augmented alphabet size and \(k\) the control count in Lemma 14. Use zero-blank base \(B=2a\), and enumerate nonterminal and terminal states with positive integer indices \(j\ge1\) and first-coordinate offsets respectively \(L_0+3j\) and \(-L_0-3j\), where \(L_0=4k+ka^3+10\). Refinement by triples gives at most \(N=ka^3\) branches. Their source prefix lengths are \((1,2)\) and image lengths \((1+e,2-e)\), so the smallest possible grid gap is \(B^{-3}\). If \(P_i,Q_i\) are lower corners, the branch map is \(Q_i+\mathop{\mathrm{diag}}(B^{-e_i},B^{e_i})(y-P_i)\).

Choose three flat ramps \(\alpha,\beta,\gamma\), active respectively in \((1/8,1/4)\), \((3/8,5/8)\) and \((3/4,7/8)\). Set \[c_i(t)=((1-\beta)P_i+\beta Q_i,\ i(\alpha-\gamma)), \quad l_i=1-\beta+\beta B^{-e_i}.\] The moving rectangle has side lengths \(l_i B^{-1}\) and \(l_i^{-1}B^{-2}\). Use a physical padding \(\varepsilon=B^{-3}/4\) around each moving rectangle in three dimensions. During ascent and descent the planar endpoint gaps separate these neighborhoods; during the affine phase the unit height gaps separate them. Use Lemma 5 for the paths \(c_i+\mathop{\mathrm{diag}}(l_i,l_i^{-1},1)(y-(P_i,0))\) and sum the disjoint fields. The first coordinate is an endpoint convex combination even though the paths are based at lower corners rather than centers. All supports lie in \([-2L_0,2L_0]^3\), by the definitions of \(L_0\) and \(N\).

Scale by \(\rho=1/(32L_0)\), so the repeated mechanism is supported in \([-1/16,1/16]^3\). Load the fixed label \((1/8,0,0)\) to the scaled rational input code by a moving translation with cutoff radius \(1/32\); its support stays in \((-1/4,1/4)^3\). For a nonterminal initial code, the entire first coordinate during loading is positive, and subsequent nonterminal branches have first coordinate greater than \(1/32\). Terminal codes have first coordinate less than \(-1/32\). Hence the stated strip detects exactly terminal entry, including initial halting, without any wraparound artifact. Periodize the compact curl fields and recompute the force at physical \(\nu\). Mean zero, bounded derivatives and comparison follow from Lemma 18. ◻

Proof. Normalize to one halt state \(h\). For \(i=(q,a)\in I=(Q\setminus\{h\})\times A\), write \(\delta(q,a)=(q_i,b_i,e_i)\). History letters are \(\mathcal H=\{\perp,\#\}\sqcup I\), with \(\mathcal H_+=\{\#\}\sqcup I\). The tracks are work, mark and history; controls are \(\mathsf R_q,\mathsf B_q,\mathsf T_i,\mathsf S_i\). Write \(\ell_j\) for the history at relative cell \(j\). Here is the complete table, with unmentioned cells unchanged: \[ \begin{array}{c|c|c|l} \text{input}&\text{output}&\text{move}&\text{condition}\\\hline \mathsf R_q(a,0,k)&\mathsf T_i(b_i,0,k)&e_i&i=(q,a),\ q\ne h\\ \mathsf T_i(c,0,k)&\mathsf S_i(c,1,k)&1&k\in\mathcal H_+\\ \mathsf S_i(c,0,k)&\mathsf S_i(c,0,k)&1&k\in\mathcal H_+\\ \mathsf S_i(c,0,\perp)&\mathsf B_{q_i}(c,0,i)&0&\ell_{+1}=\perp\\ \mathsf B_q(c,0,k)&\mathsf B_q(c,0,k)&-1&k\in\mathcal H_+\\ \mathsf B_q(c,1,k)&\mathsf R_q(c,0,k)&0&k\in\mathcal H_+. \end{array} \tag{34}\] Initialize \(\mathsf R_{q_0}\) at head zero with marks zero, history \(\#\) at every index below two and \(\perp\) elsewhere. At checkpoint \(n\), the work head \(p_n\) has \(|p_n|\le n\) and the boundary is \(r_n=2+n\). After the work move, \(p'=p_n+e_i\le n+1<r_n\). Thus the mark is placed on nonempty history, and the right scan reaches the first empty cell. Its right neighbor is also empty, so the append is allowed. A finite left scan returns to and erases the unique mark. The new boundary is \(r_n+1\); no other work symbol changes. The filler boundary at absolute index two is permanent, so it also retains the tape’s absolute alignment.

Into \(\mathsf T_i\) the tag gives the old work instruction and shift. Into \(\mathsf S_i\), both shifts are rightward, and the mark at the old cell distinguishes entry from a loop. Into \(\mathsf B_q\), an output with empty history at offset \(+1\) must be an append: its current record identifies \(i\). An output of a leftward loop has nonempty history at that offset, since it is the old current cell. This is exactly why the append guard is necessary. Into \(\mathsf R_q\) only mark erasure applies. Each case gives a unique predecessor on the full domain. In particular, entry into \(\mathsf B_q\) may have displacement zero or minus one; the neighboring history, rather than the target control alone, recovers that displacement. Apply Lemma 15 with \((a,b)=(1,2)\), refining by the full cells at \(-1,0,1\). Its full-domain injectivity hypothesis has just been proved, and every write and displacement lies in the allowed window. It therefore gives full image cylinders and separately disjoint closed rectangle families. ◻

Proof. For a source rectangle \(D\), choose a planar cutoff \(\theta_D\) with padding \(\varepsilon=\sigma_0b^{-3}/4\), equal to one near \(D\). Define \[\kappa_D(x,y)=\theta_D(x,y)-\theta_D(x,y-1/2)\] periodically. It has integral zero. On its own source it equals one, and on all other sources it equals zero; the shifted copy lies away from \(F\). Use the same construction for target rectangles \(E\). Assign branch \(i\) the height \(z_i=1/2+i/[4(N+1)]\) and choose disjoint height cutoffs \(\rho_i\) equal to one near \(z_i\), with padding \(1/[16(N+1)]\). The vertical fields with speeds \[Z_D(x,y)=\sum_i(z_i-z_*)\kappa_{D_i}(x,y),\qquad Z_E(x,y)=\sum_i(z_i-z_*)\kappa_{E_i}(x,y)\] are divergence free and mean zero, because they are independent of \(z\).

Let \(\chi\) equal one near \(F\), with planar padding \(1/32\). For branch \(i\), let \(p_i,q_i\) be the source and target centers and put \(d_i=q_i-p_i\), \(c_i(\mu)=p_i+\mu d_i\), \(l_i=1-\mu+\mu b^{-e_i}\) and \[H_i=\chi\left[d_{i,1}y-d_{i,2}x+ \frac{b^{-e_i}-1}{l_i}(x-c_{i,1})(y-c_{i,2})\right].\] The middle field \(W(\mu)=\sum_i\rho_i(z)(\partial_yH_i,-\partial_xH_i,0)\) is divergence free and mean zero. The path \[ c_i(\mu)+\mathop{\mathrm{diag}}(l_i,l_i^{-1})(\zeta-p_i) \tag{36}\] stays in \(F\): the first side is interpolated linearly and the reciprocal inequality (24) bounds the second side by the interpolated endpoint sides. Thus \(\chi=1\) along the entire path, and differentiation gives its exact velocity. For three ordered flat ramps \(\alpha_0,\alpha_1,\alpha_2\) on a unit interval, the complete field is \[ V=\alpha_0'(0,0,Z_D)+\alpha_1'W(\alpha_1) -\alpha_2'(0,0,Z_E). \tag{37}\] It lifts each source, performs its affine map in its own height layer, and lowers it at its target. The first coordinate is unchanged during vertical motion and is an endpoint convex combination in the middle. All statements hold on the full rectangles by the cutoff plateaus and ODE uniqueness. The zero-mean correction acts away from the computational rectangles and therefore changes none of these paths. ◻

Proof. The two initial tape tails are separately constant, so their radix sums are computable rationals, even though the left history tail is filled. Load the fixed label to that code at height \(z_*\) by the planar Hamiltonian \(\chi(d_x y-d_y x)\) with a flat scalar schedule. The segment stays in \(F\). A nonterminal load and every nonterminal phase have first coordinate at most \(3/8\); a terminal code lies in the stated strip. For an initially terminal machine the loading endpoint suffices. Repeat (37) after time one. Checkpoint simulation and the intermediate-time guard prove the event equivalence. The common analytic lemma proves force effectivity, uniqueness, boundedness and the asserted flow identities. In particular \(U=(\partial_t\Phi_t)\circ\Phi_t^{-1}\) and \(\partial_{tt}\Phi_t=(f-\nabla p+\nu\Delta U)\circ\Phi_t\); these are identities for the constructed flow, not additional unknown equations. ◻

Proof. The empty family is realized by zero. Otherwise put \(h_i=4i\), \(w_i=c_i'-c_i\), and \[\Theta_k(s)=\vartheta(10s-(2k-1)),\qquad 1\le k\le4.\] Only the \(k\)th motion is active on the interval \([(2k-1)/10,2k/10]\); the four intervals are disjoint. If \(a_{ij},a'_{ij}\) are the source and target half-widths, respectively, take the four swept boxes \[\begin{aligned} K_{i1}&=D_i\times[0,h_i],\\ K_{i2}&=\{c_i+(u_1,u_2):|u_j|\le\max(a_{ij},a'_{ij})\} \times\{h_i\},\\ K_{i3}&=\operatorname{box}(E_i-w_i,E_i)\times\{h_i\},\\ K_{i4}&=E_i\times[0,h_i]. \end{aligned}\] Here \(\operatorname{box}\) denotes the smallest axis-aligned rectangle containing the two displayed rectangles. Within each of the four families these boxes are disjoint: source separation handles the first, target separation the fourth, and distinct heights the other two. Choose a positive rational \(\epsilon<1\) smaller than one quarter of every pairwise distance within these families. A family with fewer than two members contributes no restriction. Its \(\epsilon\) enlargements are then still disjoint.

Choose either \[\mu_i=1+(\lambda_i-1)\Theta_2 \quad\hbox{or}\quad \mu_i=\lambda_i^{\Theta_2}, \qquad \gamma_i=\dot\mu_i/\mu_i.\] The factors \(\mu_i\) and \(\mu_i^{-1}\) stay between their respective endpoint values. Use the coordinate potentials \[\begin{aligned} B_{i1}&=(0,h_i x_1,0),& B_{i2}&=(0,0,(x_1-c_{i1})(x_2-c_{i2})),\\ B_{i3}&=(0,0,w_{i1}x_2-w_{i2}x_1),& B_{i4}&=(0,-h_i x_1,0). \end{aligned}\] For the logarithmic-scale construction we also retain the alternative translation potentials, with \(e_3=(0,0,1)\), \[B_{i1}=\tfrac12(h_ie_3)\times x,\qquad B_{i3}=\tfrac12(w_{i1},w_{i2},0)\times x,\qquad B_{i4}=\tfrac12(-h_ie_3)\times x.\] Their curls equal the same constant translation vectors. Their localized fields need not agree away from the plates. Either choice gives \[W=\sum_i\left[ \dot\Theta_1\nabla\times(\chi_{K_{i1},\epsilon}B_{i1}) +\gamma_i\nabla\times(\chi_{K_{i2},\epsilon}B_{i2}) +\dot\Theta_3\nabla\times(\chi_{K_{i3},\epsilon}B_{i3}) +\dot\Theta_4\nabla\times(\chi_{K_{i4},\epsilon}B_{i4})\right].\] This is a finite sum of spatial curls. Its spatial support is compact, its time support is contained in \([1/10,8/10]\), and the lower bound \(\mu_i\ge\min(1,\lambda_i)>0\) proves smoothness and effectivity.

For \(b\in D_i\), the successive paths are \[\begin{array}{c|c} \text{motion}&\text{position}\\\hline \text{ascent}&(b,h_i\Theta_1)\\ \text{scaling}&(c_i+\operatorname{diag}(\mu_i,\mu_i^{-1})(b-c_i),h_i)\\ \text{translation}&(A_i(b)-w_i+\Theta_3w_i,h_i)\\ \text{descent}&(A_i(b),h_i(1-\Theta_4)). \end{array}\] Each path stays in its swept box. On a neighborhood of that box the uncut curls are, in order, \[h_ie_3,\qquad (x_1-c_{i1},-(x_2-c_{i2}),0),\qquad (w_{i1},w_{i2},0),\qquad -h_ie_3.\] After multiplication by the time coefficients, these are the prescribed affine velocities. The motion matrices are positive diagonal of determinant one, the plateau margin is \(\epsilon/2\), and the support padding is \(\epsilon\). The disjoint swept boxes therefore permit the plateau argument of Lemma 5, with \(\eta=\epsilon/2\). It identifies these particular localized fields with the displayed paths on the plateaux. The normalized diagonal factors are bounded by \(L_i=\max(1,\lambda_i,\lambda_i^{-1})\), so that lemma gives the full initial neighborhood of radius \(\epsilon/(4L_i)\). ◻

Proof. Normalize to a single halt state \(q_H\) and a complete nonhalting table, as in Section 6. Apply the move-first recorder of Lemma 11. Its history frontier starts at cell \(2\); all other history cells are empty and all mark bits are zero. At checkpoint \(n\), the frontier is \(g_n=n+2\), the work head satisfies \(|h_n|\le n\), and the next work instruction ending at \(h'\) is completed in exactly \(4+2(g_n-h')\) local steps. Thus every required checkpoint is reached after finitely many steps. The incoming-direction and written-symbol inverse properties hold on the whole partial table. The records recover the successive work instructions and hence the absolute head position, even though the geometric code is head-relative.

Let \(m\) be the full alphabet size. Use odd digits \(\beta(a)\) in \(\{1,3,\ldots,2m-1\}\), with base \(B=2m+1\) in the first construction and \(B=2m+2\) in the second. Set \[C(a_0a_1\cdots)=\sum_{r\ge0}\beta(a_r)B^{-r-1},\quad P_a(v)=\frac{\beta(a)+v}{B},\quad D_a(v)=Bv-\beta(a),\quad I_a=P_a([0,1]).\] Let \(\xi\) encode the left string, nearest cell first, and \(\eta\) the string starting at the head. The complete right, stay and left branch maps are (19), with its digit \(d(a)\) equal to \(\beta(a)\) and its move \(e\) equal to \(d\) here. The left move is split over every full symbol \(c\). The shared lemma proves the full-rectangle endpoint formulas, arbitrary-tail action and linear part \(\operatorname{diag}(B^{-d},B^d)\).

Translate each state square horizontally. In the first construction give the halt state offset \(-4\) and the other states offsets \(2+2r\), \(r\ge0\); in the second use halt offset \(0\) and other offsets \(3r\), \(r\ge1\). Thus the code is \((o_s+\xi,\eta,0)\). Lemma 10 applies with exactly these odd digits and either displayed base: its incoming conditions are those supplied by the move-first recorder, and (19) is its minimal left-split branch list. Thus both complete closed rectangle families are separately disjoint, with within-state gaps at least \(B^{-2}\). This proves the hypotheses of Lemma 49; no separation between the source and target families is required.

Use its linear scale and coordinate potentials for the first construction; use its logarithmic scale and symmetric translation potentials for the second. In the latter case the scaling coefficient is exactly \((\log\lambda_i)\dot\Theta_2\) and the scale matrix is \(\operatorname{diag}(\lambda_i^{\Theta_2},\lambda_i^{-\Theta_2})\). Denote the resulting step field by \(W_{\rm step}\).

The initial code \(P_{\rm in}\) is rational. A constant full-symbol tail of digit \(\beta_0\) after \(M\) positions contributes \(B^{-M}\beta_0/(B-1)\); all other contributions are finite. For the appropriate fixed label \(a_*\), put \(v_0=P_{\rm in}-a_*\) and let \(J_0\) be the smallest box containing \([a_*,P_{\rm in}]\). The loader \[W_{\rm in}(s,x)=\frac d{ds}\vartheta(2s-1/2)\, \nabla\times\left(\chi_{J_0,1}(x)\frac{v_0\times x}{2}\right)\] has the exact marked path \(a_*+\vartheta(2s-1/2)v_0\), since the uncut curl is \(v_0\) and the entire segment lies in the plateau. Form \[W(s,x)=W_{\rm in}(s,x)+\sum_{n\ge1}W_{\rm step}(s-n,x).\] The zero time collars make this smooth, initially zero, and periodic for \(s\ge1\), with common compact spatial support and bounded mixed derivatives.

At time \(s=1+k\) the marked particle is exactly the code after \(k\) local instructions, through the first halt or indefinitely in a nonhalting run. This follows by induction from the full-rectangle maps. Each halt code lies in its stated observation box, including one reached by loading an initially halted machine. In a nonhalting run, the first loader has \(x_1\ge0\) and later nonhalt codes have \(x_1\ge2\); for the second these bounds are \(x_1\ge2\) and \(x_1\ge3\). Ascent and descent leave those planar coordinates unchanged. Every planar motion has height \(4i\ge4\), outside both observers. The intervening stationary intervals add no new positions. This proves exclusion at every real time.

For the first case take \(u=W\) and its residual force. For the second set \[s(t)=\log(1+t),\quad \theta(t)=(1+t)^{-1},\quad u(t,x)=\theta(t)W(s(t),x),\] whose force is \[f=\theta^2\{-W+\partial_sW+(W\cdot\nabla)W\}(s(t),x) -\nu\theta\Delta W(s(t),x).\] Lemma 30 applies to this fixed-support, bounded-derivative template. Its identity \(\partial_t(\theta^jY(s(t)))=\theta^{j+1}(\partial_s-j)Y(s(t))\) proves (40) for every \(k,\alpha\). Fixed support then gives the stated \(L^2_tH^m_x\) conclusion. Its clock is onto, with inverse \(t=e^s-1\), so the material path is the old path composed with \(s(t)\) and the event is unchanged.

In both cases smoothness, fixed support and bounded derivatives on finite time intervals put the reference velocity in \(CH^2\cap C^1L^2\) with bounded velocity and gradient. It is solenoidal and initially zero. Lemma 2(i) therefore gives the exact solution \((u,0)\) and uniqueness in (38). The effective profiles following (5), applied to the finite table, rational initial code, finitely many boxes and repeated template, give all derivative evaluations. A coarse time approximation encloses finitely many possible active translates, avoiding equality tests at seams. The prescription never computes the machine’s run. ◻

Proof. Use Lemma 16 with the frontier initially at cell \(1\). Its main control \(M_{q,e}\) records the last work displacement \(e\), initially zero. In one work step it writes and marks the old head, records \((q,e,a)\) at the frontier, returns to the mark, and only then makes the work displacement. All other initial history cells are blank and every mark is zero. At checkpoint \(n\) the frontier is \(n+1\), \(|h_n|\le n\), and the next checkpoint takes exactly \(2(n+1-h_n)+3\) local steps. These scans test history and marks, so the proof is unaffected by work symbols on either side of the head. Main controls with halting \(q\) have no outgoing rules. The full-domain incoming properties are those of the cited lemma.

Let \(S\) be the local state set, \(N=|S|\), let \(S_h\) be its halting main states, and enumerate \(S\) by \(n(s)=0,\ldots,N-1\). Put \[\ell=\frac1{16(N+1)},\quad z_0=\frac14,\quad x_s=\frac18+2\ell n(s)+\frac12\mathbf1_{s\in S_h},\quad T_s(\xi,\eta)=(x_s+\ell\xi,1/4+\ell\eta).\] Nonhalting squares lie in \(1/8\le x_1<1/4\) and halting squares in \(5/8\le x_1<3/4\); distinct squares have gap at least \(\ell\). Give the full alphabet, of size \(m\), odd digits in base \(\beta=2m+1\). Conjugating (19) by the source and target charts gives separately disjoint rectangle families \(R_j^0,R_j^1\) with positive diagonal factors \(\alpha_j,\alpha_j^{-1}\), \(\alpha_j\in\{\beta^{-1},1,\beta\}\). Every chart has the same scale, so this remains the physical linear part. Lemma 10, with this uniform chart scaling, gives within-family coordinate gaps at least \(\ell/\beta^2\), including the two-letter target intervals for left moves.

For \(1\le j\le L\), write the centers as \(p_j^0,p_j^1\), the source half-widths as \(a_j,b_j\), and the target half-widths as \(\alpha_ja_j,\alpha_j^{-1}b_j\). All endpoint half-widths are at most \(\ell/2\). Choose \[h_j=\frac12+\frac{j}{4(L+1)},\qquad \epsilon=\frac1{100}\min\left(1,\frac\ell{\beta^2}, \frac1{4(L+1)}\right).\] With \(\theta(s)=\vartheta(2s-1/2)\) and \(\Theta_k(\tau)=\theta(3\tau-k+1)\) for \(k=1,2,3\), define \[\begin{aligned} C_j(\tau)&=\big((1-\Theta_2)p_j^0+\Theta_2p_j^1, z_0+(\Theta_1-\Theta_3)(h_j-z_0)\big),\\ \sigma_j&=1+(\alpha_j-1)\Theta_2,\qquad \lambda_j=\dot\sigma_j/\sigma_j,\\ (w_{j1},w_{j2},w_{j3})&=(a_j\sigma_j,b_j/\sigma_j,0). \end{aligned}\] These centers and half-widths describe moving plates. Their \(\epsilon\) enlargements are disjoint: the first third uses source projections, the middle third distinct heights, and the last third target projections. The padding is at most one hundredth of every relevant gap. All plates and padding lie in a compact sub-box of \((0,1)^3\), since their centers interpolate between the stated chart centers, widths are at most \(\ell\), heights lie between \(1/4\) and \(3/4\), and \(\ell\le1/32\), \(\epsilon\le1/100\).

For \(w\ge0\) put \[\kappa_\epsilon(w,v)= \vartheta\!\left(\frac{2(w+\epsilon-v)}\epsilon\right) \vartheta\!\left(\frac{2(w+\epsilon+v)}\epsilon\right),\qquad \chi_j=\prod_{a=1}^3\kappa_\epsilon(w_{ja},x_a-C_{ja}).\] Writing \(y=x-C_j\), use \[A_j=\tfrac12\dot C_j\times y+(0,0,\lambda_jy_1y_2), \qquad V(\tau,x)=\sum_j\nabla\times(\chi_jA_j).\] Extend the potentials by zero outside the chart. The support remains away from its faces, so this gives a smooth torus field, zero near the time endpoints. On the whole plateau of plate \(j\) its velocity is \(\dot C_j+(\lambda_jy_1,-\lambda_jy_2,0)\); other summands vanish there. Hence the path from \((p_j^0+(r,s),z_0)\) is exactly \[ C_j(\tau)+(\sigma_j(\tau)r,\sigma_j(\tau)^{-1}s,0), \qquad |r|\le a_j,\quad |s|\le b_j. \tag{43}\] It ends at the prescribed affine image. The plateau argument of Lemma 5 applies with \(D_j=\operatorname{diag}(\sigma_j,\sigma_j^{-1},1)\), \(\eta=\epsilon/2\), and diagonal bound \(\max(1,\alpha_j,\alpha_j^{-1})\): the required enlarged-box separation was just proved. It gives the same motion on an initial neighborhood of radius \(\epsilon/[4\max(1,\alpha_j,\alpha_j^{-1})]\), with constant normal displacement, and therefore includes boundary codes.

The initial code \(P_{\rm in}\) is rational, since relative to constant blank tails only finitely many work and frontier digits differ. Load it from \(P_*\) along \(C_*=(1-\theta)P_*+\theta P_{\rm in}\) using \[V_0(t,x)=\nabla\times\left[ \prod_{a=1}^3\kappa_{1/64}(0,x_a-C_{*,a}(t)) \frac{\dot C_*(t)\times(x-C_*(t))}{2}\right].\] The segment and its collar stay within the chart: its first coordinate stays between \(1/8\) and \(3/4\), its second between \(1/4\) and \(1/2+\ell\), and its third is \(1/4\). Its plateau curl is \(\dot C_*\). Take \(U=V_0\) on \([0,1]\) and \(U(t)=V(t-n)\) on \([n,n+1]\) for \(n\ge1\). Time collars make this smooth and initially zero. It has fixed compact chart support, bounded mixed derivatives, and period one after loading. Define \(f=U_t+(U\cdot\nabla)U-\nu\Delta U\) using the stated physical viscosity. Both \(U\) and \(f\) have mean zero: \(U\) is a curl, the Laplacian integrates to zero, and \((U\cdot\nabla)U=\operatorname{div}(U\otimes U)\). The torus clause of Lemma 2 applies to this smooth, initially zero, solenoidal reference field and gives the unique smooth solution \((U,0)\) with mean-zero pressure.

Equations (19) and (43) identify the particle at time \(1+k\) with the \(k\)th local configuration. The recorder invariant ensures that a nonhalting computation continues indefinitely. A halt, including an initial halt reached by loading, puts the particle in \(5/8\le X_1<3/4\), proving the event. For a nonhalting computation the loader has \(X_1\le1/4\) and every later branch has both endpoint centers below \(1/4\). Its center is their convex interpolation and its first half-width is at most \(\ell/2\). Thus \(0<X_1<1/4+\ell/2<1/2\) at every intermediate time. The path stays inside the chart, so reduction modulo one cannot create a false visit.

For the separate decaying force choose the onto clock \[s(t)=(1+t)^{1/3}-1,\qquad \widehat U(t,x)=s'(t)U(s(t),x).\] Its residual is \[\widehat f=s''U+(s')^2\{\partial_sU+(U\cdot\nabla)U\} -\nu s'\Delta U,\] where the fields on the right are evaluated at \((s(t),x)\). Since \(s'=(1/3)(1+t)^{-2/3}\) and \(s''=-(2/9)(1+t)^{-5/3}\), the bounded template derivatives prove (42). Every positive-order derivative of \(s\) is bounded; repeated product and chain rules therefore bound every mixed derivative of \(\widehat U\) and \(\widehat f\). The square of \((1+t)^{-2/3}\) is integrable. Support and mean zero persist, and \(U(0)=0\) retains the zero datum. The torus clause of Lemma 2 therefore applies to \(\widehat U\) too. Finally the material path is the old one at \(s(t)\), and \(s\) has inverse \(t=(1+s)^3-1\), so the event is unchanged.

All choices are effective finite data followed by explicit smooth formulas. In particular \(\sigma_j\ge\min(1,\alpha_j)>0\); the cutoff estimates and zero collars allow derivative evaluation at approximate real arguments without tests for exact boundary equality. Only positive arguments enter the cube root. The program repeats the entire local table and never uses a computed execution. ◻

Proof. The maximum tail is \((m-1)/(B-1)=\kappa\), and \((m-1+\kappa)/B=\kappa\). Consecutive first-digit intervals have gap \((1-\kappa)/B>0\). Within one first-digit interval the second-digit gaps are this quantity divided by \(B\); different first digits already give a gap. The interval containing a code recovers its first digit, and \(v\mapsto Bv-a\) then exposes the tail. Repetition proves injectivity, including endpoint codes.

Writing \(a\) to \(b\) changes the first right digit. A right move pushes \(b\) onto the left string and removes \(a\) from the right; a left move removes \(c\) from the left and prepends \(c,b\) to the old right tail. These identities give (44), including its images and reciprocal slopes. Determinism and digit gaps separate sources by state, read digit, and \(c\) when needed. At one target state the common incoming direction determines where the written digit is visible: in \(I_b\) on the left or right, or in \(I_{c,b}\). The incoming-rule hypothesis and digit gaps then separate the closed target rectangles. ◻

Proof. Let \(d_i,b_i\) be the source and target centers, and let \(a_{ij},a'_{ij}\) be their half-widths. Put \(a^*_{ij}=\max(a_{ij},a'_{ij})\). Choose \(\epsilon\) to be one quarter of the minimum of \(1\) and all pairwise maximum-norm distances within the source and target families; omit pair entries when \(N=1\). Define \[\begin{gathered} x_{\min}=\min\{x_1:(x_1,x_2)\in D_i\cup E_i\text{ for some }i\},\\ R=\max(1,\max_i a^*_{i1}),\qquad J=4(R+1),\qquad s_i=(x_{\min}-iJ,0). \end{gathered}\] The storage rectangle at \(s_i\) with half-widths \(a^*_{ij}\) lies in \(x_1<0\). Its first-coordinate projection is separated by more than one from every source and target and from the other storage projections: the relevant gaps are at least \(J-R>1\) and \(J-2R>1\).

We need only a localized translation and a localized reciprocal scaling. On a time segment \([t_0,t_0+\delta]\), use \(g(t)=\vartheta(3(t-t_0)/\delta-1)\), whose derivative is supported in the middle third. Using the cutoff \(\kappa_\epsilon\) defined in the proof of Theorem 51, for half-widths \(a,b\) write \[\chi_{a,b}(y)=\kappa_\epsilon(a,y_1) \kappa_\epsilon(b,y_2)\kappa_\epsilon(0,y_3).\] A plate translation from center \(P\) to \(Q\) is produced by \[C(t)=(1-g)P+gQ,\qquad \nabla\times\left[ \tfrac12\chi_{a,b}(x-C(t))\dot C(t)\times(x-C(t))\right].\] On its plateau the velocity is \(\dot C\) and the exact path is \(C(t)+(r,s,0)\). At storage center \(P=(s_i,0)\), take \(q(t)=1+(\lambda_i-1)g(t)\) and the field \[\nabla\times\left(0,0, \frac{\dot q}{q}(x_1-P_1)(x_2-P_2) \chi_{a^*_{i1},a^*_{i2}}(x-P)\right).\] It has the exact path \(P+(qr,q^{-1}s,0)\). The denominator has the positive lower bound \(\min(1,\lambda_i)\), and the whole path lies in the storage rectangle, because each half-width stays between its source and target values.

Partition a period into \(7N\) equal segments and take \(Z=2\). For each \(i\) in order, use four segments for \[(d_i,0)\longrightarrow(d_i,Z)\longrightarrow(s_i,Z) \longrightarrow(s_i,0),\qquad\text{then scale at }s_i.\] Only after all these evacuations, use three segments for each \(i\) to deliver its target shape along \[(s_i,0)\longrightarrow(s_i,Z)\longrightarrow(b_i,Z) \longrightarrow(b_i,0).\] Sum these segment fields and extend periodically. The middle-third time switches give zero neighborhoods of every segment endpoint and every integer time. The result is a smooth finite sum of curls with one compact spatial support.

This is the \(7N\) schedule of Lemma 7(ii). Its separation hypotheses have explicit margins here. Unpadded source-source and target-target gaps are at least \(4\epsilon\), so their \(\epsilon\) enlargements leave gap at least \(2\epsilon\). Storage gaps exceed one, leaving more than \(1-2\epsilon>0\) after padding. At horizontal travel height the enlarged moving plate and every enlarged stationary plate have vertical gap \(2-2\epsilon>0\). Every scaling stays in its reserved storage rectangle. The lemma therefore gives simultaneous noninterference for evacuation and delivery, including when a source meets a target.

The localized potentials above have exactly these affine generators on their plateaux, so the plateau argument of Lemma 5 gives endpoint \((d_i+(r,s),0)\mapsto(b_i+(\lambda_i r,\lambda_i^{-1}s),0)\), with normal displacement unchanged. Put \(M=\max_i(1,\lambda_i,\lambda_i^{-1})\) and \(d_*=\min(2\epsilon,1-2\epsilon,2-2\epsilon)>0\). A neighborhood radius smaller than both \(\epsilon/(8M)\) and \(d_*/(4M)\) retains the plateaux and all exclusion margins throughout the concatenation. This applies the shared proofs to the displayed potentials and does not identify their fields away from the plateaux.

If \(E_i\) is negative, each translation of an individual point joins two negative first coordinates, and its linear interpolation remains negative. Scaling occurs inside the negative storage rectangle. The point is stationary during all other segments. This proves the asserted whole-period exclusion. ◻

Proof. Insert a fresh nonhalting start state that preserves the read letter, stays put and enters the original start state. Replace missing nonhalting instructions by letter-preserving stay instructions to a halt state. This finite normalization preserves the halting question, including an initially halted machine, and guarantees a nonhalting initial control and at least one rule.

Apply the six-row delayed-head recorder of Lemma 16 with initial frontier \(1\). Initially every other history cell is blank and every mark is zero. The main control records \((q,e)\), where \(e\) is the preceding displacement, and the history record is \((q,e,a)\). The old head is marked before the right scan and cleared on return, immediately before displacement into the next main state. At checkpoint \(k\) the frontier is \(k+1\) and the head \(i\) satisfies \(|i|\le k\); the next checkpoint uses \(2(k+1-i)+3\le4k+5\) local steps. The incoming-direction and written-symbol properties hold on the full domain, as required by Lemma 52.

Give the full blank symbol digit zero and the other full symbols digits \(1,\ldots,m-1\). Order nonhalting local states with the initial state first and assign offsets \(-2h\), \(h=1,2,\ldots\); assign halting main states offsets \(2h\), \(h=1,2,\ldots\). Let \(\eta_{\rm in}\) be the initial right-stack value and put \(y_*=-\eta_{\rm in}\). Only finitely many full symbols are initially nonblank, so this is a finite rational sum. Every cell left of the initial head is fully blank. The code \[(x_\sigma+\xi,y_*+\eta,0)\] therefore places the initial configuration exactly at \((-2,0,0)\). The input changes the common second-coordinate translation, while the observed label and zero initial velocity remain fixed.

Every finitely reached tape still has finite full-symbol support, so its code is rational and can be decoded by rational digit extraction until both tails are zero. At a checkpoint the leftmost nonempty history site has absolute index \(1\); its relative position also recovers the head’s absolute index. Thus this coding retains the initialized computation.

Translate the rectangles of (44) by the indicated state offsets and \(y_*\). Lemma 52 gives their full affine action and separate source and target gaps. Since \(\kappa<1\), every nonhalting state block is strictly negative in the first coordinate and every halting block strictly positive. Halting states have no outgoing rules, so all sources are negative. Apply Lemma 53 to obtain its periodic field \(v\). Zero tails can place a code on a rectangle boundary, which is included in that lemma’s full-neighborhood conclusion.

Set \[f=f_0+\nu f_1,\qquad f_0=\partial_tv+(v\cdot\nabla)v,\qquad f_1=-\Delta v.\] The field and force are smooth, periodic, uniformly compactly supported, and have all mixed derivatives bounded. Since \(v\) vanishes near integer times, its initial value is zero. Fixed support and bounded derivatives give \(v\in CH^2\cap C^1L^2\) and bounded \(v,\nabla v\) on every finite interval. Lemma 2(i) therefore supplies the exact solution \((v,0)\), uniqueness in (38), and its global material flow.

At time zero the marked particle is the initial code. If it is a nonhalting code at integer \(n\), its digits select a source rectangle and the next period applies exactly that local instruction. Induction identifies the integer-time codes through the first halt or forever. Finite completion of every recorder scan identifies the original machine at the checkpoint integers. A halt therefore reaches a positive code. If the machine never halts, each used branch has a negative target rectangle. Lemma 53 keeps its marked path negative for that whole period, ruling out an intermediate false visit.

Finally the force program uses finite normalization, the finite table, the rational input shift, a finite rectangle list, rational storage and separation data, \(7N\) time segments and explicit cutoffs. All scaling denominators have known positive lower bounds. The zero-collar and flat profile estimates following (5) permit evaluation of the repeated formula and every derivative from approximate real arguments without exact seam tests. Every period executes all branch maps, independently of the marked particle’s current state; the program never computes the input run. ◻

Proof. Use full symbols \([g,e,l]\), with work letter \(g\), mark \(e\in\{0,1\}\), and \(l\in\{E,P\}\sqcup\{J_\tau\}\). Controls are \(A_q\), \(B_\tau\) for moving instructions, \(C_\tau,S_\tau\), and \(D_q,R_q\). For \(\tau=(q,a)\) writing \(b_\tau\), the complete table is \[\begin{array}{c|c|c|c|r} \text{control}&\text{read and guard}&\text{target}&\text{write}&d\\\hline A_q&[a,1,l],\ d_\tau=0&C_\tau&[b_\tau,1,l]&0\\ A_q&[a,1,l],\ d_\tau\ne0&B_\tau&[b_\tau,0,l]&d_\tau\\ B_\tau&[g,0,l]&C_\tau&[g,1,l]&0\\ C_\tau&[g,1,l],\ l\ne P&S_\tau&[g,1,l]&1\\ S_\tau&[g,0,l],\ l\ne P&S_\tau&[g,0,l]&1\\ S_\tau&[g,0,P]&D_{q^+_\tau}&[g,0,J_\tau]&1\\ D_q&[g,0,E]&R_q&[g,0,P]&-1\\ R_q&[g,0,l],\ l\ne P&R_q&[g,0,l]&-1\\ R_q&[g,1,l],\ l\ne P&A_q&[g,1,l]&0. \end{array}\] The first two rows use nonhalting \(q\). In those rows and the \(B_\tau\) row, \(l\) is unrestricted, including \(P\); no additional history guard is imposed. Initialize in \(A_{q_0}\) at zero, with the input work tape, a unique mark at zero, pointer at two, and empty history elsewhere.

At checkpoint \(n\), the work move ends at \(w'=w_n+d_\tau\le n+1<p_n\). A stay instruction reaches \(C_\tau\) in one step with the mark retained. A moving instruction reaches it in two steps after transferring the mark. From \(C_\tau\), move right, scan the \(p_n-w'-1\) intervening nonpointer cells, record at \(p_n\), and extend the pointer to \(p_n+1\). The return scans \(p_n-w'\) unmarked cells and stops at the persistent mark, entering the correct \(A_{q^+_\tau}\). There are five other steps for a stay instruction and six for a moving instruction. This yields the stated count and restores the invariant. Every required guard holds, so the execution continues exactly until halting, including possible initial halting.

For an arbitrary allowed tape, each target determines its incoming move. It is zero into \(A_q\) or \(C_\tau\), \(d_\tau\) into \(B_\tau\), one into \(S_\tau\) or \(D_q\), and minus one into \(R_q\). The record \(\tau\) recovers the source of the work update. A fixed \(C_\tau\) receives only an \(A_q\) entry if \(d_\tau=0\), and only \(B_\tau\) entries otherwise; the written symbol recovers the old full symbol in either case. At \(S_\tau\), the written mark separates entry from scan. At \(D_q\), \(J_\tau\) identifies the recording instruction and its former pointer. At \(R_q\), the written pointer separates entry from the loop. At \(A_q\), the unique source is \(R_q\) and the marked symbol is unchanged. Undoing the indicated shift therefore gives at most one predecessor in every case, on the full stated domain. ◻

Proof. Put \(\rho(s)=\sigma(2s-1/2)\), using the effective flat switch from (5). Use the single-halt normalization and Lemma 55. For its full alphabet \(\Sigma\), choose distinct odd digits \(e(a)\in\{1,3,\ldots,2|\Sigma|-1\}\) and base \(B=2|\Sigma|+1\). For an infinite word \(\omega=\omega_0\omega_1\cdots\), define \[c(\omega)=\sum_{j\ge0}e(\omega_j)B^{-j-1}.\] Place \(A_h\) at offset zero and the other control squares at first-coordinate offsets \(3,6,9,\ldots\). The code at height zero is \[(o_s+c(\text{left stream}),\ c(\text{right stream}),0).\] Every tail value lies in \([1/(B-1),(B-2)/(B-1)]\subset(0,1)\), so a halt code lies strictly inside the observed box. The two incoming properties give the separated filled rectangles of Lemma 17, with minimal left subdivision and first factors \(1,B^{-1},B\) for stay, right and left moves.

Apply the linear version of Lemma 19 to the full cuboids with third interval \([-1,1]\), heights \(6i\), and switches \(\rho(3s),\rho(3s-1),\rho(3s-2)\). Source projections separate lifts, heights separate horizontal motion, and target projections separate descents. Equation (23) gives the exact affine flow on every swept cuboid and a neighborhood.

The initial code \(x^{\rm in}\) is rational. For example, if \(b=\max(3,\text{input length})\), the first \(b\) right-stream symbols include the entire input and the pointer at cell two, after which the stream is constant; the left stream is constant. On the loading interval use \[C_*(t)=(2,0,0)+\rho(t)(x^{\rm in}-(2,0,0))\] and \[V_*(t,x)=\nabla\times\left\{ \chi(C_*(t),(1,1,1),1;x) \tfrac12 C_*'(t)\times(x-C_*(t))\right\}.\] The cutoff is the centered one defined in Lemma 19. Its plateau velocity is \(C_*'\), so the fixed particle follows the displayed path. This field vanishes outside \([1/4,3/4]\). Extend the step field \(V\) by zero and set \(U=V_*+\sum_{k\ge1}V(t-k)\), with its residual force at the prescribed \(\nu\). The finite union of swept cuboids and the loading tube is bounded; all time joins have zero collars. Consequently \(U\) and its force are smooth, uniformly compactly supported, and have bounded mixed derivatives, with period one after loading. Effective evaluation uses the finite formulas and a finite superset of potentially active integer shifts at any approximate time, as in Lemma 21.

The precise uniqueness input is Lemma 2(i). On each finite interval the constructed \(U\) is smooth, has one compact spatial support, and has bounded velocity and gradient. Consequently \(U\in CH^2\cap C^1L^2\), \(U(0)=0\), and the force is exactly \(\mathcal F_\nu[U]\). The theorem’s competitors are smooth classical \(v\in CH^2\cap C^1L^2\) with bounded \(v\) and a representative \(p\in CH^1\), precisely case (i). No bound on \(\nabla v\) is added. That case gives \(v=U\) and fixes the \(H^1\) pressure representative to zero. The lemma also supplies the unique global material trajectories of the reference field.

Loading and the exact full-box maps put the particle at the code after \(k\) local steps at time \(1+k\), through the first halt or forever in a nonhalting execution. A halt code, including an initially halted code reached by loading, is inside the observation box. In a nonhalting run, loading has first coordinate at least two. During a later lift, the particle keeps its nonhalting source planar code, whose first coordinate exceeds three. During descent it keeps the nonhalting target planar code with the same property. Throughout the middle third its height is \(6i>1\). These three exclusions cover every real time and prove the event equivalence independently of horizontal excursions in the middle stage. The repeated processor installs every local branch from the finite table; it does not depend on the executed run or its outcome. ◻

Proof. Let \(\Gamma\) be the full alphabet of the six-rule table with \(r_0=1\). Use odd digits in base \(B=2|\Gamma|+1\) and its fully subdivided maps. Put halt lanes at \(-2,-4,\ldots\) and the other lanes at \(2,4,\ldots\), each of width one. The second coordinate lies in \([0,1]\). For the resulting rectangles \(P_i,Q_i\), centers \(p_i,q_i\), and first scale \(\lambda_i\), choose rational \(0<\varepsilon\le1/4\) smaller than one quarter of all within-source and within-target maximum-norm distances. Empty pair lists impose no condition. Thicken by \([-1,1]\) and put \(H_i=4i\).

On each third of the cycle use a flat progress function \(\vartheta\) from zero to one. Lifting uses \(z_i=H_i\vartheta\) and potential \(A_i=(0,z_i'X_1,0)\), whose curl is \((0,0,z_i')\). At height \(H_i\), set \(c_i=(1-\vartheta)p_i+\vartheta q_i\) and \(\mu_i=1-\vartheta+\vartheta\lambda_i\). Use the potential \(A_i=(0,0,\psi_i)\), where \[\psi_i=c_{i1}'(X_2-c_{i2})-c_{i2}'(X_1-c_{i1}) +\frac{\mu_i'}{\mu_i}(X_1-c_{i1})(X_2-c_{i2}).\] Its curl is \((c_{i1}'+(\mu_i'/\mu_i)(X_1-c_{i1}), c_{i2}'-(\mu_i'/\mu_i)(X_2-c_{i2}),0)\). Thus it realizes reciprocal scaling about the interpolated center. Descending uses the first potential with \(z_i=H_i(1-\vartheta)\). Multiply each potential by its current-box \(\varepsilon\) cutoff before taking the curl. Source gaps, distinct middle heights, and target gaps separate the three stages. On the whole boxes and a neighborhood the uncut formulas are exact. The first coordinate is constant during vertical motion and is the convex interpolation of its endpoints during the planar stage.

For loading put \(\theta_0(s)=\sigma(2s-1/2)\), which is zero near zero and one near one. Let \(P_*\) be the rational initial code. Load it in the first unit interval by the planar path \(m(s)=\theta_0(s)((P_*)_1,(P_*)_2)\), localized around \((m(s),0)+[-1,1]^3\), using potential \((0,0,m_1'(X_2-m_2)-m_2'(X_1-m_1))\). Repeat the three-phase instruction field afterward to obtain \(W\). Code induction and the delayed-head invariant prove the correspondence at times \(1+n\). Nonhalting codes have first coordinate at least 2, so loading and the convex-path bound exclude \(X_1<-1/2\) for every real time. A halt code has first coordinate at most \(-1\), including one reached by loading an initially halted machine. Apply Lemma 30; physical code times are \(e^{1+n}-1\). Its support, derivative, event, and uniqueness conclusions give precisely the claimed separate force choice. ◻

Proof. Choose an integer \(L>1\) larger than the absolute values of all endpoint coordinates. The storage columns have first coordinates \(U_i=4(i+1)L\) and common height \(Z=-4L\). Write \([\ell_i,r_i]\) for the first-coordinate interval of \(D_i\). Extract sources in decreasing order of \(r_i\). Translate each horizontally right to its column and then vertically to \(Z\), retaining its second coordinate. If an unextracted box meets the active box in both the second and third projections, disjointness forces their first intervals to be separated. That box cannot lie to the right: its right endpoint would then exceed \(r_i\), contrary to the extraction order. It lies strictly to the left of \(\ell_i\) and misses the entire rightward sweep. Stored boxes lie below the original boxes; vertical moves occur in distinct columns outside them.

At storage height, adjust the second center coordinate to its target value and apply \(A_i^{\theta}\). Each intermediate half-width lies between its two endpoint values, hence is less than \(L\). The columns remain disjoint throughout these operations. Deliver boxes in increasing order of the right endpoint of their target first interval: raise a box to its target height and translate it left into its target. The raising column is empty. As above, a previously delivered box whose last two intervals meet the active ones must be separated in the first coordinate. Its smaller right endpoint puts it wholly to the left of the active target interval, so the leftward sweep stops before reaching it. Undelivered boxes remain at storage height. These observations give positive rational clearances for all swept bounding boxes. Choose cutoff padding below each clearance and use Lemma 5 in successive flat slots. Induction over the slots proves the full-box assertion. A point initially in the central third-coordinate plane retains zero third-coordinate offset during each diagonal deformation. Its height is interpolated only between the two endpoint heights and \(Z\). If those endpoint heights are negative, every intermediate height is negative. ◻

Proof. Use Lemma 11 and odd digits in base \(2m+2\). Give every simulator state a distinct positive integer index \(i_s\) and put its third-coordinate center at \(z_s=i_s\) for ready halting states and \(z_s=-i_s\) otherwise. Thicken the full rectangles of Lemma 10 by \([z_s-1/4,z_s+1/4]\). The assigned third scale is one, so the maps satisfy Proposition 58. The rational initial code \(Y_0\) is loaded from the origin along the straight segment, by a localized translation during \([0,1]\). Repeat the unit-time mechanism after time one. At time \(1+k\) the particle is exactly the code after \(k\) defined simulator transitions. If the initial machine halts, the loading endpoint already has positive height. If it never halts, loading has nonpositive height and all later transitions have negative endpoint centers; the proposition excludes a positive height at every intervening time. Conversely, a halting checkpoint has positive height. Lemma 18 gives the force and uniqueness, and Lemma 30 gives the separate temporal alternative. ◻

Proof. Let \(p_i^-,p_i^+\) be the centers in the plane \(z=0\), set \(h_i=5i\), and choose a smooth \(\Theta\) equal to zero on \((-\infty,1/4]\) and one on \([3/4,\infty)\). For \(0\le t\le1\), put \[\theta=\Theta(3t-1),\quad l_i=1-\theta+\theta\lambda_i, \quad c_i=(1-\theta)p_i^-+\theta p_i^+ +h_i\{\Theta(3t)-\Theta(3t-2)\}e_3.\] The desired path is \(g_i(t,y)=c_i(t)+\mathop{\mathrm{diag}}(l_i,l_i^{-1},1)(y-p_i^-)\). Use a source padding \(\varepsilon=1/(10B^3)\), transported by this same affine map, for the cutoff neighborhood. During lifting the planar source gap is at least \(B^{-2}\). During lowering the enlarged target padding is at most \(B\varepsilon<B^{-2}/2\). During the middle phase the height difference is at least five, whereas the sum of third half-widths is \(2+2\varepsilon<5\). Thus these moving cutoff supports can be chosen disjoint. Sum the localized curls of Lemma 5. Differentiation and uniqueness give the displayed paths for the full boxes. Finally \[g_{i,1}(t,y)=(1-\theta)y_1+ \theta\{p_{i,1}^++\lambda_i(y_1-p_{i,1}^-)\},\] which proves the asserted guard. Flat time collars hold automatically. ◻

Proof. Use the frontier recorder, zero-blank base \(B=2m\), and state first-coordinate offsets \(3j\) for nonterminal controls and \(-3j\) for terminal controls, with positive distinct indices. Encode a state by \((o_s+L,R,0)\). The full branches are on the \(B^{-2}\) grid and satisfy the preceding proposition. Load the rational initial code from zero along a straight segment. Nonhalting codes have nonnegative first coordinate; the whole segment and every nonhalting transition therefore avoid \(x_1<-1\). Terminal codes have first coordinate at most \(-2\), including an initially terminal loading endpoint. Exact integer-time simulation gives both implications. The force conclusions follow from the common analytic and logarithmic-clock lemmas, with the compact support needed for all \(H^j\) norms. No claim of periodicity is made for the slowed alternative. ◻

Proof. The empty family is realized by zero. Otherwise let \(R=2+\) the largest endpoint half-width, and choose an integer \(H>20R\) so large that the expanded source centers \(A_i^*=Ha_i\) and target centers \(B_i^*=Hb_i\) are separately more than \(20R\) apart in the sup norm. Set \[Z_0=1+100R+\max_i\{\|A_i^*\|_\infty,\|B_i^*\|_\infty\}, \qquad P_i=(0,0,-Z_0-100Ri).\] Each of the unions \(\{A_i^*\}\cup\{P_i\}\) and \(\{B_i^*\}\cup\{P_i\}\) has separation greater than \(20R\). The construction has five phases: expand source centers by a factor from one to \(H\) while keeping shapes fixed; move them successively to \(P_i\); change each parked shape by the prescribed positive diagonal power; move them successively to \(B_i^*\); contract target centers to \(b_i\) while keeping the final shapes fixed.

The simultaneous expansion is collision free. For each pair of source boxes some coordinate has positive gap \(|a_{i,j}-a_{k,j}|-r_{i,j}-r_{k,j}\); multiplication of the center difference by a factor at least one preserves that gap. The same argument applies to contraction toward the disjoint targets. Choose padding less than one quarter of the minimum such gap, and less than one. During a parked deformation the \(j\)th half-width is \(r_{i,j}^{1-\theta}w_{i,j}^{\theta}\le\max(r_{i,j},w_{i,j})<R\). Thus parked deformations have all half-widths less than \(R\) and supports of half-width at most \(2R\); the storage separation suffices.

For a successive transfer, surround each stationary center by its closed sup-norm cube of radius \(4R\). These cubes are pairwise disjoint. Start with the straight segment between the moving centers. Each nontrivial intersection with such a cube can be replaced by a polygonal path on its boundary: join the entrance to a vertex of its face, follow cube edges to an exit-face vertex, and join to the exit. Fixed index orders choose the vertices and edges. All coordinates and intersection parameters are rational. Tangencies need no detour. Since different obstacle cubes are disjoint, the resulting finite polygon stays at distance at least \(4R\) from every stationary center. Move along its segments in flat slots, using a cutoff support of half-width at most \(2R\). Stationary boxes have half-width less than \(R\), so no such support touches them. Lemma 5 realizes each translation and deformation.

For the height assertion, a transfer with negative endpoints has endpoints of height at most \(-H\). Its straight segment is negative. It cannot meet an obstacle centered at positive height \(H\), whose bottom is \(H-4R>0\). Every boundary used around a negative-height obstacle lies below \(-H+4R<0\). Storage heights are negative as well. Expansion, contraction, and positive diagonal deformation preserve the central-plane condition. Thus all these paths are negative. Finitely many rational vertices, endpoint widths, and cutoff paddings give a rational support bound. ◻

Proof. Use the frontier recorder, odd base \(B=2m+1\), distinct negative integer heights for nonterminal states and height one for the terminal ready state. Thicken every branch by a third half-width \(1/4\). The reciprocal tape map is onto its target rectangle; adjoining the unchanged third interval gives an onto box map with linear part \(\mathop{\mathrm{diag}}(B^{-e},B^e,1)\) and determinant one. All source centers are negative because terminal controls have no outgoing rules. Apply the preceding proposition. The logical initial code \(Y_0\) is rational and has a uniform machine-dependent bound as the input varies: its two stack coordinates lie in \([0,1]\) and its height is fixed by the initial control. A moving translation loads the logical origin to \(Y_0\), with a fixed source half-width one and padding one. This gives a machine-dependent rational bound \(K\) for the supports of all such loaders and the repeated mechanism; for example take \(K=1+2R+M\) with \(M\) an upper bound for all routing vertices and loading coordinates and \(R\) enlarged to cover their half-widths. Choose \(\lambda=1/(4K)\) and use the chart \(x=o+\lambda Y\). Everything remains strictly inside \((1/4,3/4)^3\).

Periodize the resulting compactly supported curl fields, loading during the first unit interval and repeating thereafter. Compute the residual at physical viscosity \(\nu\), after the chart change. The fields and force have zero mean and all the bounds in Lemma 18. For a nonhalting run, loading has nonpositive logical height and each later segment remains negative by the proposition. A terminal code has physical height \(1/2+\lambda\), which lies in the event. There is no chart-boundary crossing that could create a spurious observation. An initially terminal input is detected at the loading endpoint. The initialized correspondence is exact; at checkpoints the number of history records gives \(n\), and the frontier’s relative position together with its absolute index \(2+n\) also recovers the absolute work-head position. ◻

Proof. Give rectangle \(i\) height \(h_i=4i\). A planar cutoff equal to one near its source and supported within padding \(1/(4B^2)\) has support disjoint from the other source cutoffs. Multiply it by a third-coordinate cutoff which is one near the entire interval \([0,h_i]\). The curl of the potential \((0,h_i x,0)\) times these cutoffs equals \((0,0,h_i)\) along that source column. A common flat schedule lifts all rectangles. The corresponding target cutoffs give a later descent by \(-h_i\).

In the middle phase, let \(p_i,q_i\) be the centers and \(a_i>0\) the first-coordinate scale. With phase parameter \(\mu\in[0,1]\), set \(c_i=p_i+\mu(q_i-p_i)\), \(l_i=1+\mu(a_i-1)\) and \[H_i(\mu,x,y)=d_{i,1}y-d_{i,2}x+ \frac{a_i-1}{l_i}(x-c_{i,1})(y-c_{i,2}), \qquad d_i=q_i-p_i.\] On a neighborhood of the desired path, the planar Hamiltonian field \((\partial_yH_i,-\partial_xH_i)\) generates \(c_i+\mathop{\mathrm{diag}}(l_i,l_i^{-1})(\zeta-p_i)\). The first coordinate is the convex interpolation of its endpoint values. The second half-width obeys \[\frac{r_{i,2}}{(1-\mu)+\mu a_i} \le (1-\mu)r_{i,2}+\mu r_{i,2}/a_i,\] by convexity of the reciprocal function. Consequently the swept rectangle lies in the coordinate bounding rectangle of the two endpoints. Localize \(H_i\) around that rectangle in the layer \(z=h_i\), with third padding less than one, and take the curl of \((0,0,H_i)\) times that cutoff. For a flat middle ramp \(\mu=\eta_2(t)\), multiply the resulting phase field by \(\eta_2'(t)\). Layers are disjoint, so the sum has the required action. Lift and descent preserve planar coordinates. The exact paths and ODE uniqueness establish the full-rectangle and intermediate-coordinate assertions. ◻

Proof. Put nonterminal state offsets at \(3,6,\ldots\) and terminal offsets at \(-3,-6,\ldots\) in the first coordinate. Lemma 10 and Proposition 64 give a whole-rectangle phase map. A localized planar Hamiltonian translation loads the origin to the rational initial code in \([0,1]\) time. A nonterminal loading segment has nonnegative first coordinate, and every later nonterminal segment has first coordinate at least three. Terminal codes have first coordinate below \(-1\). This proves both directions, including initial halting. Repeat the phase map and use Lemma 18. Although this route only requires planar coding rectangles, its cutoff plateaus also give a positive neighborhood on which the same affine formulas hold. ◻

Proof. Assign odd digits \(1,3,\ldots,2m-1\) in base \(B=2m+2\), with state offsets positive multiples of three except for negative terminal offsets. The tail of the blank symbol contributes \(d_{\rm blank}B^{-k}/(B-1)\) after a finite prefix, so initialization is rational even though the blank digit is nonzero. Full affine branches follow from the lemma and Lemma 10. Use the fixed-column proof with heights \(4i\) and the smaller padding \(1/(10B^2)\); schedule a slot \([a,b]\) by \(\sigma(2(t-a)/(b-a)-1/2)\), which is flat on collars of both endpoints. These choices preserve the same separation proof. A planar Hamiltonian loading segment starts at zero. The convex first- coordinate guard excludes the detector throughout every nonhalting transition and throughout a nonhalting load; a terminal code lies below \(-1\). Finite checkpoint simulation gives the equivalence. The field has one compact support and uniform pointwise bounds, hence uniformly bounded kinetic energy. Lemma 18 supplies all remaining assertions. ◻

Proof. Normalize to one halt \(h\). For \(c=(q,b)\in\mathcal D=(Q\setminus\{h\})\times\Gamma\) write the instruction as \((q^+(c),b^+(c),m(c))\). Use histories \[\Lambda_0=\{\varnothing\}\sqcup\{H_e:e\in\mathcal D\sqcup\{\star\}\}, \qquad \Lambda=\Lambda_0\sqcup\{P_e:e\in\mathcal D\sqcup\{\star\}\}.\] The full alphabet is \(\mathcal A=\Gamma\times\Lambda\times\{0,1\}\), whose tracks are data, history and return mark. Controls are \(S_q,L_q,M_c,R_c,F_c\), and the complete table is \[\begin{array}{c|c|c|l} \text{input}&\text{output}&\text{move}&\text{restriction}\\\hline S_q(b,l,0)&M_c(b^+(c),l,0)&m(c)&c=(q,b),\ l\in\Lambda_0\\ M_c(b,l,0)&R_c(b,l,1)&1&l\in\Lambda_0\\ R_c(b,l,0)&R_c(b,l,0)&1&l\in\Lambda_0\\ R_c(b,P_e,0)&F_c(b,H_e,0)&1&e\in\mathcal D\sqcup\{\star\}\\ F_c(b,\varnothing,0)&L_{q^+(c)}(b,P_c,0)&-1&\\ L_q(b,l,0)&L_q(b,l,0)&-1&l\in\Lambda_0\\ L_q(b,l,1)&S_q(b,l,0)&0&l\in\Lambda_0. \end{array}\] Initially put \(P_\star\) at absolute cell three, empty history elsewhere, marks zero, and control \(S_{q_0}\). At ready checkpoint \(k\), the pointer is at \(r=3+k\), all cells to its right are empty, and the work configuration is the original one. After the first row the work head satisfies \(j'\le k+1\le r-2\). The scan marks it, reaches the pointer, changes \(P_e\) into \(H_e\), writes \(P_c\) in the next empty cell, and returns to erase the mark. Both scans are finite and positive. This proves the initialized simulation. The initial pointer tag is retained as \(H_\star\) after the first cycle, so the absolute cell-three reference is permanent.

The inverse works on all allowed tapes. Into \(M_c\) the tag fixes the old instruction and move; into \(R_c\) mark one distinguishes entry from a loop; into \(F_c\) the written \(H_e\) identifies the old pointer \(P_e\); into \(L_q\) a pointer output identifies entry, including the old \(F_c\), whereas a loop writes a nonpointer; into \(S_q\) only mark erasure applies. Incoming shifts are fixed in every target control. The remaining written components retain all unchanged symbols, proving the two properties of Lemma 10.

Let \(m_Q\) be the number of controls, and number them bijectively by \(\iota(s)\in\{0,\ldots,m_Q-1\}\) with \(\iota(S_h)=0\). Use odd digits in base \(B=2|\mathcal A|+1\) and state offsets \(o_s=3\iota(s)\). The physical code is \((o_s+L,R,0)\). Terminal codes lie strictly in \((0,1)^2\times\{0\}\); nonterminal first coordinates are at least three. Form the full branch boxes by taking the product with \([-1,1]\). Their planar sides have length at most one. Choose \(H_x=3m_Q+1\) and parking centers \(\pi_i=(H_x+10+4i,0)\), all at height zero.

Here is a storage construction distinct from ordered low extraction. Evacuate each source in three successive operations: lift its center to height ten, translate horizontally there to \(\pi_i\), and lower it to zero. Once every box is parked, process each in four operations: deform its planar shape at the parking center, lift it to height ten, translate it there to its target center, and lower it into the target. This uses \(7N\) slots for \(N\) branches. During deformation use the positive scale \(l=1+\theta(\lambda_i-1)\) and its reciprocal, not an exponential scale. Both side lengths remain bounded by the maximum of their endpoint lengths, which is at most one. The third half-width remains one.

Let \(\delta\) be the minimum of one and all positive source-source and target-target planar distances, omitting empty sets of pairs. Use cutoff plateaus with padding \(\varepsilon=\delta/8\) and support padding \(2\varepsilon\). During a lift or descent the planar projection separates the moving box from the current sources or delivered targets. Parked boxes lie outside their entire first-coordinate range, and parking centers have spacing four. During a horizontal translation the box lies in \(9\le z\le11\), whereas all stationary boxes lie in \(-1\le z\le1\). These positive gaps also separate the padded supports. Lemma 5 therefore realizes every operation without disturbing a stationary box. The full assigned affine map follows by composition.

For a branch whose target control is nonterminal, every full point has first coordinate at least three, including the storage and transfer paths. Load the fixed label \((2,0,0)\) to the rational initial code by a localized straight translation. If the code is nonterminal, the loading segment has first coordinate at least two; if it is terminal, the endpoint is in the observation box. At later integer samples the exact recorder code gives terminal entry, and the preceding bounds exclude the observation at all other times of a nonhalting run. Repeat the finite transport field and apply Lemma 18. Finally the displayed comparison class implies \(u\in C H^2\), since \(\|u\|_{H^2}\le C(\|u\|_2+\|\Delta u\|_2)\) by the Fourier multiplier identity; its pressure assumptions give \(p\in C H^1\). Thus the stated uniqueness follows from the common whole-space comparison, without changing this route’s original hypotheses. ◻

Proof. Use the frontier recorder after normalization to one halting state, and write \(\mathcal A\) for its full cell alphabet and \(m_Q\) for its number of controls. Number the controls bijectively by \(\iota(s)\in\{0,\ldots,m_Q-1\}\), giving the terminal control index zero. Use odd base \(B=2|\mathcal A|+1\) and first-coordinate offsets \(o_s=4\iota(s)\). Code the configuration by \((o_s+L,R)\). The origin of the original tape can be recovered from the number of history records and the frontier position, so this head-relative coding loses no initialized information. The full planar branches have reciprocal positive diagonal maps.

We first realize finite separately disjoint source and target rectangle families \(S_i,T_i\) with rational endpoints and positive side lengths, equipped with assigned positive reciprocal diagonal affine maps \(F_i\) satisfying \(F_i(S_i)=T_i\), by compactly supported planar Hamiltonian motion. Write \(c_i,q_i\) for the source and target centers. With no branches use zero. With one branch, interpolate its center directly and use a positive linear first-coordinate scale and its reciprocal; there are no stationary obstacles. For \(N\ge2\), let \(R=1+\) the largest endpoint half-width and let \(m_0>0\) be the minimum sup-norm separation within the two center families. Put \(K=1+32R/m_0\). Expand all source centers by factors from one to \(K\), leaving their shapes unchanged. Let \(M=\max_i\{|(Kc_i)_1|,|(Kq_i)_1|\}\) and choose parking centers \(P_i=(M+32Ri,0)\). Each union of expanded endpoint centers and parking centers has separation at least \(32R\).

Move expanded sources successively to their parking centers, deform them there using the linear reciprocal scale, move them successively to the expanded target centers, then contract those centers to their targets. For each successive transfer put a square of sup-radius \(4R\) around every stationary center. Replace each nontrivial segment intersection with an obstacle square by a counterclockwise path along its boundary. The intersection points and corners are rational; the squares are disjoint, so the resulting finite polygon stays at distance at least \(4R\) from all stationary centers. Flat schedules on its segments give smooth center motion. The simultaneous expansion and contraction preserve a positive coordinate gap for every pair, exactly as in the solid-box argument. If \(g_0,g_1\) are their minimum positive gaps, choose padding \(\varepsilon=\min(g_0,g_1,2R)/4\) for these phases. During individual transfers use padding below \(R\); the stationary half-widths are below \(R\). Thus all required cutoff supports avoid stationary rectangles.

For a moving center \(c\) and horizontal scale \(l\), the potential \[H=\chi\left[\dot c_x(y-c_y)-\dot c_y(x-c_x) +\frac{\dot l}{l}(x-c_x)(y-c_y)\right]\] gives the required motion on the cutoff plateau. The other current rectangles are fixed. This proves every full planar branch, including its boundary, by induction through the operations. A one-square loading translation with half-width \(1/4\) takes \((-4,0)\) to the initial rational code. All these supports lie in a square \([-L,L]^2\), where one can choose \(L=10+S_0+4m_Q\) with \(S_0\) a finite bound for routing vertices and cutoff widths. The same \(L\) covers every input of a fixed machine, since initial codes are uniformly bounded.

The planar detours need not avoid the horizontal observation box. We instead make them at height three. Choose a switch \(\eta\) with collars and set \(h(s)=3\{\eta(3s)-\eta(3s-2)\}\) on a unit interval. It lifts in the first third, stays at three in the middle, and lowers in the last third. Choose \(\Gamma\) equal to one near \([-L,L]^2\times[-1,4]\), with compact support, and \(\beta(z)\) compactly supported and equal to one near three. For any one of the planar fields \(w\), zero outside its phase interval, put \[Z(s,x)=h'(s)\nabla\times(0,x_1\Gamma(x),0),\qquad V_w(s,x)=Z(s,x)+3\beta(x_3)(w(3s-1,x_1,x_2),0).\] Along each tracked path the first term is exact vertical translation. The second term acts only in the middle third, at height three, and runs the full planar phase. It is divergence free because \(w\) is planar solenoidal and \(\beta\) is independent of the planar coordinates. Thus the trajectory lifts at its exact source, performs every detour outside the vertical detector interval, and lowers at its exact target. Use this construction for the loader and for each repeated phase. While its height can lie in \((-1,1)\), a nonterminal computational particle has first coordinate at least four; the initial loading source has first coordinate \(-4\). A terminal code lies in \((0,1)^2\times\{0\}\). These facts prove the all-real-time observation equivalence, including an initially terminal input.

For the comparison assertion, the constructed field has compact support and bounded mixed derivatives, so it satisfies the reference hypotheses of Lemma 2. The displayed competitor class is exactly case (v) of that lemma. Its proof obtains \(\nabla p\in L^2\) from the equation and cancels the pressure term by Lemma 1; it does not require an integrable pressure representative. Hence the velocity is unique, and pressure is determined up to a function of time.

Finally let \(U(s)\) denote the eventually periodic field just constructed and set \(\rho(t)=(1+t)^{1/3}-1\), \(\widehat U=\rho' U(\rho(t))\). The new residual is \[\widehat f=\rho''U+(\rho')^2\{U_s+(U\cdot\nabla)U\} -\nu\rho'\Delta U, \quad \rho'=\tfrac13(1+t)^{-2/3},\quad \rho''=-\tfrac29(1+t)^{-5/3}.\] Uniform derivative bounds and the fixed compact support give the stated decay and space-time \(L^2\) estimate. Chain differentiation also gives all mixed bounds. The clock is effective on \([0,\infty)\), starts at zero, is finite at finite time and is onto; hence it preserves precisely the same material event. The initial zero collar is preserved as well. This slower force is an alternative to, not a periodic version of, the first force. ◻

Proof. Use the controls of Lemma 12. Let \(\Sigma=\Gamma\times(\{\bot,F\}\sqcup\mathcal R)\times\{0,1\}\) be its full cell alphabet after deleting the singleton passive track. Let \(S\) be the finite set of recorder controls and choose a bijection \(j:S\to\{1,\ldots,|S|\}\). Use odd digits with \(B=2|\Sigma|+1\), and offsets \(o_s=-2j(s)\) for halting main controls and \(o_s=2j(s)\) for the others. Number the resulting reciprocal branches \(i=1,\ldots,N\), and write \(k_i\) for branch \(i\)’s first-coordinate scale. Thicken each reciprocal branch by \([-1,1]\). Its horizontal family gaps are at least \(B^{-2}\). Lift branch \(i\) to \(4i\), use the linear first scale \(\kappa_i=1+\theta(k_i-1)\), and descend. Padding smaller than \((4B^2)^{-1}\) separates all phases: in the middle the half-height is one and height gaps are four; in the other phases horizontal gaps suffice. The two horizontal half-widths are at most \(1/2\). Use the true-plateau cutoff of Lemma 5; it supplies at least the full-box guarantee, together with a neighborhood extension. The exact first coordinate obeys \[ X_1(\tau)=(1-\theta)X_1(0)+\theta X_1(1). \tag{46}\]

Let \(V\) be the period-one extension of the unit-time field just constructed. All codes, paths and padded supports lie in \((-L,L)^3\) for \(L=10+2|S|+4N\). Put \(a=1/(16L)\) and define in the torus patch \((-1/2,1/2)^3\) the repeated field \(U(t,x)=aV(t,x/a)\). Its support lies in \((-1/16,1/16)^3\) and its curl potential is the scaled potential \(a^2A(t,x/a)\). In particular its mean is zero. The residual must be recomputed at the physical viscosity: \[ \mathcal F_\nu[U](t,x)=a\{V_t+(V\cdot\nabla)V-(\nu/a^2)\Delta V\}(t,x/a). \tag{47}\] This keeps the specified viscosity rather than silently rescaling it. During the first unit interval load from \((1/8,0,0)\) to the scaled initial code using a curl translation with a box of half-width \(1/32\) and padding \(1/32\). Its support lies inside \((-1/4,1/4)^3\).

For a nonhalting run both loading endpoints have positive first coordinate, and every subsequent pair of coded endpoints is positive. Equation (46) preserves that sign throughout. A halting checkpoint has negative first coordinate in \((-1/16,0)\). All paths stay in the stated patch, so passage modulo one introduces no extra visit. Checkpoint times are \(1+\sum_{l<n}(5+2(l-h_l))\). This proves the all-time equivalence, including initial halting; Proposition 6 finishes the proof. ◻

Proof. First normalize to one halting work state and use the move-first recorder of Lemma 11; let \(\Sigma\) be its full three-track cell alphabet. Initial halting is preserved by this normalization. For \(m\) controls take \(\ell=1/[64(m+1)]\). Nonhalt squares have lower left corners \((1/4+2j\ell,1/4)\), \(0\le j\le m-2\); the halt square has lower left corner \((3/4,1/4)\). Use odd digits and \(b=2|\Sigma|+1\), with coding height \(1/4\). There are \(N\) reciprocal branches with gaps at least \(\ell b^{-2}\). Thicken them by half-height \(h_0=1/[64(N+1)]\) and choose heights \(z_i=1/2+i/[4(N+1)]\). Use exponential scale \(k_i^\theta\), center interpolation, and \(\epsilon=\min(h_0/2,\ell/(8b^2))\). The first and last phases have horizontal gaps exceeding \(2\epsilon\); middle height gaps equal \(16h_0\) and exceed \(2(h_0+\epsilon)\). Widths never exceed their endpoint values. A concrete plateau cutoff for a box centered at \(C_i\), with half-widths \(w_{ij}\), is \[\chi_i=\prod_{j=1}^3\sigma\!\left( \frac{(w_{ij}+\epsilon)^2-(x_j-C_{ij})^2} {(w_{ij}+\epsilon)^2-(w_{ij}+\epsilon/2)^2}\right).\] Use its product with \(\tfrac12\dot C_i\times(x-C_i)+(0,0,(\dot a_i/a_i)(x_1-C_{i1})(x_2-C_{i2}))\), where \(a_i=k_i^\theta\). Its curl gives the exact affine motion on an open neighborhood of each box. All supports remain inside the unit cube.

Load the initial rational code along its segment from the fixed label, using a radial cutoff of radius \(1/32\) and a flat switch \(\sigma(4t-1)\). Then repeat the unit motion, with phase switches \(\sigma(8\tau-1),\sigma(8\tau-3),\sigma(8\tau-5)\). These choices give spatial zero collars of width \(1/8\) and temporal zero collars of width \(1/16\) at nonnegative integer joins, including loading. The force follows from \(f_0=U_t+(U\cdot\nabla)U\), \(f_1=-\Delta U\).

Nonhalt centers have first coordinate at most \(1/4+1/32\) and half-width at most \(\ell/2\), hence their entire trajectories satisfy \(x_1\le1/4+1/32+\ell/2<2/3\). This is a width bound, not a convexity claim for exponential scaling. Loading also avoids the slab for a nonhalting input. Halt codes enter it at a finite integer sample, including time one for an initial halt. The record block length and the known frontier site \(n+2\) recover absolute head position at each work checkpoint. All force and uniqueness assertions now follow from Proposition 6. ◻

Proof. For the cutoffs in this proof, recall \[E(r)=\begin{cases}e^{-1/r},&r>0,\\0,&r\le0,\end{cases} \qquad \Theta(r)=\frac{E(r)}{E(r)+E(1-r)}.\] Use the single-halt normalization and the seven-row recorder in Lemma 11, with frontier initially at two and all marks initially zero. In the present notation its controls are \(S_q,U_r,W_r,C_q,D_q\) (ready, work-move completed, right scan, frontier extension, left scan), and its full symbols are \((a,g,m)\) with history \(g\in\{E,F\}\sqcup\{\widehat r\}\). The first row writes and moves, the second marks and starts the scan, and the last clears the mark and stays. These are exactly the seven rows already proved, after renaming. Thus the new work head \(j'\) is less than \(n+2\), a work step takes \(2(n+2-j')+4\) local steps, and both full-domain incoming properties hold.

Let \(M\) count the partial controls, let \(H=S_h\), and put \[\ell=\frac1{100(M+1)},\qquad o_H=(3/4,1/2),\qquad o_{s_i}=(1/4+2i\ell,1/2)\quad(0\le i<M-1),\qquad z_*=1/4.\] In the state frames \(o_s+\ell[0,1]^2\), use odd digits in base \(B=2|\Sigma|+1\) for the full alphabet \(\Sigma\). The minimal left-split subdivision in Lemma 17 gives separately separated rectangles \(E_i^-,E_i^+\), centers \(c_i^-,c_i^+\), and maps \[y\longmapsto c_i^+ +\operatorname{diag}(a_i,a_i^{-1})(y-c_i^-), \qquad a_i\in\{B^{-1},1,B\}.\] The common frame scale leaves these linear parts unchanged; all rectangle side lengths are at most \(\ell\).

For \(N\) branches set \(z_i=1/2+i/(4(N+1))\), \(1\le i\le N\). Let \(\delta\) be one tenth of the minimum of \(1/100\), all pairwise source gaps, all pairwise target gaps and all \(|z_i-z_j|\); omit empty pair lists. For a box \(D=\prod_j[l_j,r_j]\), allowing zero widths, define \[\chi_D^\delta(x)=\prod_{j=1}^3 \Theta\!\left(2(x_j-l_j+\delta)/\delta\right) \Theta\!\left(2(r_j+\delta-x_j)/\delta\right).\] It equals one on the \(\delta/2\) padding and is supported in the \(\delta\) padding. Put \(\theta(s)=\Theta(3s-1)\). On the first third of a unit step use \[z_i(s)=z_*+(z_i-z_*)\theta(3s),\qquad D_i(s)=E_i^-\times\{z_i(s)\},\qquad P_i=\chi_{D_i(s)}^\delta(0,\dot z_i(s)x_1,0).\] Its curl on the plateau is \((0,0,\dot z_i)\); source gaps separate these supports. Hence every point of the whole source plate lifts with planar coordinates unchanged.

On the middle third set \[\eta=\theta(3s-1),\quad c_i=(1-\eta)c_i^-+\eta c_i^+, \quad\mu_i=1-\eta+\eta a_i,\] \[D_i(s)= \big[c_i+\operatorname{diag}(\mu_i,\mu_i^{-1})(E_i^--c_i^-)\big] \times\{z_i\}.\] With \(\gamma_i=\dot\mu_i/\mu_i\), use \[P_i=\chi_{D_i(s)}^\delta(0,0,\Psi_i),\qquad \Psi_i=\dot c_{i1}(x_2-c_{i2})-\dot c_{i2}(x_1-c_{i1}) +\gamma_i(x_1-c_{i1})(x_2-c_{i2}).\] Its plateau curl is \((\dot c_{i1}+\gamma_i(x_1-c_{i1}), \dot c_{i2}-\gamma_i(x_2-c_{i2}),0)\), which is precisely the reciprocal-scale affine velocity. The distinct heights separate the supports during this phase. In the last third use the vertical potential again, with \(E_i^+\) and \(z_i(s)=z_i+(z_*-z_i)\theta(3s-2)\); target gaps separate the descents. Summing the curls gives the required field in each phase, with zero collars at every join.

All supports lie strictly inside the unit cube: horizontal centers lie in \([1/4,3/4+\ell]\times[1/2,1/2+\ell]\), half-widths are at most \(\ell/2\), heights lie between \(1/4\) and \(3/4\), and \(\delta\le10^{-3}\). Thus the field extends smoothly and with zero mean to the torus. The construction acts on neighborhoods, not only plates. An initial perturbation of max norm less than \(\delta/(4B)\) stays within \(\delta/4\) of its affine plate path: vertical phases preserve the perturbation, and both horizontal scale factors are at most \(B\). It remains on the plateau, so uniqueness of the smooth ODE gives that same affine continuation throughout.

Write the rational initial code as \((b,z_*)\). During one unit of time load the fixed particle using \[c_0(t)=(1-\theta(t))(1/4,1/4)+\theta(t)b\] and the curl of \[\chi_{\{(c_0(t),z_*)\}}^{1/100} (0,0,\dot c_{01}(x_2-c_{02})-\dot c_{02}(x_1-c_{01})).\] This has plateau velocity \((\dot c_0,0)\), support inside the cube, and zero endpoint collars. Repeat the step field thereafter and define its residual force at \(\nu\). The finite compact phase formulas give effective mixed-derivative bounds; the only scale denominators satisfy \(\mu_i\ge\min(1,a_i)>0\). Curls have zero mean, and incompressibility gives zero mean for the residual. Lemma 21 supplies the unique zero-data PDE solution and its global trajectories.

At times \(1+k\), induction through the full branch maps gives the exact local code through the first halt, or for all \(k\) in the nonhalting case. A halt code has first coordinate in \([3/4,3/4+\ell]\subset(2/3,5/6)\), including a code reached solely by loading an initially halted machine. For a nonhalting input the loader stays below \(1/2\) in first coordinate. Lift and descent retain the nonhalting planar endpoint codes. In the middle phase the centers are convex combinations of nonhalting centers and the first half-width is at most \(\ell/2\), so \[X_1\le1/4+2M\ell+\ell/2<1/2.\] This excludes the height-independent observer at every intermediate time. Only the loader depends on the input; the repeated field depends on the finite machine table. ◻

Proof. Normalize the work machine to one halting state as in Section 5, preserving initial halting. Use four tracks: work, head mark, history, and an origin bit. The history is empty except for a frontier at cell two; the origin bit is one only at absolute cell zero. In a ready control \(S_q\), the head mark is one at the work head. At intermediate controls use the seven-row recorder unchanged, with the origin track copied in every row. The first row now requires ready mark one and writes mark zero at the old head before the work move. The final return row leaves mark one when entering \(S_q\). These are the only two changes to the seven rows.

For precision, this table is the conjugate of the frontier table on its full configuration domain by the following involution: toggle the mark at the current head exactly when the control is ready, and leave it unchanged at every other control. Also take the direct product with the passive origin track. This is a bijection of all configurations, not merely of initialized ones. It proves the full-domain inverse. The incoming-rule properties also persist: into the work-tag control the first-row output mark is zero; into the right scan the marking row and its loop still write one and zero; into the return scan frontier and nonfrontier outputs remain distinct; into the ready control only the last row applies. The initial configuration is the conjugate of the frontier initialization. Consequently its cycle length is \(2(2+n-h')+4\), its absolute frontier is \(2+n\), and its origin bit remains fixed at the absolute origin. This verifies the distinct initialization without another appeal to an existential compiler.

Denote the resulting full four-track alphabet by \(\mathcal A\), and let \(K=2|\mathcal A|+1\) be its odd-digit radix. For \(m\) controls take state-square side \(r=1/[16(m+1)]\), with nonterminal lower corners \((1/8+2(i-1)r,1/8)\) and terminal lower corner \((3/4,1/8)\). Their common coding height is \(z_0=1/4\). Lemma 10 supplies separated full planar rectangles; let \(N\) be the number of branches. Give branch \(i\) height \(h_i=1/2+i/[4(N+1)]\). During a first phase lift its center from \(z_0\) to \(h_i\), keeping its planar shape; during a middle phase interpolate its planar center, with width \(w_i=(1-\mu)w_i^-+\mu w_i^+\) and height \(v_i=w_i^-v_i^-/w_i\); during a final phase lower it to \(z_0\). The planar first coordinate is again an endpoint convex combination. Choose a fixed positive padding less than one tenth of all source and target gaps, all chart-face margins and all height separations. The moving padded rectangles are disjoint at every phase. During the middle phase this follows from the different heights; during the first and last phases it follows from endpoint planar separation.

Here is a direct potential for these moving neighborhoods. With moving center \(c=(c_x,c_y,c_z)\) and \(\lambda=\dot w/w=-\dot v/v\), set \[A=(0,\dot c_z(x-c_x), \dot c_x(y-c_y)-\dot c_y(x-c_x) +\lambda(x-c_x)(y-c_y)).\] Its curl on the plateau is \((\dot c_x+\lambda(x-c_x),\dot c_y-\lambda(y-c_y),\dot c_z)\). Multiply it by a cutoff supported in the moving padded neighborhood and sum the curls. All supports remain in the open cube. The displayed center-and-shape paths solve this field on the full planar rectangles; uniqueness proves their exact endpoint maps. The positive padding and bounded planar scales also give a uniform initial neighborhood that stays on every plateau, by the perturbation argument of Lemma 5. Each component of the total field has zero mean after periodization, since the entire field is a curl.

For initialization, choose equal small squares about the fixed label and the rational input code at height \(z_0\), with half-side less than one quarter of their minimum chart-face margin. Move one to the other by the same localized potential for translation during \([0,1]\). A nonterminal loading first coordinate lies between \(1/8\) and \(1/4\), and every subsequent nonterminal phase lies below \(1/4\). A terminal endpoint lies in the detector. The initialized symbolic equivalence, including initial halting, therefore holds at all real times. Repeating the finite middle mechanism and applying Lemma 18 gives the asserted force and comparison properties. Only the loading translation and its input code depend on the finite word. ◻

Proof. Apply Lemma 30 to each template, including its loading interval if present. Both are initially zero, have one compact support, and have uniformly bounded mixed derivatives. The lemma gives the exact residual and all mixed bounds; fixed support also gives \(\|f(t)\|_{H^j}\le C_j(1+t)^{-1}\) for every fixed \(j\).

On each finite interval the slowed reference field retains every Sobolev regularity and bounded derivative required by its original comparison class. Case (i) of Lemma 2 therefore applies to the direction shuttle’s continuous \(H^1\) pressure representative, and case (iii) applies to the guarded-history processor’s continuous \(L^2\) pressure gradient. Both constructed pressures are zero.

The new path is \(X(\log(1+t);a_0)\). The clock lemma preserves the event and the effective prescription. In particular, loaded samples \(1+k\) occur at physical times \(e^{1+k}-1\), whereas the aligned construction’s samples \(k\) occur at \(e^k-1\). These times have no finite accumulation. The new forces have the stated temporal integrability; no periodicity in physical time is asserted for them. ◻

Proof. Add the companion’s fresh nonhalting dummy start and use its seven-row compiler. The permutation of (work, mark, history) into our track order identifies its table with Lemma 11, including the full-domain inverse. Let \(K\) be the full compiler alphabet size, \(m\) the number of controls, and \(N\) the number of full branches. With odd digits in base \(B=2K+2\) and initial rational stack coordinates \(x_0,y_0\), its state-square parameters are \[ \lambda=\frac1{100(m+N+1)},\qquad a_*=1/4-\lambda x_0,\qquad b_{\rm start}=1/4-\lambda y_0. \tag{48}\] The halt control has lower height \(1/16\); enumerate the controls other than start and halt by \(\ell\) and give them lower heights \(1/3+\ell/[12(m+1)]\), as in that theorem. Thus the initial code is \((1/4,1/4)\), and the common full branch formulas of Lemma 10 are precisely its inputs. The cited theorem gives the full closed-rectangle routing, positive cutoff neighborhoods, and the all-time guard in \(J_*=[1/8,7/8]\times[1/32,7/8]\). Write \(V(s,X,Y)\) for its planar velocity, one-periodic in \(s\). Choose \(\chi_*\) smooth, compactly supported in the open unit-square chart and equal to one near \(J_*\). The autonomous field from the companion’s Section 5 is \[ W=(V_1(Z,X,Y),V_2(Z,X,Y),\partial_X(X\chi_*(X,Y))), \tag{49}\] with the compactly supported products extended periodically. It is solenoidal and mean zero; its phase speed is one along the tracked path. Its trajectory from \(P\) enters \(O\) exactly upon halting. These are the specific geometric imports, rather than an identification based only on the headline conclusion.

Take the stationary and slow profiles from (OpenAI 2026c, sec. 5 and Proposition 6.1), respectively: \[ \alpha_0=\sigma(2t-1),\qquad \alpha_1=\sigma(2t-1)/(1+t). \tag{50}\] Their exact force formula at the fixed physical viscosity is \[ U_\alpha=\alpha W,\qquad f_\alpha=\alpha'W+\alpha^2(W\cdot\nabla)W-\nu\alpha\Delta W. \tag{51}\] Lemma [lem:p2-realization] supplies the stated comparison conclusion. For \(s_\alpha(t)=\int_0^t\alpha(r)\,dr\), symmetry of \(\sigma\) gives \(s_{\alpha_0}(1)=1/4\) and \(s_{\alpha_0}(t)=t-3/4\) for \(t\ge1\). The other clock satisfies \(s_{\alpha_1}(t)=s_{\alpha_1}(1)+\log((1+t)/2)\) there. Both are continuous, finite at finite time and onto \([0,\infty)\), so both preserve the complete trajectory event, including original initial halting through the dummy start. The first force is stationary after one. In the second, the coefficient orders are respectively \((1+t)^{-2},(1+t)^{-2}\) and \((1+t)^{-1}\); each time derivative gives an additional inverse power. The spatial factors in (51) have effective bounds at every order. Thus for every \(h,r\ge0\), \[ \|\partial_t^h U_{\alpha_1}(t)\|_{C^r} +\|\partial_t^h f_{\alpha_1}(t)\|_{C^r} \le C_{h,r}(1+t)^{-1-h}. \tag{52}\] The constant is effective; increasing it over \([0,1]\) makes the estimate valid for all \(t\ge0\). The exponent is common to all spatial orders, and the square of the envelope has integral \((1+2h)^{-1}\). This proves the asserted bounded and temporal \(L^2\) norms.

For the stationary profile, Proposition 4 preserves stationarity after time one and every mixed-derivative bound. For the slow profile, (2) and (4) apply to (52) through spatial order \(r+2\). They retain the rate \((1+t)^{-1-h}\) for the projected force through order \(r\) and for its pressure through order \(r+1\), with effective constants and temporal \(L^2\) bounds. In either projection choice the velocity and event are unchanged and the mean-zero pressure is \(-\phi\). The stationary and slow profiles remain separate choices. ◻