Proof. Choose any lift \(g\in G\) of \(\sigma\) with \(c_i\)-coordinate zero, and partition \(H_0\) into \[E_0^{(i)}=\{h\in H_0:c_i(h)=0\},\qquad E_{\ne0}^{(i)}=H_0\setminus E_0^{(i)}.\] Set \[F_{i,\sigma}=E_{\ne0}^{(i)}\sqcup(g+E_0^{(i)}).\] The union is disjoint because its two parts have, respectively, nonzero and zero \(c_i\)-coordinates. Fix a target base \(b\). The unique graph point above \(b\), together with \(E_{\ne0}^{(i)}\), covers exactly those points in that fibre whose \(c_i\)-coordinate differs from \(c_i(b)\), once each. The graph point above \(b-\sigma\), together with \(g+E_0^{(i)}\), covers exactly the points whose \(c_i\)-coordinate is \(c_i(b-\sigma)\), once each. In both assertions, every choice of the other outputs and high coordinates is attained exactly once, since those offsets range freely in \(E_0^{(i)}\). The two contributions partition the target fibre precisely when the displayed equality of useful outputs holds. ◻

Proof. For each \(i\in C\), let \(\operatorname{prev}(i)\) denote the preceding channel in the cycle. Choose arbitrary maps \[d_i:P_{\operatorname{prev}(i)}\longrightarrow U_{i,j(i)}.\] For \(e=(e_i)_{i\in\mathcal I}\in P\), put \[d_i[e]= \begin{cases} d_i(e_{\operatorname{prev}(i)})&i\in C,\\ 0&i\notin C. \end{cases}\] Include a tile for every joint choice \(d=(d_i)_{i\in C}\), namely \[ F_d=\left\{ \bigl(0,\zeta,(\upsilon_i,e_i)_{i\in\mathcal I}\bigr): \begin{array}{l} e\in P,\quad \zeta\in\mathbb Z/D\mathbb Z,\\ \upsilon_i\in\mathbb Z/r_i^2\mathbb Z,\quad \upsilon_i\bmod r_i=d_i[e]\quad(i\in\mathcal I) \end{array} \right\}. \tag{7}\] This is an ordinary set: distinct \(e\) give distinct useful-output offsets, and every choice of \(\zeta\) and the lifts \(\upsilon_i\) occurs once. Both the tile and the family of all maps \(d\) are finite.

Fix a target base \((x,k)\). For a fixed \(e\), a representation using (7) must start at the graph point over \((x,(k_i-d_i[e])_i)\). The useful outputs of the target are then \[ \Phi_{x,k,d}(e) =\bigl(e_i+g_{i,x}(k_i-d_i[e])\bigr)_{i\in\mathcal I}. \tag{8}\] For any prescribed remaining coordinates of the target, the offsets \(\zeta\) and \(\upsilon_i\) are uniquely determined: each low residue is already correct, and all its lifts occur in the tile. Hence \(A\oplus F_d=G\) holds exactly when (8) is a permutation of \(P\) for every \((x,k)\). This criterion is independent of the graph’s high outputs and its \(z\)-output.

Suppose some \(j(i_0)\) is inactive at \(x\). Then the \(i_0\)-component of (8) is \(e_{i_0}+g_{i_0,x}(k_{i_0})\), independent of the preceding component of \(e\). From any target useful-output tuple, we can therefore recover \(e_{i_0}\) uniquely. The next equation around the cycle then recovers the next component, and continuing recovers all components on \(C\). The starting equation remains satisfied because it never depended on the preceding component. Off the cycle, each equation is just a fixed translation and recovers its component separately. This gives a unique preimage for every target, for all \(k\) and all maps \(d\).

Conversely, suppose every chosen label is active at some \(x\). For each \(i\in C\) choose \(w_i,w_i'\in K_i\) such that \[w_i-w_i'\in U_{i,j(i)},\qquad \varepsilon_i:=g_{i,x}(w_i)-g_{i,x}(w_i')\ne0.\] Set \(k_i=w_i\) on the cycle, and choose the other \(k_i\) arbitrarily. Extend \(\varepsilon\) by zero off \(C\) to an element of \(P\). For each \(i\in C\), choose a map with \[d_i(0)=0,\qquad d_i(\varepsilon_{\operatorname{prev}(i)})=w_i-w_i'.\] The two specified arguments are distinct, so these prescriptions extend to maps on the whole domain. They are among the maps tested in the finite family. At this base, the distinct inputs \(0\) and \(\varepsilon\) have the same image under (8): on \(C\) their outputs are respectively \(g_{i,x}(w_i)\) and \(\varepsilon_i+g_{i,x}(w_i')\), and off \(C\) they coincide as well. The permutation condition fails. This proves both directions. ◻

Proof. Fix a target base point and a batch. Every offset in that batch has the same horizontal and low coordinates, so the graph equation supplies exactly one possible source point of \(A\): the point above the base in (13). The \(i\)th low-coordinate shift is \(l\) because \(h_1\equiv l\pmod{r_i}\). Since \(x'_1=x_1-h_1\), we have \[[k_i'-x'_1]_{r_i}=[k_i-x_1]_{r_i}.\] Consequently the shear formula (5) gives \[v_i(x',k')+h_1 =x_1+[k_i-x_1]_{r_i}+r_i\beta_i(x',k') \quad\text{in }\mathbb Z/r_i^2\mathbb Z.\] Thus the offset preserves the sheared high coordinate \(\beta_i\) exactly, and the target useful output is \(c_i(x',k')+e_\delta\).

For every \(\ell\ne i\), adding the freely chosen element of \(R_\ell\) fills the entire high-coordinate fibre with low coordinate \(k_\ell\) once. Independently, adding the freely chosen element of \(P_\ell\) fills that useful-output coordinate once. In an ordinary batch the \(z\)-coordinate is also filled once; in a seed batch it remains equal to \(z(x',k')\). Hence each batch covers exactly the part of the target fibre specified by its tuple in [act:ordinary-map] or [act:seed-map], with one representation for every choice of the remaining coordinates. Distinct batches correspond to distinct tile elements, so exact coverage by the whole tile is equivalent to every specified tuple occurring once. This is precisely the asserted bijection condition. ◻

Proof. Suppose channel \(i\) has no active label at some \(x\). A function on the finite product \(K_i=\prod_{j\in J_i}\mathbb Z/r_{ij}\mathbb Z\) that is independent of every digit is constant: any two inputs can be joined by changing one digit at a time. Thus \(g_{i,x}\) has a constant value \(c_0\in P_i\). By the dependence condition (6) and the equality \(L_i(x-h)=L_i(x)\), every source in the activation test at \((x,k)\) has \(c_i(x',k')=c_0\). For each \(\delta\) there are \(r_i\) choices of \(l\). The number of batches giving any target value \(c\in P_i\) is therefore \[r_i\mu_i(c-c_0).\] The fibre criterion requires this number to be \(r_i\) for every \(c\). That is impossible because \(\mu_i\) is not constant. ◻

Proof. Fix a seed \(i\) and suppose it is nowhere active. The dependence condition and the absence of activity give a function \(q:S\to P_i\) such that \(c_i(x,k)=q(s(x))\) for every base point. Define its nonnegative integer histogram by \[Q(\alpha,\xi) =\abs{\{s\in S:q(s)=(\alpha,\xi)\}}, \qquad \sum_{(\alpha,\xi)\in P_i}Q(\alpha,\xi)=D.\] At any target base point and for a fixed \(\delta\), the source pairs \[(s(x'),k_i')=(s(x)-\tau,k_i-l)\] run once through \(S\times K_i\) as \((\tau,l)\) varies. Thus the number of seed batches with useful output \(c\in P_i\) is \(r_i\sum_{\delta\in\Delta_i}Q(c-e_\delta)\). The necessary marginal count from 9, divided by \(r_i\), yields \[(Q*\mu_i)(c)=D\qquad(c\in P_i),\] where \((Q*\mu_i)(c)=\sum_{e\in P_i}\mu_i(e)Q(c-e)\) is convolution on the finite group \(P_i\). Convolution with the constant function \(1\) already equals \(D\), so \(Q*\omega_i=0\). Written out, this says \[ Q(\alpha,\xi)-Q(\alpha-1,\xi) -Q(\alpha,\xi-1)+Q(\alpha-1,\xi-1)=0. \tag{16}\]

We spell out the integer and positivity consequences of this identity. For each \(\alpha\), the difference \(Q(\alpha,\xi)-Q(\alpha-1,\xi)\) is independent of \(\xi\), because successive differences in the second coordinate vanish. Summing such differences along the cyclic first coordinate shows that, for any \(\alpha_0\), \[Q(\alpha,\xi)-Q(\alpha_0,\xi) =Q(\alpha,0)-Q(\alpha_0,0).\] Choose \(\alpha_0\) to minimize \(Q(\alpha,0)\), and put \[u(\alpha)=Q(\alpha,0)-Q(\alpha_0,0), \qquad v(\xi)=Q(\alpha_0,\xi).\] Both functions take nonnegative integer values, and \(Q(\alpha,\xi)=u(\alpha)+v(\xi)\). Summing gives \[D=b_i\sum_{\alpha\in\mathbb Z/a_i\mathbb Z}u(\alpha) +a_i\sum_{\xi\in\mathbb Z/b_i\mathbb Z}v(\xi).\] Since \(b_i>D\) and both sums are nonnegative integers, the first sum must vanish. It follows that \(a_i\) divides \(D=p^4\). But the two seed values of \(a_i\) are \(2\) and \(3\), whereas \(p>200\) is prime. This contradiction proves the claim. ◻

Proof. Let \(A\) be any common solution. Fix a total ordering of \(\Sigma\). For each ordinary channel \(n\) and integer \(m\), define \[W(n,m)=\min\mathcal A_n((0,m)),\] where the minimum is taken in that ordering. This is defined by 10. The dependence condition implies that \(\mathcal A_n(x)\) depends only on \(L_n(x)\); since \(L_n(0,m)=m\), the selected symbol at any \(x\) is exactly \(W(n,L_n(x))\).

For any integers \(d,e\), consider \(x=(d,e)\). If the word \((W(n,dn+e))_{0\le n<N}\) were forbidden, all its selected labels would be active at \(x\). This contradicts the corresponding exclusion equations, by 8. Hence \(W\) satisfies the line rule.

By 11, for each \(t\in\{1,2\}\) there is a point \(x^{(t)}\) where the label of \(\eta_t\) is active. The seed-pair exclusions imply that \(\mathcal A_n(x^{(t)})\) contains no symbol other than \(t\). Since this set is nonempty, it equals \(\{t\}\). Therefore \[W\bigl(n,L_n(x^{(1)})\bigr)=1, \qquad W\bigl(n,L_n(x^{(2)})\bigr)=2 \quad\text{for every }0\le n<N.\] The two arguments in each column must be distinct, and every column of \(W\) is nonconstant.

Suppose now that \(A\) is invariant under a finite-index subgroup \(\Lambda\le G\). Consider the actual group element \[e_{\mathrm v}=\bigl((0,1),0_V\bigr)\in G.\] Its coset has finite order in \(G/\Lambda\), so some integer \(M>0\) satisfies \(M e_{\mathrm v}\in\Lambda\). We thus obtain a period with zero finite-coordinate component: \[A+\bigl((0,M),0_V\bigr)=A.\] Translation by this element takes the unique graph point above \((x,k)\) to the unique graph point above \((x+(0,M),k)\) and leaves all useful outputs unchanged. It follows that every active-label set is unchanged under \(x\mapsto x+(0,M)\). At \(x=(0,m)\) this gives \(W(n,m+M)=W(n,m)\) for all \(n,m\). The array is therefore vertically periodic, satisfies the line rule, and has every column nonconstant, contradicting 5. ◻

Proof. Fix a target \(((\alpha^*,\xi^*),\beta_i^*)\). The point \(\beta_i^*\) specifies one block \(B\in\mathcal B_{i,j}\) and its label \(q\). As each \(T_{i,j}(t)\) preserves its block, any preimage must have its source \(k_i'=k_i-l\) in \(B\).

Suppose first that \(B\) has size \(a_i\). Choose \(\delta\) by the two conditions \[ \begin{aligned} e_\delta^b&=\xi^* &&\text{in }\mathbb Z/b_i\mathbb Z,\\ e_\delta^a+\rho^a(\delta)&=\alpha^*+x_1-q &&\text{in }\mathbb Z/a_i\mathbb Z. \end{aligned} \tag{18}\] The first fibre ordering makes this choice unique. Set \(y=\alpha^*-e_\delta^a\), let \(k_i'\) be the unique point of \(B\) with label \(y\), and set \(l=k_i-k_i'\). The useful output in (17) is then \((\alpha^*,\xi^*)\). The high output has label \[y+x_1-h_1(l,\delta) =\alpha^*+x_1- \bigl(e_\delta^a+\rho^a(\delta)\bigr) =q \quad\text{in }\mathbb Z/a_i\mathbb Z,\] so it is \(\beta_i^*\). Conversely, both target coordinates force exactly these choices of \(\delta,y,k_i'\) and \(l\).

If \(B\) has size \(b_i\), use the second fibre ordering to choose the unique \(\delta\) satisfying \[ \begin{aligned} e_\delta^a&=\alpha^* &&\text{in }\mathbb Z/a_i\mathbb Z,\\ e_\delta^b+\rho^b(\delta)&=\xi^*+x_1-q &&\text{in }\mathbb Z/b_i\mathbb Z. \end{aligned} \tag{19}\] Now \(y=\xi^*-e_\delta^b\) specifies \(k_i'\) in \(B\), and again \(l=k_i-k_i'\). The high label is \(\xi^*+x_1-(e_\delta^b+\rho^b(\delta))=q\) in \(\mathbb Z/b_i\mathbb Z\). These choices are necessary and sufficient, proving the bijection in both cases. ◻

Proof. Use the graph, dependence, exclusion, and activation tiles constructed in [sec:encoding,sec:activation]. They are finite nonempty sets, and their finite coordinate group \(V\) is cyclic. The graph constructed in this section solves all their equations. 12 excludes every fully periodic common solution. ◻

Proof. Choose independently, for each residue \(v\), \[t(v)=v+mh_v,\qquad h_v\text{ uniform in }\{-2,-1,0,1,2\}^3.\] Fix a requested vector \(d\) on the right-hand side of [geom:difference-property]. Translation by \(d\bmod m\) on \((\mathbb Z/m\mathbb Z)^3\) is a permutation without fixed points. A cycle of length \(L\geq2\) contains \(\lfloor L/2\rfloor\geq L/3\) disjoint successive pairs. Selecting these pairs in every cycle gives at least \(m^3/3\) disjoint pairs of residues \((u,v)\) with \(v-u\equiv d\pmod m\). Discard the pair containing residue \(0\), if there is one. At least \[k_m=\lfloor m^3/3\rfloor-1\] pairs remain, all with both endpoints nonzero.

For such a pair, the equality \(t(v)-t(u)=d\) asks for \[h_v-h_u=\frac{d-(v-u)}m.\] This is an integer vector whose coordinates have absolute value at most \(2\), since \(\norm{d}_\infty\leq m\) and the coordinates of \(u,v\) lie between \(0\) and \(m-1\). Two independent uniform variables on \(\{-2,-1,0,1,2\}\) have any specified difference of absolute value at most \(2\) with probability at least \(3/25\). Hence each chosen pair realizes \(d\) with probability at least \[c=(3/25)^3=27/15625.\] These events are independent because the pairs have disjoint endpoints. Failure to realize this \(d\) has probability at most \((1-c)^{k_m}\). There are at most \((2m+1)^3\) requested vectors, so the probability that [geom:difference-property] fails is at most \[(2m+1)^3(1-c)^{k_m}.\] This tends to zero as \(m\) tends to infinity. Fix \(m\geq2\) for which it is less than one, and then fix a successful choice of the representatives. Every witnessing pair avoids residue \(0\), as required. ◻

Proof. Let \(A\oplus F=\Gamma\) be a tiling from 17, and put \(B=\pi^{-1}(A)\). Then \(B+w=B\). Also \(B\oplus U=\mathbb Z^3\): for any \(g\in\mathbb Z^3\), the unique representation of \(\pi(g)\) in \(A+F\) determines a unique \(u\in U\), and then \(g-u\in B\) is the unique possible other summand.

Place copies of \(T\) at all translations in \(mB\). Their constituent copies of \(T_0\) or \(T_1\) have centers \(m(b+u)\), for \(b\in B\) and \(u\in U\). Since \(B\oplus U=\mathbb Z^3\), there is exactly one such center at every point of \(m\mathbb Z^3\). Its type is \(1\) precisely at the centers \(mB\), because the type-\(1\) piece corresponds to \(u=0\).

The assembly with \(T_0\) at every center in \(m\mathbb Z^3\) tiles \(\mathbb Z^3\) exactly: each integer vector has a unique residue label \(v\) and a unique expression \(ma+t(v)\). Switching to type \(1\) at the centers \(mB\) removes the marked points \[t(0)+mB\] and inserts \[t(0)+m(B+w)=t(0)+mB.\] The insertion is a bijective relocation, so all multiplicities remain one. This proves \(mB\oplus T=\mathbb Z^3\). Thickening by unit cubes gives an almost-everywhere tiling by \(\Omega\) with the same translation set: away from the integer coordinate planes, every point belongs to the interior of one integer unit cube. ◻

Proof. If \(\delta\) is integral, use \(d=\delta\). Otherwise, keep every integer coordinate unchanged and round each noninteger coordinate either down or up. Both choices lie in \([-m,m]\), since the endpoints are integers, and both are at distance strictly less than one from the original coordinate. In at least one noninteger coordinate choose a rounding which is not a multiple of \(m\). Such a choice exists because two consecutive integers cannot both be multiples of \(m\geq2\). The resulting \(d\) is not in \(m\mathbb Z^3\). Coordinates of \(\delta\) equal to \(m\) or \(-m\) are left unchanged, so the strict distance estimate also holds on the boundary of the box. ◻

Proof. First, two constituent copies cannot have the same center, because both contain the positive-volume common part \(\Omega_*\). Thus the centers and their types are unambiguous. We claim that distinct centers \(c,c'\) satisfy \[ \norm{c'-c}_\infty\leq m \quad\Longrightarrow\quad c'-c\in m\mathbb Z^3. \tag{29}\] If \(\delta=c'-c\) violated this assertion, 20 would give a vector \(d\) in the difference set in (27) with \(\norm{\delta-d}_\infty<1\). Write \(d=t-t'\) for \(t,t'\in T_*\). The two common-part unit cubes \[c+t+[0,1]^3,\qquad c'+t'+[0,1]^3\] have lower corners whose coordinate differences have absolute value strictly less than one. Their intersection therefore has positive volume, contradicting the tiling property. This proves [geom:nearby-centers].

In particular, distinct centers have sup-norm distance at least \(m\): a nonzero vector in \(m\mathbb Z^3\) cannot have smaller sup norm. The set \(C\) is consequently uniformly separated and locally finite. The auxiliary closed cubes \[Q_c=c+[-m/2,m/2]^3\qquad(c\in C)\] have disjoint interiors. Each has volume \(m^3\), equal to the volume of either constituent type.

We use periodicity to show that these auxiliary cubes cover. Let \(P\) be a bounded half-open fundamental parallelepiped for \(\Lambda\). There are finitely many centers modulo \(\Lambda\), by local finiteness; choose representatives \(c_1,\ldots,c_r\). Their types \(\iota_1,\ldots,\iota_r\) are preserved under \(\Lambda\). Integrating the constituent tiling over \(P\) gives \[\begin{align*} |P| &=\int_P\sum_{j=1}^r\sum_{\lambda\in\Lambda} \mathbf{1}_{c_j+\lambda+\Omega_{\iota_j}}(x)\,dx\\ &=\sum_{j=1}^r|\Omega_{\iota_j}|=rm^3. \end{align*}\] For the second equality, translate the integration domains by \(-\lambda\) and use that \(\{P-\lambda:\lambda\in\Lambda\}\) partitions \(\mathbb R^3\) up to boundaries. Nonnegative summands justify the exchange of summation and integration.

The same calculation for the auxiliary cubes gives \[\int_P\sum_{c\in C}\mathbf{1}_{Q_c}(x)\,dx=rm^3=|P|.\] The integrand is at most one almost everywhere, because the cubes have disjoint interiors; their countably many boundaries are null. It is therefore one almost everywhere on \(P\), and periodicity gives almost-everywhere coverage of \(\mathbb R^3\). The family of closed cubes \(\{Q_c\}_{c\in C}\) is locally finite, so its union is closed. A point outside the union would have an open neighborhood of positive volume outside it, contradicting almost-everywhere coverage. Thus the cubes cover every point of \(\mathbb R^3\).

Their intersection graph is connected. Indeed, if one graph component were separated from the remaining components, the unions of their respective cubes would be disjoint nonempty closed sets covering \(\mathbb R^3\). Both unions are closed because they are subfamilies of a locally finite family of closed sets. They would disconnect \(\mathbb R^3\), which is impossible. Whenever \(Q_c\) and \(Q_{c'}\) intersect, \(\norm{c'-c}_\infty\leq m\), so [geom:nearby-centers] puts their difference in \(m\mathbb Z^3\). Following paths in the intersection graph therefore puts every center in one coset \(c_0+m\mathbb Z^3\).

No point of that coset can be missing. The interior of the side-\(m\) cube at a missing grid point is disjoint from all side-\(m\) cubes at the other grid points and would be uncovered. Hence \(C=c_0+m\mathbb Z^3\). ◻

Proof. Suppose, to the contrary, that \(\mathcal A\subset\mathbb R^3\) is such a translation set and that \(\mathcal A+\Lambda=\mathcal A\) for a full-rank lattice \(\Lambda\). Distinct vectors of \(\mathcal A\) have sup-norm distance at least one: otherwise their translates of a fixed unit cube in \(\Omega\) would overlap in positive volume. Hence \(\mathcal A\) is countable. Decompose every translated copy of \(\Omega\) using [geom:large-tile]. A copy at \(a\in\mathcal A\) has a type-\(1\) constituent centered at \(a\) and type-\(0\) constituents centered at \(a+mu\) for \(u\in U\setminus\{0\}\). The constituents form an almost-everywhere tiling and inherit the lattice periods, including their types. No two constituents have the same center, as observed in the proof of 21. Consequently the map \[ \mathcal A\times U\longrightarrow C,\qquad(a,u)\longmapsto a+mu \tag{30}\] is a bijection, and the type-\(1\) centers are exactly \(\mathcal A\).

By 21, \(C=c_0+m\mathbb Z^3\). Translate the entire assembly by \(-c_0\), which does not change its period lattice, and henceforth write \(C=m\mathbb Z^3\). There is a set \(B'\subset\mathbb Z^3\) such that \(\mathcal A=mB'\). Let \(b(a)=\mathbf{1}_{B'}(a)\); this records the type at the center \(ma\). The bijection (30), divided by \(m\), gives \[ B'\oplus U=\mathbb Z^3. \tag{31}\]

Consider the open unit cell with lower corner \(t(0)+ma\), for any \(a\in\mathbb Z^3\). Since all constituent centers and all voxel positions are integral, a unit cube can meet the interior of this cell only if it has the same lower corner. The residues modulo \(m\) then exclude every common-part point \(t(v)\) with \(v\ne0\). There are exactly two possible sources: the unshifted marked point from type \(0\) at \(ma\), or the moved marked point from type \(1\) at \(m(a-w)\). The coverage multiplicity on this open cell is therefore \[1-b(a)+b(a-w).\] The almost-everywhere tiling makes it one. Thus \(b(a)=b(a-w)\) for every \(a\), or equivalently \[ B'+w=B'. \tag{32}\] This also gives invariance under every element of \(\ker\pi=\mathbb Zw\).

The set \(B'\) is fully periodic in the discrete sense. It is nonempty, and \(\mathcal A=mB'\subset m\mathbb Z^3\). For \(a_0\in\mathcal A\) and any \(\lambda\in\Lambda\), both \(a_0\) and \(a_0+\lambda\) belong to \(m\mathbb Z^3\), so \(\lambda\in m\mathbb Z^3\). Hence \(L=m^{-1}\Lambda\) is a rank-three lattice contained in \(\mathbb Z^3\) and preserves \(B'\). Such an integer lattice has finite index: the matrix formed by an integer basis has nonzero integer determinant, whose absolute value is its index. In particular, \(L\leq\mathbb Z^3\) is a finite-index period subgroup of \(B'\).

We finally descend the exact tiling (31) to the quotient. Fix \(\gamma\in\Gamma\) and any lift \(g\in\mathbb Z^3\). For each \(u\in U\), there is a representation \[\gamma=\alpha+\pi(u),\qquad \alpha\in\pi(B'),\] if and only if \(g-u\in B'\). One direction follows by applying \(\pi\). For the other, if \(\alpha=\pi(b_0)\) with \(b_0\in B'\), then \(g-u-b_0\in\ker\pi\); [geom:kernel-invariance] therefore puts \(g-u\) in \(B'\). By [geom:upstairs-tiling], exactly one \(u\) has this property, and \(\pi\) is injective on \(U\). Thus \[\pi(B')\oplus\pi(U)=\Gamma.\] This quotient tiling is fully periodic. Indeed, \(\pi(L)\) preserves \(\pi(B')\), and the surjection \(\mathbb Z^3/L\to\Gamma/\pi(L)\) shows that \(\pi(L)\) has finite index. Since \(\pi(U)=F\), this contradicts 17. ◻

Completion of the proof of 1. The finite nonempty tile \(T\) from [geom:large-tile] and its closed unit-cube thickening \(\Omega\) admit the tilings in 19. The Euclidean tiling cannot be fully periodic by 22. If a discrete tiling of \(T\) had a finite-index period subgroup in \(\mathbb Z^3\), thickening its unit cells would give an almost-everywhere tiling of \(\Omega\) invariant under the same subgroup, viewed as a full-rank lattice in \(\mathbb R^3\). This is also excluded by 22, proving both claims. ◻