---

Source. Sections 4.1–4.9, physical pp. 9–18 of the selected author version. This page expands the displayed arrows as unbounded exponent regions; it is an exact symbolic check, rather than a finite cutoff sample.

Put

$$(a,b,c,d,e,f,g,h) =(v_2(m),v_3(m),v_5(m),v_7(m),v_{11}(m), v_{13}(m),v_{17}(m),v_{19}(m)). \tag{1}$$

An expression such as $8^\uparrow$ gives the interval $a\ge3$; $9^\uparrow(1,2)$ gives $b\ge2$ with $a\in\{0,1\}$. Ordinary nodes and multiplication add fixed exponent coordinates, while an arrow changes one coordinate through a half-infinite interval. Consequently every displayed regular package is a finite union of Cartesian exponent boxes. The comparisons below establish when that union is disjoint.

Initial prime-7 boundary

Before prime $7$, the exact prime-$2,3,5$ expressions and the explicit deleted classes have disjoint signatures. Expanding the prime-$7$ display gives the following new signatures.

The rows are disjoint. The unrestricted complement at a positive $7$-exponent also contains the five excluded signatures $7,14,21,28,35$. Among moduli at least $40$, the complement is exactly the two unused families recorded at the end of Section 4.4,

$$7^{u+1}5^{v+3}3^{w+2}2^a,\qquad 7^{u+2}5^{v+3}3^b2^a \quad(u,v,w,a,b\ge0). \tag{2}$$

Thus the initial expressions add no duplicate regular modulus and leave precisely the stated unused regions.

The prime-11 template

For $d=0$, the ten packages $A_1,\ldots,A_{10}$ have the following exact partition.

For $d\ge1$, $A_9$ has exactly

$$\begin{aligned} &c=0,b=0,\ a\ge0;\\ &c=0,b=1,\ a=0;\\ &c\ge1,b=0,\ a\in\{0,1,2\}. \end{aligned} \tag{3}$$

The $C$-part of $A_{10}$ is the disjoint complement:

$$\begin{aligned} &c=0,b=1,\ a\ge1;\qquad c=0,b\ge2,\ a\ge0;\\ &c\ge1,b=0,\ a\ge3;\qquad c\ge1,b\ge1,\ a\ge0. \end{aligned} \tag{4}$$

These assertions follow directly by expanding the six displayed inputs of $A_9$ and $C$. For example, the fourth $C$-input gives $c\ge1,b=0,a\ge3$, $c\ge1,b=1,a=0$, and $c\ge1,b=2,a\le2$; its sixth input supplies the complementary $a$'s for $b=1,2$, and its fifth supplies all $a$ for $b\ge3$. At $c=0$, its first, second, third, fifth, and sixth inputs give exactly (4).

Therefore $A_1,\ldots,A_{10}$ partition every quadruple $(a,b,c,d)\in\mathbb Z_{\ge0}^4$ except the signatures of $1,2,3$:

$$(0,0,0,0),\quad(1,0,0,0),\quad(0,1,0,0). \tag{5}$$

In particular, the ten input packages are pairwise signature-disjoint at every exponent, not merely up to a finite bound. Attaching $e\ge1$ proves regular injectivity of $\mathcal T_{11}$.

The prime-13 reflection

The first ten $D_i$ have the same partition (5). In $D_{11}$, deleting every structural suffix with $a=0$ or $a=1$ leaves exactly the prime-$11$ profiles with $a\ge2$. The structural replacements $4\mapsto1$ and $8^\uparrow\mapsto2$ in $D_{12}$ map those two source rays bijectively to $a=0$ and $a=1$. Thus

$$\begin{array}{c|c} D_{11}&e\ge1,\ a\ge2,\quad b,c,d\ge0,\\ D_{12}&e\ge1,\ a\in\{0,1\},\quad b,c,d\ge0. \end{array} \tag{6}$$

The transform is applied to the finite arrow syntax before expansion; it does not collapse separately enumerated leaves. The two regions in (6) are disjoint, and $D_1,\ldots,D_{10}$ have $e=0$. Hence all twelve inputs of $\mathcal T_{13}$ are disjoint.

The partial prime-19 template

The first ten packages have no primes $7,11,13,17$. Their union is

$$\begin{cases} c=0,b=0,a\ge2,\\ c=0,b\ge1,a\ge0,\\ c\ge1,b\in\{0,1\},a\ge0. \end{cases} \tag{7}$$

The individual lines in (7) split exactly as the ten displayed $F_i$: $F_1,F_2$ split the first line by $a=2$ and $a\ge3$; $F_3,\ldots,F_8$ split the second by $b=1$ or $b\ge2$ and the four $2$-adic ranges; $F_9,F_{10}$ are the two $b$-values in the last line.

Adding the temporary packages $1,2$ makes the exact pool

$$c=0\quad\text{or}\quad(c\ge1\ \text{and}\ b\le1), \qquad a\ge0. \tag{8}$$

The explicit two six-element blocks on the template page partition (8). Consequently $F_{11},F_{12}$, which add $d\ge1$, are disjoint. The package $G_1$ has $c\ge1,b\ge2,d=e=f=g=0$, precisely the unused $5$-region in (7), and $G_2$ has $d\ge1,c\ge1,b\ge2,e=f=g=0$, the unused $7$-region complementary to $F_{11},F_{12}$.

At $e\ge1,f=g=0$, $F_{13},F_{14}$ together occupy

$$\begin{aligned} &d=0,\quad[c=0\ \text{or}\ (c\ge1,\ b\le1)],\\ &d\ge1,\quad c=0,\ b=0, \end{aligned} \tag{9}$$

for every $a\ge0$. The structural enlargement $1\mapsto4$, $2\mapsto8^\uparrow$ splits these regions between $a\in\{0,1\}$ and $a\ge2$. Expanding the four nonblank inputs of $G_3$ gives the disjoint complement of (9),

$$d=0,c\ge1,b\ge2 \quad\text{or}\quad d\ge1,c+b\ge1, \tag{10}$$

again for every $a$. Thus $F_{13},F_{14},G_3$ partition every $(a,b,c,d)$ at $e\ge1$.

The exact $F_{15}$ selection has $f\ge1,e=g=0$ and

$$d=c=0 \quad\text{or}\quad d\ge1\ \text{with}\ (c=0\ \text{or}\ b\le1). \tag{11}$$

The first five inputs of $G_4$ have $f\ge1,e=0,d=0,c\ge1$, split by $b=0,1,\ge2$ and then by $a\le1$ or $a\ge2$. They miss (11).

Number the seven full $J(u,v,w)$ inputs on the prime-$19$ page by $J_1,\ldots,J_7$. All have $e,f\ge1$, and their exact inner partition is

Thus the last seven inputs have $e\ge1$, miss $F_{15}$, and are internally disjoint. Finally $F_{16}$ alone has $g\ge1$.

It follows that $F_1,\ldots,F_{16}$ and all four components of $F_{17}$ are pairwise signature-disjoint. In the surrounding $19^\uparrow$, each has $h\ge1$ and a nontrivial inner signature. The selected-input tail $(19^2)^\uparrow\!\cdot1$ has $h\ge2$ and all other exponents zero, so it does not meet them. Its missing first-level signature is $19$, exactly the regular hole later filled at prime $47$.

The prime-23 template

The first sixteen displayed packages have the following disjoint union; the omitted regions are supplied by $H_{17}$.

The reserve $9^\uparrow(16^\uparrow,\_)$ in $H_{17}$ supplies $d=c=0,b\ge2,a\ge4$. Its $7^\uparrow$-part supplies

$$d\ge1,c=0,b\ge2,a\ge0, \quad\text{and}\quad d\ge1,c\ge1,b\ge2,a\le3. \tag{12}$$

Thus $H_1,\ldots,H_{17}$ are disjoint and their union is

$$\begin{array}{ll} d=0,c=0:&\text{every }a,b\text{ except }(a,b)=(0,0),\\ d=0,c\ge1:&b\le1\text{ with every }a,\text{ or }b\ge2,a\le3,\\ d\ge1,c=0:&\text{every }a,b,\\ d\ge1,c\ge1:&b=0\text{ with every }a,\ b=1,a\le2,\ \text{or }b\ge2,a\le3. \end{array} \tag{13}$$

The exact prefix selections for $H_{18},H_{19},H_{20}$ add respectively new $13,17,19$ factors, so these three packages are mutually disjoint and disjoint from $H_1,\ldots,H_{17}$. None of the first twenty packages has an $11$-factor. The two final $11^\uparrow$ packages use the disjoint blocks $H_1,\ldots,H_{10}$ and $H_{11},\ldots,H_{20}$, so they are disjoint from each other and from the first twenty. No input has a $23$-factor. Therefore every regular modulus in $\mathcal T_{23}$ occurs once.

Consequence

The four reusable templates are regular-signature injective over their full unbounded exponent ranges. The argument does not use terminal primes and does not shift a regular arrow's starting level. Their finite terminal realizations may therefore be applied afterward without concealing a regular-modulus collision.

Bears on. Problem 2.