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Source. Section 3, especially physical pp. 5–8, and the completion remarks on physical p. 23 of the selected author version. The fresh-prime realization below is the paper's explicitly permitted alternative on physical p. 7. The proof spells out the nested finite closure that the source leaves informal.

Statement

Let $q$ be prime and $m\ge1$ an integer. Fix an explicit ancestor context $C=b\pmod L$, with $\gcd(L,q)=1$, and an explicit $q$-coordinate modulo $q^{m-1}$. Represent that coordinate by the unique

$$1\le c\le q^{m-1}, \tag{1}$$

where $c=q^{m-1}$ represents zero. Let the target set $T$ lie in the intersection of $C$ with this $q$-coordinate. The context $C$ consists only of congruence conditions actually present in the expression; unrelated conditions defining $T$ are not inherited.

Suppose each $A_j$ is either a finite package or a finite acyclic arrow expression and, when intersected with the successive explicit $q$-children below, covers the corresponding portions of $T$. Then

$$(q^m)^\uparrow(A_1,\ldots,A_{q-1}) \tag{2}$$

has a finite realization covering $T$.

For a finite dependency expression made from such arrows, assume separately that the regular modulus signatures generated by its finite descents are distinct. Then all arrows can be realized so that the resulting family is finite and all moduli remain distinct. Terminal moduli may be forced above any prescribed bound.

One normalized arrow

For $k\ge m$ and $1\le j<q$, define the $j$th nonmarked $q$-coordinate at level $k$ by

$$Q_{k,j}:\quad x\equiv c+jq^{k-1}-q^{m-1}\pmod {q^k}. \tag{3}$$

For $k\ge m-1$, define the marked coordinate by

$$Q_k^*:\quad x\equiv c-q^{m-1}\pmod {q^k}. \tag{4}$$

The normalization (1) makes (3), in increasing $j$, exactly the first $q-1$ compatible children in increasing least-positive order; (4) is the last child. At $k=m$, these $q$ classes partition the parent $q$-coordinate. At the next level, the classes in (3) partition the first $q-1$ children of $Q_{k-1}^*$, and $Q_k^*$ is its last child. Induction proves the same statement at every level.

At each level $k$, intersect every explicit class of $A_j$ with $C\cap Q_{k,j}$. If that class has modulus $d$, the output modulus is

$$\operatorname{lcm}(L,q^k,d), \tag{5}$$

with no additional factor coming from the target $T$. By hypothesis these packages cover all nonmarked pieces of $T$. After levels $m,m+1,\ldots,n$, only $T\cap C\cap Q_n^*$ remains.

Closing the marked tail

Before choosing the cutoff, reserve a fresh prime $r$. It is chosen outside the finite regular-prime alphabet, different from $q$, coprime to $L$, and different from every terminal prime reserved earlier. Now choose

$$n\ge m+r-2 \tag{6}$$

and as large as any requested terminal-modulus bound requires. For $1\le i\le r$, take the explicit class determined by

$$C,\qquad x\equiv i\pmod r,\qquad Q_{n+1-i}^*. \tag{7}$$

Every integer in $T\cap C\cap Q_n^*$ has one residue $i\pmod r$ and satisfies $Q_{n+1-i}^*$, so (7) covers the remaining target. Its $r$ moduli have $q$-adic exponents

$$n,n-1,\ldots,n+1-r,$$

which are distinct and, by (6), at least $m-1$. Hence they preserve the explicit parent coordinate.

For $q=2,m=1,c=1,C=\mathbb Z$, formula (7) is the source's family

$$x\equiv i\pmod r,\qquad x\equiv2^{\,n+1-i}\pmod {2^{\,n+1-i}}, \qquad1\le i\le r. \tag{8}$$

Thus an arrow is a finite cover macro. The recurrence

$$q^\uparrow(A_1,\ldots,A_{q-1}) =q(A_1,\ldots,A_{q-1},q^\uparrow(A_1,\ldots,A_{q-1}))$$

describes its regular levels, but an infinite literal expansion would not be a finite covering system.

Domination of the ideal arrow

The finite realization contains, in the sense of unions of integer sets, the entire ideal regular union represented by the infinite arrow. Indeed, every regular class at a level at most the cutoff is included explicitly. Every deeper regular class lies inside $Q_n^*$, and (7) covers all of that marked tail. If an input package itself contains arrows, apply the same statement inductively to its realized child occurrences.

The same domination holds for a partial arrow with blank inputs: it contains all ideal regular classes arising from its nonblank inputs, while making no claim about a blank input. Therefore a later $x$ may use coverage supplied by an earlier ideal-arrow calculation even when the two arrow occurrences are ultimately assigned different cutoffs. The proof relies on containment of unions, not on the finite realization retaining every deeper ideal class as a literal member.

A selected input's arrow portion

If the $j$th displayed input is omitted from $q^\uparrow(A_1,\ldots,A_{q-1})$, then its classes $C\cap Q_{k,j}\cap A_j$ are missing for every $k\ge m$. A contextually selected shifted package beginning at $q^{m+1}$, denoted in the source by $(q^{m+1})^\uparrow\!\cdot A_j$, restores precisely the copies with $k\ge m+1$. The first class $C\cap Q_{m,j}\cap A_j$ remains a hole.

Indeed, the outer marked child $Q_m^*$ has normalized representative

$$c'=c-q^{m-1}+q^m.$$

The $j$th regular coordinate of the shifted arrow at level $k\ge m+1$ is

$$c'+jq^{k-1}-q^m=c+jq^{k-1}-q^{m-1},$$

which is exactly $Q_{k,j}$. Thus the shifted arrow begins on the old marked child and reproduces the later copies; it does not descend inside the first-level $j$th child.

This “arrow portion” is an input tail and is distinct from the marked spine $Q_k^*$. It explains the two deleted prime-$17$ inputs and the single unfilled prime-$19$ input.

Nested arrows and collisions

Realize a finite acyclic dependency expression recursively.

1. At the current arrow occurrence, reserve its fresh terminal prime.
2. Choose a finite cutoff satisfying (6), and expand its finitely many regular levels.
3. Recursively realize the finitely many child-arrow occurrences created by that expansion. The dependency depth decreases, so this terminates.
4. Add the current occurrence's terminal classes (7).

All terminal primes are chosen outside the regular alphabet. A terminal prime is introduced only in the terminal leaves owned by its occurrence; it never becomes part of another occurrence's fixed regular context. Descendants created under an ancestor's regular level retain their own fresh primes, while the ancestor contributes only its explicit regular coordinate.

A terminal leaf contains the unique prime assigned to its owner. It cannot share a modulus with a regular leaf or with a terminal leaf owned by another occurrence. Within one occurrence, the distinct $q$-adic exponents in (7) separate the terminal moduli. The assumed regular-signature injectivity separately handles every nonterminal leaf.

That assumption is essential. Choosing a deeper cutoff or a fresh terminal prime changes tail classes only; it cannot repair two equal regular moduli. The construction ledger checks regular signatures before this lemma is applied. The proof also does not certify the optional assertion that a single prime $107$ suffices for every occurrence; fresh terminal primes already prove the existence theorem.

Bears on. Problem 2.