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Source. Section 4.7, physical pp. 15–16 of the selected author version.

Available partial covers

This stage works simultaneously on the deleted classes $8\pmod {16}$ and $16\pmod {32}$. On each branch, the prime-$5$ stage already covers the fifth child and reduces the third child to one input of a $9^\uparrow$. The last prime-$7$ branch needs only three inputs of a $49^\uparrow$; its third child needs one class modulo $3$, and its fourth needs the class $4\pmod5$ and two inputs of a $25^\uparrow$ inside $1\pmod5$.

Put

$$P=3^\uparrow(16,32^\uparrow)+64^\uparrow. \tag{1}$$

Here the first summand fills the modulus-$16$ branch and the second fills the modulus-$32$ branch. This contextual meaning is fixed throughout this page.

The sixteen ordered packages

The first eleven packages are

$$\begin{aligned} E_1={}&1,& E_2={}&2,& E_3={}&4,& E_4={}&8,\\ E_5={}&16+32,& E_6={}&3(4,2,1),& E_7={}&3(8,3^\uparrow(8,4),3^\uparrow(2,1)),\\ E_8={}&P,\\ E_9={}&5(1,2,9^\uparrow\!\cdot1,4,x),\\ E_{10}={}&5(8,16+32,9^\uparrow\!\cdot2,P,x),\\ E_{11}={}&5\bigl(5^\uparrow(1,2,4,8),3(1,2,4), 9^\uparrow\!\cdot4,\\ &\hspace{34mm}3^\uparrow(8,\_) +5^\uparrow(3^\uparrow\!\cdot1,3^\uparrow\!\cdot2, 3^\uparrow\!\cdot4,3^\uparrow\!\cdot8),x\bigr). \tag{2} \end{aligned}$$

Keep in reserve the partial package

$$R=5\bigl(5^\uparrow(\_,\_,16+32,P),\_,\_,\_,x\bigr). \tag{3}$$

The first ten packages in (2) fill a complete $11^\uparrow$; call the resulting twelfth package $E_{12}=11^\uparrow(*)$. The twelve complete packages now available fill a complete $13^\uparrow$; call it $E_{13}=13^\uparrow(*)$. The star always refers to these exact ordered inputs, not an arbitrary complete package.

The next two packages are

$$\begin{aligned} E_{14}={}&7(1,2,3\cdot1,5\cdot1+25^\uparrow(x,x,1,2),4,8, 7^\uparrow(x,1,x,x,2,4)),\\ E_{15}={}&7(16+32,P,3\cdot2,5\cdot2+25^\uparrow(x,x,4,8), 3^\uparrow(4,8),11^\uparrow(*), 7^\uparrow(x,8,x,x,16+32,P)). \tag{4} \end{aligned}$$

In the last arrow of $E_{15}$, the old $49$-input list has positions $1,3,4$ already covered on both restricted targets. The six displayed inputs therefore record the exact mask $7^\uparrow(x,8,x,x,16+32,P)$. Finally, define

$$\begin{aligned} F=7\bigl(&13^\uparrow(*), 5(\_,4,9^\uparrow\!\cdot1,8,x), 9^\uparrow(1,2),\\ &5\cdot3(1,2,4)+25^\uparrow(x,x,x,x),\\ &5(5^\uparrow(16+32,P,x,x),16+32, 9^\uparrow\!\cdot2,P,x),\\ &5(5^\uparrow(3^\uparrow(1,2),3^\uparrow(4,8),x,x), 3^\uparrow(8,\_),9^\uparrow\!\cdot4,\_,x)\\ &\quad+7^\uparrow\bigl( 5(x,3^\uparrow(x,1),x,1,x), 5(x,3^\uparrow(x,2),x,2,x),\\ &\hspace{29mm}5(x,3^\uparrow(x,4),x,4,x), 5(x,3^\uparrow(x,8),x,8,x),\\ &\hspace{29mm}5(x,16+32,x,P,x),3^\uparrow(1,2)\bigr),\\ &7^\uparrow(x,3^\uparrow(4,8),x,x,11^\uparrow(*), 13^\uparrow(*))\bigr),\\ E_{16}={}&R+F. \tag{5} \end{aligned}$$

Complete template verification

Packages $E_1$ through $E_8$ supply the atomic $2$- and $3$-profiles needed on both target branches. In $E_9,E_{10},E_{11}$, the fifth $5$-child was already covered and the displayed $9^\uparrow$ entries fill the sole surviving third child. Thus these eleven packages are complete. Their first ten have no prime $11$ and are pairwise signature-disjoint, so they fill $11^\uparrow(*)$; adding that new package gives twelve disjoint inputs for $13^\uparrow(*)$.

For $E_{14}$ and $E_{15}$, the earlier nested $25$-package covers the first two positions in the fourth $7$-child, so the exact remaining masks are $25^\uparrow(x,x,1,2)$ and $25^\uparrow(x,x,4,8)$. Their final arrow entries fill the still-open $7$-children. In $E_{16}$, the reserve $R$ also supplies positions three and four, so $25^\uparrow(x,x,x,x)$ records four contextually precovered inputs and contributes no new regular class. Reading (5) from left to right, the remaining seven children are supplied respectively by the completed $13$ package, the $5$-package with its sole $9$ input, the direct $9$ package, the $3$ plus $25$ package, the combined two-branch $5$ package, the displayed $5+7^\uparrow$ package, and the last $7^\uparrow$. Every $x$ is a child already covered by the initial prime-$5$ or prime-$7$ package. The source records one small child still open in $F$'s second outer prime-$7$ input, namely the first input of the $5$-node displayed there. This is deliberately retained for prime $103$.

At this point $E_3,\ldots,E_{15}$ are thirteen complete inputs and $E_{16}$ is the stated partial input. The first-level classes represented by $E_1=1$ and $E_2=2$ would have moduli $17$ and $34$, so delete them. The shifted packages $(17^2)^\uparrow\!\cdot1$ and $(17^2)^\uparrow\!\cdot2$ retain their copies at all prime-$17$ levels $k\ge2$. Thus only the first-level regular classes are empty; these are selected-input tails, not the marked continuation of the surrounding $17^\uparrow$. Primes $41$ and $43$ handle those two first-level holes. The partial last branch is handled by prime $103$.

No package in (2)–(5) has a regular factor $17$ before it is placed here. Their inner signature sets are disjoint by the construction order and the explicit $x$ deletions. Hence attaching the prime-$17$ level preserves regular injectivity. Arrow terminal classes are made finite separately by the finite realization lemma.

Used by. Later Nielsen templates.

Bears on. Problem 2.