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Source. Section 4.8, physical pp. 16–17 of the selected author version.

The first sixteen inputs

Work in the deleted class $5\pmod {12}$, which lies on its $1\pmod4$ branch. Begin with the ten ordered packages

$$\begin{aligned} F_1,\ldots,F_{10}={}&\bigl(4,8^\uparrow,3\cdot1,3\cdot2, 3\cdot4,3\cdot8^\uparrow,9^\uparrow(1,2),\\ &9^\uparrow(4,8^\uparrow),5^\uparrow(1,2,4,8^\uparrow),\\ &5^\uparrow(3\cdot1,3\cdot2,3\cdot4, 3\cdot8^\uparrow)\bigr). \tag{1} \end{aligned}$$

Put

$$(B_1,\ldots,B_{12})=(1,2,F_1,F_2,\ldots,F_{10}). \tag{2}$$

Use the first and last blocks of six, in that order, to define

$$F_{11}=7^\uparrow(B_1,\ldots,B_6),\qquad F_{12}=7^\uparrow(B_7,\ldots,B_{12}). \tag{3}$$

Thus $F_1,\ldots,F_{12}$, rather than the temporary atomic inputs $1,2$, are the first twelve inputs of the outer $19$-arrow.

The prime-$11$ stage already partially covers this branch. Define

$$\begin{aligned} F_{13}=11^\uparrow\bigl(&x,x,1,2,3\cdot1, 5^\uparrow(x,x,1,x),\\ &5^\uparrow(x,2,3\cdot1,3\cdot2),3\cdot2, 7^\uparrow(x,x,1,2,x,x),9^\uparrow(1,2)\bigr). \tag{4} \end{aligned}$$

Let $F_{14}$ be the same package after replacing every atomic $1$ by $4$ and every atomic $2$ by $8^\uparrow$. These two substitutions fill the two complementary remaining prime-$11$ profiles.

The source allows the earlier packages to be selected in any compatible order. Fix the reproducible choices

$$\begin{aligned} F_{15}&=13^\uparrow(1,2,F_1,\ldots,F_8,F_{11},F_{12}),\\ F_{16}&=17^\uparrow(1,2,F_1,\ldots,F_{14}). \tag{5} \end{aligned}$$

The first list has twelve entries and excludes $F_9,F_{10}$, which begin with $5^\uparrow$, and $F_{13},F_{14}$, which begin with $11^\uparrow$, exactly as the source requires. The second list has sixteen entries. Neither list contains its new outer prime, and each is drawn without repetition from a signature-disjoint ordered pool.

The seventeenth input

It is the union of four packages. Put

$$G_1=5^\uparrow(\_,9^\uparrow(1,2), 9^\uparrow(4,8^\uparrow),\_), \tag{6}$$ $$G_2=7^\uparrow(\_,\_,\_, 5^\uparrow(9^\uparrow(1,2),x,x,9^\uparrow(4,8^\uparrow)), \_,\_), \tag{7}$$

and

$$\begin{aligned} G_3=11^\uparrow\bigl(&x,x, 5^\uparrow(9^\uparrow(1,2),x,x,9^\uparrow(4,8^\uparrow)),\\ &7^\uparrow(3\cdot1,3\cdot2,3\cdot4,x, 3\cdot8^\uparrow,9^\uparrow(1,2)),\\ &7^\uparrow(9^\uparrow(4,8^\uparrow), 5^\uparrow(1,x,x,2),5^\uparrow(4,x,x,8^\uparrow),x,\\ &\hspace{24mm}5^\uparrow(3\cdot1,x,x,3\cdot2), 5^\uparrow(3\cdot4,x,x,3\cdot8^\uparrow)),\\ &x,\_,\_, 7^\uparrow(x,x, 5^\uparrow(9^\uparrow(1,2),x,x,9^\uparrow(4,8^\uparrow)), x,x,x),\_\bigr). \tag{8} \end{aligned}$$

The source abbreviates several $11^\uparrow$ packages below to three displayed entries. Precisely, write

$$J(u,v,w)=11^\uparrow \bigl(x,x,x,x,x,x,5^\uparrow(x,x,x,u),v,x,w\bigr). \tag{9}$$

The earlier prime-$11$ package together with $G_1,G_2,G_3$ covers prime-$11$ inputs $1,\ldots,6,9$. In input $7$, the first three regular inputs of the displayed $5^\uparrow$ are covered, so $u$ occupies its fourth input; $v,w$ occupy prime-$11$ inputs $8,10$. Every compressed expression $11^\uparrow(5^\uparrow\!\cdot u,v,w)$ below means this full $J(u,v,w)$. With that exact expansion, the fourth package is

$$\begin{aligned} G_4=13^\uparrow\bigl(& 5^\uparrow(1,x,x,2), 5^\uparrow(4,x,x,8^\uparrow),\\ &5^\uparrow(3\cdot1,x,x,3\cdot2), 5^\uparrow(3\cdot4,x,x,3\cdot8^\uparrow),\\ &5^\uparrow(9^\uparrow(1,2),x,x,9^\uparrow(4,8^\uparrow)),\\ &J(1,1,2), J(2,4,8^\uparrow),\\ &J(4,3\cdot1,3\cdot2),\\ &J(8^\uparrow,3\cdot4,3\cdot8^\uparrow),\\ &J(3\cdot1,9^\uparrow(1,2),9^\uparrow(4,8^\uparrow)),\\ &J(3\cdot2, 5^\uparrow(3\cdot4,x,x,3\cdot8^\uparrow), 7^\uparrow(1,2,4,x,8^\uparrow,3\cdot1)),\\ &J(9^\uparrow(1,2),\\ &\qquad7^\uparrow(3\cdot2,3\cdot4,3\cdot8^\uparrow,x, 9^\uparrow(1,2),9^\uparrow(4,8^\uparrow)),\\ &\qquad7^\uparrow(5^\uparrow(1,x,x,2), 5^\uparrow(4,x,x,8^\uparrow), 5^\uparrow(3\cdot1,x,x,3\cdot2),x,\\ &\hspace{39mm}5^\uparrow(3\cdot4,x,x,3\cdot8^\uparrow), 5^\uparrow(9^\uparrow(1,2),x,x,9^\uparrow(4,8^\uparrow)))\bigr)\bigr). \tag{10} \end{aligned}$$

Set

$$F_{17}=G_1+G_2+G_3+G_4. \tag{11}$$

The unfilled input, its shifted tail, and proof

The construction leaves the eighteenth regular input of the outer $19^\uparrow$ empty at its first level. This is the hole completed at prime $47$. Its copies at all higher prime-$19$ levels are covered by the contextually selected input tail

$$(19^2)^\uparrow\!\cdot1. \tag{12}$$

For (1), every package is complete on the target branch by the initial prime-$3$ and prime-$5$ coverage. The two six-package blocks therefore fill the two $7$ arrows. In (4), the black and gray children supplied by the earlier $11$ template are exactly the $x$ positions; the remaining displayed inputs fill every white child. The substitution defining $F_{14}$ switches to the other unused $2$-profiles. The source's exclusions in $F_{15}$ prevent reuse of a $5$- or $11$-signature.

For (6)–(11), $G_1$ supplies the sixth child of the partially filled $11^\uparrow$, and $G_2$ supplies the fourth child of the needed $7^\uparrow$. Together with prior coverage, $G_3$ leaves exactly prime-$11$ positions $7,8,10$ unresolved. Each full $J(u,v,w)$ in $G_4$ fills those three positions with the stated fourth prime-$5$ input and the two complete packages $v,w$. The twelve resulting $J$-based and direct packages fill the children of the outer $13^\uparrow$. Reading the displayed inputs in order leaves no blank except an explicitly precovered $x$. Hence $F_{17}$ is complete. Formula (12) covers the eighteenth input's copies at levels $k\ge2$; it does not fill that input at level $k=1$, and it is not the outer arrow's marked spine.

The exact exponent-region calculation on the signature-certificate page proves that the seventeen inner signature sets are disjoint. It also checks the selected-input tail against their full unbounded $19$-exponent ranges. None contains $19$ before being placed in the outer arrow. Thus

$$19^\uparrow(F_1,\ldots,F_{17},\_) +(19^2)^\uparrow\!\cdot1 \tag{13}$$

is a regular-pattern-injective partial package for the modulus-$12$ hole. It retains all repeated copies of $F_1,\ldots,F_{17}$ and the outer marked spine. After (12), exactly the first-level eighteenth child remains unresolved among the regular input targets. Prime $47$ fills that class and thereby completes the hole; the outer marked spine is later terminated by the general finite-arrow lemma.

Used by. The prime-23 template, later Nielsen stages, and Owens's construction.

Bears on. Problem 2.