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Source. Section 4.9, physical pp. 17–18 of the selected author version.

The first seventeen inputs

Work in the deleted class $11\pmod {24}$. The first sixteen ordered packages are

$$\begin{aligned} H_1,\ldots,H_{16}=\bigl(&2,4,8,16^\uparrow, 3\cdot1,3\cdot2,3\cdot4,3\cdot8,3\cdot16^\uparrow,\\ &9^\uparrow(1,2),9^\uparrow(4,8), 5^\uparrow(1,2,4,8),\\ &5^\uparrow(16^\uparrow,3\cdot1,3\cdot2,3\cdot4),\\ &5^\uparrow(3\cdot8,3\cdot16^\uparrow, 9^\uparrow(1,2),9^\uparrow(4,8)),\\ &7^\uparrow(1,2,4,8,16^\uparrow,3\cdot1),\\ &7^\uparrow(3\cdot2,3\cdot4,3\cdot8,3\cdot16^\uparrow, 5^\uparrow(1,2,4,8), 5^\uparrow(16^\uparrow,3\cdot1,3\cdot2,3\cdot4))\bigr). \tag{1} \end{aligned}$$

The prime-power package $9^\uparrow(16^\uparrow,\_)$ was kept in reserve. Put

$$\begin{aligned} H_{17}={}&9^\uparrow(16^\uparrow,\_)\\ &+7^\uparrow\bigl(9^\uparrow(x,1),9^\uparrow(x,2), 9^\uparrow(x,4),9^\uparrow(x,8),9^\uparrow(x,16^\uparrow),\\ &\hspace{28mm}5^\uparrow(9^\uparrow(x,1),9^\uparrow(x,2), 9^\uparrow(x,4),9^\uparrow(x,8))\bigr). \tag{2} \end{aligned}$$

The reserve supplies the one previously missing $9$-child and the six $7$-inputs in (2) fill its remaining descendants.

The last five inputs

The source permits any suitable earlier packages, adding the atomic package $1$ if needed. Fix the ordered pool

$$(K_1,\ldots,K_{18})=(1,H_1,H_2,\ldots,H_{17}) \tag{3}$$

and the reproducible selections

$$\begin{aligned} H_{18}&=13^\uparrow(K_1,\ldots,K_{12}),\\ H_{19}&=17^\uparrow(K_1,\ldots,K_{16}),\\ H_{20}&=19^\uparrow(K_1,\ldots,K_{18}). \tag{4} \end{aligned}$$

These have exactly the required $12,16,18$ regular inputs. They are new complete arrow packages on the present branch; they do not assert that the earlier partial prime-$17$ or prime-$19$ target packages have become complete in their original contexts.

None of $H_1,\ldots,H_{20}$ has a regular factor $11$. Partition these twenty ordered packages into the first ten and last ten, and use the two blocks to fill two copies of $11^\uparrow$. Call the results $H_{21}$ and $H_{22}$.

The complete prime-$23$ package is

$$\boxed{\mathcal T_{23}=23^\uparrow(H_1,\ldots,H_{22}).} \tag{5}$$

Complete proof

Each of the first fourteen packages in (1) is complete by the initial $2,3,5$ construction. The last two are complete $7$-packages with six displayed inputs. Formula (2) closes the one reserved $9$-branch. The ordinary arrow rule and the exact prefixes in (4) then give $H_{18},H_{19},H_{20}$ on the stated restricted branch. Their new primes $13,17,19$ distinguish them from the earlier pool and from one another. The absence of $11$ makes the final two ten-package blocks valid inputs for two new $11$ arrows.

The exact region partition on the signature-certificate page shows that $H_1,\ldots,H_{17}$ are disjoint. The next three acquire, respectively, new $13,17,19$ factors, and the last two attach $11$ to disjoint exact ten-package blocks. None has prime $23$ before being placed in (5). Thus (5) covers every child of the modulus-$24$ hole without repeating a regular modulus. The separate finite-arrow lemma closes every marked tail.

Used by. Later Nielsen templates and Owens's construction.

Bears on. Problem 2.