\section{The coefficient in (I4$'$)}\label{app:js}

We use the parameters of (I4): $(Y_0,\alpha_1,\alpha_2,C)=(7.9,7,2,3.98)$,
$u_0=400$, $\delta=1.3$, $\alpha=1/4$, and $M=40$.
We may take $\eps=0.0005543$, $\xi\le801$, and $C_1(\eps),C_2(\eps)\le114$ \cite[proof of Theorem~1.3]{JS}.
For each even $N\ge X_2'$, apply \cite[Theorem 4.2]{JS} separately with $X_2=N$.
Write
\begin{gather*}
 L=\log N\ge L_0:=2736.86,\quad w=L-5\log L,\\
 j(w)=\frac{1.3841\cdot7\log w}{\log(7\log w)},\quad
 s_J(L)=40\left(\frac14-\frac{14\log L}{L}\right).
\end{gather*}
Set $\mathfrak S_2=\prod_{p>2}(1-(p-1)^{-2})$ and $T=2e^{-\gamma}\mathfrak S_2$.

\emph{Correction to the definition of $a$.}
In \cite[\S2.1, p.~5]{JS} (arXiv v3), the denominator $\log^{1+\delta}n$
in $a(X_2)$ should read $\log^\delta n$, as in the definition (138) of \cite{BJS}. We use
\[
 a(X_2)=a_1(X_2)\max_{n\ge X_2}\left[
 \frac{\log\log\log n}{\log^\delta n}\mathfrak S_2^{-1}
 \left(e^\gamma\log\log(\log^{\alpha_1}x_1(n))
 +\frac{2.5}{\log\log(\log^{\alpha_1}x_1(n))}\right)\right].
\]
This is supported by the proof of \cite[Theorem 44(b)]{BJS} imported in
\cite[Theorem 4.2(b)]{JS}: the error bound
$a_1NV^{(\ell)}\log\log\log N/\log^{1+\delta}N$ becomes
$a_1|A|V^{(\ell)}\log\log\log N/\log^\delta N$ using $|A|>N/\log N$.
The theorem expresses the error relative to $|A|V(z)$.
Neither $a_1$ nor the outside factor changes. The authors' Mathematica notebook for the unconditional case
(\url{https://github.com/DJmath1729/prime_and_almost_prime}, commit \texttt{b2330c99d8}, definition of $a$) also uses $\delta$.

\emph{Hypotheses.}
The condition $\log\log x_1(X_2)\ge7.9$ reads $w\ge e^{7.9}$.
At $L=L_0$, $w-e^{7.9}>0.00483894$, and the other conditions have margins
\begin{gather*}
 \frac w2-7\log w-7\log L>1237.94,\quad
 7\log L-\log10^9>34.67,\quad 1-801/L^2>0.999893,\\
 \frac1{74}-\eps>0.012959,\quad
 \frac L4-7\log w-2.5\log L-400(1+9\cdot10^{-7}/\log400)>209.12,\\
 0.2L-14\log L>436.56,\qquad w^{1.3}>28853.95>3022.
\end{gather*}
These margins persist: $w$ increases, $\log L/L$ decreases, and the support
and level expressions have positive derivatives.
Also $N\ge4\cdot10^{18}$, $z_0=N^{1/40}>10^{12}$, and $w^{1.3}>15015$,
so $\eps_1(N,\delta)\le1/11$. Finally $s_J(L)\ge8.3805685>7$, giving
$f(s_J(L)),f(10)\ge1-\rho$ and $\overline m\le\rho$, where $\rho=0.000105$.
Here $f$ increases and $F$ decreases to $1$ (see \cite[\S2]{JS}), so these follow from $1-f(6)\le1.05\cdot10^{-4}$ and
$F(7)-1\le5\cdot10^{-6}$; ball arithmetic on the closed forms quoted there gives $1-f(6)=1.0493998\ldots\cdot10^{-4}$ and
$F(7)-1=4.9945225\ldots\cdot10^{-6}$ (\texttt{author/fF\_enclose.py}).
We use $h(s_J(L))<0.000082075421$; the bound $h\le10^{-5}$ printed in
the proof of \cite[Theorem 1.3]{JS} is too small
($h(s_J(L))>8.4\cdot10^{-5}$ at their $X_2$).

\emph{Suprema and decreasing majorants.}
In the supremum defining $p_2$, $Q(x_1(N))$ and $\nu(x_1(N))$ are fixed
while $t=\log y\ge w$ varies. The $C$ contribution is constant, and
$t^2e^{-\nu t}$ decreases since $\nu w>11.92>2$.
The remaining terms decrease, so the supremum is attained at $y=x_1(N)$.
The arXiv v3 definition of $p_2$ in \cite[\S2.1]{JS} contains a stray line
``$+18.78(\log y)^{2.515}\exp(-0.8274\sqrt{\log y})+c_1p_3(y)$'' with $p_3$ undefined.
It is not in the authors' notebook, and we omit it; its first term is below $2\cdot10^{-9}$ here.
For $c_1$, the exponent $E_P(t)=t/(2\cdot2.0452\cdot7\log t)$ satisfies
$dE_P/d\log t>10.41497>2$; its denominator improves and its other terms decrease.

Put $r(L)=\log\log L$ and $q(L)=\log(7\log w)$, and define
\begin{align*}
 M_3(L)&=\left(\frac{1.5}{L}+\frac{L+7\log L}{L-7\log L}\right)
 \left(\frac Lw\right)^{1.3}
 \left[e^\gamma\left(1+\frac{\log1.3}{r(L)}\right)
 +\frac{2.5}{\log(1.3\log w)r(L)}\right]
 +\frac{1.3841L^{3.3}e^{-L}}{\log L\,r(L)},\\
 \Gamma(L)&=\frac{1.3841\cdot7\log L}{L^{0.7}r(L)q(L)},\\
 A_*(L)&=\frac{r(L)}{\mathfrak S_2L^{1.3}}
 \left[e^\gamma(\log7+r(L))+\frac{2.5}{q(L)}\right].
\end{align*}
Then $c_3(N)\le M_3(L)$, $a_1(N)\le c_2(N)\Gamma(L)+M_3(L)$, and
$a(N)\le[c_2(N)\Gamma(L)+M_3(L)]A_*(L)$.
All three majorants decrease for $L\ge L_0$.
For $M_3$, use the decrease of $\log L/L$ and the displayed positive factors.
For $\Gamma$, use the decrease of $\log L/L^{0.7}$ and the increasing denominators.
For the first term of $A_*$, the numerator's logarithmic derivative is
\[
 \frac1{\log L}\left(\frac1{r(L)}+\frac1{\log7+r(L)}\right)<1.3;
\]
for its second term, $r(L)/L^{1.3}$ decreases and $q(L)$ increases.
The terms of $c_2$ decrease as well. In particular $a(N)<0.00379644$ for $L\ge L_0$.

\emph{Finite interval computation.}
Let $E_i(L)=\eps C_i(\eps)e^2h(s_J(L))$ and set
\[
 \mathfrak B(L)=1-\rho-\rho/11-(1+1/11)E_2(L)
 -(3/11+a(N))(\rho+E_1(L))-a(N)-1602/L^2.
\]
The factor $40(1+\xi/L^2)$ in case (b) cancels its reciprocal in the final
remainder term. Once $\mathfrak B(L)>0$, dropping the excess positive factor gives
\[
 C_b(L)=40\mathfrak B(L)-\frac{c_4^*(N)j(w)}{LT}.
\]
We enclose this expression on $[L_0,3000]$, retaining the denominator $L$.
For each interval, evaluate $M_3,\Gamma,A_*$ at its left endpoint and enclose
$c_2(N)$ on the whole interval.

The certificate \texttt{audit/audit\_js.py}, with output \texttt{expected/audit/audit\_js.json}, uses
256-bit Arb arithmetic and exact rational endpoints. It verifies a cover by
twelve adjacent intervals with no gaps. With $C_1=C_2=114$ and the certified
bound $\mathfrak S_2>0.6601617536$, the minimum lower enclosures are
$>38.41700825$ in case (a) and $>9.80367826$ in case (b).
The brace is positive on every interval. The output retains every interval
and its bounds; the first is
$[136843/50,280267621/102400]=[2736.86,2736.988486328125]$.
The certificate also checks $C_1=113$, as given in \cite{JS}.
An mpmath interval implementation, \texttt{third/js\_cert/js\_cert.py}, confirms
the conservative conclusions $C_a>38.417$ and $C_b>9.8036$.

\emph{Infinite tail.}
For $L\ge3000$, replace $w/L$ by 1 in the positive remainder terms and freeze
$1/\beta$ at its upper bound at $L=3000$.
Including $j$, the delicate exponential terms have adverse logarithmic
derivatives at most $1+2/\log w_0<1.254$.
The exponent derivatives satisfy
\[
 \frac{dE_P}{d\log w}>10.41497,\qquad
 \frac{dE_S}{d\log w}>1.45786,\qquad
 E_S(w)=\frac{100w^{0.35}}{1.3^2\log^2w},
\]
and both increase. Thus these terms decrease. The $C$ term and remaining
power and exponential terms decrease even after multiplication by $j$.
The partial-summation multiplier decreases. For the deletion term times $j/L$,
an upper logarithmic derivative is $-1/2+0.5/L+1/(w\log w)<0$.
Also $M_3,\Gamma,A_*$ decrease, while $s_J(L)$ increases and $h(s_J(L))$ decreases.
Evaluation at 3000 gives coefficients $>39.28$ in case (a) and $>26.218$ in case (b): both implementations give more than
$26.2182$ in case (b), and \texttt{audit/audit\_js.py}, with $C_1=C_2=114$, encloses it above $26.21829064$.

The coefficients therefore exceed $38.417$ and $9.8036$ on $[L_0,3000]$, and $39.28$ and $26.218$ for $L\ge3000$.
Their positivity and $|A|>N/L$ give
$S(\cA,N^{1/40})>9.8036\,U_NN/L^2$ for every even $N$ with $L_0\le\log N\le3000$, and
$S(\cA,N^{1/40})>26.218\,U_NN/L^2$ for every even $N$ with $\log N\ge3000$, proving (I4$'$).