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Source. Section 4.1, printed pp. 386–387 (PDF pp. 6–7). This is the full pruning deduction used by the original upper proof. Use $X,\ell,B,T,f(x)$ from the common definitions.

Statement. There is a nonnegative function $\eta(x)\to0$ such that, for every sufficiently large real $x$, every admissible family $\mathcal Q\subseteq[1,x]$ with $S=|\mathcal Q|=f(x)$, and every choice of its disjoint residue classes, a subset $\mathcal Q'$ exists with $S'=|\mathcal Q'|$ satisfying

1. $S'\ge S e^{-\eta(x)T}$;
2. $x e^{-2T}\le q\le x$ for every $q\in\mathcal Q'$;
3. $h(q)\le e^{\sqrt X}$ for every $q\in\mathcal Q'$;
4. there is one integer $K$, $1\le K\le3B$, with $\omega(q)=K$ for every $q\in\mathcal Q'$;
5. the kernels $\operatorname{ker}(q)$, $q\in\mathcal Q'$, are distinct.

The same $\eta$ and sufficiently-large threshold work for all such families and residues. In particular, for every fixed $\epsilon>0$ one may replace $\eta(x)$ by $\epsilon$ eventually. The retained progressions remain disjoint because only members are removed.

Full proof

The lower construction gives $S\ge x e^{-(1+o(1))T}$, independently of the extremal family chosen. Remove all moduli $q<x e^{-2T}$, all those with $h(q)>e^{\sqrt X}$, and all those with $\omega(q)\ge3B$. The numbers removed are at most, respectively,

$$x e^{-2T},\qquad x e^{-\sqrt X\ell/5},\qquad x e^{(-3/2+o(1))T}.$$

These follow from elementary counting, Lemma 3.2, and Lemma 3.1. Each is $o(S)$; for the middle expression, $\sqrt X\ell/T=\sqrt\ell$ tends to infinity. Consequently at least $S/2$ members remain for all sufficiently large $x$, uniformly over the original family.

Also $x e^{-2T}\to\infty$, so these remaining moduli are greater than one and have at least one prime divisor. Their integer values of $\omega(q)$ lie in $[1,3B]$, giving at most $3B$ possibilities. One value $K$ occurs at least $S/(6B)$ times. This treats the endpoints without assuming that $3B$ is an integer.

For this group, apply Lemma 3.3 with the real value $H=e^{\sqrt X}$. At most $H^2 2^K$ members have any one kernel. Keeping one representative of every kernel leaves

$$S'\ge\frac{S}{6B\,2^K e^{2\sqrt X}} \ge S\exp\!\left(-\log(6B)-3B\log2-2\sqrt X\right). \tag{1}$$

The expression subtracted in (1) is $o(T)$. Its ratio to $T$ defines a choice of $\eta(x)$ valid for every retained $K$ and every original family. All five properties follow. The construction never changes a residue, so no dependence on a favorable residue choice has entered the estimates.

Scope. Modulus one is removed by the lower cutoff, not silently allowed in an intersecting-support argument. This page proves the same-paper reduction once; later arguments may import these five properties and their uniform loss without repeating the proof.

Bears on. Problem 202, as an input to the original upper bound. This is a counting lemma, not a current-status claim or a formal verification.