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Statement. Let $P,Q$ be finite planar sets with $2\leq m=|P|\leq n=|Q|$, and use the line families of Proposition 27. Every regulus $R$ satisfies

$$|L\cap R|\leq\max\{4m,\;4D(P,Q)+2m\}. \tag{1}$$

Consequently, for each fixed $c>0$, either $D(P,Q)\geq c\sqrt{mn}$ or every regulus contains at most $(4c+4)\sqrt{mn}$ lines of $L$.

Source. Mathialagan, published 2021 PDF, pp. 12, 23, Lemma 26. The source states the consequence for sufficiently large $c$. The explicit estimate (1) supplies its required constant and fills in the counts for the second ruling and the second color.

Proof. There are $2m$ fixed-endpoint families $L_p^i$. If each supplies at most two lines in $R$, then $|L\cap R|\leq4m$. Otherwise one supplies three distinct lines. Reflection across $z=0$ interchanges the colors and preserves $L$, so suppose they are $\ell_{p,a_1},\ell_{p,a_2}, \ell_{p,a_3}$ with $p\in P$ and distinct $a_i\in Q$.

By Proposition 27 these three lines are projectively disjoint. The projective ruling classification in Proposition 36 puts them in the same ruling and says they determine the unique quadratic surface $R$. The three endpoints are either collinear, or lie on a unique circle of positive radius. The latter follows by intersecting two perpendicular bisectors; their directions are independent for a noncollinear triple.

In the circle case, Proposition 40 says that all lines in $R$ have one of the two forms

$$A=\{\ell_{p,a}:a\in C(q,r)\},\qquad B=\{\ell_{b,q}:b\in C(p,r)\}.$$

Uniqueness of ordered endpoints in Proposition 27 gives the following bounds, including the membership conditions which can make a count zero.

The $2D(P,Q)$ bounds are precisely the circle case of Lemma 34 with $A=P$. Summing is an upper bound even when a line has both colors. The two rulings themselves have no common line. This yields $|L\cap R|\leq4D(P,Q)+2m$.

In the collinear case Proposition 42 gives one ruling $\{\ell_{p,a}:a\in\lambda\}$ and a wholly horizontal opposite ruling. No line of $L$ is horizontal. The first ruling contributes at most $|Q\cap\lambda|\leq2D(P,Q)$ lines of color 1 and at most $m$ of color 2, so this case is bounded by $2D(P,Q)+m$, which is sufficient for (1).

Finally $m\leq\sqrt{mn}$. If $D(P,Q)<c\sqrt{mn}$, each term in the maximum (1) is at most $(4c+4)\sqrt{mn}$. This proves the alternative.

Thresholds and dependencies. The source begins with at most four lines per family and otherwise selects three same-ruling lines from five. Its Propositions 40 and 42 refer to the externally cited Lemma 39, whose printed threshold is seven. This proof does not infer a seven-line hypothesis from five lines. Instead Proposition 36 proves the ruling classification and uniqueness directly; the three fixed-endpoint lines are already projectively disjoint and hence in the same ruling. Propositions 40 and 42 prove the complete rulings without Lemma 39. Only the line/circle specialization of Lemma 34 is needed.

Verification scope. Verified within the independently reviewed Theorem 3 chain, retained in the final review; the strengthened bookkeeping estimate, both-color/both-ruling counts and geometric dependencies belong to the living Theorem 3 record. This is a compilation-supplied elaboration of the published argument, not an author-issued correction.

Bears on. Problem 661.