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Statement

Setting (p. 252). Fix a graph $G$ and write $\nu=\nu(G)$, with $\nu(G)$ and $\tau(G)$ the packing and transversal numbers of triangles defined on p. 251 (see theorem_5). Fix an independent family $\mathscr B$ of $\nu$ triangles of $G$, and write $E[\mathscr B]$ for the set of edges of its triangles. A triangle of $G$ is of type $(\mathscr B,i)$ when exactly $i$ of its edges lie in $E[\mathscr B]$; since $\mathscr B$ is a maximum independent family, every triangle has type $(\mathscr B,i)$ for some $i\in\{1,2,3\}$. Let $\mathscr B_1$ be an independent family of type-$(\mathscr B,1)$ triangles of maximum size in $G$, and define $\gamma$ by $|\mathscr B_1|=\gamma\nu$.

Lemma 1 (p. 252, quoted). "We have $\tau(G)\le(3-\gamma)\nu(G)$."

Proof sketch

Proof on p. 252. Each triangle of $\mathscr B_1$ shares its single $E[\mathscr B]$-edge with exactly one triangle of $\mathscr B$, and by the maximality of $\mathscr B$ different triangles of $\mathscr B_1$ are paired with different triangles of $\mathscr B$; each pair spans a $K_4$ minus an edge. The transversal keeps all edges of the unpaired triangles of $\mathscr B$ and, for each pair, the shared edge and the missing edge of the $K_4$ when it is present in $G$: at most $3(1-\gamma)\nu+2\gamma\nu$ edges. Swapping either member of each pair into $\mathscr B$ keeps a maximum independent family, so a triangle avoiding the unpaired triangles must meet both members of some pair, and then it contains the shared edge or the missing one.

Dependencies

None outside the paper: only the maximality of $\mathscr B$.

Bears on

• Problem 167: one of the four inequalities whose weighted sum proves Theorem 5, $\tau(G)\le\frac{66}{23}\nu(G)$, toward Tuza's conjecture $\tau(G)\le2\nu(G)$; it is also one of the four lemmas that the 2026 preprint of Yi restates for its claimed constant $\frac{63}{22}$ (Corollary 1). It settles nothing the problem page leaves open.