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Source. Ruiliang Li, On an Erdős--Lovász problem: 3-critical 3-graphs of minimum degree 7, arXiv:2512.24850v1 (31 December 2025), Lemma 4.2 and proof, printed p. 8 (PDF p. 8).

Setup. The construction (5) of Theorem 4.1.

Used in. Theorem 1.2.

Bears on. #834: a step in the example behind the yes answer under the chromatic reading of "33-critical".

Statement

For the hypergraph HH of Theorem 4.1,

dH(1)=10,dH(v)=7(2≤v≤9).d_H(1)=10,\qquad d_H(v)=7\quad(2\leq v\leq9).

In particular, δ(H)=7\delta(H)=7.

Rewritten proof

Reading the edge list and grouping the edges by vertex gives

1123,129,138,146,148,149,157,158,159,1672123,129,236,237,249,259,2673123,138,236,237,348,358,3674146,148,149,249,348,468,4695157,158,159,259,358,578,5796146,167,236,267,367,468,4697157,167,237,267,367,578,5798138,148,158,348,358,468,5789129,149,159,249,259,469,579.\begin{array}{c|l} 1&123,129,138,146,148,149,157,158,159,167\\ 2&123,129,236,237,249,259,267\\ 3&123,138,236,237,348,358,367\\ 4&146,148,149,249,348,468,469\\ 5&157,158,159,259,358,578,579\\ 6&146,167,236,267,367,468,469\\ 7&157,167,237,267,367,578,579\\ 8&138,148,158,348,358,468,578\\ 9&129,149,159,249,259,469,579. \end{array}

The first row contains ten edges and every other row contains seven, proving the claim. As a consistency check, the degree sum is 10+8⋅7=6610+8\cdot7=66, equal to three times the 22 edges.