GPT 6 Astra Ultra attempt on Math-694

Erdős Problem #151: independent continuation

Date: 23 September 2026. Source versions read in full: 151.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

Let $\tau(G)$ hit every maximal clique with at least two vertices, and let $H(n)$ be the minimum independence number among triangle-free $n$-vertex graphs. The target is $\tau(G)\le n-H(n)$. The source codegree bound is correct, but its proof only uses one endpoint: the stronger bound $\tau(G)\le n-\Delta(G)$ follows immediately. The asserted new codegree obstruction is therefore redundant with this simpler bound.

2) WORK: A LOCAL-PLUS-REMOTE CERTIFICATE

For a vertex $v$ put $U=N(v)$, and let $W$ consist of vertices outside $U\cup\{v\}$ which have no neighbour in $U$. Then \[ \tau(G)\le n-|U|-\alpha(G[W]). \tag{1} \] Take an independent set $I\subset W$ of size $\alpha(G[W])$ and set $T=V(G)\setminus(U\cup I)$. There are no edges between $U$ and $I$ and no edges inside $I$. Consequently a clique of size at least two contained in $U\cup I$ must lie entirely in $U$. It extends by adding $v$, so is not maximal in $G$. Thus $T$ is a transversal, proving $1$. The convention excluding singleton cliques is important here.

A computable weaker form is \[ \tau(G)\le n-d(v)-\frac{|W|}{\Delta(G[W])+1}, \] with the last term zero when $W$ is empty. Indeed, greedily select a vertex of $G[W]$ and remove it and its neighbours; at most $\Delta(G[W])+1$ vertices are removed at each step, leaving an independent set of the claimed size (with an integer ceiling if desired).

Hence a counterexample must satisfy, for every $v$, \[ d(v)+\alpha(G[W_v])<H(n), \] as well as $\alpha(G)<H(n)$. This strengthens the supplied local certificate by adding a compatible remote independent set. It is a sufficient condition, not a characterization of all clique transversals.

3) ADVERSARIAL VERIFICATION

One cannot add an arbitrary independent set outside $N(v)$: it may have edges into $N(v)$ and create a maximal clique entirely inside the proposed avoided set. The definition of $W$ is exactly what prevents this failure. In a complete graph $1$ gives $\tau\le1$; with an isolated $v$ it reduces to an independence-based bound. Neither boundary case creates singleton clique obligations excluded by the problem.

7) FINAL AND REFERENCE

UNRESOLVED. The certificate improves on the source codegree argument, but I cannot prove its left side always reaches $H(n)$, or construct a graph violating the full conjecture. The definition of $H(n)$ and the clique convention were checked at https://www.erdosproblems.com/latex/151.

8) COMPLETION ESTIMATE

COMPLETION: 10% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.