Perfect divisibility conjecture for odd hole-free graphs

Let $G$ be a graph with no induced odd cycle of length at least five. A graph is perfectly divisible if, for every induced subgraph $H$ with at least one edge, its vertex set can be partitioned into $A$ and $B$ such that $H[A]$ is perfect and $\omega(H[B])<\omega(H)$. Perfect divisibility conjecture. Every odd hole-free graph is perfectly divisible. This would extend several known perfect-divisibility results for odd hole-free graphs under additional forbidden-subgraph conditions, including the absence of a claw, cochair, dart, bull, odd balloon, fork, or banner. The general case remains unresolved in the supplied source.