One job, $s₁ = 1$, $d₁ = 2$, two machines, $q_max = 1$. The shift at the only job is $δ₁ = 1 = q_max$, so the printed recursion's $hshift$ is vacuous and the taking branch derives $T₁[(2), 3] ≤ 0$ — two jobs due at $2$ — in an instance with one job. The instance has positive processing times and positive weights, so it is not excluded by any standing assumption.