will add these medium jobs to the extra added bags with large jobs and add them to the schedule such that there is no conflict among any of these jobs. To do so we will use a result of [@bagptas]. As it might not be obvious that we can apply the lemma of [@bagptas] we will give a rephrase of this lemma and also a proof.

\[lem:mod2\] Let $I$ be an instance of the machine scheduling with bag-constraint problem and let $I'$ be the modified instance. Let $S'$ be a solution of the instance $I'$. We can expand $S'$ by adding and scheduling all medium jobs from non-priority bags of instance $I$ such that no medium and large job from the same bag $B_l$ in $I$ are on the same machine in $S'$. Additionally this schedule will increase the makespan of the solution $S'$ by at most $2*\epsilon \in O(\epsilon)$.

With this lemma we gain a solution $S'$ for Instance $I'$ and this solution also contains all jobs from instance $I$. The only difference is that $S'$ may have put a small job and either a medium or a large job of the same bag in $I$ on the same 