Proof. Consider an optimal schedule in which job precedes job , job is fully or partially late and job is fully early. Assume that in a set of jobs is scheduled prior to , and a set of jobs is scheduled after job . Create a new schedule by moving job to be scheduled at time zero. Note that: ( ) ; ( ) (the jobs in are delayed in by an amount of ); ) (job is delayed in by an amount of ); ( )(the jobs in are not affected when moving job ). It follows that is optimal as well. By repeating this procedure for all pairs of jobs where the first is not fully early and the second is fully early, we obtain an optimal schedule consisting of a first set of fully early jobs followed by a set of fully or partially late jobs.