Proof.
(1) Consider an optimal schedule S satisfying property 1, in which job k is fully early and is not assigned to the first position. Create a schedule S' by moving job k to the first position. It is clear that job k in S' remains fully early and all other jobs are either delayed or not moved. Thus, the total early work of S' is not larger than that of , implying that S' is optimal as well.
(2) Consider a schedule satisfying property 1, in which the set of fully late or partially late jobs are not sequenced according to EDD. Let and denote a pair of consecutive, fully or partially late jobs in ( precedes ) with . Let denote the starting time of job . Obtain schedule by swapping jobs and . Note first that since job in is fully or partially late and , job in must be fully late, i.e
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