Prove
Lemma 16. Every pruned list produced by this construction has at most five…Lemma 15. For either update, assuming in the switch case, the pre-pruning…Proposition
Track the original entries through later reindexings. As long as the first entry survives, a suffix operation cannot trim or discard the second entry: a cutoff at that later index would already discard the first, permanently. Thus the second entry is either still carried untrimmed, or was pruned because it was empty on some intermediate node containing U . In the latter case D_ a_0 ∩U=∅ . On U∩D_ a_0 , the second entry, when this set is nonempty, therefore selects every job. Every job of U D_ a_0 lies in G_0 , because U⊆Q_0 . On that remainder the first entry selects exactly the rank suffix ≥k . This proves equality of the two unions on U . Both factors in are upsets, so their union is globally upward closed, although it need not agree with the carried pair outside U .