 Shortest-duration interval scheduling counterexample

In this separate three-job example, labels $A,B,C$ denote new jobs:Their durations are five, five and two. Consider the rule "accept a shortest remaining job, discard all jobs conflicting with it, and repeat". It selects $C$ first. But $C$ overlaps $A$ on $[4,5)$ and $B$ on $[5,6)$, so both are discarded. The rule returns one job, while the compatible subset $\{A,B\}$ contains two. The common endpoint five is allowed, and no feasible solution contains all three jobs, so two is the maximum.

Duration alone ignores where a job sits. Here the shortest job straddles the useful handover time. Earliest finish instead chooses $A$, then $B$, and attains two. A counterexample can refute a proposed greedy-choice property even when the rule succeeds on many other inputs.