Two five-cycles forced by a three-cycle
= Two five-cycles forced by a three-cycle
The covering words $I J J J J$ and $I J I J J$ both occur under the interval coverings in <period three implies all periods>. Their points cannot be the boundary three-cycle. Each has least period five, because five is prime and each itinerary visits both intervals. Their respective one and two visits to the interior of $I$ distinguish the orbits even after cyclic changes of starting point.