 Solution

Replace the one-place marker $\mathrm{Empty}$ by three nullary planning fluents $q_0,q_1,q_2$, meaning that zero, one or two items are aboard. Initially only $q_0$ is true. Exactly one of these facts is maintained. No additional object constants are needed.

For compactness, define the common loading preconditionsand common unloading preconditionsThe four STRIPS actions areHere the first unloading removes one of two items, and the last unloading removes the sole remaining item; items do not require fixed seat identities. The movement action remains as repaired in part (a) and changes none of the occupancy facts.

The planning state invariant is $q_j$ true exactly when $j$ items are aboard. Every action changes the marker by the same amount as the actual onboard count, so it follows by mathematical induction from the initial empty boat. In particular, a third load is impossible.