 Solution

Label the balls $a,b,c,d$ and buckets $1,2,3$. Ignoring unchanged static type facts, the initial state's true planning fluents areExpanding it produces exactly three successors. The held ball makes $\mathrm{Empty}$ false, so neither ball in bucket $1$ can be picked up yet.

For $\mathrm{Drop}(d,1)$, replace $\mathrm{Holding}(d)$ by $\mathrm{In}(d,1)$ and $\mathrm{Empty}$; the arm stays at bucket $1$. All four balls are then in buckets, with $a,b,d$ in $1$ and $c$ in $2$.

For $\mathrm{Move}(1,2)$, replace $\mathrm{At}(1)$ by $\mathrm{At}(2)$ and leave every ball planning fluent unchanged, including $\mathrm{Holding}(d)$.

For $\mathrm{Move}(1,3)$, similarly replace only the arm-location planning fluent by $\mathrm{At}(3)$. The arm is now over the destination while still holding $d$; it has not yet dropped that ball.

No other action is applicable: the other drop locations fail their arm-location precondition, no pick has an empty hand, and moves from another source bucket fail their source-location precondition. None of these three successors satisfies ${\bigwedge }_{z\in \{a,b,c,d\}}\mathrm{In}(z,3)$. This describes one forward state-space planning expansion, not an assertion that the search has already found a complete plan.