The decidability of the special case of the won-position problem, restricted to positions having only short-range pieces, remains open to my knowledge. Nevertheless, as you suspected, one can use the methods of the mate-in-$n$ analysis to provide a much lower upper bound on the complexity. Whereas we had conjectured that the general won-position problem might not be arithmetic, in the case of your restricted positions, the problem is at worst computably enumerable. **Theorem.** A position with only short-range pieces is a won position for white if and only if it is mate-in-n for white for some $n$. Proof: Since the game position is finitely branching, the recursive game values on positions with black-to-move will always be taking the supremum of a finite set, and so inductively we can see that all game values will be finite. This is a general fact: in any open game, where black has only finitely many moves at any stage, then all the game values are finite. In particular, if the initial position has a value, which is to say, if white can force a win, then the value must be finite. Thus, if white can force a win at all, then white will be able to force a win in $n$ moves for some specific $n$. QED In particular, the phenomenon of [transfinite game values in infinite chess](http://jdh.hamkins.org/game-values-in-infinite-chess/) does not arise with positions having only short-range pieces. **Corollary.** The won-position problem for short-range-piece positions is computably enumerable. Proof: Given any finite position having only short-range pieces, we can search for an $n$ such that it is mate-in-$n$, and those questions are decidable. By the theorem, this is equivalent to the original position begin winning for white. QED One can similarly enumerate computably the won-positions for black, and also enumerate the positions for which white or black can force a draw by means of forcing the position into a closed finite space of positions. But this is not the same as forcing a draw, since perhaps black can force the play to continue indefinitely, without forcing it into a finite closed space of positions. So this possibility prevents us from having a partition of the positions into finitely many c.e. classes, and so undecidability still seems possible.