As long as you have a method for assigning any board state a well-defined score if the game must end in that board state, it’s pretty easy to handle the lack of a superko rule, as well as draws. So for Tromp Taylor but only a simple ko rule and a fixed (possibly integer) komi, and assuming each player prefers a win over a draw or infinite cycle, and prefers a draw or infinite cycle over a loss, then we have the possibilities:
- Black can force a win under optimal play
- White can force a win under optimal play
- Neither player can force a win if the other responds optimally, in which case the result will be either a draw or an infinite cycle, black’s choice of which one.
- Neither player can force a win if the other responds optimally, in which case the result will be either a draw or an infinite cycle, white’s choice of which one.
- Neither player can force a win if the other responds optimally, in which case the result will be either a draw or an infinite cycle, either player can opt for a draw, but if both players prefer the infinite cycle, then there can be an infinite cycle.
- Neither player can force a win if the other responds optimally, in which case the result will be a draw. (infinite cycles only occur in branches where one of the players plays a losing move, in which case the other player just forces a win rather than going for an infinite cycle).
- Neither player can force a win if the other responds optimally, in which case the result will be an infinite cycle. (draws only occur in branches where one of the players plays a losing move, in which case the other player just forces a win rather than going for a draw).
I believe these are all the possibilities. Of course, the draw outcomes require an integer komi. For non-integer komi, then the only possibilities are:
- Black can force a win under optimal play
- White can force a win under optimal play
- Neither player can force a win if the other responds optimally, in which case the result will be an infinite cycle.
I might have missed a case, but I think I got them all, in which case the proof proceeds basically the same way.