Breton adverse: When a piece is captured, another piece – and only one – of same nature belonging to captured side (if any) must also disappear…if there are more than one such pieces, the capturing side chooses the one to remove.
No.1334Pierre Tritten France original – 20.10.2018
Pseudo four-captures cyclic Zilahi, each white officer is captured once
Three first half-moves are specific captures in each solution:
1.g×f6(xh3) allow wSh5 play on f6 (orthodox motivation) but also to remove guard of f6 (specific effect) underlining that 1.B×f6(×h3) S×f6(×f1) 2.b×a6(×g3) Sc6+? doesn’t work because f6 is guarded by bPg7;
1…S×f6(×b7) the capture on f6 is used to clear b7 (specific effect);
2.B×g3(×a6) unguard f6, otherwise bBh4 would prevent mate (specific effect);
1.B×h3(×f6) is used to clear f6 (specific effect) and remove the guard of b5 (specific effect);
1...B×b5(×f4) clears a line for wBg3 (specific effect);
2.B×e7(×h5) captures both white Knights (specific effect);
1.b×a6(×g3) clears a line for wRh3(specific effect);
1...R×e3(×f5) clears a line for bRb5 (specific effect);
No.1334 
Interesting condition. Who decides decides which piece/pawn shall vanish (when there are more than two)?
The capturing side
К сожалению в третьем решении версии №1334.1 правило Breton adverse не нужно.
Thank you Alexander for your comment, you are right, we did not pay attention to this!