This method of showing various cycles (here a cyclic le Grand = Ukrainian cycle), with different guards produced by promotion…
A nice foreplan to turn Rb7 from an unwanted mate to just a check - which can indeed be seen…
My comment meant to be tongue-in-cheek only 😀
there are magazines that would not publish this task
Thank you very much for the dedication!
Yes, top class: agreed! But after 1.c8=Q? ... Gf8! (2.Nd4xf8[+bGb4]+? also fails to 1.- Gb4-d4!)
The solution is 1.Rd4+ Ke5 2.Rxg4 Kd5 3.Rd4 Ke5 4.Rxh4 Kd5 5.Kd4 Ke5 6.Rc4 Kd5 7.Rc3 g1=Q 8.Rd3 Qd4 9.Rxd4…
The problem by O. Schmitt can be done without black Bishop, as follows: 8/5p2/3p4/pK1k1pN1/P1R3pp/1p6/1B2P1p1/4rn1r #15 1.Rd4+
Probably all hs= , ser.s= and auto= can be designated as reaching dead positions.
In the try in Phase (b) 2...Rf4+ also impossible due to the fairy condition