Josef Breuer
1.S4d6 Ke6 2.Sf6 K×e5 3.e8S Ke6 4.Re4;
2...Kxe7 3.Rc8 Ke6 4.Re8;
1...K×e5 2.Re4+ Kd5 3.Sf6+ K×d6 4.e8S.
Jacob Hoover: Each mate seen in the solution is a model mate. In addition, two of the mates had an
aesthetically pleasing symmetrical layout. I was actually surprised at just how easy this was once I thought
about it.
Guy Meissonnier: What a beautiful Meredith with underpromotion!