BDS

feenschach, 1990

1K6/1Pp5/6*3q1/2r1Bk2/3*3S*3bP2/2*3r1pP2/2P5/8

Helpmate in 2; 2 solutions

LEO g6

PAO c3

VAO e4

MAO d4

1.PAd3 MAb3 2.VAxc2 MAd4#
1.Rc6  MAb5 2.VAxb7 MAd4#

If e4 were empty, Black would be mated in the diagram position. So, in each solution, the play involves unpinning the bVA and allowing it to move to a square from which it cannot return to e4, following which the wMA returns to d4.