theorem Th17:
  x<>y & X misses Y implies (X-->x)+*(Y-->y) in Choose(X\/Y,card X ,x,y)
proof
  assume that
A1: x<>y and
A2: X misses Y;
  dom (X-->x)=X & dom (Y-->y)=Y;
  then (X-->x)+*(Y-->y)=(Y-->y)+*(X-->x) by A2,FUNCT_4:35;
  hence thesis by A1,Th16;
end;
