theorem
  (f+*g)+~(x,y) = f+~(x,y)+*(g+~(x,y))
proof
  set A = rng f \/ rng g;
A1: g+~(x,y) =(id A+*(x,y))*g by Th115,XBOOLE_1:7;
A2: dom(id A+*(x,y)) = dom id A by Th29
    .= A;
  rng(f+*g) c= A by FUNCT_4:17;
  hence (f+*g)+~(x,y) = (id A+*(x,y))*(f+*g) by Th115
    .= (id A+*(x,y))*f+*((id A+*(x,y))*g) by A2,Th9,XBOOLE_1:7
    .= f+~(x,y)+*(g+~(x,y)) by A1,Th115,XBOOLE_1:7;
end;
