theorem
  (for x holds f.x = bD(g,h).x) implies
  [!f,x0,x1!] = [!g,x0,x1!]-[!g,x0-h,x1-h!]
proof
  assume
A1: for x holds f.x = bD(g,h).x;
  [!f,x0,x1!] = (bD(g,h).x0-f.x1)/(x0-x1) by A1
    .= (bD(g,h).x0-bD(g,h).x1)/(x0-x1) by A1
    .= ((g.x0-g.(x0-h))-bD(g,h).x1)/(x0-x1) by DIFF_1:4
    .= ((g.x0-g.(x0-h))-(g.x1-g.(x1-h)))/(x0-x1) by DIFF_1:4
    .= [!g,x0,x1!]-[!g,x0-h,x1-h!];
  hence thesis;
end;
