theorem
  (for x holds f.x = fD(g,h).x) implies
  [!f,x0,x1!] = [!g,x0+h,x1+h!]-[!g,x0,x1!]
proof
  assume
A1: for x holds f.x = fD(g,h).x;
  [!f,x0,x1!] = (fD(g,h).x0-f.x1)/(x0-x1) by A1
    .= (fD(g,h).x0-fD(g,h).x1)/(x0-x1) by A1
    .= ((g.(x0+h)-g.x0)-fD(g,h).x1)/(x0-x1) by DIFF_1:3
    .= ((g.(x0+h)-g.x0)-(g.(x1+h)-g.x1))/(x0-x1) by DIFF_1:3
    .= [!g,x0+h,x1+h!]-[!g,x0,x1!];
  hence thesis;
end;
