theorem
  (for x holds f.x = cD(g,h).x) implies
  [!f,x0,x1!] = [!g,x0+h/2,x1+h/2!]-[!g,x0-h/2,x1-h/2!]
proof
  assume
A1: for x holds f.x = cD(g,h).x;
  [!f,x0,x1!] = (cD(g,h).x0-f.x1)/(x0-x1) by A1
    .= (cD(g,h).x0-cD(g,h).x1)/(x0-x1) by A1
    .= ((g.(x0+h/2)-g.(x0-h/2))-cD(g,h).x1)/(x0-x1) by DIFF_1:5
    .= ((g.(x0+h/2)-g.(x0-h/2))-(g.(x1+h/2)-g.(x1-h/2)))/(x0-x1) by DIFF_1:5
    .= [!g,x0+h/2,x1+h/2!]-[!g,x0-h/2,x1-h/2!];
  hence thesis;
end;
