theorem
  (for x holds f.x = sqrt x) & x0<>x1 & x0>0 & x1>0 implies
  [!f,x0,x1!] = 1/(sqrt x0 + sqrt x1)
proof
  assume that
A1:for x holds f.x = sqrt x and
A2:x0<>x1 and
A3:x0>0 & x1>0;
  [!f,x0,x1!] = (sqrt x0 - f.x1)/(x0-x1) by A1
    .= (sqrt x0 - sqrt x1)/(x0-x1) by A1
    .= 1/(sqrt x0 + sqrt x1) by A2,A3,SQUARE_1:36;
  hence thesis;
end;
