
theorem
for R being ordered domRing,
    O being Ordering of R,
    a being square Element of R,
    b being Element of R
holds b is Sqrt of a iff (b = sqrt(O,a) or b = -sqrt(O,a))
proof
let R be ordered domRing, O be Ordering of R;
let a be square Element of R, b be Element of R;
hereby assume b is Sqrt of a;
   then b^2 = a by defsqrt .= sqrt(O,a)^2 by defq;
   hence b = sqrt(O,a) or b = -sqrt(O,a) by sq00;
   end;
assume b = sqrt(O,a) or b = -sqrt(O,a);
then per cases;
suppose b = sqrt(O,a);
  hence b is Sqrt of a;
  end;
suppose B: b = -sqrt(O,a);
  (-sqrt(O,a))^2 = sqrt(O,a)^2 by VECTSP_1:10;
  hence b is Sqrt of a by B,defsqrt;
  end;
end;
