theorem Th1:
  0 <= r & 0 <= s & r^2 = s^2 implies r = s
  proof
    assume that
A1: 0 <= r and
A2: 0 <= s and
A3: r^2 = s^2 and
A4: r <> s;
    -Q >= --s by A1,A3,A4,SQUARE_1:40;
    hence contradiction by A4,A2,A3,SQUARE_1:40;
  end;
