
theorem sq5:
for R being preordered Ring,
    P being Preordering of R,
    a being Element of R st a in -P \ {0.R} holds a is non square
proof
let R be preordered Ring, P be Preordering of R, a be Element of R;
assume a in -P \ {0.R};
then A: a in -P & not a in {0.R} by XBOOLE_0:def 5;
B: SQ R c= P & P /\ -P = {0.R} by REALALG1:def 14;
assume a is square;
  then a in SQ R;
  hence contradiction by A,B;
end;
