
theorem
for R being preordered Ring,
    P being Preordering of R holds (-P) /\ (SQ R) = {0.R}
proof
let R be preordered Ring, P be Preordering of R;
0.R in P by REALALG1:25; then
C: -0.R in -P;
A: now let o be object;
   assume o in {0.R};
   then o = 0.R by TARSKI:def 1;
   then o in -P & o in SQ R by C;
   hence o in (-P) /\ (SQ R);
   end;
now let o be object;
  assume o in (-P) /\ (SQ R);
   then H0: o in -P & o in SQ R by XBOOLE_0:def 4;
   then consider a being Element of R such that H1: o = a & a is square;
   not(a in -P \ {0.R}) by H1,sq5;
   hence o in {0.R} by H0,H1,XBOOLE_0:def 5;
  end;
hence thesis by A,TARSKI:2;
end;
