
theorem
for R being preordered non degenerated Ring,
    P being Preordering of R,
    a being Element of R holds abs(P,a) = 0.R iff a = 0.R
proof
let R be preordered non degenerated Ring, O be Preordering of R,
    a be Element of R;
hereby assume B: abs(O,a) = 0.R;
   now assume abs(O,a) = -1.R;
     then -1.R + 1.R = 0.R + 1.R by B;
     hence contradiction by RLVECT_1:5;
     end;
   then C: a is O-ordered by av00;
   per cases;
   suppose a in O;
     hence a = 0.R by B,defa;
     end;
   suppose not a in O;
     then a in -O by C,XBOOLE_0:def 3;
     then -a = 0.R by B,defa;
     then --a = 0.R;
     hence a = 0.R;
     end;
   end;
thus thesis by REALALG1:25,defa;
end;
