
theorem
for R being preordered Ring,
    P being Preordering of R,
    a being Element of R holds abs(P,a) = abs(P,-a)
proof
let R be preordered Ring, O be Preordering of R, a be Element of R;
per cases;
    suppose D: a in -O;
      then E: abs(O,a) = -a by defa;
      -a in --O by D;
      hence abs(O,a) = abs(O,-a) by E,defa;
      end;
    suppose D: a in O;
      then E: abs(O,a) = a by defa;
      -a in -O & --a = a by D;
      hence abs(O,a) = abs(O,-a) by E,defa;
      end;
    suppose D: not(a in O) & not(a in -O);
      then E: abs(O,a) = -1.R by defa;
      not(-a in O) & not(-a in -O)
      proof
        assume -a in O or -a in -O;
        then --a in -O or --a in --O;
        hence contradiction by D;
      end;
      hence abs(O,a) = abs(O,-a) by E,defa;
      end;
end;
