
theorem
for R being preordered Ring,
    P being Preordering of R,
    a,b being Element of R holds abs(P,a-b) = abs(P,b-a)
proof
let R be preordered Ring, O be Preordering of R, a,b be Element of R;
per cases;
suppose A: a-b in O;
  then -(a-b) in -O;
  then b-a in -O by RLVECT_1:33;
  hence abs(O,b-a) = -(b-a) by defa
                  .= a-b by RLVECT_1:33
                  .= abs(O,a-b) by A,defa;
  end;
suppose A: a-b in -O;
  then -(a-b) in --O;
  then B: b-a in O by RLVECT_1:33;
  thus abs(O,a-b) = -(a-b) by A,defa
                 .= b-a by RLVECT_1:33
                 .= abs(O,b-a) by B,defa;
  end;
suppose A: not(a-b in O) & not(a-b in -O);
  then B: abs(O,a-b) = -1.R by defa;
  not(b-a in O) & not(b-a in -O)
    proof
      assume b-a in O or b-a in -O;
      then -(b-a) in -O or -(b-a) in --O;
      hence contradiction by A,RLVECT_1:33;
    end;
  hence abs(O,a-b) = abs(O,b-a) by B,defa;
  end;
end;
