
theorem sq2:
for F being preordered Field,
    P being Preordering of F,
    a being non zero P-ordered Element of F holds abs(P,a") = abs(P,a)"
proof
let F be preordered Field, P be Preordering of F,
    a be non zero P-ordered Element of F;
AS: a in P \/ -P by defppp;
Y: a <> 0.F; -a <> -0.F; then
X: a" is non zero & -a" is non zero & -a is non zero;
abs(P,a") * abs(P,a) = 1_F
  proof
  per cases by AS,XBOOLE_0:def 3;
  suppose K: a in P;
    then 0.F <=P, a;
    then L: abs(P,a) = a by av2;
    0.F <=P, a" by K,REALALG1:27;
    then abs(P,a") = a" by av2;
    hence abs(P,a") * abs(P,a) = 1_F by Y,L,VECTSP_2:9;
    end;
  suppose a in -P;
    then K: -a in --P;
    then --a <=P, 0.F;
    then L: abs(P,a) = -a by av3;
    -((-a)") <=P, 0.F by K,X,REALALG1:27;
    then a" <=P, 0.F by YZ;
    then abs(P,a") = -(a") by av3;
    hence abs(P,a")*abs(P,a) = a" * a by L,VECTSP_1:10
                            .= 1_F by Y,VECTSP_2:9;

    end;
  end;
hence thesis by VECTSP_2:10;
end;
