
theorem
for O being Ordering of F_Rat holds O = Positives(F_Rat)
proof
let O be Ordering of F_Rat;
defpred P[Nat] means $1 in O;
A: 0.(F_Rat) = 0 & 1.(F_Rat) = 1 by GAUSSINT:def 14;
B: 1.F_Rat in O & 0.F_Rat in O by ord3;
IA: P[0] by A,ord3;
E: O + O c= O & O * O c= O by defpc;
IS: now let k be Nat;
    assume C: P[k];
    then consider a being Element of F_Rat such that D: a = k;
    a + 1.F_Rat in {x + y where x,y is Element of F_Rat :
         x in O & y in O} by D,C,B;
    then k + 1 in O + O by GAUSSINT:def 14,D,IDEAL_1:def 19;
    hence P[k+1] by E;
    end;
I: for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
Positives(F_Rat) c= O
proof let o be object;
  assume o in Positives(F_Rat);
  then consider r being Element of RAT such that A1: o = r & 0 <= r;
  consider m,n being Integer such that B1: n > 0 & r = m/n by RAT_1:1;
  reconsider a=n,b=m as Element of F_Rat by RAT_1:def 2,GAUSSINT:def 14;
  per cases;
  suppose m = 0;
    hence o in O by B1,A1,A,ord3;
    end;
  suppose C1: 0 < m;
    0 <= n by B1;
    then reconsider n1 = n, m1 = m as Element of NAT by C1,INT_1:3;
    C: m1 in O & n1 in O by I;
    a is non zero by B1,GAUSSINT:def 14;
    then a" in O by C,ord5;
    then b * a" in O * O by C;
    then b * a" in O by E;
    hence o in O by A1,B1,GAUSSINT:def 14,GAUSSINT:15;
    end;
  suppose C1: m < 0;
    then 0 >= n by A1,B1;
    then reconsider n1 = -n, m1 = -m as Element of NAT by C1,INT_1:3;
    C: m1 in O & n1 in O by I;
    K: -a <> 0.F_Rat by B1,GAUSSINT:def 14;
    -a is non zero by B1,GAUSSINT:def 14;
    then (-a)" in O by C,ord5;
    then F: (-b) * ((-a)") in O * O by C;
    H: -(n") = -(a") by B1,GAUSSINT:def 14,GAUSSINT:15;
    a <> 0.F_Rat by B1,GAUSSINT:def 14; then
    1.F_Rat = a * a" by VECTSP_1:def 10 .= (-(a")) * (-a);
    then (-a)" = -(a") by VECTSP_1:def 10,K;
    hence o in O by A1,B1,F,E,H;
    end;
   end;
hence thesis by ordsub;
end;
