theorem Th41:
  K is having_valuation implies
  (0 <= v.a iff 0 <= normal-valuation(v).a)
  proof
    set f = normal-valuation(v);
    assume
A1: K is having_valuation;
    hereby
      assume 0 <= v.a;
      then v.a is positive or 0 = v.a;
      hence 0 <= f.a by A1,Th40,Th37;
    end;
    assume 0 <= f.a;
    then f.a is positive or 0 = f.a;
    hence thesis by A1,Th40,Th37;
  end;
