theorem Th61:
  K is having_valuation implies
  (a in vp v iff 0 < v.a)
  proof
    assume K is having_valuation;
    then
A1: vp v = {x where x is Element of K: 0 < v.x} by Def13;
    hereby
      assume a in vp v;
      then ex b being Element of K st b = a & 0 < v.b by A1;
      hence 0 < v.a;
    end;
    thus thesis by A1;
  end;
