theorem Th63:
  K is having_valuation implies not 1.K in vp v
  proof
    assume
A1: K is having_valuation;
    then
A2: vp v = {x where x is Element of K: 0 < v.x} by Def13;
    assume 1.K in vp v;
    then ex x being Element of K st x = 1.K & 0 < v.x by A2;
    hence thesis by A1,Th17;
  end;
