theorem
  K is having_valuation implies vp(v) is maximal
  proof
    assume
A1: K is having_valuation;
    thus vp(v) is proper
    proof
      1.ValuatRing v = 1.K by A1,Def12;
      hence vp(v) <> the carrier of ValuatRing v by A1,Th63;
    end;
    let J be Ideal of ValuatRing v;
    assume
A2: vp(v) c= J;
    J is non proper or J = vp(v)
    by A1,Th79,A2;
    hence thesis;
  end;
