theorem
  K is having_valuation & 0 <= v.(1.K-a) implies 0 <= v.a
  proof
    assume that
A1: K is having_valuation and
A2: 0 <= v.(1.K-a);
    1.K-a = 1.K + -a & v.-a = v.a by A1,Th20;
    hence thesis by A1,A2,Th30;
  end;
