theorem Th30:
  K is having_valuation & 0 <= v.(1.K+a) implies 0 <= v.a
  proof
    assume that
A1: K is having_valuation & 0 <= v.(1.K+a) and
A2: v.a < 0;
    0 = v.1.K by A1,Th17;
    hence contradiction by A1,A2,Th28;
  end;
