theorem
  K is having_valuation & a <> 0.K & v.a <= v.b implies 0 <= v.(b/a)
  proof
    assume that
A1: K is having_valuation and
A2: a <> 0.K;
    assume v.a <= v.b;
    then 0 <= v.b - v.a by Lm4;
    hence thesis by A1,A2,Th22;
  end;
