theorem Th26:
  K is having_valuation & b <> 0.K & v.(a/b) <= 0 implies v.a <= v.b
  proof
    assume that
A1: K is having_valuation and
A2: b <> 0.K and
A3: v.(a/b) <= 0;
A4: a <> 0.K by A1,Def8,A3;
    then 0 <= v.(b/a) by A1,A2,A3,Th25;
    hence v.a <= v.b by A1,A4,Th24;
  end;
