theorem Th25:
  K is having_valuation & a <> 0.K & b <> 0.K & v.(a/b) <= 0 implies
  0 <= v.(b/a)
  proof
    assume K is having_valuation & a <> 0.K & b <> 0.K;
    then v.(a/b) = -v.(b/a) by Th23;
    hence thesis;
  end;
