theorem Th22:
  K is having_valuation & b <> 0.K implies v.(a/b) = v.a - v.b
  proof
    assume
A1: K is having_valuation;
    assume
A2: b <> 0.K;
    thus v.(a/b) = v.a + v.(b") by A1,Def8
    .= v.a - v.b by A1,A2,Th21;
  end;
