theorem Th23:
  K is having_valuation & a <> 0.K & b <> 0.K implies v.(a/b) = -v.(b/a)
  proof
    assume
A1: K is having_valuation;
    assume
A2: a <> 0.K;
    assume b <> 0.K;
    hence v.(a/b) = v.a - v.b by A1,Th22
    .= -(v.b-v.a) by XXREAL_3:26
    .= -v.(b/a) by A1,A2,Th22;
  end;
