theorem Th20:
  K is having_valuation implies v.-a = v.a
  proof
    assume
A1: K is having_valuation;
    (-1.K) * a = -a by VECTSP_2:29;
    hence v.-a = v.(-1.K) + v.a by A1,Def8
    .= 0 + v.a by A1,Th19
    .= v.a by XXREAL_3:4;
  end;
