theorem Th19:
  K is having_valuation implies v.-1.K = 0
  proof
    assume
A1: K is having_valuation;
    (-1.K) * (-1.K) = 1.K by Th12;
    then v.(-1.K) + v.(-1.K) = v.(1.K) by A1,Def8
    .= 0 by A1,Th17;
    hence thesis by Th2;
  end;
