theorem Th17:
  K is having_valuation implies v.1.K = 0
  proof
    assume
A1: K is having_valuation;
A2: v.1.K = v.(1.K*1.K)
    .= v.1.K + v.1.K by A1,Def8;
    v.1.K in INT by A1,Def8;
    then reconsider x = v.1.K as Element of REAL by XREAL_0:def 1;
    x + 0 = x + x by A2,XXREAL_3:def 2;
    hence thesis;
  end;
