theorem Th7:
  n>0 implies aseq(k).n=1-(k/n)
proof
  assume
A1: n>0;
  thus aseq(k).n = (n-k)/n by Def1
    .= (n/n)-(k/n)
    .= 1-(k/n) by A1,XCMPLX_1:60;
end;
