theorem Th13:
  k<n implies 0<aseq(k).n & aseq(k).n<=1
proof
A1: aseq(k).n=(n-k)/n by Def1;
  assume
A2: k<n;
  then n-k>0 by XREAL_1:50;
  hence aseq(k).n>0 by A2,A1,XREAL_1:139;
  1-(k/n)<=1-0 by XREAL_1:6;
  hence thesis by A2,Th7;
end;
