reserve i,j,k,n,m for Nat,
  x,y,z,y1,y2 for object, X,Y,D for set,
  p,q for XFinSequence;

theorem Th9:
  rng (p/^n) c= rng p
proof
  thus rng (p/^n) c= rng p
  proof
    let z be object;
    assume z in rng (p/^n);
    then consider x being object such that
A1: x in dom (p/^n) and
A2: z=(p/^n).x by FUNCT_1:def 3;
    reconsider nx=x as Element of NAT by A1;
    nx<len (p/^n) by A1,AFINSQ_1:86;
    then
A3: nx < len p -' n by Def2;
    per cases;
    suppose
      n<len p;
      then len p-'n=len p-n by XREAL_1:233; then
A4:   nx+n in dom p by AFINSQ_1:86,A3,XREAL_1:20;
      (p/^n).nx=p.(nx+n) by A1,Def2;
      hence thesis by A2,A4,FUNCT_1:def 3;
    end;
    suppose
      n>=len p;
      then (p/^n)={} by Th6;
      hence thesis by A1;
    end;
  end;
end;
