 reserve x for Real,
    p,k,l,m,n,s,h,i,j,k1,t,t1 for Nat,
    X for Subset of REAL;
reserve x for object, X,Y,Z for set;
 reserve M,N for Cardinal;
reserve X for non empty set,
  s for sequence of X;

theorem
  for N being sequence of NAT holds (s * N) ^\k = s * (N ^\k)
proof
  let N be sequence of NAT;
  now
    let n be Element of NAT;
    thus ((s * N) ^\k).n = (s * N).(n + k) by Def2
      .= s.(N.(n + k)) by FUNCT_2:15,ORDINAL1:def 12
      .= s.((N ^\k).n) by Def2
      .= (s * (N ^\k)).n by FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;
