theorem
  for p being n-element FinSequence,
      q being m-element FinSequence
   holds (p^q).(n+(1 qua Nat)) = q.1 & ... & (p^q).(n+m) = q.m
proof let p be n-element FinSequence, q be m-element FinSequence;
A1: len p = n by Th151;
A2: len q = m by Th151;
 let k be Nat;
 assume 1 <= k & k <= m;
  then
A3: n+(1 qua Nat) <= n+k & n+k <= n+m by XREAL_1:6;
 thus (p^q).(n+k)= (p^q).(n+k)
     .= q.(n+k-n) by A3,A1,A2,FINSEQ_1:23
     .= q.k;
end;
