theorem Th113:
  for n,m being Nat,f being FinSequence st
  1<= m & m+n<=len f holds (f/^n).m=f.(m+n)
proof
  let n,m be Nat,f be FinSequence;
  assume that
A1: 1<= m and
A2: m+n<=len f;
A3: m+n-n<=len f -n by A2,XREAL_1:9;
  n<=n+m by NAT_1:11;
  then
A4: n<=len f by A2,XXREAL_0:2;
  then len (f/^n) = len f -n by RFINSEQ:def 1;
  then m in dom (f/^n) by A1,A3,FINSEQ_3:25;
  hence thesis by A4,RFINSEQ:def 1;
end;
