theorem :: GRAPH_2:14
  for f1, f2 being FinSequence of D holds 1<=n & n<=len f1 implies
  (f1^'f2)/.n = f1/.n
proof
  let f1, f2 be FinSequence of D;
  assume that
A1: 1<=n and
A2: n<=len f1;
  per cases;
  suppose
    f2 = {};
    hence thesis by Th55;
  end;
  suppose
A3: f2 <> {};
    then
A4: len f1 + 0 < len f1 + len f2 by XREAL_1:6;
    len (f1^'f2) + 1 = len f1 + len f2 by A3,Th13;
    then n < len (f1^'f2) + 1 by A2,A4,XXREAL_0:2;
    then n <= len (f1^'f2) by NAT_1:13;
    hence (f1^'f2)/.n = (f1^'f2).n by A1,FINSEQ_4:15
      .= f1.n by A1,A2,Th14
      .= f1/.n by A1,A2,FINSEQ_4:15;
  end;
end;
