
theorem Th2:
for D be set, Y be FinSequenceSet of D, F be FinSequence of Y,
    k be Nat st k < len F holds
  Length(F|(k+1)) = Length(F|k) ^ <*len(F.(k+1))*>
proof
   let D be set, Y be FinSequenceSet of D, F be FinSequence of Y, k be Nat;
   assume A1: k < len F; then
   k+1 <= len F by NAT_1:13; then
A3:len(F|(k+1)) = k+1 by FINSEQ_1:59;
A6:len(F|k) = k by A1,FINSEQ_1:59;
A5:dom (Length(F|(k+1))) = dom (F|(k+1)) & dom (Length(F|k)) = dom(F|k)
     by Def1; then
A7:len(Length(F|(k+1))) = k+1 & len(Length(F|k)) = k by A3,A6,FINSEQ_3:29; then
A8:len( Length(F|k) ^ <*len(F.(k+1))*> )
     = k + len <*len(F.(k+1))*> by FINSEQ_1:22
    .= k+1 by FINSEQ_1:40;
    now let n be Nat;
     assume A9: 1 <= n & n <= len(Length(F|(k+1))); then
     n in dom(Length(F|(k+1))) by FINSEQ_3:25; then
A10: (Length(F|(k+1))).n = len((F|(k+1)).n) by Def1
      .= len(F.n) by A7,A9,FINSEQ_3:112;
     per cases;
     suppose n = len(Length(F|(k+1)));
      hence (Length(F|(k+1))).n = (Length(F|k)^<*len(F.(k+1))*>).n
        by A7,A10,FINSEQ_1:42;
     end;
     suppose n <> len(Length(F|(k+1))); then
      n < k+1 by A7,A9,XXREAL_0:1; then
A11:  n <= k by NAT_1:13; then
      (Length(F|k) ^ <*len(F.(k+1))*>).n = (Length(F|k)).n by A9,A7,FINSEQ_1:64
        .= len((F|k).n) by A11,Def1,A9,A5,A6,FINSEQ_3:25
        .= len(F.n) by A11,FINSEQ_3:112;
      hence (Length(F|(k+1))).n = (Length(F|k) ^ <*len(F.(k+1))*>).n by A10;
     end;
    end;
    hence thesis by A5,A8,A3,FINSEQ_3:29;
end;
