reserve S for 1-sorted,
  i for Element of NAT,
  p for FinSequence,
  X for set;

theorem Th33:
  for X being non empty set, s being FinSequence of bspace(X), f
being Element of bspace(X), x being Element of X holds (s ^ <*f*>)@x = (s@x) ^
  <*f@x*>
proof
  let X be non empty set, s be FinSequence of bspace(X), f be Element of
  bspace(X), x be Element of X;
  set L = (s ^ <*f*>)@x;
  set R = (s@x) ^ <*f@x*>;
A1: len L = len (s ^ <*f*>) by Def9
    .= (len s) + (len <*f*>) by FINSEQ_1:22
    .= (len s) + 1 by FINSEQ_1:39;
A2: for k being Nat st 1 <= k & k <= len L holds L.k = R.k
  proof
    let k be Nat such that
A3: 1 <= k and
A4: k <= len L;
    per cases by A1,A4,NAT_1:8;
    suppose
A5:   k <= len s;
      dom (s@x) = Seg (len (s@x)) by FINSEQ_1:def 3
        .= Seg (len s) by Def9;
      then k in dom (s@x) by A3,A5;
      then
A6:   R.k = (s@x).k by FINSEQ_1:def 7
        .= (s.k)@x by A3,A5,Def9;
      dom s = Seg (len s) by FINSEQ_1:def 3;
      then
A7:   k in dom s by A3,A5;
      k <= len (s ^ <*f*>) by A4,Def9;
      then L.k = ((s ^ <*f*>).k)@x by A3,Def9;
      hence thesis by A6,A7,FINSEQ_1:def 7;
    end;
    suppose
A8:   k = len L;
      dom (<*f@x*>) = {1} by FINSEQ_1:2,def 8;
      then
A9:  1 in dom (<*f@x*>) by TARSKI:def 1;
      len (s@x) = len s by Def9;
      then
A10:  R.k = <*f@x*>.1 by A1,A8,A9,FINSEQ_1:def 7
        .= f@x;
      dom (<*f*>) = {1} by FINSEQ_1:2,def 8;
      then 1 in dom (<*f*>) by TARSKI:def 1;
      then
A11:  (s ^ <*f*>).k = <*f*>.1 by A1,A8,FINSEQ_1:def 7
        .= f;
      k <= len (s ^ <*f*>) by A4,Def9;
      hence thesis by A3,A10,A11,Def9;
    end;
  end;
  len ((s@x) ^ <*f@x*>) = (len (s@x)) + (len <*f@x*>) by FINSEQ_1:22
    .= (len s) + (len <*f@x*>) by Def9
    .= (len s) + 1 by FINSEQ_1:39;
  hence thesis by A1,A2;
end;
