theorem Th40:
  for qr being State of rtfsm st rtfsm = the_reduction_of tfsm & q
  in qr holds for k being Nat st k in Seg (len w +1) holds (q,w)
  -admissible.k in (qr,w)-admissible.k
proof
  let qr be State of rtfsm;
  set TR = the Tran of tfsm;
  assume that
A1: rtfsm = the_reduction_of tfsm and
A2: q in qr;
  defpred P[Nat] means $1 in Seg(len w +1) implies (q,w)-admissible
  .$1 in (qr,w)-admissible.$1;
A3: for k being Nat st P[k] holds P[k + 1]
  proof
    let k be Nat;
    assume
A4: k in Seg (len w + 1) implies (q,w)-admissible.k in (qr,w) -admissible.k;
    assume
A5: (k+1) in Seg (len w + 1);
A6: k = 0 or 0 < k & 0+1 = 1;
    per cases by A6,NAT_1:13;
    suppose
A7:   k = 0;
      then (q,w)-admissible.(k+1) = q by Def2;
      hence thesis by A2,A7,Def2;
    end;
    suppose
A8:   1 <= k;
A9:   k+1 <= len w + 1 by A5,FINSEQ_1:1;
      then
A10:  k <= len w by XREAL_1:6;
      then consider
      w1i being Element of IAlph, q1i, q1i1 being Element of tfsm
      such that
A11:  w1i = w.k & q1i = (q,w)-admissible.k and
A12:  q1i1 = (q,w)-admissible.(k+1) & w1i-succ_of q1i = q1i1 by A8,Def2;
      consider w2i being Element of IAlph, q2i, q2i1 being Element of rtfsm
      such that
A13:  w2i = w.k & q2i = (qr,w)-admissible.k and
A14:  q2i1 = (qr,w)-admissible.(k+1) & w2i-succ_of q2i = q2i1 by A8,A10,Def2;
      k <= k+1 by NAT_1:11;
      then k <= len w + 1 by A9,XXREAL_0:2;
      then TR.(q1i,w1i) in (the Tran of rtfsm).(q2i,w2i) by A1,A4,A8,A11,A13
,Def18,FINSEQ_1:1;
      hence thesis by A12,A14;
    end;
  end;
A15: P[0] by FINSEQ_1:1;
  thus for k being Nat holds P[k] from NAT_1:sch 2(A15,A3 );
end;
