
theorem Th11:
for m,n be non zero Nat, X be non-empty m-element FinSequence,
 P be Function of CarProduct SubFin(X,n),product SubFin(X,n) st
  n <= m & P = (Pt2FinSeq X).n holds P is bijective
proof
    let m,n be non zero Nat, X be non-empty m-element FinSequence,
    P be Function of CarProduct SubFin(X,n),product SubFin(X,n);
    assume that
A1: n <= m and
A2: P = (Pt2FinSeq X).n;

    defpred P[Nat] means 1 <= $1 & $1 <= n implies
     ex i be non zero Nat,
     F be Function of CarProduct SubFin(X,i),product SubFin(X,i)
      st $1 = i & F = (Pt2FinSeq X).i & F is bijective;
A3: P[0];
A4: for k be Nat st P[k] holds P[k+1]
    proof
     let k be Nat;
     assume
A5:  P[k];
     assume
A6:  1 <= k+1 & k+1 <= n;
     per cases;
     suppose
A7:   k = 0;
      ex id1 be Function of CarProduct SubFin(X,1),product SubFin(X,1) st
      (Pt2FinSeq X).1 = id1 & id1 is bijective
    & for x be object st x in CarProduct SubFin(X,1) holds id1.x = <*x*>
        by Def5;
      hence thesis by A7;
     end;
     suppose k <> 0; then
      consider i0 be non zero Nat,
       F0 be Function of CarProduct SubFin(X,i0),product SubFin(X,i0)
      such that
A8:  k = i0 & F0 = (Pt2FinSeq X).i0 & F0 is bijective by A5,A6,NAT_1:13,14;

      k < n by A6,NAT_1:13; then
A9:  i0 < m by A1,A8,XXREAL_0:2; then
      consider F be Function of CarProduct SubFin(X,i0),product SubFin(X,i0),
       IK be Function of [: CarProduct SubFin(X,i0),ElmFin(X,i0+1) :],
          product SubFin(X,i0+1) such that
A10:  F = (Pt2FinSeq X).i0 & IK = (Pt2FinSeq X).(i0+1) & F is bijective
    & IK is bijective
    & for x,y be object st x in CarProduct SubFin(X,i0) & y in ElmFin(X,i0+1)
       ex s be FinSequence st F.x = s & IK.(x,y) = s^ <*y*> by Def5;

      reconsider i = i0+1 as non zero Nat;
      CarProduct SubFin(X,i0+1) = [: CarProduct SubFin(X,i0),ElmFin(X,i0+1) :]
        by A9,MEASUR13:9; then
      reconsider IK as Function of CarProduct SubFin(X,i),product SubFin(X,i);
      take i,IK;
      thus thesis by A8,A10;
     end;
    end;
    for k be Nat holds P[k] from NAT_1:sch 2(A3,A4); then
    P[n];
    hence P is bijective by A2,NAT_1:14;
end;
