reserve n, k, r, m, i, j for Nat;

theorem Th46:
  for k being Nat st k > 1 holds k < n implies Fib (k) < Fib (n)
proof
  let k be Nat such that
A1: k > 1;
  assume
A2: k < n;
  then consider m being Nat such that
A3: n = k + m by NAT_1:10;
  reconsider k as non zero Element of NAT by A1,ORDINAL1:def 12;
  reconsider m as non zero Element of NAT by A2,A3,ORDINAL1:def 12;
  for k, m being non zero Element of NAT st k > 1 holds Fib (k) < Fib (k + m)
  proof
    let k, m be non zero Element of NAT such that
A4: k > 1;
    defpred P[Nat] means Fib (k) < Fib (k + $1);
A5: for r being non zero Nat st P[r] holds P[r+1]
    proof
      let r be non zero Nat;
      k + r > 0 + 1 by A4,XREAL_1:8;
      then
A6:   Fib (k + r) < Fib ((k + r) + 1) by Th44;
      assume P[r];
      hence thesis by A6,XXREAL_0:2;
    end;
A7: P[1] by A4,Th44;
    for k being non zero Nat holds P[k] from NAT_1:sch 10(A7,A5);
    hence thesis;
  end;
  then Fib (k) < Fib (k + m) by A1;
  hence thesis by A3;
end;
