
theorem LemmaGe1:
  for m,n being Nat st m < n & m > 3 holds
    Fib n - Fib m > 1
  proof
    let m,n be Nat;
    assume
A1: m < n & m > 3; then
    m + 1 <= n by NAT_1:13; then
A3: Fib (m+1) - Fib m <= Fib n - Fib m by XREAL_1:9,FIB_NUM2:45;
    reconsider m1 = m - 1 as Element of NAT by INT_1:5,A1,XXREAL_0:2;
a2: Fib (m1+1+1) = Fib (m1+1) + Fib m1 by PRE_FF:1;
    m - 1 > 3 - 1 by A1,XREAL_1:14; then
    Fib m1 >= 2 by FibGe2a; then
    Fib n - Fib m >= 2 by A3,a2,XXREAL_0:2;
    hence thesis by XXREAL_0:2;
  end;
