
theorem LemmaGe2a:
  for m,n being Nat st m < n & m > 4 holds
    Fib n - Fib m > 2
  proof
    let m,n be Nat;
    assume
A1: m < n & m > 4; 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 > 4 - 1 by A1,XREAL_1:14; then
    Fib m1 >= 3 by FibGe3; then
    Fib n - Fib m >= 3 by A3,XXREAL_0:2,A2;
    hence thesis by XXREAL_0:2;
  end;
