A1: for k, n being Element of NAT holds Fib (n + k) >= Fib n
proof
defpred S1[ Nat] means for n being Element of NAT holds Fib (n + $1) >= Fib n;
A2: for k being Nat st S1[k] holds
S1[k + 1]
proof
let k be Nat; :: thesis: ( S1[k] implies S1[k + 1] )
assume A3: S1[k] ; :: thesis: S1[k + 1]
let n be Element of NAT ; :: thesis: Fib (n + (k + 1)) >= Fib n
n + (k + 1) = (n + k) + 1 ;
then A4: Fib (n + (k + 1)) >= Fib (n + k) by Lm4;
Fib (n + k) >= Fib n by A3;
hence Fib (n + (k + 1)) >= Fib n by A4, XXREAL_0:2; :: thesis: verum
end;
let k, n be Element of NAT ; :: thesis: Fib (n + k) >= Fib n
A5: S1[ 0 ] ;
for k being Nat holds S1[k] from NAT_1:sch 2(A5, A2);
 hence Fib (n + k) >= Fib n ; :: thesis: verum
end;
let m, n be Element of NAT ; :: thesis: ( m >= n implies Fib m >= Fib n )
assume m >= n ; :: thesis: Fib m >= Fib n
then consider k being Nat such that
A6: m = n + k by NAT_1:10;
reconsider k = k as Element of NAT by ORDINAL1:def 12;
m = n + k by A6;
hence Fib m >= Fib n by A1; :: thesis: verum