A1: for k, n being Nat holds Fib (n + k) >= Fib n

proof

defpred S₁[ Nat] means for n being 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 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 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 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