Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma

Lemma for Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers
Let $\map f n$ be an arbitrary arithmetic function.

Let $\sequence {a_n}$ be the sequence defined as:
 * $a_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + a_{n - 2} + \map f {n - 2} & : n > 1 \end{cases}$

Then:
 * $a_n = F_n + \ds \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map f k$

Proof
Trying out a few values:

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $a_n = F_n + \ds \sum_{k \mathop = 1}^{n - 1} F_{n - k - 1} \map f k$

$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

$\map P 2$ is the case:

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $a_r = F_r + \ds \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} \map f k$

and:
 * $a_{r - 1} = F_{r - 1} + \ds \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} \map f k$

from which it is to be shown that:
 * $a_{r + 1} = F_{r + 1} + \ds \sum_{k \mathop = 0}^{r - 1} F_{r - k} \map f k$

Induction Step
This is the induction step:

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $a_n = F_n + \ds \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map f k$