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

Lemma for Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers
Let $f \left({n}\right)$ be an arbitrary arithmetic function.

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

0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + a_{n - 2} + f \left({n - 2}\right) & : n > 1 \end{cases}$

Then:
 * $a_n = F_n + \displaystyle \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} f \left({k}\right)$

Proof
Trying out a few values:

The proof proceeds by induction.

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

$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

$P \left({2}\right)$ is the case:

Thus $P \left({2}\right)$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $a_r = F_r + \displaystyle \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} f \left({k}\right)$

and:
 * $a_{r - 1} = F_{r - 1} + \displaystyle \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} f \left({k}\right)$

from which it is to be shown that:
 * $a_{r + 1} = F_{r + 1} + \displaystyle \sum_{k \mathop = 0}^{r - 1} F_{r - k} f \left({k}\right)$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $a_n = F_n + \displaystyle \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} f \left({k}\right)$