Sum of Non-Consecutive Fibonacci Numbers

Theorem
Let $S$ be a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.

Let the largest element of $S$ be $F_j$.

Then:
 * $\displaystyle \sum_{F_i \mathop \in S} F_i < F_{j + 1}$

That is, the sum of all the elements of $S$ is strictly less than the next largest Fibonacci number.

That is, given some increasing sequence $\left\langle {c_i}\right\rangle$ satisfying $c_i \ge 2$ and $c_{i + 1} \ge c_i + 1$:
 * $\displaystyle F_{c_k + 1} > \sum_{i \mathop = 0}^k F_{c_i}$

Proof
The proof proceeds by induction on $j$ for $j \ge 2$.

For all $j \in \N_{> 0}$, let $P \left({j}\right)$ be the proposition:
 * $\displaystyle \sum_{F_i \mathop \in S} F_i < F_{j + 1}$

Let the term allowable set be used to mean a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.

Basis for the Induction
The only possible allowable set whose largest member is $F_2 = 1$ is the set:
 * $\left\{{F_2}\right\} = \left\{{1}\right\}$

This has sum $1$, which is strictly less than $F_3 = 2$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{F_i \mathop \in S} F_i < F_{k + 1}$

where $S$ is an allowable set whose largest element is $F_k$.

Then we need to show:
 * $\displaystyle \sum_{F_i \mathop \in S} F_i < F_{k + 2}$

where $S$ is an allowable set whose largest element is $F_{k + 1}$.

Induction Step
This is our induction step:

Let $S$ be an allowable set whose greatest element is $F_{k + 1}$.

Let $S' := S \setminus \left\{ {F_{k + 1} }\right\}$.

That is, $S'$ is $S$ without its largest element $F_{k + 1}$.

We have that $S$ cannot contain consecutive Fibonacci numbers.

Hence the largest element of $S'$ is at most $F_{k - 1}$.

So by the induction hypothesis:
 * $\displaystyle \sum S' < F_k$

Then:

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

Therefore:
 * $\displaystyle \forall j \in \N_{> 0}: \sum_{F_i \mathop \in S} F_i < F_{j + 1}$