A closed form for nth Fibonacci number

I have made an attempt to arrive at a closed form by setting up a bijection in a counting process.

Consider the set . In how many ways can we select non null subset of such that no two elements in the subset give difference one ?

Let be the number of ways of selecting such a subset. Depending upon whether is included in the subset or not , we get


with and

It is clear that = (n+1)th Fibonacci number,

Looking the other way, let our subset contain elements.
{min, max}= {, }

I arrange all the elements in ascending order from left to right. Except for the last element, I place a ball just right of each element. Now I am left with balls with places available. You are getting what I am telling ?

This can be done in ways.
So,

=