Problem Solving Part 3

I will attempt to prove the following claim by contradiction:

For each natural number n there exists a natural number m such that F(m,n) is a power of 2.

Using the definitions found in Part 1, originally from here.

Assume there exists some n such that F(m,n) is not a power of 2 for all m.

Then the set of natural numbers n that satisfy "F(m,n) is not a power of 2 for all m" exists and is non-empty. Call this set S.

By the Well-ordering principle, S has a smallest element. Call this number x.

Note that, if x is in S, then F(m,x) is also in x for any m.

The case that x is even can easily be ruled out. If x is even, then f(x) = x/2. Since we are not considering 0, Its clear x/2 < x. We also know that f(x) is in S.  But x was chosen as the smallest element of S, f(x) cannot be in S. This is a contradiction; x can therefore not be an even number.

So x is an odd number. Proving that this leads to a contradiction would be a much more difficult task. I will assume that one of the following is true, and try to reach contradictions in both cases:

F(m,x) will eventually reach some cycle, which does not contain the number 1. 

or

S has infinitely many elements.

Where to go from here stumps me for now. Tomorrow, I will try to advance at least 1 of the above.