Would love any feedback for this proposed proof of the Collatz Conjecture.
(For more details explaining each step, I made a video here: https://www.youtube.com/watch?v=P0F4zbNdbTU )
Ignoring all even number (which trivially lead to an odd number), we create a "Collatz Odd-Only Variant" as such, where $n$ is a positive odd integer:
$f(n) = \left\{\begin{matrix} \frac{n-1}{4} & \mbox{if }n\equiv 5 \mbox{ }(\mbox{mod }8)\\ \frac{3n+1}{4} & \mbox{if }n\equiv 1 \mbox{ }(\mbox{mod }8)\\ \frac{3n+1}{2} & \mbox{if }n\equiv 3 \mbox{ }(\mbox{mod }4)\\ \end{matrix}\right.$
This creates a tree that looks like this:
We then must show that all odd numbers in this variant are connected to 1. (That is, they do not create loops or diverge elsewhere.)
To do this, we create a new simpler function, again where $n$ is a positive odd integer:
$f(n) = \left\{\begin{matrix}\ \frac{n-1}{4} & \mbox{if }n\equiv 5 \mbox{ }(\mbox{mod }8)\\ \frac{n+3}{4} & \mbox{if }n\equiv 1 \mbox{ }(\mbox{mod }8)\\ \frac{n-1}{2} & \mbox{if }n\equiv 3 \mbox{ }(\mbox{mod }4)\\ \end{matrix}\right.$
In this function, we know that all $n$'s will lead to 1, because $n$ must shrink with every iteration, and it cannot shrink below 1.
This creates a tree like this:
Since we know this tree must contain all positive odd integers, all we have to do is disconnect and reattach some nodes so that this tree matches the "Collatz Odd-Only Variant" tree. As long as we do not reconnect a node to one of its own children (a node somewhere above it, representing the iteration of the function that preceded it), we are sure that we will never get loops and that the odd numbers will remain connected to 1.
When we disconnected the $\frac{n+3}{4}$ nodes and reattach them to $\frac{3n+1}{4}$ nodes, we know we will never connect it to a child, because we must connect it to a node that is less than $n$ (and all of $n$'s children will, by definition, be greater than $n$).
Disconnecting $\frac{n-1}{2}$ nodes to $\frac{3n+1}{2}$ is also certain to never connect to one of its own children, because iterating backwards through our function can never give us $\frac{3n+1}{2}$. (We can see this by checking the inverses of the equations in our function, even when we combine them to create composite functions.)
Since we know all nodes can be disconnected and reconnected while remaining attached to the tree connecting them all to 1, doesn't this prove the conjecture? Or what am I missing?
Would love to know your thoughts! Thank you!