Let's say we've a function similar to the function in Collatz conjecture. $$ f(n)= \begin{cases} 1 \ \ \ \ \ \ \ \ \ \text{if $n=1$}\\ \tfrac12n \ \ \ \ \ \ \text{if $n \equiv 0 \ \ $ (mod 2)}\\ n-1 \ \ \ \text{if $n \equiv 1 \ $ (mod 2) }\\ \end{cases} \\ , \forall \ \ n \in \mathbb{Z}^+ $$
Now, Can we prove that this will go to $1$ , or maybe more formally:
$$ f^m(n) = 1, m \to \infty $$
If yes, what steps would we take?
This will always go to $1$.
Assume $n$ is even; then $f(n)=\tfrac12n<n$. Assume $n>1$ is odd; then $f(n)=n-1<n$. Thus, $f(n)<n$ for all $n>1$. This makes sure that we decrease every time we apply $f$, thus, we must reach $1$ or less at some point, and since $f(n)\ge 1$, we know this must be $1$.