We have that for a partial function $f : \mathbb{N}^{k+1} \to \mathbb{N}$, the $\mu-$ operator is defined as the following:
$$\mu (f)(x_1, ..., x_k)=\begin{cases} n, & f(n,x_1,...,x_k)=0 \land f(n',x_1,...,x_k) \neq 0 \land \text{is defined for all} \ n' < n, \\ \bot, & \text{else} \end{cases}$$
We are then given the following primitive recursive functions:
$$\text{modsub}: \mathbb{N}^2 \to \mathbb{N}, \text{modsub}(x,y) = \text{max}\{0,x-y\},$$ $$\text{exp}: \mathbb{N} \to \mathbb{N}, \text{exp}(x)=2^x$$
Finally, we have that $f: \mathbb{N}^2 \to \mathbb{N}$ is defined as $f(x,y):= \text{modsub}(y, \text{exp}(x))$.
How can I show that $\mu(f)(4) = 2$ and $\mu(f)(9) =4$?
$(\mu f)(4) =$ smallest $n$ such that $f(n,4) = \max\{0,4-2^n\}$ is zero, which is $n=2$.
$(\mu f)(9) =$ smallest $n$ such that $f(n,9) = \max\{0,9-2^n\}$ is zero, which is $n=4$. Here $n=3$ would lead to $\max\{0,9-2^3\}=1$.