In Introduction to metamathematics (Kleene), The function ${\mu}y_{y<z}R(x_1,...x_n,y)$ is shown to be primitive recursive in $R$ (where ${\mu}y_{y<z}R(x_1,...x_n,y)$ is the smallest $y<z$ such that $R(x_1,...x_n,y)$ is $t$ for any n-tuple $x_1,...,x_n$).
My question is if $z$ is $x_k$ ($k = 0,1,2,...,n$) then is the function still primitive recursive in $R$.
i.e.
Is the function ${\mu}y_{y<x_k}R(x_1,...,x_n,y)$ ($k=0,1,2,...,n$) primitive recursive in $R$?
Yes, this is still primitive recursive. Intuitively, it's a composition of $f$ with a "rewriting" of the input tuple; formally, we use the composition and projection clauses.
We have the initial function $$f(x_1,...,x_n, z)=\mu y_{y<z}R(x_1,...,x_n,y),$$ and we want to build the function $$h(x_1,...,x_n)=f(x_1,...,x_n,x_k)$$ (for $k$ fixed). To do this we'll apply the composition clause of the definition of primitive recursion. Let $g_i(x_1,...,x_n)=x_i$ for $1\le i\le n$ and let $g_{n+1}(x_1,...,x_n)=x_k$. Each $g_i$ is primitive recursive (by the projection clause), and we already know $f$ is primitive recursive, so the function $$f(g_1(x_1,...,x_n), g_2(x_1,...,x_n), ..., g_n(x_1,...,x_n), g_{n+1}(x_1,...,x_n))$$ is also primitive recursive. But this simplifies to just $$f(x_1,x_2,...,x_n, x_k).$$ And that's our desired $h$.