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bounded minimization
One useful way of generating more primitive recursive functions from existing ones is through what is known as bounded summation and bounded product. Given a primitive recursive function $f:\mathbb{N}^{{m+1}}\to\mathbb{N}$, define two functions $f_{s},f_{p}:\mathbb{N}^{{m+1}}\to\mathbb{N}$ as follows: for $\boldsymbol{x}\in\mathbb{N}^{m}$ and $y\in\mathbb{N}$:
$f_{s}(\boldsymbol{x},y):=\sum_{{i=0}}^{y}f(\boldsymbol{x},i)$ 
$f_{p}(\boldsymbol{x},y):=\prod_{{i=0}}^{y}f(\boldsymbol{x},i)$ 
These are easily seen to be primitive recursive, because they are defined by primitive recursion. For example,
$f_{s}(\boldsymbol{x},0)=f(\boldsymbol{x},0),\quad\mbox{and}\quad f_{s}(% \boldsymbol{x},n+1)=g(\boldsymbol{x},n,f_{s}(\boldsymbol{x},n)),$ 
where $g(\boldsymbol{x},n,y)=\operatorname{add}(f(\boldsymbol{x},n),y)$, which is primitive recursive by functional composition.
Definition. We call $f_{s}$ and $f_{p}$ functions obtained from $f$ by bounded sum and bounded product respectively.
Using bounded summation and bounded product, another useful class of primitive recursive functions can be generated:
Definition. Let $f:\mathbb{N}^{{m+1}}\to\mathbb{N}$ be a function. For each $y\in\mathbb{N}$, set
$A_{f}(\boldsymbol{x},y):=\{z\in\mathbb{N}\mid z\leq y\mbox{ and }f(\boldsymbol% {x},z)=0\}.$ 
Define
$f_{{bmin}}(\boldsymbol{x},y):=\left\{\begin{array}[]{ll}\min A_{f}(\boldsymbol% {x},y)&\textrm{if }A_{f}(\boldsymbol{x},y)\neq\varnothing,\\ s(y)&\textrm{otherwise.}\end{array}\right.$ 
$f_{{bmin}}$ is called the function obtained from $f$ by bounded minimization, and is usually denoted
$\mu z\leq y(f(\boldsymbol{x},z)=0).$ 
Proposition 1.
If $f:\mathbb{N}^{{m+1}}\to\mathbb{N}$ is primitive recursive, so is $f_{{bmin}}$.
Proof.
Define $g:=\operatorname{sgn}\circ f$. Then
$g(\boldsymbol{x},y):=\left\{\begin{array}[]{ll}0&\textrm{if }f(\boldsymbol{x},% y)=0,\\ 1&\textrm{otherwise.}\end{array}\right.$ 
As $f$ is primitive recursive, so is $g$, since the sign function $\operatorname{sgn}$ is primitive recursive (see this entry).
Next, the function $g_{p}$ obtained from $g$ by bounded product has the following properties:

if $g_{p}(\boldsymbol{x},y)=1$, then $g_{p}(\boldsymbol{x},z)=1$ for all $z<y$,

if $g_{p}(\boldsymbol{x},y)=0$, then $g_{p}(\boldsymbol{x},z)=0$ for all $z\geq y$.
Finally, the function $(g_{p})_{s}$ obtained from $g_{p}$ by bounded sum has the property that, when applied to $(\boldsymbol{x},y)$, counts the number of $z\leq y$ such that $g_{p}(\boldsymbol{x},z)=1$. Based on the property of $g_{p}$, this count is then exactly the least $z\leq y$ such that $g_{p}(\boldsymbol{x},z)=1$. This means that $(g_{p})_{s}=f_{{bmin}}$ for all $(\boldsymbol{x},y)\in\mathbb{N}^{{m+1}}$. Since $g_{p}$ is primitive recursive, so is $(g_{p})_{s}$, or that $f_{{bmin}}$ is primitive recursive. ∎
In fact, if $f$ is a (total) recursive function, so is $f_{{bmin}}$, because all of the derived functions in the proof above preserve primitive recursiveness as well as totalness.
Remarks.

In the definition of bounded minimization, if we take the $y$ out, then we arrive at the notion of unbounded minimization, or just minimization. The proposition above shows that the set $\mathcal{PR}$ of primitive recursive functions is closed under bounded minimization. However, $\mathcal{PR}$ is not closed under minimization. The closure of $\mathcal{PR}$ under minimization is the set $\mathcal{R}$ of recursive functions (total or not).

It is not hard to show that $\mathcal{ER}$, the set of all elementary recursive functions, is closed under bounded minimization.
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