I am trying to show, that the set $PR$ of primitive recursive functions is a subset of $R$, the recursive functions. Could someone help me, complete the proof of that assertion ?
My idea: Since $PR$ is defined as the smallest set of all sets $PR'\subseteq \cup _{k\in \mathbb{N}} \{f:\mathbb{N}^k \rightarrow \mathbb{N}\}$, satisfying the properties
1) $PR'$ contains the constant zero function, the successor function and the projection functions
2) $PR'$ is closed under composition by functions
3) $PR'$ is closed under primitive recursion
and $R$ is the smallest set of all sets $R'\subseteq \cup _{k\in \mathbb{N}} \{f:\mathbb{N}^k \rightarrow \mathbb{N}\}$, satisfying the properties 1)-3) and additionaly the property
4) $R'$ is closed under $\mu$-recursion
we obviously have $R' \subseteq PR'$, since there are more constraints on the functions in $R'$ than in $PR'$. But why does the "$\subseteq$" sign gets reversed, if we take the minimum ob both sets, meaning $\min R'=R\supseteq PR=\min PR'$ ?
(m idea was nonsense)