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)