Since the Union of Primitive Recursive Sets is itself primitive recursive, all we need to do is show that each of $\operatorname{Zinstr}$, $\operatorname{Sinstr}$, $\operatorname{Cinstr}$ and $\operatorname{Jinstr}$ are primitive recursive. First we consider $\operatorname{Zinstr}$. :$\operatorname{Zinstr} = \left\{{\beta \left({Z \left({n}\right)}\right): n \in \N^*}\right\} = \left\{{6 n - 3: n \in \N^*}\right\}$. So $\operatorname{Zinstr}$ is the set of natural numbers which are divisible by $3$ but not $6$. Thus its characteristic function is given by: :$\chi_{\operatorname{Zinstr}} \left({k}\right) = \operatorname{div} \left({k, 3}\right) \times \overline{\operatorname{sgn}}\left({\operatorname{div} \left({k, 6}\right)}\right)$ where: : $\operatorname{div}$ is primitive recursive : $\overline{\operatorname{sgn}}$ is primitive recursive : Multiplication is Primitive Recursive : $3$ and $6$ are constants Hence $\operatorname{Zinstr}$ is primitive recursive. Next we consider $\operatorname{Sinstr}$. :$\operatorname{Sinstr} = \left\{{\beta \left({S \left({n}\right)}\right): n \in \N^*}\right\} = \left\{{6 n: n \in \N^*}\right\}$. So $\operatorname{Sinstr}$ is the set of natural numbers which are divisible by $6$. Thus its characteristic function is given by: :$\chi_{\operatorname{Sinstr}} \left({k}\right) = \operatorname{sgn} \left({k}\right) \times \operatorname{div} \left({k, 6}\right)$ where: : $\operatorname{div}$ is primitive recursive : $\operatorname{sgn}$ is primitive recursive : Multiplication is Primitive Recursive : $6$ is constant Hence $\operatorname{Sinstr}$ is primitive recursive. Next we consider $\operatorname{Cinstr}$. We have that $\beta \left({C \left({n, m}\right)}\right) = 2^m 3^n + 1$. Hence $k \in \operatorname{Cinstr}$ iff $k \equiv 1 \pmod 3$ and $k - 1$ codes a pair of positive integers. That is: {{begin-eqn}} {{eqn | l = k \in \operatorname{Cinstr}       | o = \iff       | r = \operatorname{rem} \left({k, 3}\right) = 1       | c =  }} {{eqn | o = \land       | r = \chi_{\operatorname{Seq} } \left({k \, \dot - \, 1}\right) = 1       | c =  }} {{eqn | o = \land       | r = \operatorname{len} \left({k \, \dot - \, 1}\right) = 2       | c =  }} {{end-eqn}} We can introduce two properties: : $P \left({k}\right) \iff \operatorname{eq} \left({\operatorname{rem} \left({k, 3}\right), 1}\right)$ : $Q \left({k}\right) \iff \operatorname{eq} \left({\operatorname{len} \left({k \, \dot - \, 1}\right), 2}\right)$ $\chi_{P}$ is primitive recursive since it is obtained by substitution from: : $\operatorname{eq}$ is primitive recursive : $\operatorname{rem}$ is primitive recursive : $3$ and $1$ are constants $\chi_{Q}$ is primitive recursive since it is obtained by substitution from: : $\operatorname{eq}$ is primitive recursive : $\operatorname{len}$ is primitive recursive : $k \, \dot - \, 1$ is primitive recursive : $2$ and $1$ are constants Thus the characteristic function of $\operatorname{Cinstr}$ is given by: :$\chi_{\operatorname{Cinstr}} = \chi_{P} \left({k}\right) \chi_{Q} \left({k}\right) \chi_{\operatorname{Seq}} \left({k \, \dot - \, 1}\right)$ where: : $\chi_{P}$ is primitive recursive : $\chi_{Q}$ is primitive recursive : Multiplication is Primitive Recursive : $\chi_{\operatorname{Seq}}$ is primitive recursive : $k \, \dot - \, 1$ is primitive recursive : $1$ is constant Hence $\operatorname{Cinstr}$ is primitive recursive. Finally we consider $\operatorname{Jinstr}$. We have that $\beta \left({J \left({n, m, q}\right)}\right) = 2^m 3^n 5^n + 2$. Hence $k \in \operatorname{Jinstr}$ iff $k \equiv 2 \pmod 3$ and $k - 1$ codes a triad of positive integers. That is: {{begin-eqn}} {{eqn | l = k \in \operatorname{Jinstr}       | o = \iff       | r = \operatorname{rem} \left({k, 3}\right) = 2       | c =  }} {{eqn | o = \land       | r = \chi_{\operatorname{Seq} } \left({k \, \dot - \, 2}\right) = 1       | c =  }} {{eqn | o = \land       | r = \operatorname{len} \left({k \, \dot - \, 2}\right) = 3       | c =  }} {{end-eqn}} We can introduce two properties: : $R \left({k}\right) \iff \operatorname{eq} \left({\operatorname{rem} \left({k, 3}\right), 2}\right)$ : $S \left({k}\right) \iff \operatorname{eq} \left({\operatorname{len} \left({k \, \dot - \, 2}\right), 3}\right)$ $\chi_R$ is primitive recursive since it is obtained by substitution from: : $\operatorname{eq}$ is primitive recursive : $\operatorname{rem}$ is primitive recursive : $3$ and $2$ are constants $\chi_S$ is primitive recursive since it is obtained by substitution from: : $\operatorname{eq}$ is primitive recursive : $\operatorname{len}$ is primitive recursive : $k \, \dot - \, 2$ is primitive recursive : $2$ and $3$ are constants Thus the characteristic function of $\operatorname{Jinstr}$ is given by: :$\chi_{\operatorname{Jinstr}} = \chi_R \left({k}\right) \chi_S \left({k}\right) \chi_{\operatorname{Seq}} \left({k \, \dot - \, 2}\right)$ where: : $\chi_R$ is primitive recursive : $\chi_S$ is primitive recursive : Multiplication is Primitive Recursive : $\chi_{\operatorname{Seq}}$ is primitive recursive : $k \, \dot - \, 2$ is primitive recursive : $2$ is constant Hence $\operatorname{Jinstr}$ is primitive recursive. The result follows. {{qed}} Category:URM Programs Category:Primitive Recursive Functions lx6sqfd2svyrgvh97vgzh6pvnffq772