First, observe that condition 1 is satisfied by both $\mathcal{PR}^{{\prime}}$ and $\mathcal{D}$. To see that $\mathcal{D}\subseteq\mathcal{PR}^{{\prime}}$, note that condition 3 is just the definition in the parent entry, and conditions 2 and 4 are discussed and proved, also in the parent entry. So we have one inclusion. To see the other inclusion $\mathcal{PR}^{{\prime}}\subseteq\mathcal{D}$, we need to verify the two closure properties:

1. 1.

   functional composition (in $\mathcal{S}$): suppose $g_{1},\ldots,g_{m}\in\mathcal{V}(k,1)$ and $h\in\mathcal{V}(m,1)$ are in $\mathcal{D}$, we want to show that $f:=h(g_{1},\ldots,g_{m})\in\mathcal{V}(k,1)$ is in $\mathcal{D}$ too. First, form $g=(g_{1},\ldots,g_{m})$. Then $g\in\mathcal{D}$ by extension of coordinates. Then $f=h\circ g\in\mathcal{D}$ by functional composition (in $\mathcal{V}$).

2. 2.

   primitive recursion: suppose $g\in\mathcal{V}(k,1)$ and $h\in\mathcal{V}(k+2,1)$ are both in $\mathcal{D}$, we want to show that $f\in\mathcal{V}(k+1,1)$ given by $f(\boldsymbol{x},0)=g(\boldsymbol{x})$ and $f(\boldsymbol{x},n+1)=h(\boldsymbol{x},n,f(\boldsymbol{x},n))$ is in $\mathcal{D}$ too. First, define a function $H\in\mathcal{V}(k+2,k+2)$ by

   $$H(\boldsymbol{x},y,z):=(\boldsymbol{x},s(y),h(\boldsymbol{x},y,z)).$$

   Then $H$ is formed by extension of coordinates via the projection functions $p_{i}^{{k+2}}$ (with $i=1,\ldots,k$) producing the first $k$ coordinates (the coordinates of $\boldsymbol{x}$), the function $s\circ p_{{k+1}}^{{k+2}}$ producing the $k+1$st coordinate $s(y)$, and $h$ producing the last coordinate. Since each of the coordinate functions is in $D$, so is $H$.

   Next, the function $F\in\mathcal{V}(k+3,k+2)$ given by $F(\boldsymbol{x},y,z,m)=H^{m}(\boldsymbol{x},y,z)$ is in $D$ by iterated composition. We now verify by induction on $n$ that

   $$F(\boldsymbol{x},0,g(\boldsymbol{x}),n)=(\boldsymbol{x},n,f(\boldsymbol{x},n)).$$

   • $F(\boldsymbol{x},0,g(\boldsymbol{x}),0)=(\boldsymbol{x},0,g(\boldsymbol{x})$, and its third coordinate is $f(\boldsymbol{x},0)$, as desired.

   • Suppose now that $F(\boldsymbol{x},0,g(\boldsymbol{x}),n)=(\boldsymbol{x},n,f(\boldsymbol{x},n))$. Then

     	$\displaystyle F(\boldsymbol{x},0,g(\boldsymbol{x}),n+1)$	$\displaystyle=$	$\displaystyle H(\boldsymbol{x},n,f(\boldsymbol{x},n))$
     		$\displaystyle=$	$\displaystyle(\boldsymbol{x},s(n),h(\boldsymbol{x},n,f(\boldsymbol{x},n)))$
     		$\displaystyle=$	$\displaystyle(\boldsymbol{x},n+1,f(\boldsymbol{x},n+1)).$

   As a result, $f(\boldsymbol{x},n)=p_{{k+2}}^{{k+2}}\circ F(\boldsymbol{x},0,g(\boldsymbol{x}),n)$ is in $\mathcal{D}$ also.

Therefore, $\mathcal{PR}^{{\prime}}\subseteq\mathcal{D}$, and the proof is complete. ∎