How does primitive recursion handle mutual recursion?

Suppose you have two mutually recursive maps $f, g : \mathbb{N} \to \mathbb{N}$ defined by \begin{align*} f(n) &= \Phi(f, g, n), \\ g(n) &= \Psi(f, g, n) \end{align*} We may replace this with a single recursive map $h : \mathbb{N} \to \mathbb{N} \times \mathbb{N}$, where $h(n) = (f(n), g(n))$, so that $f(n) = \pi_1(h(n))$, $g(n) = \pi_2(h(n))$. Then the above recursive definition may be rewriten as $$ h(n) = (\Phi(\pi_1 \circ h, \pi_2 \circ h, n), \Psi(\pi_1 \circ h, \pi_2 \circ h, n)) $$ The above trick is very general and works in many situations. Regarding primitive recursion, we just have to verify that the passage between $f, g$ and $h$ preserves the primitive recursive nature of functions, which it does because pairing, projections and compositions are primitive recursive.