One way is to use the primitive recursion (catalog item 13). Let $d(n,0,b)=g(b)$ , and $d(n,a+1,b)=h((n)_{a},a,b)$ . Then, $f(0,b)=g(b)=d(\langle\rangle,0,b)$ , and $f(a+1,b)=h(f(a,b),a,b)=d(\overline{f}(a+1,b),a+1,b)$ , therefore, $f$ is the function defined by $d$ by primitive recursion. Therefore, it is sufficient to show that $d$ is representable. $d(n,0,b)=g(I_{3}^{3}(n,0,b))$ is representable, and for $a>0$ , $d(n,a,b)=h((I_{1}^{3}(n,a,b))_{I_{2}^{3}(n,a,b)-1},I_{2}^{3}(n,a,b),I_{3}^{3}(n,a,b))$ is representable. Hence, $g$ is representable (a more general case of “gluing” representable functions together is presented in Exercise 10).