1. 1.

   Induct on $x$. First, $A(0,y)=y+1$ is well-defined, so $(0,y)\in\operatorname{dom}(A)$ for all $y$. Next, suppose that for a given $x$, $(x,y)\in\operatorname{dom}(A)$ for all $y$. We want to show that $(x+1,y)\in\operatorname{dom}(A)$ for all $y$. To do this, induct on $y$. First, $(x+1,0)\in\operatorname{dom}(A)$, since $A(x+1,0)=A(x,1)$ is well-defined. Next, assume that $(x+1,y)\in\operatorname{dom}(A)$. Then $A(x,A(x+1,y))=A(x+1,y+1)$ is well-defined. so $(x+1,y+1)\in\operatorname{dom}(A)$ as well.

2. 2.

   Induct on $y$. First, $A(1,0)=A(0,1)=2$. Next, assume $A(1,y)=y+2$. Then $A(1,y+1)=A(0,y+2)=y+3=(y+1)+2$.

3. 3.

   Induct on $y$. First, $A(2,0)=A(1,1)=1+2=3$. Next, assume $A(2,y)=2y+3$. Then $A(2,y+1)=A(1,A(2,y))=A(2,y)+2=(2y+3)+2=2(y+1)+3$.

4. 4.

   Induct on $x$. First, $y<y+1=A(0,y)$. Next, assume $y<A(x,y)$, where $x>0$. Then $y+1\leq A(x,y)<A(x-1,A(x,y))=A(x,y+1)$.

5. 5.

   Induct on $x$. First, $A(0,y)=y+1<y+2=A(0,y+1)$. Next, assume that $A(x,y)<A(x,y+1)$. Then $A(x+1,y)<A(x,A(x+1,y))=A(x+1,y+1)$.

6. 6.

   Induct on $y$. First, $A(x,1)=A(x+1,0)$. Next, assume that $A(x,y+1)\leq A(x+1,y)$. Then $A(x,y+2)\leq A(x,A(x,y+1))\leq A(x,A(x+1,y))=A(x+1,y+1)$.

7. 7.

   Induct on $x$. First, $A(0,y)=y+1<y+2=A(1,y)$. Next, assume that $A(x,y)<A(x+1,y)$. There are two cases: $y=0$. Then $A(x+1,0)=A(x,1)<A(x+1,1)$. Otherwise, $y=z+1$, so that $A(x+1,y)=A(x+1,z+1)=A(x,A(x+1,z))\leq A(x,A(x,z+1))=A(x,A(x,y))<A(x,A(x+1,y))=A(x+1,y+1)$.

8. 8.

   $A(r,A(s,y))<A(r+s,A(s,y))<A(r+s,A(r+s+1,y))=A(r+s+1,y+1)\leq A(r+s+2,y)$.

9. 9.

   Let $z=\max\{r,s\}$. Then $A(r,y)+A(s,y)\leq 2A(z,y)<2A(z,y)+3=A(2,A(z,y))<A(4+z,y)$. The proof is completed by setting $t=4+z$.

∎

$A_{0}(m)=m+1=s(m)$, so is primitive recursive. Now, assume $A_{n}$ is primitive recursive, then $A_{{n+1}}(0)=A(n+1,0)=A(n,1)=A_{n}(1)=k$, and $A_{{n+1}}(m+1)=A(n+1,m+1)=A(n,A(n+1,m))=A_{n}(A(n+1,m))=A_{n}(A_{{n+1}}(m))$, so that $A_{{n+1}}$ is defined by primitive recursion via the constant function $\operatorname{const}_{k}$, and $A_{n}$, which is primitive recursive by the induction hypothesis. Therefore $A_{{n+1}}$ is primitive recursive also. ∎