properties of Ackermann function

In this entry, we derive some basic properties of the Ackermann function $A(x,y)$, defined by double recursion, as follows:

$$A(0,y)=y+1,A(x+1,0)=A(x,1),A(x+1,y+1)=A(x,A(x+1,y)).$$

These properties will be useful in proving that $A$ is not primitive recursive.

1. $A$ is total ($\mathrm{dom}(A)={\mathbb{N}}^2$).

2. $A(1,y)=y+2$.

3. $A(2,y)=2y+3$.

4. .

5. .

6. $A(x,y+1)\le A(x+1,y)$.

7. .

9. For any $r,s$, for some $t$ not depending on $y$.

Most of the proofs are done by induction.

Proof.

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

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. 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. Induct on $x$. First, . Next, assume , where $x>0$. Then .

5. Induct on $x$. First, . Next, assume that . Then .

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

7. Induct on $x$. First, . Next, assume that . There are two cases: $y=0$. Then . Otherwise, $y=z+1$, so that .

8. .

9. Let $z=\max \{r,s\}$. Then . The proof is completed by setting $t=4+z$.

∎

With respect to the recursive property of $A$, we see that $A$ is recursive, since, by Church’s Thesis, $A$ can be effectively computed (in fact, a program can be easily written to compute $A(x,y)$). We also have the following:

Proposition 1.

Define $A_n\mathrm{:}\mathrm{N}\mathrm{\to }\mathrm{N}$ by $A_n\mathit{}\mathrm{(}m\mathrm{)}\mathrm{=}A\mathit{}\mathrm{(}n\mathrm{,}m\mathrm{)}$. Then $A_n$ is primitive recursive for each $n$.

Proof.

$A_0(m)=m+1=s(m)$, 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 ${\mathrm{const}}_k$, and $A_n$, which is primitive recursive by the induction hypothesis. Therefore $A_{n+1}$ is primitive recursive also.∎

The most important fact about $A$ concerning recursiveness is that $A$ is not primitive recursive. Due to the length of its proof, it is demonstrated in this entry (http://planetmath.org/AckermannFunctionIsNotPrimitiveRecursive).