Ackermann function

In computability theory, the Ackermann function, named after Wilhelm Ackermann, is one of the simplest^[1] and earliest-discovered examples of a total computable function that is not primitive recursive. All primitive recursive functions are total and computable, but the Ackermann function illustrates that not all total computable functions are primitive recursive.

After Ackermann's publication^[2] of his function (which had three non-negative integer arguments), many authors modified it to suit various purposes, so that today "the Ackermann function" may refer to any of numerous variants of the original function. One common version is the two-argument Ackermann–Péter function developed by Rózsa Péter and Raphael Robinson. This function is defined from the recurrence relation ${\displaystyle \operatorname {A} (m+1,n+1)=\operatorname {A} (m,\operatorname {A} (m+1,n))}$ with appropriate base cases. Its value grows very rapidly; for example, ${\displaystyle \operatorname {A} (4,2)}$ results in ${\displaystyle 2^{65536}-3}$, an integer with 19,729 decimal digits.^[3]

In the late 1920s, the mathematicians Gabriel Sudan and Wilhelm Ackermann, students of David Hilbert, were studying the foundations of computation. Both Sudan and Ackermann are credited^[4] with discovering total computable functions (termed simply "recursive" in some references) that are not primitive recursive. Sudan published the lesser-known Sudan function, then shortly afterwards and independently, in 1928, Ackermann published his function ${\displaystyle \varphi }$ (from Greek, the letter phi). Ackermann's three-argument function, ${\displaystyle \varphi (m,n,p)}$, is defined such that for ${\displaystyle p=0,1,2}$, it reproduces the basic operations of addition, multiplication, and exponentiation as

$${\displaystyle {\begin{aligned}\varphi (m,n,0)&=m+n\\\varphi (m,n,1)&=m\times n\\\varphi (m,n,2)&=m^{n}\end{aligned}}}$$

and for ${\displaystyle p>2}$ it extends these basic operations in a way that can be compared to the hyperoperations:

$${\displaystyle {\begin{aligned}\varphi (m,n,3)&=m[4](n+1)\\\varphi (m,n,p)&\gtrapprox m[p+1](n+1)&&{\text{for }}p>3\end{aligned}}}$$

(Aside from its historic role as a total-computable-but-not-primitive-recursive function, Ackermann's original function is seen to extend the basic arithmetic operations beyond exponentiation, although not as seamlessly as do variants of Ackermann's function that are specifically designed for that purpose—such as Goodstein's hyperoperation sequence.)

In On the Infinite,^[5] David Hilbert hypothesized that the Ackermann function was not primitive recursive, but it was Ackermann, Hilbert's personal secretary and former student, who actually proved the hypothesis in his paper On Hilbert's Construction of the Real Numbers.^[2][6]

Rózsa Péter^[7] and Raphael Robinson^[8] later developed a two-variable version of the Ackermann function that became preferred by almost all authors.

The generalized hyperoperation sequence, e.g. ${\displaystyle G(m,a,b)=a[m]b}$, is a version of the Ackermann function as well.^[9]

In 1963 R.C. Buck based an intuitive two-variable ^{[n 1]} variant ${\displaystyle \operatorname {F} }$ on the hyperoperation sequence:^[10][11]

$${\displaystyle \operatorname {F} (m,n)=2[m]n.}$$

Compared to most other versions, Buck's function has no unessential offsets:

$${\displaystyle {\begin{aligned}\operatorname {F} (0,n)&=2[0]n=n+1\\\operatorname {F} (1,n)&=2[1]n=2+n\\\operatorname {F} (2,n)&=2[2]n=2\times n\\\operatorname {F} (3,n)&=2[3]n=2^{n}\\\operatorname {F} (4,n)&=2[4]n=2^{2^{2^{{}^{.^{.^{{}_{.}2}}}}}}\\&\quad \vdots \end{aligned}}}$$

Many other versions of Ackermann function have been investigated.^[12][13]

Ackermann's original three-argument function ${\displaystyle \varphi (m,n,p)}$ is defined recursively as follows for nonnegative integers ${\displaystyle m,n,}$ and ${\displaystyle p}$:

$${\displaystyle {\begin{aligned}\varphi (m,n,0)&=m+n\\\varphi (m,0,1)&=0\\\varphi (m,0,2)&=1\\\varphi (m,0,p)&=m&&{\text{for }}p>2\\\varphi (m,n,p)&=\varphi (m,\varphi (m,n-1,p),p-1)&&{\text{for }}n,p>0\end{aligned}}}$$

Of the various two-argument versions, the one developed by Péter and Robinson (called "the" Ackermann function by most authors) is defined for nonnegative integers ${\displaystyle m}$ and ${\displaystyle n}$ as follows:

$${\displaystyle {\begin{array}{lcl}\operatorname {A} (0,n)&=&n+1\\\operatorname {A} (m+1,0)&=&\operatorname {A} (m,1)\\\operatorname {A} (m+1,n+1)&=&\operatorname {A} (m,\operatorname {A} (m+1,n))\end{array}}}$$

The Ackermann function has also been expressed in relation to the hyperoperation sequence:^[14][15]

$${\displaystyle A(m,n)={\begin{cases}n+1&m=0\\2[m](n+3)-3&m>0\\\end{cases}}}$$

or, written in Knuth's up-arrow notation (extended to integer indices ${\displaystyle \geq -2}$):

$${\displaystyle A(m,n)={\begin{cases}n+1&m=0\\2\uparrow ^{m-2}(n+3)-3&m>0\\\end{cases}}}$$

or, equivalently, in terms of Buck's function F:^[10]

$${\displaystyle A(m,n)={\begin{cases}n+1&m=0\\F(m,n+3)-3&m>0\\\end{cases}}}$$

Define ${\displaystyle f^{n}}$ as the n-th iterate of ${\displaystyle f}$:

$${\displaystyle {\begin{array}{rll}f^{0}(x)&=&x\\f^{n+1}(x)&=&f(f^{n}(x))\end{array}}}$$

Iteration is the process of composing a function with itself a certain number of times. Function composition is an associative operation, so ${\displaystyle f(f^{n}(x))=f^{n}(f(x))}$.

Conceiving the Ackermann function as a sequence of unary functions, one can set ${\displaystyle \operatorname {A} _{m}(n)=\operatorname {A} (m,n)}$.

The function then becomes a sequence ${\displaystyle \operatorname {A} _{0},\operatorname {A} _{1},\operatorname {A} _{2},...}$ of unary^{[n 2]} functions, defined from iteration:

$${\displaystyle {\begin{array}{lcl}\operatorname {A} _{0}(n)&=&n+1\\\operatorname {A} _{m+1}(n)&=&\operatorname {A} _{m}^{n+1}(1)\\\end{array}}}$$

The recursive definition of the Ackermann function can naturally be transposed to a term rewriting system (TRS).

The definition of the 2-ary Ackermann function leads to the obvious reduction rules^[16][17]

$${\displaystyle {\begin{array}{lll}{\text{(r1)}}&A(0,n)&\rightarrow &S(n)\\{\text{(r2)}}&A(S(m),0)&\rightarrow &A(m,S(0))\\{\text{(r3)}}&A(S(m),S(n))&\rightarrow &A(m,A(S(m),n))\end{array}}}$$

Example

Compute ${\displaystyle A(1,2)\rightarrow _{*}4}$

The reduction sequence is ^{[n 3]}

Leftmost-outermost (one-step) strategy:	Leftmost-innermost (one-step) strategy:
${\displaystyle {\underline {A(S(0),S(S(0)))}}}$	${\displaystyle {\underline {A(S(0),S(S(0)))}}}$
${\displaystyle \rightarrow _{r3}{\underline {A(0,A(S(0),S(0))}})}$	${\displaystyle \rightarrow _{r3}A(0,{\underline {A(S(0),S(0))}})}$
${\displaystyle \rightarrow _{r1}S({\underline {A(S(0),S(0))}})}$	${\displaystyle \rightarrow _{r3}A(0,A(0,{\underline {A(S(0),0)}}))}$
${\displaystyle \rightarrow _{r3}S({\underline {A(0,A(S0,0))}})}$	${\displaystyle \rightarrow _{r2}A(0,A(0,{\underline {A(0,S(0))}}))}$
${\displaystyle \rightarrow _{r1}S(S({\underline {A(S(0),0)}}))}$	${\displaystyle \rightarrow _{r1}A(0,{\underline {A(0,S(S(0)))}})}$
${\displaystyle \rightarrow _{r2}S(S({\underline {A(0,S(0))}}))}$	${\displaystyle \rightarrow _{r1}{\underline {A(0,S(S(S(0))))}}}$
${\displaystyle \rightarrow _{r1}S(S(S(S(0))))}$	${\displaystyle \rightarrow _{r1}S(S(S(S(0))))}$

To compute ${\displaystyle \operatorname {A} (m,n)}$ one can use a stack, which initially contains the elements ${\displaystyle \langle m,n\rangle }$.

Then repeatedly the two top elements are replaced according to the rules^{[n 4]}

$${\displaystyle {\begin{array}{lllllllll}{\text{(r1)}}&0&,&n&\rightarrow &(n+1)\\{\text{(r2)}}&(m+1)&,&0&\rightarrow &m&,&1\\{\text{(r3)}}&(m+1)&,&(n+1)&\rightarrow &m&,&(m+1)&,&n\end{array}}}$$

Schematically, starting from ${\displaystyle \langle m,n\rangle }$:

WHILE stackLength <> 1 {    POP 2 elements;    PUSH 1 or 2 or 3 elements, applying the rules r1, r2, r3 } 

The pseudocode is published in Grossman & Zeitman (1988).

For example, on input ${\displaystyle \langle 2,1\rangle }$,

the stack configurations	reflect the reduction[n 5]
${\displaystyle {\underline {2,1}}}$	${\displaystyle {\underline {A(2,1)}}}$
${\displaystyle \rightarrow 1,{\underline {2,0}}}$	${\displaystyle \rightarrow _{r1}A(1,{\underline {A(2,0)}})}$
${\displaystyle \rightarrow 1,{\underline {1,1}}}$	${\displaystyle \rightarrow _{r2}A(1,{\underline {A(1,1)}})}$
${\displaystyle \rightarrow 1,0,{\underline {1,0}}}$	${\displaystyle \rightarrow _{r3}A(1,A(0,{\underline {A(1,0)}}))}$
${\displaystyle \rightarrow 1,0,{\underline {0,1}}}$	${\displaystyle \rightarrow _{r2}A(1,A(0,{\underline {A(0,1)}}))}$
${\displaystyle \rightarrow 1,{\underline {0,2}}}$	${\displaystyle \rightarrow _{r1}A(1,{\underline {A(0,2)}})}$
${\displaystyle \rightarrow {\underline {1,3}}}$	${\displaystyle \rightarrow _{r1}{\underline {A(1,3)}}}$
${\displaystyle \rightarrow 0,{\underline {1,2}}}$	${\displaystyle \rightarrow _{r3}A(0,{\underline {A(1,2)}})}$
${\displaystyle \rightarrow 0,0,{\underline {1,1}}}$	${\displaystyle \rightarrow _{r3}A(0,A(0,{\underline {A(1,1)}}))}$
${\displaystyle \rightarrow 0,0,0,{\underline {1,0}}}$	${\displaystyle \rightarrow _{r3}A(0,A(0,A(0,{\underline {A(1,0)}})))}$
${\displaystyle \rightarrow 0,0,0,{\underline {0,1}}}$	${\displaystyle \rightarrow _{r2}A(0,A(0,A(0,{\underline {A(0,1)}})))}$
${\displaystyle \rightarrow 0,0,{\underline {0,2}}}$	${\displaystyle \rightarrow _{r1}A(0,A(0,{\underline {A(0,2)}}))}$
${\displaystyle \rightarrow 0,{\underline {0,3}}}$	${\displaystyle \rightarrow _{r1}A(0,{\underline {A(0,3)}})}$
${\displaystyle \rightarrow {\underline {0,4}}}$	${\displaystyle \rightarrow _{r1}{\underline {A(0,4)}}}$
${\displaystyle \rightarrow 5}$	${\displaystyle \rightarrow _{r1}5}$

Remarks

• The leftmost-innermost strategy is implemented in 225 computer languages on Rosetta Code.
• For all ${\displaystyle m,n}$ the computation of ${\displaystyle A(m,n)}$ takes no more than ${\displaystyle (A(m,n)+1)^{m}}$ steps.^[18]
• Grossman & Zeitman (1988) pointed out that in the computation of ${\displaystyle \operatorname {A} (m,n)}$ the maximum length of the stack is ${\displaystyle \operatorname {A} (m,n)}$, as long as ${\displaystyle m>0}$.

  Their own algorithm, inherently iterative, computes ${\displaystyle \operatorname {A} (m,n)}$ within ${\displaystyle {\mathcal {O}}(m\operatorname {A} (m,n))}$ time and within ${\displaystyle {\mathcal {O}}(m)}$ space.

The definition of the iterated 1-ary Ackermann functions leads to different reduction rules

$${\displaystyle {\begin{array}{lll}{\text{(r4)}}&A(S(0),0,n)&\rightarrow &S(n)\\{\text{(r5)}}&A(S(0),S(m),n)&\rightarrow &A(S(n),m,S(0))\\{\text{(r6)}}&A(S(S(x)),m,n)&\rightarrow &A(S(0),m,A(S(x),m,n))\end{array}}}$$

As function composition is associative, instead of rule r6 one can define

$${\displaystyle {\begin{array}{lll}{\text{(r7)}}&A(S(S(x)),m,n)&\rightarrow &A(S(x),m,A(S(0),m,n))\end{array}}}$$

Like in the previous section the computation of ${\displaystyle \operatorname {A} _{m}^{1}(n)}$ can be implemented with a stack.

Initially the stack contains the three elements ${\displaystyle \langle 1,m,n\rangle }$.

Then repeatedly the three top elements are replaced according to the rules^{[n 4]}

$${\displaystyle {\begin{array}{lllllllll}{\text{(r4)}}&1&,0&,n&\rightarrow &(n+1)\\{\text{(r5)}}&1&,(m+1)&,n&\rightarrow &(n+1)&,m&,1\\{\text{(r6)}}&(x+2)&,m&,n&\rightarrow &1&,m&,(x+1)&,m&,n\\\end{array}}}$$

Schematically, starting from ${\displaystyle \langle 1,m,n\rangle }$:

WHILE stackLength <> 1 {    POP 3 elements;    PUSH 1 or 3 or 5 elements, applying the rules r4, r5, r6; } 

Example

On input ${\displaystyle \langle 1,2,1\rangle }$ the successive stack configurations are

$${\displaystyle {\begin{aligned}&{\underline {1,2,1}}\rightarrow _{r5}{\underline {2,1,1}}\rightarrow _{r6}1,1,{\underline {1,1,1}}\rightarrow _{r5}1,1,{\underline {2,0,1}}\rightarrow _{r6}1,1,1,0,{\underline {1,0,1}}\\&\rightarrow _{r4}1,1,{\underline {1,0,2}}\rightarrow _{r4}{\underline {1,1,3}}\rightarrow _{r5}{\underline {4,0,1}}\rightarrow _{r6}1,0,{\underline {3,0,1}}\rightarrow _{r6}1,0,1,0,{\underline {2,0,1}}\\&\rightarrow _{r6}1,0,1,0,1,0,{\underline {1,0,1}}\rightarrow _{r4}1,0,1,0,{\underline {1,0,2}}\rightarrow _{r4}1,0,{\underline {1,0,3}}\rightarrow _{r4}{\underline {1,0,4}}\rightarrow _{r4}5\end{aligned}}}$$

The corresponding equalities are

$${\displaystyle {\begin{aligned}&A_{2}(1)=A_{1}^{2}(1)=A_{1}(A_{1}(1))=A_{1}(A_{0}^{2}(1))=A_{1}(A_{0}(A_{0}(1)))\\&=A_{1}(A_{0}(2))=A_{1}(3)=A_{0}^{4}(1)=A_{0}(A_{0}^{3}(1))=A_{0}(A_{0}(A_{0}^{2}(1)))\\&=A_{0}(A_{0}(A_{0}(A_{0}(1))))=A_{0}(A_{0}(A_{0}(2)))=A_{0}(A_{0}(3))=A_{0}(4)=5\end{aligned}}}$$

When reduction rule r7 is used instead of rule r6, the replacements in the stack will follow

$${\displaystyle {\begin{array}{lllllllll}{\text{(r7)}}&(x+2)&,m&,n&\rightarrow &(x+1)&,m&,1&,m&,n\end{array}}}$$

The successive stack configurations will then be

$${\displaystyle {\begin{aligned}&{\underline {1,2,1}}\rightarrow _{r5}{\underline {2,1,1}}\rightarrow _{r7}1,1,{\underline {1,1,1}}\rightarrow _{r5}1,1,{\underline {2,0,1}}\rightarrow _{r7}1,1,1,0,{\underline {1,0,1}}\\&\rightarrow _{r4}1,1,{\underline {1,0,2}}\rightarrow _{r4}{\underline {1,1,3}}\rightarrow _{r5}{\underline {4,0,1}}\rightarrow _{r7}3,0,{\underline {1,0,1}}\rightarrow _{r4}{\underline {3,0,2}}\\&\rightarrow _{r7}2,0,{\underline {1,0,2}}\rightarrow _{r4}{\underline {2,0,3}}\rightarrow _{r7}1,0,{\underline {1,0,3}}\rightarrow _{r4}{\underline {1,0,4}}\rightarrow _{r4}5\end{aligned}}}$$

The corresponding equalities are

$${\displaystyle {\begin{aligned}&A_{2}(1)=A_{1}^{2}(1)=A_{1}(A_{1}(1))=A_{1}(A_{0}^{2}(1))=A_{1}(A_{0}(A_{0}(1)))\\&=A_{1}(A_{0}(2))=A_{1}(3)=A_{0}^{4}(1)=A_{0}^{3}(A_{0}(1))=A_{0}^{3}(2)\\&=A_{0}^{2}(A_{0}(2))=A_{0}^{2}(3)=A_{0}(A_{0}(3))=A_{0}(4)=5\end{aligned}}}$$

Remarks

• On any given input the TRSs presented so far converge in the same number of steps. They also use the same reduction rules (in this comparison the rules r1, r2, r3 are considered "the same as" the rules r4, r5, r6/r7 respectively). For example, the reduction of ${\displaystyle A(2,1)}$ converges in 14 steps: 6 × r1, 3 × r2, 5 × r3. The reduction of ${\displaystyle A_{2}(1)}$ converges in the same 14 steps: 6 × r4, 3 × r5, 5 × r6/r7. The TRSs differ in the order in which the reduction rules are applied.
• When ${\displaystyle A_{i}(n)}$ is computed following the rules {r4, r5, r6}, the maximum length of the stack stays below ${\displaystyle 2\times A(i,n)}$. When reduction rule r7 is used instead of rule r6, the maximum length of the stack is only ${\displaystyle 2(i+2)}$. The length of the stack reflects the recursion depth. As the reduction according to the rules {r4, r5, r7} involves a smaller maximum depth of recursion,^{[n 6]} this computation is more efficient in that respect.

As Sundblad (1971) — or Porto & Matos (1980) — showed explicitly, the Ackermann function can be expressed in terms of the hyperoperation sequence:

$${\displaystyle A(m,n)={\begin{cases}n+1&m=0\\2[m](n+3)-3&m>0\\\end{cases}}}$$

or, after removal of the constant 2 from the parameter list, in terms of Buck's function

${\displaystyle A(m,n)={\begin{cases}n+1&m=0\\F(m,n+3)-3&m>0\\\end{cases}}}$

Buck's function ${\displaystyle \operatorname {F} (m,n)=2[m]n}$,^[10] a variant of Ackermann function by itself, can be computed with the following reduction rules:

$${\displaystyle {\begin{array}{lll}{\text{(b1)}}&F(S(0),0,n)&\rightarrow &S(n)\\{\text{(b2)}}&F(S(0),S(0),0)&\rightarrow &S(S(0))\\{\text{(b3)}}&F(S(0),S(S(0)),0)&\rightarrow &0\\{\text{(b4)}}&F(S(0),S(S(S(m))),0)&\rightarrow &S(0)\\{\text{(b5)}}&F(S(0),S(m),S(n))&\rightarrow &F(S(n),m,F(S(0),S(m),0))\\{\text{(b6)}}&F(S(S(x)),m,n)&\rightarrow &F(S(0),m,F(S(x),m,n))\end{array}}}$$ Instead of rule b6 one can define the rule

$${\displaystyle {\begin{array}{lll}{\text{(b7)}}&F(S(S(x)),m,n)&\rightarrow &F(S(x),m,F(S(0),m,n))\end{array}}}$$ To compute the Ackermann function it suffices to add three reduction rules

$${\displaystyle {\begin{array}{lll}{\text{(r8)}}&A(0,n)&\rightarrow &S(n)\\{\text{(r9)}}&A(S(m),n)&\rightarrow &P(F(S(0),S(m),S(S(S(n)))))\\{\text{(r10)}}&P(S(S(S(m))))&\rightarrow &m\\\end{array}}}$$

These rules take care of the base case A(0,n), the alignment (n+3) and the fudge (-3).

Example

Compute ${\displaystyle A(2,1)\rightarrow _{*}5}$

using reduction rule ${\displaystyle {\text{b7}}}$:[n 5]	using reduction rule ${\displaystyle {\text{b6}}}$:[n 5]
${\displaystyle {\underline {A(2,1)}}}$	${\displaystyle {\underline {A(2,1)}}}$
${\displaystyle \rightarrow _{r9}P({\underline {F(1,2,4)}})}$	${\displaystyle \rightarrow _{r9}P({\underline {F(1,2,4)}})}$
${\displaystyle \rightarrow _{b5}P(F(4,1,{\underline {F(1,2,0)}}))}$	${\displaystyle \rightarrow _{b5}P(F(4,1,{\underline {F(1,2,0)}}))}$
${\displaystyle \rightarrow _{b3}P({\underline {F(4,1,0)}})}$	${\displaystyle \rightarrow _{b3}P({\underline {F(4,1,0)}})}$
${\displaystyle \rightarrow _{b7}P(F(3,1,{\underline {F(1,1,0)}}))}$	${\displaystyle \rightarrow _{b6}P(F(1,1,{\underline {F(3,1,0)}}))}$
${\displaystyle \rightarrow _{b2}P({\underline {F(3,1,2)}})}$	${\displaystyle \rightarrow _{b6}P(F(1,1,F(1,1,{\underline {F(2,1,0)}})))}$
${\displaystyle \rightarrow _{b7}P(F(2,1,{\underline {F(1,1,2)}}))}$	${\displaystyle \rightarrow _{b6}P(F(1,1,F(1,1,F(1,1,{\underline {F(1,1,0)}}))))}$
${\displaystyle \rightarrow _{b5}P(F(2,1,F(2,0,{\underline {F(1,1,0)}})))}$	${\displaystyle \rightarrow _{b2}P(F(1,1,F(1,1,{\underline {F(1,1,2)}})))}$
${\displaystyle \rightarrow _{b2}P(F(2,1,{\underline {F(2,0,2)}}))}$	${\displaystyle \rightarrow _{b5}P(F(1,1,F(1,1,F(2,0,{\underline {F(1,1,0)}}))))}$
${\displaystyle \rightarrow _{b7}P(F(2,1,F(1,0,{\underline {F(1,0,2)}})))}$	${\displaystyle \rightarrow _{b2}P(F(1,1,F(1,1,{\underline {F(2,0,2)}})))}$
${\displaystyle \rightarrow _{b1}P(F(2,1,{\underline {F(1,0,3)}}))}$	${\displaystyle \rightarrow _{b6}P(F(1,1,F(1,1,F(1,0,{\underline {F(1,0,2)}}))))}$
${\displaystyle \rightarrow _{b1}P({\underline {F(2,1,4)}})}$	${\displaystyle \rightarrow _{b1}P(F(1,1,F(1,1,{\underline {F(1,0,3)}})))}$
${\displaystyle \rightarrow _{b7}P(F(1,1,{\underline {F(1,1,4)}}))}$	${\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(1,1,4)}}))}$
${\displaystyle \rightarrow _{b5}P(F(1,1,F(4,0,{\underline {F(1,1,0)}})))}$	${\displaystyle \rightarrow _{b5}P(F(1,1,F(4,0,{\underline {F(1,1,0)}})))}$
${\displaystyle \rightarrow _{b2}P(F(1,1,{\underline {F(4,0,2)}}))}$	${\displaystyle \rightarrow _{b2}P(F(1,1,{\underline {F(4,0,2)}}))}$
${\displaystyle \rightarrow _{b7}P(F(1,1,F(3,0,{\underline {F(1,0,2)}})))}$	${\displaystyle \rightarrow _{b6}P(F(1,1,F(1,0,{\underline {F(3,0,2)}})))}$
${\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(3,0,3)}}))}$	${\displaystyle \rightarrow _{b6}P(F(1,1,F(1,0,F(1,0,{\underline {F(2,0,2)}}))))}$
${\displaystyle \rightarrow _{b7}P(F(1,1,F(2,0,{\underline {F(1,0,3)}})))}$	${\displaystyle \rightarrow _{b6}P(F(1,1,F(1,0,F(1,0,F(1,0,{\underline {F(1,0,2)}})))))}$
${\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(2,0,4)}}))}$	${\displaystyle \rightarrow _{b1}P(F(1,1,F(1,0,F(1,0,{\underline {F(1,0,3)}}))))}$
${\displaystyle \rightarrow _{b7}P(F(1,1,F(1,0,{\underline {F(1,0,4)}})))}$	${\displaystyle \rightarrow _{b1}P(F(1,1,F(1,0,{\underline {F(1,0,4)}})))}$
${\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(1,0,5)}}))}$	${\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(1,0,5)}}))}$
${\displaystyle \rightarrow _{b1}P({\underline {F(1,1,6)}})}$	${\displaystyle \rightarrow _{b1}P({\underline {F(1,1,6)}})}$
${\displaystyle \rightarrow _{b5}P(F(6,0,{\underline {F(1,1,0)}}))}$	${\displaystyle \rightarrow _{b5}P(F(6,0,{\underline {F(1,1,0)}}))}$
${\displaystyle \rightarrow _{b2}P({\underline {F(6,0,2)}})}$	${\displaystyle \rightarrow _{b2}P({\underline {F(6,0,2)}})}$
${\displaystyle \rightarrow _{b7}P(F(5,0,{\underline {F(1,0,2)}}))}$	${\displaystyle \rightarrow _{b6}P(F(1,0,{\underline {F(5,0,2)}}))}$
${\displaystyle \rightarrow _{b1}P({\underline {F(5,0,3)}})}$	${\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,{\underline {F(4,0,2)}})))}$
${\displaystyle \rightarrow _{b7}P(F(4,0,{\underline {F(1,0,3)}}))}$	${\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,F(1,0,{\underline {F(3,0,2)}}))))}$
${\displaystyle \rightarrow _{b1}P({\underline {F(4,0,4)}})}$	${\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,F(1,0,F(1,0,{\underline {F(2,0,2)}})))))}$
${\displaystyle \rightarrow _{b7}P(F(3,0,{\underline {F(1,0,4)}}))}$	${\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,F(1,0,F(1,0,F(1,0,{\underline {F(1,0,2)}}))))))}$
${\displaystyle \rightarrow _{b1}P({\underline {F(3,0,5)}})}$	${\displaystyle \rightarrow _{b1}P(F(1,0,F(1,0,F(1,0,F(1,0,{\underline {F(1,0,3)}})))))}$
${\displaystyle \rightarrow _{b7}P(F(2,0,{\underline {F(1,0,5)}}))}$	${\displaystyle \rightarrow _{b1}P(F(1,0,F(1,0,F(1,0,{\underline {F(1,0,4)}}))))}$
${\displaystyle \rightarrow _{b1}P({\underline {F(2,0,6)}})}$	${\displaystyle \rightarrow _{b1}P(F(1,0,F(1,0,{\underline {F(1,0,5)}})))}$
${\displaystyle \rightarrow _{b7}P(F(1,0,{\underline {F(1,0,6)}}))}$	${\displaystyle \rightarrow _{b1}P(F(1,0,{\underline {F(1,0,6)}}))}$
${\displaystyle \rightarrow _{b1}P({\underline {F(1,0,7)}})}$	${\displaystyle \rightarrow _{b1}P({\underline {F(1,0,7)}})}$
${\displaystyle \rightarrow _{b1}{\underline {P(8)}}}$	${\displaystyle \rightarrow _{b1}{\underline {P(8)}}}$
${\displaystyle \rightarrow _{r10}5}$	${\displaystyle \rightarrow _{r10}5}$

The matching equalities are

• when the TRS with the reduction rule ${\displaystyle {\text{b6}}}$ is applied:

$${\displaystyle {\begin{aligned}&A(2,1)+3=F(2,4)=\dots =F^{6}(0,2)=F(0,F^{5}(0,2))=F(0,F(0,F^{4}(0,2)))\\&=F(0,F(0,F(0,F^{3}(0,2))))=F(0,F(0,F(0,F(0,F^{2}(0,2)))))=F(0,F(0,F(0,F(0,F(0,F(0,2))))))\\&=F(0,F(0,F(0,F(0,F(0,3)))))=F(0,F(0,F(0,F(0,4))))=F(0,F(0,F(0,5)))=F(0,F(0,6))=F(0,7)=8\end{aligned}}}$$

• when the TRS with the reduction rule ${\displaystyle {\text{b7}}}$ is applied:

$${\displaystyle {\begin{aligned}&A(2,1)+3=F(2,4)=\dots =F^{6}(0,2)=F^{5}(0,F(0,2))=F^{5}(0,3)=F^{4}(0,F(0,3))=F^{4}(0,4)\\&=F^{3}(0,F(0,4))=F^{3}(0,5)=F^{2}(0,F(0,5))=F^{2}(0,6)=F(0,F(0,6))=F(0,7)=8\end{aligned}}}$$ Remarks