We consider the system 02Ackermann.

  Alphabet:

    ack : [N * N] --> N 
    s : [N] --> N 
    z : [] --> N 

  Rules:

    ack(z, x) => s(x) 
    ack(s(x), z) => ack(x, s(z)) 
    ack(s(x), s(y)) => ack(x, ack(s(x), y)) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  ack(z, X) >? s(X) 
  ack(s(X), z) >? ack(X, s(z)) 
  ack(s(X), s(Y)) >? ack(X, ack(s(X), Y)) 

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {ack} and Mul = {s, z}, and the following precedence: ack > s > z

With these choices, we have:

  1] ack(z, X) >= s(X)  because [2], by (Star) 
  2] ack*(z, X) >= s(X)  because ack > s and [3], by (Copy) 
  3] ack*(z, X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] ack(s(X), z) > ack(X, s(z))  because [6], by definition 
  6] ack*(s(X), z) >= ack(X, s(z))  because [7], [10] and [12], by (Stat) 
  7] s(X) > X  because [8], by definition 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] ack*(s(X), z) >= X  because [11], by (Select) 
  11] s(X) >= X  because [8], by (Star) 
  12] ack*(s(X), z) >= s(z)  because ack > s and [13], by (Copy) 
  13] ack*(s(X), z) >= z  because [14], by (Select) 
  14] s(X) >= z  because [15], by (Star) 
  15] s*(X) >= z  because s > z, by (Copy) 

  16] ack(s(X), s(Y)) >= ack(X, ack(s(X), Y))  because [17], by (Star) 
  17] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y))  because [18], [21] and [23], by (Stat) 
  18] s(X) > X  because [19], by definition 
  19] s*(X) >= X  because [20], by (Select) 
  20] X >= X  by (Meta) 
  21] ack*(s(X), s(Y)) >= X  because [22], by (Select) 
  22] s(X) >= X  because [19], by (Star) 
  23] ack*(s(X), s(Y)) >= ack(s(X), Y)  because [24], [26], [29] and [30], by (Stat) 
  24] s(X) >= s(X)  because s in Mul and [25], by (Fun) 
  25] X >= X  by (Meta) 
  26] s(Y) > Y  because [27], by definition 
  27] s*(Y) >= Y  because [28], by (Select) 
  28] Y >= Y  by (Meta) 
  29] ack*(s(X), s(Y)) >= s(X)  because [24], by (Select) 
  30] ack*(s(X), s(Y)) >= Y  because [31], by (Select) 
  31] s(Y) >= Y  because [27], by (Star) 

We can thus remove the following rules:

  ack(s(X), z) => ack(X, s(z)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  ack(z, X) >? s(X) 
  ack(s(X), s(Y)) >? ack(X, ack(s(X), Y)) 

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {ack} and Mul = {s, z}, and the following precedence: z > ack > s

With these choices, we have:

  1] ack(z, X) >= s(X)  because [2], by (Star) 
  2] ack*(z, X) >= s(X)  because ack > s and [3], by (Copy) 
  3] ack*(z, X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] ack(s(X), s(Y)) > ack(X, ack(s(X), Y))  because [6], by definition 
  6] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y))  because [7], [10] and [12], by (Stat) 
  7] s(X) > X  because [8], by definition 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] ack*(s(X), s(Y)) >= X  because [11], by (Select) 
  11] s(X) >= X  because [8], by (Star) 
  12] ack*(s(X), s(Y)) >= ack(s(X), Y)  because [13], [15], [18] and [19], by (Stat) 
  13] s(X) >= s(X)  because s in Mul and [14], by (Fun) 
  14] X >= X  by (Meta) 
  15] s(Y) > Y  because [16], by definition 
  16] s*(Y) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 
  18] ack*(s(X), s(Y)) >= s(X)  because [13], by (Select) 
  19] ack*(s(X), s(Y)) >= Y  because [20], by (Select) 
  20] s(Y) >= Y  because [16], by (Star) 

We can thus remove the following rules:

  ack(s(X), s(Y)) => ack(X, ack(s(X), Y)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  ack(z, X) >? s(X) 

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {ack, s, z}, and the following precedence: ack = s > z

With these choices, we have:

  1] ack(z, X) > s(X)  because [2], by definition 
  2] ack*(z, X) >= s(X)  because ack = s, ack in Mul and [3], by (Stat) 
  3] X >= X  by (Meta) 

We can thus remove the following rules:

  ack(z, X) => s(X) 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.