/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  ack : [o * o] --> o 
  s : [o] --> o 

  ack(0, X) => s(X) 
  ack(s(X), 0) => ack(X, s(0)) 
  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(0, X) >? s(X) 
  ack(s(X), 0) >? ack(X, s(0)) 
  ack(s(X), s(Y)) >? ack(X, ack(s(X), Y)) 

about to try horpo

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

We choose Lex = {ack} and Mul = {0, s}, and the following precedence: ack > s > 0

With these choices, we have:

  1] ack(0, X) >= s(X)  because [2], by (Star) 
  2] ack*(0, X) >= s(X)  because ack > s and [3], by (Copy) 
  3] ack*(0, X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] ack(s(X), 0) >= ack(X, s(0))  because [6], by (Star) 
  6] ack*(s(X), 0) >= ack(X, s(0))  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), 0) >= X  because [11], by (Select) 
  11] s(X) >= X  because [8], by (Star) 
  12] ack*(s(X), 0) >= s(0)  because ack > s and [13], by (Copy) 
  13] ack*(s(X), 0) >= 0  because [14], by (Select) 
  14] s(X) >= 0  because [15], by (Star) 
  15] s*(X) >= 0  because s > 0, by (Copy) 

  16] ack(s(X), s(Y)) > ack(X, ack(s(X), Y))  because [17], by definition 
  17] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y))  because [7], [18] and [19], by (Stat) 
  18] ack*(s(X), s(Y)) >= X  because [11], by (Select) 
  19] ack*(s(X), s(Y)) >= ack(s(X), Y)  because [20], [22], [24] and [25], by (Stat) 
  20] s(X) >= s(X)  because s in Mul and [21], by (Fun) 
  21] X >= X  by (Meta) 
  22] s(Y) > Y  because [23], by definition 
  23] s*(Y) >= Y  because [4], by (Select) 
  24] ack*(s(X), s(Y)) >= s(X)  because [20], by (Select) 
  25] ack*(s(X), s(Y)) >= Y  because [26], by (Select) 
  26] s(Y) >= Y  because [23], 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(0, X) >? s(X) 
  ack(s(X), 0) >? ack(X, s(0)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  ack = \y0y1.3 + y0 + y1 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[ack(0, _x0)]] = 3 + x0 > x0 = [[s(_x0)]] 
  [[ack(s(_x0), 0)]] = 3 + x0 >= 3 + x0 = [[ack(_x0, s(0))]] 

We can thus remove the following rules:

  ack(0, X) => s(X) 

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(s(X), 0) >? ack(X, s(0)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  ack = \y0y1.2y1 + 3y0 
  s = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[ack(s(_x0), 0)]] = 6 + 3x0 > 4 + 3x0 = [[ack(_x0, s(0))]] 

We can thus remove the following rules:

  ack(s(X), 0) => ack(X, s(0)) 

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.