/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 
  activate : [o] --> o 
  cons : [o * o] --> o 
  f : [o] --> o 
  n!6220!62200 : [] --> o 
  n!6220!6220f : [o] --> o 
  n!6220!6220s : [o] --> o 
  p : [o] --> o 
  s : [o] --> o 

  f(0) => cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) 
  f(s(0)) => f(p(s(0))) 
  p(s(X)) => X 
  f(X) => n!6220!6220f(X) 
  s(X) => n!6220!6220s(X) 
  0 => n!6220!62200 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(n!6220!62200) => 0 
  activate(X) => 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]):

  f(0) >? cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) 
  f(s(0)) >? f(p(s(0))) 
  p(s(X)) >? X 
  f(X) >? n!6220!6220f(X) 
  s(X) >? n!6220!6220s(X) 
  0 >? n!6220!62200 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220s(X)) >? s(activate(X)) 
  activate(n!6220!62200) >? 0 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.1 + 2y0 
  cons = \y0y1.y0 + y1 
  f = \y0.y0 
  n!6220!62200 = 0 
  n!6220!6220f = \y0.y0 
  n!6220!6220s = \y0.y0 
  p = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(0)]] = 0 >= 0 = [[cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200)))]] 
  [[f(s(0))]] = 0 >= 0 = [[f(p(s(0)))]] 
  [[p(s(_x0))]] = x0 >= x0 = [[_x0]] 
  [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[0]] = 0 >= 0 = [[n!6220!62200]] 
  [[activate(n!6220!6220f(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[s(activate(_x0))]] 
  [[activate(n!6220!62200)]] = 1 > 0 = [[0]] 
  [[activate(_x0)]] = 1 + 2x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  activate(n!6220!62200) => 0 
  activate(X) => 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]):

  f(0) >? cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) 
  f(s(0)) >? f(p(s(0))) 
  p(s(X)) >? X 
  f(X) >? n!6220!6220f(X) 
  s(X) >? n!6220!6220s(X) 
  0 >? n!6220!62200 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220s(X)) >? s(activate(X)) 

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

The following interpretation satisfies the requirements:

  0 = 1 
  activate = \y0.y0 
  cons = \y0y1.y0 + y1 
  f = \y0.y0 
  n!6220!62200 = 0 
  n!6220!6220f = \y0.y0 
  n!6220!6220s = \y0.y0 
  p = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(0)]] = 1 >= 1 = [[cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200)))]] 
  [[f(s(0))]] = 1 >= 1 = [[f(p(s(0)))]] 
  [[p(s(_x0))]] = x0 >= x0 = [[_x0]] 
  [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[0]] = 1 > 0 = [[n!6220!62200]] 
  [[activate(n!6220!6220f(_x0))]] = x0 >= x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(activate(_x0))]] 

We can thus remove the following rules:

  0 => n!6220!62200 

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

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

  f(0) >? cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) 
  f(s(0)) >? f(p(s(0))) 
  p(s(X)) >? X 
  f(X) >? n!6220!6220f(X) 
  s(X) >? n!6220!6220s(X) 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220s(X)) >? s(activate(X)) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.2 + 2y0 
  cons = \y0y1.y0 + y1 
  f = \y0.2 + y0 
  n!6220!62200 = 0 
  n!6220!6220f = \y0.1 + y0 
  n!6220!6220s = \y0.1 + y0 
  p = \y0.y0 
  s = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[f(0)]] = 2 >= 2 = [[cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200)))]] 
  [[f(s(0))]] = 4 >= 4 = [[f(p(s(0)))]] 
  [[p(s(_x0))]] = 2 + x0 > x0 = [[_x0]] 
  [[f(_x0)]] = 2 + x0 > 1 + x0 = [[n!6220!6220f(_x0)]] 
  [[s(_x0)]] = 2 + x0 > 1 + x0 = [[n!6220!6220s(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 4 + 2x0 >= 4 + 2x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = 4 + 2x0 >= 4 + 2x0 = [[s(activate(_x0))]] 

We can thus remove the following rules:

  p(s(X)) => X 
  f(X) => n!6220!6220f(X) 
  s(X) => n!6220!6220s(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]):

  f(0) >? cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) 
  f(s(0)) >? f(p(s(0))) 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220s(X)) >? s(activate(X)) 

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

The following interpretation satisfies the requirements:

  0 = 2 
  activate = \y0.2 + 3y0 
  cons = \y0y1.y0 + y1 
  f = \y0.3 + 2y0 
  n!6220!62200 = 0 
  n!6220!6220f = \y0.2 + 2y0 
  n!6220!6220s = \y0.1 + 3y0 
  p = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(0)]] = 7 > 6 = [[cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200)))]] 
  [[f(s(0))]] = 7 >= 7 = [[f(p(s(0)))]] 
  [[activate(n!6220!6220f(_x0))]] = 8 + 6x0 > 7 + 6x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = 5 + 9x0 > 2 + 3x0 = [[s(activate(_x0))]] 

We can thus remove the following rules:

  f(0) => cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] f#(s(0)) =#> f#(p(s(0)))   

  Rules R_0:

    f(s(0)) => f(p(s(0))) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.