/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.

  activate : [o] --> o 
  c : [o] --> o 
  d : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  h : [o] --> o 
  n!6220!6220d : [o] --> o 
  n!6220!6220f : [o] --> o 
  n!6220!6220g : [o] --> o 

  f(f(X)) => c(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) 
  c(X) => d(activate(X)) 
  h(X) => c(n!6220!6220d(X)) 
  f(X) => n!6220!6220f(X) 
  g(X) => n!6220!6220g(X) 
  d(X) => n!6220!6220d(X) 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(n!6220!6220g(X)) => g(X) 
  activate(n!6220!6220d(X)) => d(X) 
  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(f(X)) >? c(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) 
  c(X) >? d(activate(X)) 
  h(X) >? c(n!6220!6220d(X)) 
  f(X) >? n!6220!6220f(X) 
  g(X) >? n!6220!6220g(X) 
  d(X) >? n!6220!6220d(X) 
  activate(n!6220!6220f(X)) >? f(activate(X)) 
  activate(n!6220!6220g(X)) >? g(X) 
  activate(n!6220!6220d(X)) >? d(X) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  activate = \y0.y0 
  c = \y0.y0 
  d = \y0.y0 
  f = \y0.y0 
  g = \y0.y0 
  h = \y0.3 + 3y0 
  n!6220!6220d = \y0.y0 
  n!6220!6220f = \y0.y0 
  n!6220!6220g = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(f(_x0))]] = x0 >= x0 = [[c(n!6220!6220f(n!6220!6220g(n!6220!6220f(_x0))))]] 
  [[c(_x0)]] = x0 >= x0 = [[d(activate(_x0))]] 
  [[h(_x0)]] = 3 + 3x0 > x0 = [[c(n!6220!6220d(_x0))]] 
  [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] 
  [[g(_x0)]] = x0 >= x0 = [[n!6220!6220g(_x0)]] 
  [[d(_x0)]] = x0 >= x0 = [[n!6220!6220d(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = x0 >= x0 = [[f(activate(_x0))]] 
  [[activate(n!6220!6220g(_x0))]] = x0 >= x0 = [[g(_x0)]] 
  [[activate(n!6220!6220d(_x0))]] = x0 >= x0 = [[d(_x0)]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  h(X) => c(n!6220!6220d(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#(f(X)) =#> c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(X))))   
    1] c#(X) =#> d#(activate(X))   
    2] c#(X) =#> activate#(X)   
    3] activate#(n!6220!6220f(X)) =#> f#(activate(X))   
    4] activate#(n!6220!6220f(X)) =#> activate#(X)   
    5] activate#(n!6220!6220g(X)) =#> g#(X)   
    6] activate#(n!6220!6220d(X)) =#> d#(X)   

  Rules R_0:

    f(f(X)) => c(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) 
    c(X) => d(activate(X)) 
    f(X) => n!6220!6220f(X) 
    g(X) => n!6220!6220g(X) 
    d(X) => n!6220!6220d(X) 
    activate(n!6220!6220f(X)) => f(activate(X)) 
    activate(n!6220!6220g(X)) => g(X) 
    activate(n!6220!6220d(X)) => d(X) 
    activate(X) => X 

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 : 1, 2 
    * 1 :  
    * 2 : 3, 4, 5, 6 
    * 3 : 0 
    * 4 : 3, 4, 5, 6 
    * 5 :  
    * 6 :  

This graph has the following strongly connected components:

  P_1:

    f#(f(X)) =#> c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(X))))   
    c#(X) =#> activate#(X)   
    activate#(n!6220!6220f(X)) =#> f#(activate(X))   
    activate#(n!6220!6220f(X)) =#> activate#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

The formative rules of (P_1, R_0) are R_1 ::=

  f(X) => n!6220!6220f(X) 
  activate(n!6220!6220f(X)) => f(activate(X)) 
  activate(X) => X 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(f(X)) >? c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) 
  c#(X) >? activate#(X) 
  activate#(n!6220!6220f(X)) >? f#(activate(X)) 
  activate#(n!6220!6220f(X)) >? activate#(X) 
  f(X) >= n!6220!6220f(X) 
  activate(n!6220!6220f(X)) >= f(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  activate = \y0.y0 
  activate# = \y0.y0 
  c# = \y0.y0 
  f = \y0.1 + 2y0 
  f# = \y0.1 
  n!6220!6220f = \y0.1 + 2y0 
  n!6220!6220g = \y0.0 

Using this interpretation, the requirements translate to:

  [[f#(f(_x0))]] = 1 >= 1 = [[c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(_x0))))]] 
  [[c#(_x0)]] = x0 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220f(_x0))]] = 1 + 2x0 >= 1 = [[f#(activate(_x0))]] 
  [[activate#(n!6220!6220f(_x0))]] = 1 + 2x0 > x0 = [[activate#(_x0)]] 
  [[f(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[n!6220!6220f(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[f(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of:

  f#(f(X)) =#> c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(X))))   
  c#(X) =#> activate#(X)   
  activate#(n!6220!6220f(X)) =#> f#(activate(X))   

Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(f(X)) >? c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) 
  c#(X) >? activate#(X) 
  activate#(n!6220!6220f(X)) >? f#(activate(X)) 
  f(X) >= n!6220!6220f(X) 
  activate(n!6220!6220f(X)) >= f(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  activate = \y0.y0 
  activate# = \y0.3 + y0 
  c# = \y0.3 + 2y0 
  f = \y0.1 + 3y0 
  f# = \y0.3 + 3y0 
  n!6220!6220f = \y0.1 + 3y0 
  n!6220!6220g = \y0.0 

Using this interpretation, the requirements translate to:

  [[f#(f(_x0))]] = 6 + 9x0 > 5 = [[c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(_x0))))]] 
  [[c#(_x0)]] = 3 + 2x0 >= 3 + x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220f(_x0))]] = 4 + 3x0 > 3 + 3x0 = [[f#(activate(_x0))]] 
  [[f(_x0)]] = 1 + 3x0 >= 1 + 3x0 = [[n!6220!6220f(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 1 + 3x0 >= 1 + 3x0 = [[f(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_1, minimal, formative) by (P_3, R_1, minimal, formative), where P_3 consists of:

  c#(X) =#> activate#(X)   

Thus, the original system is terminating if (P_3, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_1, 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.