/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.

  a : [] --> o 
  activate : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  n!6220!6220a : [] --> o 
  n!6220!6220f : [o] --> o 
  n!6220!6220g : [o] --> o 

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

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

The following interpretation satisfies the requirements:

  a = 3 
  activate = \y0.2y0 
  f = \y0.2y0 
  g = \y0.y0 
  n!6220!6220a = 2 
  n!6220!6220f = \y0.2y0 
  n!6220!6220g = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(n!6220!6220f(n!6220!6220a))]] = 8 >= 8 = [[f(n!6220!6220g(n!6220!6220f(n!6220!6220a)))]] 
  [[f(_x0)]] = 2x0 >= 2x0 = [[n!6220!6220f(_x0)]] 
  [[a]] = 3 > 2 = [[n!6220!6220a]] 
  [[g(_x0)]] = x0 >= x0 = [[n!6220!6220g(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 4x0 >= 2x0 = [[f(_x0)]] 
  [[activate(n!6220!6220a)]] = 4 > 3 = [[a]] 
  [[activate(n!6220!6220g(_x0))]] = 2x0 >= 2x0 = [[g(activate(_x0))]] 
  [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  a => n!6220!6220a 
  activate(n!6220!6220a) => a 

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(n!6220!6220f(n!6220!6220a)) >? f(n!6220!6220g(n!6220!6220f(n!6220!6220a))) 
  f(X) >? n!6220!6220f(X) 
  g(X) >? n!6220!6220g(X) 
  activate(n!6220!6220f(X)) >? f(X) 
  activate(n!6220!6220g(X)) >? g(activate(X)) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[f(n!6220!6220f(n!6220!6220a))]] = 3 >= 3 = [[f(n!6220!6220g(n!6220!6220f(n!6220!6220a)))]] 
  [[f(_x0)]] = 2 + x0 > 1 + x0 = [[n!6220!6220f(_x0)]] 
  [[g(_x0)]] = x0 >= x0 = [[n!6220!6220g(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 2 + 2x0 >= 2 + x0 = [[f(_x0)]] 
  [[activate(n!6220!6220g(_x0))]] = 2x0 >= 2x0 = [[g(activate(_x0))]] 
  [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

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

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

The following interpretation satisfies the requirements:

  activate = \y0.3 + y0 
  f = \y0.2y0 
  g = \y0.y0 
  n!6220!6220a = 0 
  n!6220!6220f = \y0.2y0 
  n!6220!6220g = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f(n!6220!6220f(n!6220!6220a))]] = 0 >= 0 = [[f(n!6220!6220g(n!6220!6220f(n!6220!6220a)))]] 
  [[g(_x0)]] = x0 >= x0 = [[n!6220!6220g(_x0)]] 
  [[activate(n!6220!6220f(_x0))]] = 3 + 2x0 > 2x0 = [[f(_x0)]] 
  [[activate(n!6220!6220g(_x0))]] = 3 + x0 >= 3 + x0 = [[g(activate(_x0))]] 
  [[activate(_x0)]] = 3 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  activate(n!6220!6220f(X)) => f(X) 
  activate(X) => 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#(n!6220!6220f(n!6220!6220a)) =#> f#(n!6220!6220g(n!6220!6220f(n!6220!6220a)))   
    1] activate#(n!6220!6220g(X)) =#> g#(activate(X))   
    2] activate#(n!6220!6220g(X)) =#> activate#(X)   

  Rules R_0:

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

This graph has the following strongly connected components:

  P_1:

    activate#(n!6220!6220g(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).

We apply the subterm criterion with the following projection function:

  nu(activate#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(activate#(n!6220!6220g(X))) = n!6220!6220g(X) |> X = nu(activate#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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.