/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 
  a!6220!6220f : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  mark : [o] --> o 

  a!6220!6220f(f(a)) => a!6220!6220f(g(f(a))) 
  mark(f(X)) => a!6220!6220f(mark(X)) 
  mark(a) => a 
  mark(g(X)) => g(X) 
  a!6220!6220f(X) => f(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]):

  a!6220!6220f(f(a)) >? a!6220!6220f(g(f(a))) 
  mark(f(X)) >? a!6220!6220f(mark(X)) 
  mark(a) >? a 
  mark(g(X)) >? g(X) 
  a!6220!6220f(X) >? f(X) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(f(a))]] = 6 >= 6 = [[a!6220!6220f(g(f(a)))]] 
  [[mark(f(_x0))]] = 8 + 2x0 > 5 + 2x0 = [[a!6220!6220f(mark(_x0))]] 
  [[mark(a)]] = 2 > 0 = [[a]] 
  [[mark(g(_x0))]] = 2 + 2x0 > x0 = [[g(_x0)]] 
  [[a!6220!6220f(_x0)]] = 3 + x0 >= 3 + x0 = [[f(_x0)]] 

We can thus remove the following rules:

  mark(f(X)) => a!6220!6220f(mark(X)) 
  mark(a) => a 
  mark(g(X)) => g(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]):

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(f(a))]] = 3 >= 3 = [[a!6220!6220f(g(f(a)))]] 
  [[a!6220!6220f(_x0)]] = 3 + 3x0 > x0 = [[f(_x0)]] 

We can thus remove the following rules:

  a!6220!6220f(X) => f(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] a!6220!6220f#(f(a)) =#> a!6220!6220f#(g(f(a)))   

  Rules R_0:

    a!6220!6220f(f(a)) => a!6220!6220f(g(f(a))) 

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.