/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.

  a : [] --> o 
  b : [] --> o 
  c : [o] --> o 
  d : [o] --> o 
  e : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 

  f(a) => f(c(a)) 
  f(c(X)) => X 
  f(c(a)) => f(d(b)) 
  f(a) => f(d(a)) 
  f(d(X)) => X 
  f(c(b)) => f(d(a)) 
  e(g(X)) => e(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(a) >? f(c(a)) 
  f(c(X)) >? X 
  f(c(a)) >? f(d(b)) 
  f(a) >? f(d(a)) 
  f(d(X)) >? X 
  f(c(b)) >? f(d(a)) 
  e(g(X)) >? e(X) 

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

The following interpretation satisfies the requirements:

  a = 0 
  b = 0 
  c = \y0.2y0 
  d = \y0.y0 
  e = \y0.y0 
  f = \y0.3 + 2y0 
  g = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[f(a)]] = 3 >= 3 = [[f(c(a))]] 
  [[f(c(_x0))]] = 3 + 4x0 > x0 = [[_x0]] 
  [[f(c(a))]] = 3 >= 3 = [[f(d(b))]] 
  [[f(a)]] = 3 >= 3 = [[f(d(a))]] 
  [[f(d(_x0))]] = 3 + 2x0 > x0 = [[_x0]] 
  [[f(c(b))]] = 3 >= 3 = [[f(d(a))]] 
  [[e(g(_x0))]] = 3 + 3x0 > x0 = [[e(_x0)]] 

We can thus remove the following rules:

  f(c(X)) => X 
  f(d(X)) => X 
  e(g(X)) => e(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#(a) =#> f#(c(a))   
    1] f#(c(a)) =#> f#(d(b))   
    2] f#(a) =#> f#(d(a))   
    3] f#(c(b)) =#> f#(d(a))   

  Rules R_0:

    f(a) => f(c(a)) 
    f(c(a)) => f(d(b)) 
    f(a) => f(d(a)) 
    f(c(b)) => f(d(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 : 1 
    * 1 :  
    * 2 :  
    * 3 :  

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.