/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 
  f : [o * o] --> o 
  h : [o] --> o 

  f(X, f(Y, a)) => f(f(f(f(a, a), Y), h(a)), 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#(X, f(Y, a)) =#> f#(f(f(f(a, a), Y), h(a)), X)   
    1] f#(X, f(Y, a)) =#> f#(f(f(a, a), Y), h(a))   
    2] f#(X, f(Y, a)) =#> f#(f(a, a), Y)   
    3] f#(X, f(Y, a)) =#> f#(a, a)   

  Rules R_0:

    f(X, f(Y, a)) => f(f(f(f(a, a), Y), h(a)), 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 : 0, 1, 2, 3 
    * 1 :  
    * 2 : 0, 1, 2, 3 
    * 3 :  

This graph has the following strongly connected components:

  P_1:

    f#(X, f(Y, a)) =#> f#(f(f(f(a, a), Y), h(a)), X)   
    f#(X, f(Y, a)) =#> f#(f(a, a), Y)   

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 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#(X, f(Y, a)) >? f#(f(f(f(a, a), Y), h(a)), X) 
  f#(X, f(Y, a)) >? f#(f(a, a), Y) 
  f(X, f(Y, a)) >= f(f(f(f(a, a), Y), h(a)), X) 

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

The following interpretation satisfies the requirements:

  a = 1 
  f = \y0y1.2y1 + 3y0y1 
  f# = \y0y1.y0 + 3y1 + y0y1 
  h = \y0.0 

Using this interpretation, the requirements translate to:

  [[f#(_x0, f(_x1, a))]] = 6 + 3x0 + 3x0x1 + 9x1 > 3x0 = [[f#(f(f(f(a, a), _x1), h(a)), _x0)]] 
  [[f#(_x0, f(_x1, a))]] = 6 + 3x0 + 3x0x1 + 9x1 > 5 + 8x1 = [[f#(f(a, a), _x1)]] 
  [[f(_x0, f(_x1, a))]] = 4 + 6x0 + 6x1 + 9x0x1 >= 2x0 = [[f(f(f(f(a, a), _x1), h(a)), _x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0).  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.