/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.

  ap : [o * o] --> o 
  cons : [] --> o 
  ff : [] --> o 
  nil : [] --> o 

  ap(ap(ff, X), X) => ap(ap(X, ap(ff, X)), ap(ap(cons, X), nil)) 

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, all):

  Dependency Pairs P_0:

    0] ap#(ap(ff, X), X) =#> ap#(ap(X, ap(ff, X)), ap(ap(cons, X), nil))   
    1] ap#(ap(ff, X), X) =#> ap#(X, ap(ff, X))   
    2] ap#(ap(ff, X), X) =#> ap#(ff, X)   
    3] ap#(ap(ff, X), X) =#> ap#(ap(cons, X), nil)   
    4] ap#(ap(ff, X), X) =#> ap#(cons, X)   

  Rules R_0:

    ap(ap(ff, X), X) => ap(ap(X, ap(ff, X)), ap(ap(cons, X), nil)) 

Thus, the original system is terminating if (P_0, R_0, minimal, all) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, all).

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

This graph has the following strongly connected components:

  P_1:

    ap#(ap(ff, X), X) =#> ap#(ap(X, ap(ff, X)), ap(ap(cons, X), nil))   
    ap#(ap(ff, X), X) =#> ap#(X, ap(ff, X))   
    ap#(ap(ff, X), X) =#> ap#(ap(cons, X), nil)   

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, all) is finite.

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

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:

  ap#(ap(ff, X), X) >? ap#(ap(X, ap(ff, X)), ap(ap(cons, X), nil)) 
  ap#(ap(ff, X), X) >? ap#(X, ap(ff, X)) 
  ap#(ap(ff, X), X) >? ap#(ap(cons, X), nil) 
  ap(ap(ff, X), X) >= ap(ap(X, ap(ff, X)), ap(ap(cons, X), nil)) 

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

The following interpretation satisfies the requirements:

  ap = \y0y1.y0 + y0y1 
  ap# = \y0y1.2y0 + 2y0y1 
  cons = 0 
  ff = 2 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[ap#(ap(ff, _x0), _x0)]] = 4 + 4x0x0 + 8x0 > 4x0x0 + 6x0 = [[ap#(ap(_x0, ap(ff, _x0)), ap(ap(cons, _x0), nil))]] 
  [[ap#(ap(ff, _x0), _x0)]] = 4 + 4x0x0 + 8x0 > 4x0x0 + 6x0 = [[ap#(_x0, ap(ff, _x0))]] 
  [[ap#(ap(ff, _x0), _x0)]] = 4 + 4x0x0 + 8x0 > 0 = [[ap#(ap(cons, _x0), nil)]] 
  [[ap(ap(ff, _x0), _x0)]] = 2 + 2x0x0 + 4x0 >= 2x0x0 + 3x0 = [[ap(ap(_x0, ap(ff, _x0)), ap(ap(cons, _x0), nil))]] 

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.