/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] --> o 
  b : [o] --> o 
  p : [o * o] --> o 

  p(a(X), p(b(Y), p(a(Z), U))) => p(Z, p(a(a(X)), p(b(Y), U))) 

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] p#(a(X), p(b(Y), p(a(Z), U))) =#> p#(Z, p(a(a(X)), p(b(Y), U)))   
    1] p#(a(X), p(b(Y), p(a(Z), U))) =#> p#(a(a(X)), p(b(Y), U))   
    2] p#(a(X), p(b(Y), p(a(Z), U))) =#> p#(b(Y), U)   

  Rules R_0:

    p(a(X), p(b(Y), p(a(Z), U))) => p(Z, p(a(a(X)), p(b(Y), U))) 

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 
    * 1 : 0, 1, 2 
    * 2 :  

This graph has the following strongly connected components:

  P_1:

    p#(a(X), p(b(Y), p(a(Z), U))) =#> p#(Z, p(a(a(X)), p(b(Y), U)))   
    p#(a(X), p(b(Y), p(a(Z), U))) =#> p#(a(a(X)), p(b(Y), U))   

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:

  p#(a(X), p(b(Y), p(a(Z), U))) >? p#(Z, p(a(a(X)), p(b(Y), U))) 
  p#(a(X), p(b(Y), p(a(Z), U))) >? p#(a(a(X)), p(b(Y), U)) 
  p(a(X), p(b(Y), p(a(Z), U))) >= p(Z, p(a(a(X)), p(b(Y), U))) 

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

The following interpretation satisfies the requirements:

  a = \y0.0 
  b = \y0.0 
  p = \y0y1.3 + y1 
  p# = \y0y1.2y1 

Using this interpretation, the requirements translate to:

  [[p#(a(_x0), p(b(_x1), p(a(_x2), _x3)))]] = 12 + 2x3 >= 12 + 2x3 = [[p#(_x2, p(a(a(_x0)), p(b(_x1), _x3)))]] 
  [[p#(a(_x0), p(b(_x1), p(a(_x2), _x3)))]] = 12 + 2x3 > 6 + 2x3 = [[p#(a(a(_x0)), p(b(_x1), _x3))]] 
  [[p(a(_x0), p(b(_x1), p(a(_x2), _x3)))]] = 9 + x3 >= 9 + x3 = [[p(_x2, p(a(a(_x0)), p(b(_x1), _x3)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_2, R_0, minimal, formative), where P_2 consists of:

  p#(a(X), p(b(Y), p(a(Z), U))) =#> p#(Z, p(a(a(X)), p(b(Y), U)))   

Thus, the original system is terminating if (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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:

  p#(a(X), p(b(Y), p(a(Z), U))) >? p#(Z, p(a(a(X)), p(b(Y), U))) 
  p(a(X), p(b(Y), p(a(Z), U))) >= p(Z, p(a(a(X)), p(b(Y), U))) 

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

The following interpretation satisfies the requirements:

  a = \y0.2 + 3y0 
  b = \y0.1 
  p = \y0y1.2y0 + 2y0y1 + 2y1 
  p# = \y0y1.y1 + 2y0 + y0y1 

Using this interpretation, the requirements translate to:

  [[p#(a(_x0), p(b(_x1), p(a(_x2), _x3)))]] = 22 + 2*18 + 224x2 + 224x2x3 + 224x3 + 6x0 + 24x2 + 24x2x3 + 24x3 + 3x0*18 + 3x024x2 + 3x024x2x3 + 3x024x3 >= 52 + 54x0 + 54x0x2 + 54x2 + 72x0x2x3 + 72x0x3 + 72x2x3 + 72x3 = [[p#(_x2, p(a(a(_x0)), p(b(_x1), _x3)))]] 
  [[p(a(_x0), p(b(_x1), p(a(_x2), _x3)))]] = 112 + 114x0 + 144x0x2 + 144x0x2x3 + 144x0x3 + 144x2 + 144x2x3 + 144x3 >= 104 + 106x2 + 108x0 + 108x0x2 + 144x0x2x3 + 144x0x3 + 144x2x3 + 144x3 = [[p(_x2, p(a(a(_x0)), p(b(_x1), _x3)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, 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.