/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.

  app : [o * o] --> o 
  cons : [] --> o 
  fcons : [] --> o 
  fmap : [] --> o 
  fnil : [] --> o 
  nil : [] --> o 

  app(app(fmap, fnil), X) => nil 
  app(app(fmap, app(app(fcons, X), Y)), Z) => app(app(cons, app(X, Z)), app(app(fmap, Y), Z)) 

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] app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(app(cons, app(X, Z)), app(app(fmap, Y), Z))   
    1] app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(cons, app(X, Z))   
    2] app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(X, Z)   
    3] app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(app(fmap, Y), Z)   
    4] app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(fmap, Y)   

  Rules R_0:

    app(app(fmap, fnil), X) => nil 
    app(app(fmap, app(app(fcons, X), Y)), Z) => app(app(cons, app(X, Z)), app(app(fmap, Y), Z)) 

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

This graph has the following strongly connected components:

  P_1:

    app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(app(cons, app(X, Z)), app(app(fmap, Y), Z))   
    app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(X, Z)   
    app#(app(fmap, app(app(fcons, X), Y)), Z) =#> app#(app(fmap, Y), Z)   

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).

The formative rules of (P_1, R_0) are R_1 ::=

  app(app(fmap, app(app(fcons, X), Y)), Z) => app(app(cons, app(X, Z)), app(app(fmap, Y), Z)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, 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:

  app#(app(fmap, app(app(fcons, X), Y)), Z) >? app#(app(cons, app(X, Z)), app(app(fmap, Y), Z)) 
  app#(app(fmap, app(app(fcons, X), Y)), Z) >? app#(X, Z) 
  app#(app(fmap, app(app(fcons, X), Y)), Z) >? app#(app(fmap, Y), Z) 
  app(app(fmap, app(app(fcons, X), Y)), Z) >= app(app(cons, app(X, Z)), app(app(fmap, Y), Z)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + 2y0y1 
  app# = \y0y1.2y0 
  cons = 0 
  fcons = 3 
  fmap = 2 

Using this interpretation, the requirements translate to:

  [[app#(app(fmap, app(app(fcons, _x0), _x1)), _x2)]] = 28 + 48x0 + 48x1 + 96x0x1 > 0 = [[app#(app(cons, app(_x0, _x2)), app(app(fmap, _x1), _x2))]] 
  [[app#(app(fmap, app(app(fcons, _x0), _x1)), _x2)]] = 28 + 48x0 + 48x1 + 96x0x1 > 2x0 = [[app#(_x0, _x2)]] 
  [[app#(app(fmap, app(app(fcons, _x0), _x1)), _x2)]] = 28 + 48x0 + 48x1 + 96x0x1 > 4 + 8x1 = [[app#(app(fmap, _x1), _x2)]] 
  [[app(app(fmap, app(app(fcons, _x0), _x1)), _x2)]] = 14 + 24x0 + 24x1 + 28x2 + 48x0x1 + 48x0x2 + 48x1x2 + 96x0x1x2 >= 0 = [[app(app(cons, app(_x0, _x2)), app(app(fmap, _x1), _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1).  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.