/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 
  map : [] --> o 
  nil : [] --> o 

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

  Rules R_0:

    app(app(map, X), nil) => nil 
    app(app(map, X), app(app(cons, Y), Z)) => app(app(cons, app(X, Y)), app(app(map, X), 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(map, X), app(app(cons, Y), Z)) =#> app#(app(cons, app(X, Y)), app(app(map, X), Z))   
    app#(app(map, X), app(app(cons, Y), Z)) =#> app#(X, Y)   
    app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(map, X), 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(map, X), app(app(cons, Y), Z)) => app(app(cons, app(X, Y)), app(app(map, X), 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(map, X), app(app(cons, Y), Z)) >? app#(app(cons, app(X, Y)), app(app(map, X), Z)) 
  app#(app(map, X), app(app(cons, Y), Z)) >? app#(X, Y) 
  app#(app(map, X), app(app(cons, Y), Z)) >? app#(app(map, X), Z) 
  app(app(map, X), app(app(cons, Y), Z)) >= app(app(cons, app(X, Y)), app(app(map, X), Z)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.2 + 2y0y1 
  app# = \y0y1.y0 
  cons = 0 
  map = 1 

Using this interpretation, the requirements translate to:

  [[app#(app(map, _x0), app(app(cons, _x1), _x2))]] = 2 + 2x0 >= 2 = [[app#(app(cons, app(_x0, _x1)), app(app(map, _x0), _x2))]] 
  [[app#(app(map, _x0), app(app(cons, _x1), _x2))]] = 2 + 2x0 > x0 = [[app#(_x0, _x1)]] 
  [[app#(app(map, _x0), app(app(cons, _x1), _x2))]] = 2 + 2x0 >= 2 + 2x0 = [[app#(app(map, _x0), _x2)]] 
  [[app(app(map, _x0), app(app(cons, _x1), _x2))]] = 10 + 8x0 + 16x0x2 + 16x2 >= 10 + 16x0x2 + 16x2 = [[app(app(cons, app(_x0, _x1)), app(app(map, _x0), _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of:

  app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(cons, app(X, Y)), app(app(map, X), Z))   
  app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(map, X), Z)   

Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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(map, X), app(app(cons, Y), Z)) >? app#(app(cons, app(X, Y)), app(app(map, X), Z)) 
  app#(app(map, X), app(app(cons, Y), Z)) >? app#(app(map, X), Z) 
  app(app(map, X), app(app(cons, Y), Z)) >= app(app(cons, app(X, Y)), app(app(map, X), Z)) 

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

We consider usable_rules with respect to the following argument filtering:

  app(x_1,x_2) = app(x_1) 
  app#(x_1,x_2) = app#(x_1) 

This leaves the following ordering requirements:

  app#(app(map, X), app(app(cons, Y), Z)) > app#(app(cons, app(X, Y)), app(app(map, X), Z)) 
  app#(app(map, X), app(app(cons, Y), Z)) >= app#(app(map, X), Z) 

The following interpretation satisfies the requirements:

  app = \y0y1.y0 
  app# = \y0y1.y0 
  cons = 1 
  map = 2 

Using this interpretation, the requirements translate to:

  [[app#(app(map, _x0), app(app(cons, _x1), _x2))]] = 2 > 1 = [[app#(app(cons, app(_x0, _x1)), app(app(map, _x0), _x2))]] 
  [[app#(app(map, _x0), app(app(cons, _x1), _x2))]] = 2 >= 2 = [[app#(app(map, _x0), _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_1, minimal, formative) by (P_3, R_1, minimal, formative), where P_3 consists of:

  app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(map, X), Z)   

Thus, the original system is terminating if (P_3, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_1, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(app#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(app#(app(map, X), app(app(cons, Y), Z))) = app(app(cons, Y), Z) |> Z = nu(app#(app(map, X), Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_1, minimal, f) by ({}, R_1, minimal, f).  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.