/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.

  0 : [] --> o 
  app : [o * o] --> o 
  cons : [] --> o 
  nil : [] --> o 
  plus : [] --> o 
  s : [] --> o 
  sumwith : [] --> o 

  app(app(plus, 0), X) => X 
  app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) 
  app(app(sumwith, X), nil) => nil 
  app(app(sumwith, X), app(app(cons, Y), Z)) => app(app(plus, app(X, Y)), app(app(sumwith, 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(plus, app(s, X)), Y) =#> app#(s, app(app(plus, X), Y))   
    1] app#(app(plus, app(s, X)), Y) =#> app#(app(plus, X), Y)   
    2] app#(app(plus, app(s, X)), Y) =#> app#(plus, X)   
    3] app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(app(plus, app(X, Y)), app(app(sumwith, X), Z))   
    4] app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(plus, app(X, Y))   
    5] app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(X, Y)   
    6] app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(app(sumwith, X), Z)   
    7] app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(sumwith, X)   

  Rules R_0:

    app(app(plus, 0), X) => X 
    app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) 
    app(app(sumwith, X), nil) => nil 
    app(app(sumwith, X), app(app(cons, Y), Z)) => app(app(plus, app(X, Y)), app(app(sumwith, 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 :  
    * 1 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 2 :  
    * 3 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 4 :  
    * 5 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 6 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 7 :  

This graph has the following strongly connected components:

  P_1:

    app#(app(plus, app(s, X)), Y) =#> app#(app(plus, X), Y)   
    app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(app(plus, app(X, Y)), app(app(sumwith, X), Z))   
    app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(X, Y)   
    app#(app(sumwith, X), app(app(cons, Y), Z)) =#> app#(app(sumwith, 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(plus, 0), X) => X 
  app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) 
  app(app(sumwith, X), app(app(cons, Y), Z)) => app(app(plus, app(X, Y)), app(app(sumwith, 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(plus, app(s, X)), Y) >? app#(app(plus, X), Y) 
  app#(app(sumwith, X), app(app(cons, Y), Z)) >? app#(app(plus, app(X, Y)), app(app(sumwith, X), Z)) 
  app#(app(sumwith, X), app(app(cons, Y), Z)) >? app#(X, Y) 
  app#(app(sumwith, X), app(app(cons, Y), Z)) >? app#(app(sumwith, X), Z) 
  app(app(plus, 0), X) >= X 
  app(app(plus, app(s, X)), Y) >= app(s, app(app(plus, X), Y)) 
  app(app(sumwith, X), app(app(cons, Y), Z)) >= app(app(plus, app(X, Y)), app(app(sumwith, X), Z)) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  app = \y0y1.1 + 2y0 + y0y1 
  app# = \y0y1.2y1 
  cons = 3 
  plus = 0 
  s = 0 
  sumwith = 0 

Using this interpretation, the requirements translate to:

  [[app#(app(plus, app(s, _x0)), _x1)]] = 2x1 >= 2x1 = [[app#(app(plus, _x0), _x1)]] 
  [[app#(app(sumwith, _x0), app(app(cons, _x1), _x2))]] = 30 + 6x1x2 + 12x1 + 14x2 > 6 + 2x2 = [[app#(app(plus, app(_x0, _x1)), app(app(sumwith, _x0), _x2))]] 
  [[app#(app(sumwith, _x0), app(app(cons, _x1), _x2))]] = 30 + 6x1x2 + 12x1 + 14x2 > 2x1 = [[app#(_x0, _x1)]] 
  [[app#(app(sumwith, _x0), app(app(cons, _x1), _x2))]] = 30 + 6x1x2 + 12x1 + 14x2 > 2x2 = [[app#(app(sumwith, _x0), _x2)]] 
  [[app(app(plus, 0), _x0)]] = 3 + x0 >= x0 = [[_x0]] 
  [[app(app(plus, app(s, _x0)), _x1)]] = 3 + x1 >= 1 = [[app(s, app(app(plus, _x0), _x1))]] 
  [[app(app(sumwith, _x0), app(app(cons, _x1), _x2))]] = 18 + 3x1x2 + 6x1 + 7x2 >= 6 + x2 = [[app(app(plus, app(_x0, _x1)), app(app(sumwith, _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(plus, app(s, X)), Y) =#> app#(app(plus, X), Y)   

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(plus, app(s, X)), Y) >? app#(app(plus, X), Y) 
  app(app(plus, 0), X) >= X 
  app(app(plus, app(s, X)), Y) >= app(s, app(app(plus, X), Y)) 
  app(app(sumwith, X), app(app(cons, Y), Z)) >= app(app(plus, app(X, Y)), app(app(sumwith, 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:


This leaves the following ordering requirements:

  app#(app(plus, app(s, X)), Y) > app#(app(plus, X), Y) 

The following interpretation satisfies the requirements:

  0 = 0 
  app = \y0y1.1 + 2y1 
  app# = \y0y1.2y0 
  cons = 3 
  plus = 0 
  s = 0 
  sumwith = 0 

Using this interpretation, the requirements translate to:

  [[app#(app(plus, app(s, _x0)), _x1)]] = 6 + 8x0 > 2 + 4x0 = [[app#(app(plus, _x0), _x1)]] 

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