/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 
  double : [o] --> o 
  minus : [o * o] --> o 
  plus : [o * o] --> o 
  s : [o] --> o 

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  double(0) => 0 
  double(s(X)) => s(s(double(X))) 
  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  plus(s(X), Y) => plus(X, s(Y)) 
  plus(s(X), Y) => s(plus(minus(X, Y), double(Y))) 

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] minus#(s(X), s(Y)) =#> minus#(X, Y)   
    1] double#(s(X)) =#> double#(X)   
    2] plus#(s(X), Y) =#> plus#(X, Y)   
    3] plus#(s(X), Y) =#> plus#(X, s(Y))   
    4] plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y))   
    5] plus#(s(X), Y) =#> minus#(X, Y)   
    6] plus#(s(X), Y) =#> double#(Y)   

  Rules R_0:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    double(0) => 0 
    double(s(X)) => s(s(double(X))) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    plus(s(X), Y) => plus(X, s(Y)) 
    plus(s(X), Y) => s(plus(minus(X, Y), double(Y))) 

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 : 1 
    * 2 : 2, 3, 4, 5, 6 
    * 3 : 2, 3, 4, 5, 6 
    * 4 : 2, 3, 4, 5, 6 
    * 5 : 0 
    * 6 : 1 

This graph has the following strongly connected components:

  P_1:

    minus#(s(X), s(Y)) =#> minus#(X, Y)   

  P_2:

    double#(s(X)) =#> double#(X)   

  P_3:

    plus#(s(X), Y) =#> plus#(X, Y)   
    plus#(s(X), Y) =#> plus#(X, s(Y))   
    plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y))   

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), (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_3, R_0) are:

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  double(0) => 0 
  double(s(X)) => s(s(double(X))) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(s(X), Y) >? plus#(X, Y) 
  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  double(0) >= 0 
  double(s(X)) >= s(s(double(X))) 

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

We consider usable_rules with respect to the following argument filtering:

  plus#(x_1,x_2) = plus#(x_1) 

This leaves the following ordering requirements:

  plus#(s(X), Y) >= plus#(X, Y) 
  plus#(s(X), Y) >= plus#(X, s(Y)) 
  plus#(s(X), Y) > plus#(minus(X, Y), double(Y)) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 

The following interpretation satisfies the requirements:

  0 = 0 
  double = \y0.0 
  minus = \y0y1.y0 
  plus# = \y0y1.2y0 
  s = \y0.2 + 2y0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 4 + 4x0 > 2x0 = [[plus#(_x0, _x1)]] 
  [[plus#(s(_x0), _x1)]] = 4 + 4x0 > 2x0 = [[plus#(_x0, s(_x1))]] 
  [[plus#(s(_x0), _x1)]] = 4 + 4x0 > 2x0 = [[plus#(minus(_x0, _x1), double(_x1))]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 2 + 2x0 >= x0 = [[minus(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(double#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(double#(s(X))) = s(X) |> X = nu(double#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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 apply the subterm criterion with the following projection function:

  nu(minus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, 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.