/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.

  !dot : [o * o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  nil : [] --> o 

  f(nil) => nil 
  f(!dot(nil, X)) => !dot(nil, f(X)) 
  f(!dot(!dot(X, Y), Z)) => f(!dot(X, !dot(Y, Z))) 
  g(nil) => nil 
  g(!dot(X, nil)) => !dot(g(X), nil) 
  g(!dot(X, !dot(Y, Z))) => g(!dot(!dot(X, Y), Z)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  f(nil) >? nil 
  f(!dot(nil, X)) >? !dot(nil, f(X)) 
  f(!dot(!dot(X, Y), Z)) >? f(!dot(X, !dot(Y, Z))) 
  g(nil) >? nil 
  g(!dot(X, nil)) >? !dot(g(X), nil) 
  g(!dot(X, !dot(Y, Z))) >? g(!dot(!dot(X, Y), Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.y0 + y1 
  f = \y0.y0 
  g = \y0.1 + y0 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[f(nil)]] = 0 >= 0 = [[nil]] 
  [[f(!dot(nil, _x0))]] = x0 >= x0 = [[!dot(nil, f(_x0))]] 
  [[f(!dot(!dot(_x0, _x1), _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[f(!dot(_x0, !dot(_x1, _x2)))]] 
  [[g(nil)]] = 1 > 0 = [[nil]] 
  [[g(!dot(_x0, nil))]] = 1 + x0 >= 1 + x0 = [[!dot(g(_x0), nil)]] 
  [[g(!dot(_x0, !dot(_x1, _x2)))]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[g(!dot(!dot(_x0, _x1), _x2))]] 

We can thus remove the following rules:

  g(nil) => nil 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  f(nil) >? nil 
  f(!dot(nil, X)) >? !dot(nil, f(X)) 
  f(!dot(!dot(X, Y), Z)) >? f(!dot(X, !dot(Y, Z))) 
  g(!dot(X, nil)) >? !dot(g(X), nil) 
  g(!dot(X, !dot(Y, Z))) >? g(!dot(!dot(X, Y), Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.y0 + y1 
  f = \y0.2 + y0 
  g = \y0.2y0 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[f(nil)]] = 2 > 0 = [[nil]] 
  [[f(!dot(nil, _x0))]] = 2 + x0 >= 2 + x0 = [[!dot(nil, f(_x0))]] 
  [[f(!dot(!dot(_x0, _x1), _x2))]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[f(!dot(_x0, !dot(_x1, _x2)))]] 
  [[g(!dot(_x0, nil))]] = 2x0 >= 2x0 = [[!dot(g(_x0), nil)]] 
  [[g(!dot(_x0, !dot(_x1, _x2)))]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[g(!dot(!dot(_x0, _x1), _x2))]] 

We can thus remove the following rules:

  f(nil) => nil 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  f(!dot(nil, X)) >? !dot(nil, f(X)) 
  f(!dot(!dot(X, Y), Z)) >? f(!dot(X, !dot(Y, Z))) 
  g(!dot(X, nil)) >? !dot(g(X), nil) 
  g(!dot(X, !dot(Y, Z))) >? g(!dot(!dot(X, Y), Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.y0 + y1 
  f = \y0.2 + 2y0 
  g = \y0.y0 
  nil = 2 

Using this interpretation, the requirements translate to:

  [[f(!dot(nil, _x0))]] = 6 + 2x0 > 4 + 2x0 = [[!dot(nil, f(_x0))]] 
  [[f(!dot(!dot(_x0, _x1), _x2))]] = 2 + 2x0 + 2x1 + 2x2 >= 2 + 2x0 + 2x1 + 2x2 = [[f(!dot(_x0, !dot(_x1, _x2)))]] 
  [[g(!dot(_x0, nil))]] = 2 + x0 >= 2 + x0 = [[!dot(g(_x0), nil)]] 
  [[g(!dot(_x0, !dot(_x1, _x2)))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[g(!dot(!dot(_x0, _x1), _x2))]] 

We can thus remove the following rules:

  f(!dot(nil, X)) => !dot(nil, f(X)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  f(!dot(!dot(X, Y), Z)) >? f(!dot(X, !dot(Y, Z))) 
  g(!dot(X, nil)) >? !dot(g(X), nil) 
  g(!dot(X, !dot(Y, Z))) >? g(!dot(!dot(X, Y), Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.1 + y0 + y1 
  f = \y0.y0 
  g = \y0.1 + 2y0 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[f(!dot(!dot(_x0, _x1), _x2))]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[f(!dot(_x0, !dot(_x1, _x2)))]] 
  [[g(!dot(_x0, nil))]] = 3 + 2x0 > 2 + 2x0 = [[!dot(g(_x0), nil)]] 
  [[g(!dot(_x0, !dot(_x1, _x2)))]] = 5 + 2x0 + 2x1 + 2x2 >= 5 + 2x0 + 2x1 + 2x2 = [[g(!dot(!dot(_x0, _x1), _x2))]] 

We can thus remove the following rules:

  g(!dot(X, nil)) => !dot(g(X), nil) 

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] f#(!dot(!dot(X, Y), Z)) =#> f#(!dot(X, !dot(Y, Z)))   
    1] g#(!dot(X, !dot(Y, Z))) =#> g#(!dot(!dot(X, Y), Z))   

  Rules R_0:

    f(!dot(!dot(X, Y), Z)) => f(!dot(X, !dot(Y, Z))) 
    g(!dot(X, !dot(Y, Z))) => g(!dot(!dot(X, 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 : 1 

This graph has the following strongly connected components:

  P_1:

    f#(!dot(!dot(X, Y), Z)) =#> f#(!dot(X, !dot(Y, Z)))   

  P_2:

    g#(!dot(X, !dot(Y, Z))) =#> g#(!dot(!dot(X, 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) and (P_2, R_0, m, f).

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).

This combination (P_2, R_0) has no formative rules!  We will name the empty set of rules:R_1.

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative).

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

  g#(!dot(X, !dot(Y, Z))) >? g#(!dot(!dot(X, Y), Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.3 + 3y1 
  g# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[g#(!dot(_x0, !dot(_x1, _x2)))]] = 24 + 18x2 > 6 + 6x2 = [[g#(!dot(!dot(_x0, _x1), _x2))]] 

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.

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).

This combination (P_1, R_0) has no formative rules!  This is exactly R_1.

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:

  f#(!dot(!dot(X, Y), Z)) >? f#(!dot(X, !dot(Y, Z))) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.3 + 3y0 
  f# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[f#(!dot(!dot(_x0, _x1), _x2))]] = 24 + 18x0 > 6 + 6x0 = [[f#(!dot(_x0, !dot(_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.