/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.

  a : [] --> o 
  active : [o] --> o 
  b : [] --> o 
  f : [o * o * o] --> o 
  mark : [o] --> o 

  active(f(a, X, X)) => mark(f(X, b, b)) 
  active(b) => mark(a) 
  mark(f(X, Y, Z)) => active(f(X, mark(Y), Z)) 
  mark(a) => active(a) 
  mark(b) => active(b) 
  f(mark(X), Y, Z) => f(X, Y, Z) 
  f(X, mark(Y), Z) => f(X, Y, Z) 
  f(X, Y, mark(Z)) => f(X, Y, Z) 
  f(active(X), Y, Z) => f(X, Y, Z) 
  f(X, active(Y), Z) => f(X, Y, Z) 
  f(X, Y, active(Z)) => f(X, 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, all):

  Dependency Pairs P_0:

    0] active#(f(a, X, X)) =#> mark#(f(X, b, b))   
    1] active#(f(a, X, X)) =#> f#(X, b, b)   
    2] active#(b) =#> mark#(a)   
    3] mark#(f(X, Y, Z)) =#> active#(f(X, mark(Y), Z))   
    4] mark#(f(X, Y, Z)) =#> f#(X, mark(Y), Z)   
    5] mark#(f(X, Y, Z)) =#> mark#(Y)   
    6] mark#(a) =#> active#(a)   
    7] mark#(b) =#> active#(b)   
    8] f#(mark(X), Y, Z) =#> f#(X, Y, Z)   
    9] f#(X, mark(Y), Z) =#> f#(X, Y, Z)   
    10] f#(X, Y, mark(Z)) =#> f#(X, Y, Z)   
    11] f#(active(X), Y, Z) =#> f#(X, Y, Z)   
    12] f#(X, active(Y), Z) =#> f#(X, Y, Z)   
    13] f#(X, Y, active(Z)) =#> f#(X, Y, Z)   

  Rules R_0:

    active(f(a, X, X)) => mark(f(X, b, b)) 
    active(b) => mark(a) 
    mark(f(X, Y, Z)) => active(f(X, mark(Y), Z)) 
    mark(a) => active(a) 
    mark(b) => active(b) 
    f(mark(X), Y, Z) => f(X, Y, Z) 
    f(X, mark(Y), Z) => f(X, Y, Z) 
    f(X, Y, mark(Z)) => f(X, Y, Z) 
    f(active(X), Y, Z) => f(X, Y, Z) 
    f(X, active(Y), Z) => f(X, Y, Z) 
    f(X, Y, active(Z)) => f(X, Y, Z) 

Thus, the original system is terminating if (P_0, R_0, minimal, all) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, all).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 3, 4, 5 
    * 1 : 8, 11 
    * 2 : 6 
    * 3 : 0, 1 
    * 4 : 8, 9, 10, 11, 12, 13 
    * 5 : 3, 4, 5, 6, 7 
    * 6 :  
    * 7 : 2 
    * 8 : 8, 9, 10, 11, 12, 13 
    * 9 : 8, 9, 10, 11, 12, 13 
    * 10 : 8, 9, 10, 11, 12, 13 
    * 11 : 8, 9, 10, 11, 12, 13 
    * 12 : 8, 9, 10, 11, 12, 13 
    * 13 : 8, 9, 10, 11, 12, 13 

This graph has the following strongly connected components:

  P_1:

    active#(f(a, X, X)) =#> mark#(f(X, b, b))   
    mark#(f(X, Y, Z)) =#> active#(f(X, mark(Y), Z))   
    mark#(f(X, Y, Z)) =#> mark#(Y)   

  P_2:

    f#(mark(X), Y, Z) =#> f#(X, Y, Z)   
    f#(X, mark(Y), Z) =#> f#(X, Y, Z)   
    f#(X, Y, mark(Z)) =#> f#(X, Y, Z)   
    f#(active(X), Y, Z) =#> f#(X, Y, Z)   
    f#(X, active(Y), Z) =#> f#(X, Y, Z)   
    f#(X, Y, active(Z)) =#> f#(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, all) and (P_2, R_0, minimal, all) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(f#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(f#(mark(X), Y, Z)) = mark(X) |> X = nu(f#(X, Y, Z)) 
  nu(f#(X, mark(Y), Z)) = X = X = nu(f#(X, Y, Z)) 
  nu(f#(X, Y, mark(Z))) = X = X = nu(f#(X, Y, Z)) 
  nu(f#(active(X), Y, Z)) = active(X) |> X = nu(f#(X, Y, Z)) 
  nu(f#(X, active(Y), Z)) = X = X = nu(f#(X, Y, Z)) 
  nu(f#(X, Y, active(Z))) = X = X = nu(f#(X, Y, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains:

  f#(X, mark(Y), Z) =#> f#(X, Y, Z)   
  f#(X, Y, mark(Z)) =#> f#(X, Y, Z)   
  f#(X, active(Y), Z) =#> f#(X, Y, Z)   
  f#(X, Y, active(Z)) =#> f#(X, Y, Z)   

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

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

We apply the subterm criterion with the following projection function:

  nu(f#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(f#(X, mark(Y), Z)) = mark(Y) |> Y = nu(f#(X, Y, Z)) 
  nu(f#(X, Y, mark(Z))) = Y = Y = nu(f#(X, Y, Z)) 
  nu(f#(X, active(Y), Z)) = active(Y) |> Y = nu(f#(X, Y, Z)) 
  nu(f#(X, Y, active(Z))) = Y = Y = nu(f#(X, Y, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains:

  f#(X, Y, mark(Z)) =#> f#(X, Y, Z)   
  f#(X, Y, active(Z)) =#> f#(X, Y, Z)   

Thus, the original system is terminating if each of (P_1, R_0, minimal, all) and (P_4, R_0, minimal, all) is finite.

We consider the dependency pair problem (P_4, R_0, minimal, all).

We apply the subterm criterion with the following projection function:

  nu(f#) = 3 

Thus, we can orient the dependency pairs as follows:

  nu(f#(X, Y, mark(Z))) = mark(Z) |> Z = nu(f#(X, Y, Z)) 
  nu(f#(X, Y, active(Z))) = active(Z) |> Z = nu(f#(X, Y, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, 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, all) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, all).

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:

  active#(f(a, X, X)) >? mark#(f(X, b, b)) 
  mark#(f(X, Y, Z)) >? active#(f(X, mark(Y), Z)) 
  mark#(f(X, Y, Z)) >? mark#(Y) 
  active(f(a, X, X)) >= mark(f(X, b, b)) 
  active(b) >= mark(a) 
  mark(f(X, Y, Z)) >= active(f(X, mark(Y), Z)) 
  mark(a) >= active(a) 
  mark(b) >= active(b) 
  f(mark(X), Y, Z) >= f(X, Y, Z) 
  f(X, mark(Y), Z) >= f(X, Y, Z) 
  f(X, Y, mark(Z)) >= f(X, Y, Z) 
  f(active(X), Y, Z) >= f(X, Y, Z) 
  f(X, active(Y), Z) >= f(X, Y, Z) 
  f(X, Y, active(Z)) >= f(X, Y, Z) 

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

We consider usable_rules with respect to the following argument filtering:

  active#(x_1) = active#() 

This leaves the following ordering requirements:

  active#(f(a, X, X)) >= mark#(f(X, b, b)) 
  mark#(f(X, Y, Z)) >= active#(f(X, mark(Y), Z)) 
  mark#(f(X, Y, Z)) > mark#(Y) 
  f(mark(X), Y, Z) >= f(X, Y, Z) 
  f(X, mark(Y), Z) >= f(X, Y, Z) 
  f(X, Y, mark(Z)) >= f(X, Y, Z) 
  f(active(X), Y, Z) >= f(X, Y, Z) 
  f(X, active(Y), Z) >= f(X, Y, Z) 
  f(X, Y, active(Z)) >= f(X, Y, Z) 

The following interpretation satisfies the requirements:

  a = 0 
  active = \y0.y0 
  active# = \y0.1 
  b = 0 
  f = \y0y1y2.1 + 2y1 
  mark = \y0.y0 
  mark# = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active#(f(a, _x0, _x0))]] = 1 >= 1 = [[mark#(f(_x0, b, b))]] 
  [[mark#(f(_x0, _x1, _x2))]] = 1 + 2x1 >= 1 = [[active#(f(_x0, mark(_x1), _x2))]] 
  [[mark#(f(_x0, _x1, _x2))]] = 1 + 2x1 > x1 = [[mark#(_x1)]] 
  [[f(mark(_x0), _x1, _x2)]] = 1 + 2x1 >= 1 + 2x1 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, mark(_x1), _x2)]] = 1 + 2x1 >= 1 + 2x1 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, _x1, mark(_x2))]] = 1 + 2x1 >= 1 + 2x1 = [[f(_x0, _x1, _x2)]] 
  [[f(active(_x0), _x1, _x2)]] = 1 + 2x1 >= 1 + 2x1 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, active(_x1), _x2)]] = 1 + 2x1 >= 1 + 2x1 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, _x1, active(_x2))]] = 1 + 2x1 >= 1 + 2x1 = [[f(_x0, _x1, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, all) by (P_5, R_0, minimal, all), where P_5 consists of:

  active#(f(a, X, X)) =#> mark#(f(X, b, b))   
  mark#(f(X, Y, Z)) =#> active#(f(X, mark(Y), Z))   

Thus, the original system is terminating if (P_5, R_0, minimal, all) is finite.

We consider the dependency pair problem (P_5, R_0, minimal, all).

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:

  active#(f(a, X, X)) >? mark#(f(X, b, b)) 
  mark#(f(X, Y, Z)) >? active#(f(X, mark(Y), Z)) 
  active(f(a, X, X)) >= mark(f(X, b, b)) 
  active(b) >= mark(a) 
  mark(f(X, Y, Z)) >= active(f(X, mark(Y), Z)) 
  mark(a) >= active(a) 
  mark(b) >= active(b) 
  f(mark(X), Y, Z) >= f(X, Y, Z) 
  f(X, mark(Y), Z) >= f(X, Y, Z) 
  f(X, Y, mark(Z)) >= f(X, Y, Z) 
  f(active(X), Y, Z) >= f(X, Y, Z) 
  f(X, active(Y), Z) >= f(X, Y, Z) 
  f(X, Y, active(Z)) >= f(X, Y, Z) 

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

We consider usable_rules with respect to the following argument filtering:

  f(x_1,x_2,x_3) = f(x_1x_3) 

This leaves the following ordering requirements:

  active#(f(a, X, X)) > mark#(f(X, b, b)) 
  mark#(f(X, Y, Z)) >= active#(f(X, mark(Y), Z)) 
  f(mark(X), Y, Z) >= f(X, Y, Z) 
  f(X, mark(Y), Z) >= f(X, Y, Z) 
  f(X, Y, mark(Z)) >= f(X, Y, Z) 
  f(active(X), Y, Z) >= f(X, Y, Z) 
  f(X, active(Y), Z) >= f(X, Y, Z) 
  f(X, Y, active(Z)) >= f(X, Y, Z) 

The following interpretation satisfies the requirements:

  a = 2 
  active = \y0.3y0 
  active# = \y0.y0 
  b = 0 
  f = \y0y1y2.y0 + 3y2 
  mark = \y0.y0 
  mark# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[active#(f(a, _x0, _x0))]] = 2 + 3x0 > 2x0 = [[mark#(f(_x0, b, b))]] 
  [[mark#(f(_x0, _x1, _x2))]] = 2x0 + 6x2 >= x0 + 3x2 = [[active#(f(_x0, mark(_x1), _x2))]] 
  [[f(mark(_x0), _x1, _x2)]] = x0 + 3x2 >= x0 + 3x2 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, mark(_x1), _x2)]] = x0 + 3x2 >= x0 + 3x2 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, _x1, mark(_x2))]] = x0 + 3x2 >= x0 + 3x2 = [[f(_x0, _x1, _x2)]] 
  [[f(active(_x0), _x1, _x2)]] = 3x0 + 3x2 >= x0 + 3x2 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, active(_x1), _x2)]] = x0 + 3x2 >= x0 + 3x2 = [[f(_x0, _x1, _x2)]] 
  [[f(_x0, _x1, active(_x2))]] = x0 + 9x2 >= x0 + 3x2 = [[f(_x0, _x1, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, minimal, all) by (P_6, R_0, minimal, all), where P_6 consists of:

  mark#(f(X, Y, Z)) =#> active#(f(X, mark(Y), Z))   

Thus, the original system is terminating if (P_6, R_0, minimal, all) is finite.

We consider the dependency pair problem (P_6, R_0, minimal, all).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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.