/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.

  0 : [] --> o 
  activate : [o] --> o 
  cons : [o * o] --> o 
  first : [o * o] --> o 
  from : [o] --> o 
  n!6220!6220first : [o * o] --> o 
  n!6220!6220from : [o] --> o 
  n!6220!6220s : [o] --> o 
  nil : [] --> o 
  s : [o] --> o 

  first(0, X) => nil 
  first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
  from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
  first(X, Y) => n!6220!6220first(X, Y) 
  from(X) => n!6220!6220from(X) 
  s(X) => n!6220!6220s(X) 
  activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) => from(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(X) => X 

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] first#(s(X), cons(Y, Z)) =#> activate#(Z)   
    1] activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
    2] activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
    3] activate#(n!6220!6220first(X, Y)) =#> activate#(Y)   
    4] activate#(n!6220!6220from(X)) =#> from#(activate(X))   
    5] activate#(n!6220!6220from(X)) =#> activate#(X)   
    6] activate#(n!6220!6220s(X)) =#> s#(activate(X))   
    7] activate#(n!6220!6220s(X)) =#> activate#(X)   

  Rules R_0:

    first(0, X) => nil 
    first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
    from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
    first(X, Y) => n!6220!6220first(X, Y) 
    from(X) => n!6220!6220from(X) 
    s(X) => n!6220!6220s(X) 
    activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
    activate(n!6220!6220from(X)) => from(activate(X)) 
    activate(n!6220!6220s(X)) => s(activate(X)) 
    activate(X) => X 

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, 2, 3, 4, 5, 6, 7 
    * 1 : 0 
    * 2 : 1, 2, 3, 4, 5, 6, 7 
    * 3 : 1, 2, 3, 4, 5, 6, 7 
    * 4 :  
    * 5 : 1, 2, 3, 4, 5, 6, 7 
    * 6 :  
    * 7 : 1, 2, 3, 4, 5, 6, 7 

This graph has the following strongly connected components:

  P_1:

    first#(s(X), cons(Y, Z)) =#> activate#(Z)   
    activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
    activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
    activate#(n!6220!6220first(X, Y)) =#> activate#(Y)   
    activate#(n!6220!6220from(X)) =#> activate#(X)   
    activate#(n!6220!6220s(X)) =#> activate#(X)   

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 ::=

  first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
  from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
  first(X, Y) => n!6220!6220first(X, Y) 
  from(X) => n!6220!6220from(X) 
  s(X) => n!6220!6220s(X) 
  activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) => from(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(X) => X 

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:

  first#(s(X), cons(Y, Z)) >? activate#(Z) 
  activate#(n!6220!6220first(X, Y)) >? first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? activate#(Y) 
  activate#(n!6220!6220from(X)) >? activate#(X) 
  activate#(n!6220!6220s(X)) >? activate#(X) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  from(X) >= n!6220!6220from(X) 
  s(X) >= n!6220!6220s(X) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) >= from(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  activate = \y0.y0 
  activate# = \y0.y0 
  cons = \y0y1.y1 
  first = \y0y1.y1 + 2y0 
  first# = \y0y1.y1 
  from = \y0.1 + 2y0 
  n!6220!6220first = \y0y1.y1 + 2y0 
  n!6220!6220from = \y0.1 + 2y0 
  n!6220!6220s = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[first#(s(_x0), cons(_x1, _x2))]] = x2 >= x2 = [[activate#(_x2)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x1 + 2x0 >= x1 = [[first#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x1 + 2x0 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x1 + 2x0 >= x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220from(_x0))]] = 1 + 2x0 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220s(_x0))]] = x0 >= x0 = [[activate#(_x0)]] 
  [[first(s(_x0), cons(_x1, _x2))]] = x2 + 2x0 >= x2 + 2x0 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] 
  [[from(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] 
  [[first(_x0, _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[n!6220!6220first(_x0, _x1)]] 
  [[from(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[n!6220!6220from(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[activate(n!6220!6220first(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[first(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220from(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[from(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

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:

  first#(s(X), cons(Y, Z)) =#> activate#(Z)   
  activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
  activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
  activate#(n!6220!6220first(X, Y)) =#> activate#(Y)   
  activate#(n!6220!6220s(X)) =#> activate#(X)   

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:

  first#(s(X), cons(Y, Z)) >? activate#(Z) 
  activate#(n!6220!6220first(X, Y)) >? first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? activate#(Y) 
  activate#(n!6220!6220s(X)) >? activate#(X) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  from(X) >= n!6220!6220from(X) 
  s(X) >= n!6220!6220s(X) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) >= from(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  activate = \y0.y0 
  activate# = \y0.y0 
  cons = \y0y1.2y1 
  first = \y0y1.y0 + 2y1 
  first# = \y0y1.y1 
  from = \y0.0 
  n!6220!6220first = \y0y1.y0 + 2y1 
  n!6220!6220from = \y0.0 
  n!6220!6220s = \y0.1 + 2y0 
  s = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[first#(s(_x0), cons(_x1, _x2))]] = 2x2 >= x2 = [[activate#(_x2)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x0 + 2x1 >= x1 = [[first#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x0 + 2x1 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x0 + 2x1 >= x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220s(_x0))]] = 1 + 2x0 > x0 = [[activate#(_x0)]] 
  [[first(s(_x0), cons(_x1, _x2))]] = 1 + 2x0 + 4x2 >= 2x0 + 4x2 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] 
  [[from(_x0)]] = 0 >= 0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] 
  [[first(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[n!6220!6220first(_x0, _x1)]] 
  [[from(_x0)]] = 0 >= 0 = [[n!6220!6220from(_x0)]] 
  [[s(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[n!6220!6220s(_x0)]] 
  [[activate(n!6220!6220first(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[first(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220from(_x0))]] = 0 >= 0 = [[from(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[s(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

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:

  first#(s(X), cons(Y, Z)) =#> activate#(Z)   
  activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
  activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
  activate#(n!6220!6220first(X, Y)) =#> activate#(Y)   

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 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:

  first#(s(X), cons(Y, Z)) >? activate#(Z) 
  activate#(n!6220!6220first(X, Y)) >? first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? activate#(Y) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  from(X) >= n!6220!6220from(X) 
  s(X) >= n!6220!6220s(X) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) >= from(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  activate = \y0.y0 
  activate# = \y0.y0 
  cons = \y0y1.y1 
  first = \y0y1.2 + 2y0 + 2y1 
  first# = \y0y1.2y1 
  from = \y0.0 
  n!6220!6220first = \y0y1.2 + 2y0 + 2y1 
  n!6220!6220from = \y0.0 
  n!6220!6220s = \y0.2y0 
  s = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[first#(s(_x0), cons(_x1, _x2))]] = 2x2 >= x2 = [[activate#(_x2)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2 + 2x0 + 2x1 > 2x1 = [[first#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2 + 2x0 + 2x1 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2 + 2x0 + 2x1 > x1 = [[activate#(_x1)]] 
  [[first(s(_x0), cons(_x1, _x2))]] = 2 + 2x2 + 4x0 >= 2 + 2x0 + 2x2 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] 
  [[from(_x0)]] = 0 >= 0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] 
  [[first(_x0, _x1)]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[n!6220!6220first(_x0, _x1)]] 
  [[from(_x0)]] = 0 >= 0 = [[n!6220!6220from(_x0)]] 
  [[s(_x0)]] = 2x0 >= 2x0 = [[n!6220!6220s(_x0)]] 
  [[activate(n!6220!6220first(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[first(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220from(_x0))]] = 0 >= 0 = [[from(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = 2x0 >= 2x0 = [[s(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of:

  first#(s(X), cons(Y, Z)) =#> activate#(Z)   

Thus, the original system is terminating if (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_1, 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 :  

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.