/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 
  2ndsneg : [o * o] --> o 
  2ndspos : [o * o] --> o 
  activate : [o] --> o 
  cons : [o * o] --> o 
  cons2 : [o * o] --> o 
  from : [o] --> o 
  n!6220!6220from : [o] --> o 
  n!6220!6220s : [o] --> o 
  negrecip : [o] --> o 
  pi : [o] --> o 
  plus : [o * o] --> o 
  posrecip : [o] --> o 
  rcons : [o * o] --> o 
  rnil : [] --> o 
  s : [o] --> o 
  square : [o] --> o 
  times : [o * o] --> o 

  from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
  2ndspos(0, X) => rnil 
  2ndspos(s(X), cons(Y, Z)) => 2ndspos(s(X), cons2(Y, activate(Z))) 
  2ndspos(s(X), cons2(Y, cons(Z, U))) => rcons(posrecip(Z), 2ndsneg(X, activate(U))) 
  2ndsneg(0, X) => rnil 
  2ndsneg(s(X), cons(Y, Z)) => 2ndsneg(s(X), cons2(Y, activate(Z))) 
  2ndsneg(s(X), cons2(Y, cons(Z, U))) => rcons(negrecip(Z), 2ndspos(X, activate(U))) 
  pi(X) => 2ndspos(X, from(0)) 
  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  times(0, X) => 0 
  times(s(X), Y) => plus(Y, times(X, Y)) 
  square(X) => times(X, X) 
  from(X) => n!6220!6220from(X) 
  s(X) => n!6220!6220s(X) 
  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] 2ndspos#(s(X), cons(Y, Z)) =#> 2ndspos#(s(X), cons2(Y, activate(Z)))   
    1] 2ndspos#(s(X), cons(Y, Z)) =#> s#(X)   
    2] 2ndspos#(s(X), cons(Y, Z)) =#> activate#(Z)   
    3] 2ndspos#(s(X), cons2(Y, cons(Z, U))) =#> 2ndsneg#(X, activate(U))   
    4] 2ndspos#(s(X), cons2(Y, cons(Z, U))) =#> activate#(U)   
    5] 2ndsneg#(s(X), cons(Y, Z)) =#> 2ndsneg#(s(X), cons2(Y, activate(Z)))   
    6] 2ndsneg#(s(X), cons(Y, Z)) =#> s#(X)   
    7] 2ndsneg#(s(X), cons(Y, Z)) =#> activate#(Z)   
    8] 2ndsneg#(s(X), cons2(Y, cons(Z, U))) =#> 2ndspos#(X, activate(U))   
    9] 2ndsneg#(s(X), cons2(Y, cons(Z, U))) =#> activate#(U)   
    10] pi#(X) =#> 2ndspos#(X, from(0))   
    11] pi#(X) =#> from#(0)   
    12] plus#(s(X), Y) =#> s#(plus(X, Y))   
    13] plus#(s(X), Y) =#> plus#(X, Y)   
    14] times#(s(X), Y) =#> plus#(Y, times(X, Y))   
    15] times#(s(X), Y) =#> times#(X, Y)   
    16] square#(X) =#> times#(X, X)   
    17] activate#(n!6220!6220from(X)) =#> from#(activate(X))   
    18] activate#(n!6220!6220from(X)) =#> activate#(X)   
    19] activate#(n!6220!6220s(X)) =#> s#(activate(X))   
    20] activate#(n!6220!6220s(X)) =#> activate#(X)   

  Rules R_0:

    from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
    2ndspos(0, X) => rnil 
    2ndspos(s(X), cons(Y, Z)) => 2ndspos(s(X), cons2(Y, activate(Z))) 
    2ndspos(s(X), cons2(Y, cons(Z, U))) => rcons(posrecip(Z), 2ndsneg(X, activate(U))) 
    2ndsneg(0, X) => rnil 
    2ndsneg(s(X), cons(Y, Z)) => 2ndsneg(s(X), cons2(Y, activate(Z))) 
    2ndsneg(s(X), cons2(Y, cons(Z, U))) => rcons(negrecip(Z), 2ndspos(X, activate(U))) 
    pi(X) => 2ndspos(X, from(0)) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    times(0, X) => 0 
    times(s(X), Y) => plus(Y, times(X, Y)) 
    square(X) => times(X, X) 
    from(X) => n!6220!6220from(X) 
    s(X) => n!6220!6220s(X) 
    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 : 3, 4 
    * 1 :  
    * 2 : 17, 18, 19, 20 
    * 3 : 5, 6, 7, 8, 9 
    * 4 : 17, 18, 19, 20 
    * 5 : 8, 9 
    * 6 :  
    * 7 : 17, 18, 19, 20 
    * 8 : 0, 1, 2, 3, 4 
    * 9 : 17, 18, 19, 20 
    * 10 : 0, 1, 2 
    * 11 :  
    * 12 :  
    * 13 : 12, 13 
    * 14 : 12, 13 
    * 15 : 14, 15 
    * 16 : 14, 15 
    * 17 :  
    * 18 : 17, 18, 19, 20 
    * 19 :  
    * 20 : 17, 18, 19, 20 

This graph has the following strongly connected components:

  P_1:

    2ndspos#(s(X), cons(Y, Z)) =#> 2ndspos#(s(X), cons2(Y, activate(Z)))   
    2ndspos#(s(X), cons2(Y, cons(Z, U))) =#> 2ndsneg#(X, activate(U))   
    2ndsneg#(s(X), cons(Y, Z)) =#> 2ndsneg#(s(X), cons2(Y, activate(Z)))   
    2ndsneg#(s(X), cons2(Y, cons(Z, U))) =#> 2ndspos#(X, activate(U))   

  P_2:

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

  P_3:

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

  P_4:

    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), (P_2, R_0, m, f), (P_3, R_0, m, f) and (P_4, 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), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(activate#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(activate#(n!6220!6220from(X))) = n!6220!6220from(X) |> X = nu(activate#(X)) 
  nu(activate#(n!6220!6220s(X))) = n!6220!6220s(X) |> X = nu(activate#(X)) 

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 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 apply the subterm criterion with the following projection function:

  nu(times#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, 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 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(plus#) = 1 

Thus, we can orient the dependency pairs as follows:

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

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(2ndsneg#) = 1 
  nu(2ndspos#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(2ndspos#(s(X), cons(Y, Z))) = s(X) = s(X) = nu(2ndspos#(s(X), cons2(Y, activate(Z)))) 
  nu(2ndspos#(s(X), cons2(Y, cons(Z, U)))) = s(X) |> X = nu(2ndsneg#(X, activate(U))) 
  nu(2ndsneg#(s(X), cons(Y, Z))) = s(X) = s(X) = nu(2ndsneg#(s(X), cons2(Y, activate(Z)))) 
  nu(2ndsneg#(s(X), cons2(Y, cons(Z, U)))) = s(X) |> X = nu(2ndspos#(X, activate(U))) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by (P_5, R_0, minimal, f), where P_5 contains:

  2ndspos#(s(X), cons(Y, Z)) =#> 2ndspos#(s(X), cons2(Y, activate(Z)))   
  2ndsneg#(s(X), cons(Y, Z)) =#> 2ndsneg#(s(X), cons2(Y, activate(Z)))   

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

We consider the dependency pair problem (P_5, 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 :  

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.