/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 
  A : [] --> o 
  B : [] --> o 
  C : [] --> o 
  f : [o * o] --> o 
  f!450 : [o * o] --> o 
  f!450!450 : [o] --> o 
  fold : [o * o * o] --> o 
  foldB : [o * o] --> o 
  foldC : [o * o] --> o 
  g : [o] --> o 
  s : [o] --> o 
  triple : [o * o * o] --> o 

  g(A) => A 
  g(B) => A 
  g(B) => B 
  g(C) => A 
  g(C) => B 
  g(C) => C 
  foldB(X, 0) => X 
  foldB(X, s(Y)) => f(foldB(X, Y), B) 
  foldC(X, 0) => X 
  foldC(X, s(Y)) => f(foldC(X, Y), C) 
  f(X, Y) => f!450(X, g(Y)) 
  f!450(triple(X, Y, Z), C) => triple(X, Y, s(Z)) 
  f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) 
  f!450(triple(X, Y, Z), A) => f!450!450(foldB(triple(s(X), 0, Z), Y)) 
  f!450!450(triple(X, Y, Z)) => foldC(triple(X, Y, 0), Z) 
  fold(X, Y, 0) => X 
  fold(X, Y, s(Z)) => f(fold(X, Y, Z), 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] foldB#(X, s(Y)) =#> f#(foldB(X, Y), B)   
    1] foldB#(X, s(Y)) =#> foldB#(X, Y)   
    2] foldC#(X, s(Y)) =#> f#(foldC(X, Y), C)   
    3] foldC#(X, s(Y)) =#> foldC#(X, Y)   
    4] f#(X, Y) =#> f!450#(X, g(Y))   
    5] f#(X, Y) =#> g#(Y)   
    6] f!450#(triple(X, Y, Z), B) =#> f#(triple(X, Y, Z), A)   
    7] f!450#(triple(X, Y, Z), A) =#> f!450!450#(foldB(triple(s(X), 0, Z), Y))   
    8] f!450#(triple(X, Y, Z), A) =#> foldB#(triple(s(X), 0, Z), Y)   
    9] f!450!450#(triple(X, Y, Z)) =#> foldC#(triple(X, Y, 0), Z)   
    10] fold#(X, Y, s(Z)) =#> f#(fold(X, Y, Z), Y)   
    11] fold#(X, Y, s(Z)) =#> fold#(X, Y, Z)   

  Rules R_0:

    g(A) => A 
    g(B) => A 
    g(B) => B 
    g(C) => A 
    g(C) => B 
    g(C) => C 
    foldB(X, 0) => X 
    foldB(X, s(Y)) => f(foldB(X, Y), B) 
    foldC(X, 0) => X 
    foldC(X, s(Y)) => f(foldC(X, Y), C) 
    f(X, Y) => f!450(X, g(Y)) 
    f!450(triple(X, Y, Z), C) => triple(X, Y, s(Z)) 
    f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) 
    f!450(triple(X, Y, Z), A) => f!450!450(foldB(triple(s(X), 0, Z), Y)) 
    f!450!450(triple(X, Y, Z)) => foldC(triple(X, Y, 0), Z) 
    fold(X, Y, 0) => X 
    fold(X, Y, s(Z)) => f(fold(X, Y, Z), 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 : 4, 5 
    * 1 : 0, 1 
    * 2 : 4, 5 
    * 3 : 2, 3 
    * 4 : 6, 7, 8 
    * 5 :  
    * 6 : 4, 5 
    * 7 : 9 
    * 8 : 0, 1 
    * 9 : 2, 3 
    * 10 : 4, 5 
    * 11 : 10, 11 

This graph has the following strongly connected components:

  P_1:

    foldB#(X, s(Y)) =#> f#(foldB(X, Y), B)   
    foldB#(X, s(Y)) =#> foldB#(X, Y)   
    foldC#(X, s(Y)) =#> f#(foldC(X, Y), C)   
    foldC#(X, s(Y)) =#> foldC#(X, Y)   
    f#(X, Y) =#> f!450#(X, g(Y))   
    f!450#(triple(X, Y, Z), B) =#> f#(triple(X, Y, Z), A)   
    f!450#(triple(X, Y, Z), A) =#> f!450!450#(foldB(triple(s(X), 0, Z), Y))   
    f!450#(triple(X, Y, Z), A) =#> foldB#(triple(s(X), 0, Z), Y)   
    f!450!450#(triple(X, Y, Z)) =#> foldC#(triple(X, Y, 0), Z)   

  P_2:

    fold#(X, Y, s(Z)) =#> fold#(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).

We apply the subterm criterion with the following projection function:

  nu(fold#) = 3 

Thus, we can orient the dependency pairs as follows:

  nu(fold#(X, Y, s(Z))) = s(Z) |> Z = nu(fold#(X, Y, Z)) 

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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_0) are:

  g(A) => A 
  g(B) => A 
  g(B) => B 
  g(C) => A 
  g(C) => B 
  g(C) => C 
  foldB(X, 0) => X 
  foldB(X, s(Y)) => f(foldB(X, Y), B) 
  foldC(X, 0) => X 
  foldC(X, s(Y)) => f(foldC(X, Y), C) 
  f(X, Y) => f!450(X, g(Y)) 
  f!450(triple(X, Y, Z), C) => triple(X, Y, s(Z)) 
  f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) 
  f!450(triple(X, Y, Z), A) => f!450!450(foldB(triple(s(X), 0, Z), Y)) 
  f!450!450(triple(X, Y, Z)) => foldC(triple(X, Y, 0), Z) 

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

  foldB#(X, s(Y)) >? f#(foldB(X, Y), B) 
  foldB#(X, s(Y)) >? foldB#(X, Y) 
  foldC#(X, s(Y)) >? f#(foldC(X, Y), C) 
  foldC#(X, s(Y)) >? foldC#(X, Y) 
  f#(X, Y) >? f!450#(X, g(Y)) 
  f!450#(triple(X, Y, Z), B) >? f#(triple(X, Y, Z), A) 
  f!450#(triple(X, Y, Z), A) >? f!450!450#(foldB(triple(s(X), 0, Z), Y)) 
  f!450#(triple(X, Y, Z), A) >? foldB#(triple(s(X), 0, Z), Y) 
  f!450!450#(triple(X, Y, Z)) >? foldC#(triple(X, Y, 0), Z) 
  g(A) >= A 
  g(B) >= A 
  g(B) >= B 
  g(C) >= A 
  g(C) >= B 
  g(C) >= C 
  foldB(X, 0) >= X 
  foldB(X, s(Y)) >= f(foldB(X, Y), B) 
  foldC(X, 0) >= X 
  foldC(X, s(Y)) >= f(foldC(X, Y), C) 
  f(X, Y) >= f!450(X, g(Y)) 
  f!450(triple(X, Y, Z), C) >= triple(X, Y, s(Z)) 
  f!450(triple(X, Y, Z), B) >= f(triple(X, Y, Z), A) 
  f!450(triple(X, Y, Z), A) >= f!450!450(foldB(triple(s(X), 0, Z), Y)) 
  f!450!450(triple(X, Y, Z)) >= foldC(triple(X, Y, 0), Z) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  A = 0 
  B = 0 
  C = 0 
  f = \y0y1.1 + y0 
  f!450 = \y0y1.1 + y0 
  f!450!450 = \y0.y0 
  f!450!450# = \y0.y0 
  f!450# = \y0y1.y0 
  f# = \y0y1.y0 
  foldB = \y0y1.y0 + 2y1 
  foldB# = \y0y1.y0 + 2y1 
  foldC = \y0y1.y0 + y1 
  foldC# = \y0y1.y0 + y1 
  g = \y0.0 
  s = \y0.1 + y0 
  triple = \y0y1y2.y2 + 2y1 

Using this interpretation, the requirements translate to:

  [[foldB#(_x0, s(_x1))]] = 2 + x0 + 2x1 > x0 + 2x1 = [[f#(foldB(_x0, _x1), B)]] 
  [[foldB#(_x0, s(_x1))]] = 2 + x0 + 2x1 > x0 + 2x1 = [[foldB#(_x0, _x1)]] 
  [[foldC#(_x0, s(_x1))]] = 1 + x0 + x1 > x0 + x1 = [[f#(foldC(_x0, _x1), C)]] 
  [[foldC#(_x0, s(_x1))]] = 1 + x0 + x1 > x0 + x1 = [[foldC#(_x0, _x1)]] 
  [[f#(_x0, _x1)]] = x0 >= x0 = [[f!450#(_x0, g(_x1))]] 
  [[f!450#(triple(_x0, _x1, _x2), B)]] = x2 + 2x1 >= x2 + 2x1 = [[f#(triple(_x0, _x1, _x2), A)]] 
  [[f!450#(triple(_x0, _x1, _x2), A)]] = x2 + 2x1 >= x2 + 2x1 = [[f!450!450#(foldB(triple(s(_x0), 0, _x2), _x1))]] 
  [[f!450#(triple(_x0, _x1, _x2), A)]] = x2 + 2x1 >= x2 + 2x1 = [[foldB#(triple(s(_x0), 0, _x2), _x1)]] 
  [[f!450!450#(triple(_x0, _x1, _x2))]] = x2 + 2x1 >= x2 + 2x1 = [[foldC#(triple(_x0, _x1, 0), _x2)]] 
  [[g(A)]] = 0 >= 0 = [[A]] 
  [[g(B)]] = 0 >= 0 = [[A]] 
  [[g(B)]] = 0 >= 0 = [[B]] 
  [[g(C)]] = 0 >= 0 = [[A]] 
  [[g(C)]] = 0 >= 0 = [[B]] 
  [[g(C)]] = 0 >= 0 = [[C]] 
  [[foldB(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[foldB(_x0, s(_x1))]] = 2 + x0 + 2x1 >= 1 + x0 + 2x1 = [[f(foldB(_x0, _x1), B)]] 
  [[foldC(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[foldC(_x0, s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[f(foldC(_x0, _x1), C)]] 
  [[f(_x0, _x1)]] = 1 + x0 >= 1 + x0 = [[f!450(_x0, g(_x1))]] 
  [[f!450(triple(_x0, _x1, _x2), C)]] = 1 + x2 + 2x1 >= 1 + x2 + 2x1 = [[triple(_x0, _x1, s(_x2))]] 
  [[f!450(triple(_x0, _x1, _x2), B)]] = 1 + x2 + 2x1 >= 1 + x2 + 2x1 = [[f(triple(_x0, _x1, _x2), A)]] 
  [[f!450(triple(_x0, _x1, _x2), A)]] = 1 + x2 + 2x1 >= x2 + 2x1 = [[f!450!450(foldB(triple(s(_x0), 0, _x2), _x1))]] 
  [[f!450!450(triple(_x0, _x1, _x2))]] = x2 + 2x1 >= x2 + 2x1 = [[foldC(triple(_x0, _x1, 0), _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, formative) by (P_3, R_0, minimal, formative), where P_3 consists of:

  f#(X, Y) =#> f!450#(X, g(Y))   
  f!450#(triple(X, Y, Z), B) =#> f#(triple(X, Y, Z), A)   
  f!450#(triple(X, Y, Z), A) =#> f!450!450#(foldB(triple(s(X), 0, Z), Y))   
  f!450#(triple(X, Y, Z), A) =#> foldB#(triple(s(X), 0, Z), Y)   
  f!450!450#(triple(X, Y, Z)) =#> foldC#(triple(X, Y, 0), Z)   

Thus, the original system is terminating if (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, 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 
    * 1 : 0 
    * 2 : 4 
    * 3 :  
    * 4 :  

This graph has the following strongly connected components:

  P_4:

    f#(X, Y) =#> f!450#(X, g(Y))   
    f!450#(triple(X, Y, Z), B) =#> f#(triple(X, Y, Z), A)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_3, R_0, m, f) by (P_4, R_0, m, f).

Thus, the original system is terminating if (P_4, R_0, minimal, formative) is finite.

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

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_4, R_0) are:

  g(A) => A 
  g(B) => A 
  g(B) => B 
  g(C) => A 
  g(C) => B 
  g(C) => C 

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

  f#(X, Y) >? f!450#(X, g(Y)) 
  f!450#(triple(X, Y, Z), B) >? f#(triple(X, Y, Z), A) 
  g(A) >= A 
  g(B) >= A 
  g(B) >= B 
  g(C) >= A 
  g(C) >= B 
  g(C) >= C 

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

The following interpretation satisfies the requirements:

  A = 0 
  B = 3 
  C = 3 
  f!450# = \y0y1.1 + y1 
  f# = \y0y1.2 + y1 
  g = \y0.y0 
  triple = \y0y1y2.0 

Using this interpretation, the requirements translate to:

  [[f#(_x0, _x1)]] = 2 + x1 > 1 + x1 = [[f!450#(_x0, g(_x1))]] 
  [[f!450#(triple(_x0, _x1, _x2), B)]] = 4 > 2 = [[f#(triple(_x0, _x1, _x2), A)]] 
  [[g(A)]] = 0 >= 0 = [[A]] 
  [[g(B)]] = 3 >= 0 = [[A]] 
  [[g(B)]] = 3 >= 3 = [[B]] 
  [[g(C)]] = 3 >= 0 = [[A]] 
  [[g(C)]] = 3 >= 3 = [[B]] 
  [[g(C)]] = 3 >= 3 = [[C]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0).  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.