/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!6220!6220add : [o * o] --> o 
  a!6220!6220from : [o] --> o 
  a!6220!6220fst : [o * o] --> o 
  a!6220!6220len : [o] --> o 
  add : [o * o] --> o 
  cons : [o * o] --> o 
  from : [o] --> o 
  fst : [o * o] --> o 
  len : [o] --> o 
  mark : [o] --> o 
  nil : [] --> o 
  s : [o] --> o 

  a!6220!6220fst(0, X) => nil 
  a!6220!6220fst(s(X), cons(Y, Z)) => cons(mark(Y), fst(X, Z)) 
  a!6220!6220from(X) => cons(mark(X), from(s(X))) 
  a!6220!6220add(0, X) => mark(X) 
  a!6220!6220add(s(X), Y) => s(add(X, Y)) 
  a!6220!6220len(nil) => 0 
  a!6220!6220len(cons(X, Y)) => s(len(Y)) 
  mark(fst(X, Y)) => a!6220!6220fst(mark(X), mark(Y)) 
  mark(from(X)) => a!6220!6220from(mark(X)) 
  mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) 
  mark(len(X)) => a!6220!6220len(mark(X)) 
  mark(0) => 0 
  mark(s(X)) => s(X) 
  mark(nil) => nil 
  mark(cons(X, Y)) => cons(mark(X), Y) 
  a!6220!6220fst(X, Y) => fst(X, Y) 
  a!6220!6220from(X) => from(X) 
  a!6220!6220add(X, Y) => add(X, Y) 
  a!6220!6220len(X) => len(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] a!6220!6220fst#(s(X), cons(Y, Z)) =#> mark#(Y)   
    1] a!6220!6220from#(X) =#> mark#(X)   
    2] a!6220!6220add#(0, X) =#> mark#(X)   
    3] mark#(fst(X, Y)) =#> a!6220!6220fst#(mark(X), mark(Y))   
    4] mark#(fst(X, Y)) =#> mark#(X)   
    5] mark#(fst(X, Y)) =#> mark#(Y)   
    6] mark#(from(X)) =#> a!6220!6220from#(mark(X))   
    7] mark#(from(X)) =#> mark#(X)   
    8] mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y))   
    9] mark#(add(X, Y)) =#> mark#(X)   
    10] mark#(add(X, Y)) =#> mark#(Y)   
    11] mark#(len(X)) =#> a!6220!6220len#(mark(X))   
    12] mark#(len(X)) =#> mark#(X)   
    13] mark#(cons(X, Y)) =#> mark#(X)   

  Rules R_0:

    a!6220!6220fst(0, X) => nil 
    a!6220!6220fst(s(X), cons(Y, Z)) => cons(mark(Y), fst(X, Z)) 
    a!6220!6220from(X) => cons(mark(X), from(s(X))) 
    a!6220!6220add(0, X) => mark(X) 
    a!6220!6220add(s(X), Y) => s(add(X, Y)) 
    a!6220!6220len(nil) => 0 
    a!6220!6220len(cons(X, Y)) => s(len(Y)) 
    mark(fst(X, Y)) => a!6220!6220fst(mark(X), mark(Y)) 
    mark(from(X)) => a!6220!6220from(mark(X)) 
    mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) 
    mark(len(X)) => a!6220!6220len(mark(X)) 
    mark(0) => 0 
    mark(s(X)) => s(X) 
    mark(nil) => nil 
    mark(cons(X, Y)) => cons(mark(X), Y) 
    a!6220!6220fst(X, Y) => fst(X, Y) 
    a!6220!6220from(X) => from(X) 
    a!6220!6220add(X, Y) => add(X, Y) 
    a!6220!6220len(X) => len(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, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 1 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 2 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 3 : 0 
    * 4 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 5 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 6 : 1 
    * 7 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 8 : 2 
    * 9 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 10 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 11 :  
    * 12 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 
    * 13 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 

This graph has the following strongly connected components:

  P_1:

    a!6220!6220fst#(s(X), cons(Y, Z)) =#> mark#(Y)   
    a!6220!6220from#(X) =#> mark#(X)   
    a!6220!6220add#(0, X) =#> mark#(X)   
    mark#(fst(X, Y)) =#> a!6220!6220fst#(mark(X), mark(Y))   
    mark#(fst(X, Y)) =#> mark#(X)   
    mark#(fst(X, Y)) =#> mark#(Y)   
    mark#(from(X)) =#> a!6220!6220from#(mark(X))   
    mark#(from(X)) =#> mark#(X)   
    mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y))   
    mark#(add(X, Y)) =#> mark#(X)   
    mark#(add(X, Y)) =#> mark#(Y)   
    mark#(len(X)) =#> mark#(X)   
    mark#(cons(X, Y)) =#> mark#(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).

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:

  a!6220!6220fst#(s(X), cons(Y, Z)) >? mark#(Y) 
  a!6220!6220from#(X) >? mark#(X) 
  a!6220!6220add#(0, X) >? mark#(X) 
  mark#(fst(X, Y)) >? a!6220!6220fst#(mark(X), mark(Y)) 
  mark#(fst(X, Y)) >? mark#(X) 
  mark#(fst(X, Y)) >? mark#(Y) 
  mark#(from(X)) >? a!6220!6220from#(mark(X)) 
  mark#(from(X)) >? mark#(X) 
  mark#(add(X, Y)) >? a!6220!6220add#(mark(X), mark(Y)) 
  mark#(add(X, Y)) >? mark#(X) 
  mark#(add(X, Y)) >? mark#(Y) 
  mark#(len(X)) >? mark#(X) 
  mark#(cons(X, Y)) >? mark#(X) 
  a!6220!6220fst(0, X) >= nil 
  a!6220!6220fst(s(X), cons(Y, Z)) >= cons(mark(Y), fst(X, Z)) 
  a!6220!6220from(X) >= cons(mark(X), from(s(X))) 
  a!6220!6220add(0, X) >= mark(X) 
  a!6220!6220add(s(X), Y) >= s(add(X, Y)) 
  a!6220!6220len(nil) >= 0 
  a!6220!6220len(cons(X, Y)) >= s(len(Y)) 
  mark(fst(X, Y)) >= a!6220!6220fst(mark(X), mark(Y)) 
  mark(from(X)) >= a!6220!6220from(mark(X)) 
  mark(add(X, Y)) >= a!6220!6220add(mark(X), mark(Y)) 
  mark(len(X)) >= a!6220!6220len(mark(X)) 
  mark(0) >= 0 
  mark(s(X)) >= s(X) 
  mark(nil) >= nil 
  mark(cons(X, Y)) >= cons(mark(X), Y) 
  a!6220!6220fst(X, Y) >= fst(X, Y) 
  a!6220!6220from(X) >= from(X) 
  a!6220!6220add(X, Y) >= add(X, Y) 
  a!6220!6220len(X) >= len(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220add = \y0y1.3 + y0 + y1 
  a!6220!6220add# = \y0y1.1 + y1 
  a!6220!6220from = \y0.1 + 2y0 
  a!6220!6220from# = \y0.1 + 2y0 
  a!6220!6220fst = \y0y1.2y0 + 2y1 
  a!6220!6220fst# = \y0y1.2y0 + 2y1 
  a!6220!6220len = \y0.2 + y0 
  add = \y0y1.3 + y0 + y1 
  cons = \y0y1.2y0 
  from = \y0.1 + 2y0 
  fst = \y0y1.2y0 + 2y1 
  len = \y0.2 + y0 
  mark = \y0.y0 
  mark# = \y0.1 + y0 
  nil = 0 
  s = \y0.2 

Using this interpretation, the requirements translate to:

  [[a!6220!6220fst#(s(_x0), cons(_x1, _x2))]] = 4 + 4x1 > 1 + x1 = [[mark#(_x1)]] 
  [[a!6220!6220from#(_x0)]] = 1 + 2x0 >= 1 + x0 = [[mark#(_x0)]] 
  [[a!6220!6220add#(0, _x0)]] = 1 + x0 >= 1 + x0 = [[mark#(_x0)]] 
  [[mark#(fst(_x0, _x1))]] = 1 + 2x0 + 2x1 > 2x0 + 2x1 = [[a!6220!6220fst#(mark(_x0), mark(_x1))]] 
  [[mark#(fst(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + x0 = [[mark#(_x0)]] 
  [[mark#(fst(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + x1 = [[mark#(_x1)]] 
  [[mark#(from(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[a!6220!6220from#(mark(_x0))]] 
  [[mark#(from(_x0))]] = 2 + 2x0 > 1 + x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = 4 + x0 + x1 > 1 + x1 = [[a!6220!6220add#(mark(_x0), mark(_x1))]] 
  [[mark#(add(_x0, _x1))]] = 4 + x0 + x1 > 1 + x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = 4 + x0 + x1 > 1 + x1 = [[mark#(_x1)]] 
  [[mark#(len(_x0))]] = 3 + x0 > 1 + x0 = [[mark#(_x0)]] 
  [[mark#(cons(_x0, _x1))]] = 1 + 2x0 >= 1 + x0 = [[mark#(_x0)]] 
  [[a!6220!6220fst(0, _x0)]] = 2x0 >= 0 = [[nil]] 
  [[a!6220!6220fst(s(_x0), cons(_x1, _x2))]] = 4 + 4x1 >= 2x1 = [[cons(mark(_x1), fst(_x0, _x2))]] 
  [[a!6220!6220from(_x0)]] = 1 + 2x0 >= 2x0 = [[cons(mark(_x0), from(s(_x0)))]] 
  [[a!6220!6220add(0, _x0)]] = 3 + x0 >= x0 = [[mark(_x0)]] 
  [[a!6220!6220add(s(_x0), _x1)]] = 5 + x1 >= 2 = [[s(add(_x0, _x1))]] 
  [[a!6220!6220len(nil)]] = 2 >= 0 = [[0]] 
  [[a!6220!6220len(cons(_x0, _x1))]] = 2 + 2x0 >= 2 = [[s(len(_x1))]] 
  [[mark(fst(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220fst(mark(_x0), mark(_x1))]] 
  [[mark(from(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[a!6220!6220from(mark(_x0))]] 
  [[mark(add(_x0, _x1))]] = 3 + x0 + x1 >= 3 + x0 + x1 = [[a!6220!6220add(mark(_x0), mark(_x1))]] 
  [[mark(len(_x0))]] = 2 + x0 >= 2 + x0 = [[a!6220!6220len(mark(_x0))]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2 >= 2 = [[s(_x0)]] 
  [[mark(nil)]] = 0 >= 0 = [[nil]] 
  [[mark(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[cons(mark(_x0), _x1)]] 
  [[a!6220!6220fst(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[fst(_x0, _x1)]] 
  [[a!6220!6220from(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[from(_x0)]] 
  [[a!6220!6220add(_x0, _x1)]] = 3 + x0 + x1 >= 3 + x0 + x1 = [[add(_x0, _x1)]] 
  [[a!6220!6220len(_x0)]] = 2 + x0 >= 2 + x0 = [[len(_x0)]] 

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_2, R_0, minimal, formative), where P_2 consists of:

  a!6220!6220from#(X) =#> mark#(X)   
  a!6220!6220add#(0, X) =#> mark#(X)   
  mark#(fst(X, Y)) =#> mark#(X)   
  mark#(fst(X, Y)) =#> mark#(Y)   
  mark#(cons(X, Y)) =#> mark#(X)   

Thus, the original system is terminating if (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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 : 2, 3, 4 
    * 1 : 2, 3, 4 
    * 2 : 2, 3, 4 
    * 3 : 2, 3, 4 
    * 4 : 2, 3, 4 

This graph has the following strongly connected components:

  P_3:

    mark#(fst(X, Y)) =#> mark#(X)   
    mark#(fst(X, Y)) =#> mark#(Y)   
    mark#(cons(X, Y)) =#> mark#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_2, R_0, m, f) by (P_3, R_0, m, f).

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 apply the subterm criterion with the following projection function:

  nu(mark#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(mark#(fst(X, Y))) = fst(X, Y) |> X = nu(mark#(X)) 
  nu(mark#(fst(X, Y))) = fst(X, Y) |> Y = nu(mark#(Y)) 
  nu(mark#(cons(X, Y))) = cons(X, Y) |> X = nu(mark#(X)) 

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.

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.