/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 
  active : [o] --> o 
  cons : [o * o] --> o 
  mark : [o] --> o 
  tail : [o] --> o 
  zeros : [] --> o 

  active(zeros) => mark(cons(0, zeros)) 
  active(tail(cons(X, Y))) => mark(Y) 
  mark(zeros) => active(zeros) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(0) => active(0) 
  mark(tail(X)) => active(tail(mark(X))) 
  cons(mark(X), Y) => cons(X, Y) 
  cons(X, mark(Y)) => cons(X, Y) 
  cons(active(X), Y) => cons(X, Y) 
  cons(X, active(Y)) => cons(X, Y) 
  tail(mark(X)) => tail(X) 
  tail(active(X)) => tail(X) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  active(zeros) >? mark(cons(0, zeros)) 
  active(tail(cons(X, Y))) >? mark(Y) 
  mark(zeros) >? active(zeros) 
  mark(cons(X, Y)) >? active(cons(mark(X), Y)) 
  mark(0) >? active(0) 
  mark(tail(X)) >? active(tail(mark(X))) 
  cons(mark(X), Y) >? cons(X, Y) 
  cons(X, mark(Y)) >? cons(X, Y) 
  cons(active(X), Y) >? cons(X, Y) 
  cons(X, active(Y)) >? cons(X, Y) 
  tail(mark(X)) >? tail(X) 
  tail(active(X)) >? tail(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  active = \y0.y0 
  cons = \y0y1.y0 + y1 
  mark = \y0.y0 
  tail = \y0.1 + 2y0 
  zeros = 0 

Using this interpretation, the requirements translate to:

  [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] 
  [[active(tail(cons(_x0, _x1)))]] = 1 + 2x0 + 2x1 > x1 = [[mark(_x1)]] 
  [[mark(zeros)]] = 0 >= 0 = [[active(zeros)]] 
  [[mark(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(tail(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[active(tail(mark(_x0)))]] 
  [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[tail(mark(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[tail(_x0)]] 
  [[tail(active(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  active(tail(cons(X, Y))) => mark(Y) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  active(zeros) >? mark(cons(0, zeros)) 
  mark(zeros) >? active(zeros) 
  mark(cons(X, Y)) >? active(cons(mark(X), Y)) 
  mark(0) >? active(0) 
  mark(tail(X)) >? active(tail(mark(X))) 
  cons(mark(X), Y) >? cons(X, Y) 
  cons(X, mark(Y)) >? cons(X, Y) 
  cons(active(X), Y) >? cons(X, Y) 
  cons(X, active(Y)) >? cons(X, Y) 
  tail(mark(X)) >? tail(X) 
  tail(active(X)) >? tail(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  active = \y0.y0 
  cons = \y0y1.y0 + y1 
  mark = \y0.2y0 
  tail = \y0.2 + 2y0 
  zeros = 0 

Using this interpretation, the requirements translate to:

  [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] 
  [[mark(zeros)]] = 0 >= 0 = [[active(zeros)]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(tail(_x0))]] = 4 + 4x0 > 2 + 4x0 = [[active(tail(mark(_x0)))]] 
  [[cons(mark(_x0), _x1)]] = x1 + 2x0 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = x0 + 2x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[tail(mark(_x0))]] = 2 + 4x0 >= 2 + 2x0 = [[tail(_x0)]] 
  [[tail(active(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  mark(tail(X)) => active(tail(mark(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] active#(zeros) =#> mark#(cons(0, zeros))   
    1] active#(zeros) =#> cons#(0, zeros)   
    2] mark#(zeros) =#> active#(zeros)   
    3] mark#(cons(X, Y)) =#> active#(cons(mark(X), Y))   
    4] mark#(cons(X, Y)) =#> cons#(mark(X), Y)   
    5] mark#(cons(X, Y)) =#> mark#(X)   
    6] mark#(0) =#> active#(0)   
    7] cons#(mark(X), Y) =#> cons#(X, Y)   
    8] cons#(X, mark(Y)) =#> cons#(X, Y)   
    9] cons#(active(X), Y) =#> cons#(X, Y)   
    10] cons#(X, active(Y)) =#> cons#(X, Y)   
    11] tail#(mark(X)) =#> tail#(X)   
    12] tail#(active(X)) =#> tail#(X)   

  Rules R_0:

    active(zeros) => mark(cons(0, zeros)) 
    mark(zeros) => active(zeros) 
    mark(cons(X, Y)) => active(cons(mark(X), Y)) 
    mark(0) => active(0) 
    cons(mark(X), Y) => cons(X, Y) 
    cons(X, mark(Y)) => cons(X, Y) 
    cons(active(X), Y) => cons(X, Y) 
    cons(X, active(Y)) => cons(X, Y) 
    tail(mark(X)) => tail(X) 
    tail(active(X)) => tail(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 
    * 1 :  
    * 2 : 0, 1 
    * 3 :  
    * 4 : 7, 8, 9, 10 
    * 5 : 2, 3, 4, 5, 6 
    * 6 :  
    * 7 : 7, 8, 9, 10 
    * 8 : 7, 8, 9, 10 
    * 9 : 7, 8, 9, 10 
    * 10 : 7, 8, 9, 10 
    * 11 : 11, 12 
    * 12 : 11, 12 

This graph has the following strongly connected components:

  P_1:

    active#(zeros) =#> mark#(cons(0, zeros))   
    mark#(zeros) =#> active#(zeros)   
    mark#(cons(X, Y)) =#> mark#(X)   

  P_2:

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

  P_3:

    tail#(mark(X)) =#> tail#(X)   
    tail#(active(X)) =#> tail#(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) and (P_3, 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) 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(tail#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(tail#(mark(X))) = mark(X) |> X = nu(tail#(X)) 
  nu(tail#(active(X))) = active(X) |> X = nu(tail#(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.

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(cons#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains:

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

Thus, the original system is terminating if each of (P_1, 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(cons#) = 2 

Thus, we can orient the dependency pairs as follows:

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

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, 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 ::=

  active(zeros) => mark(cons(0, zeros)) 
  mark(zeros) => active(zeros) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(0) => active(0) 
  cons(mark(X), Y) => cons(X, Y) 
  cons(X, mark(Y)) => cons(X, Y) 
  cons(active(X), Y) => cons(X, Y) 
  cons(X, active(Y)) => cons(X, Y) 

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 with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_1) are:

  cons(mark(X), Y) => cons(X, Y) 
  cons(X, mark(Y)) => cons(X, Y) 
  cons(active(X), Y) => cons(X, Y) 
  cons(X, active(Y)) => cons(X, Y) 

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

  active#(zeros) >? mark#(cons(0, zeros)) 
  mark#(zeros) >? active#(zeros) 
  mark#(cons(X, Y)) >? mark#(X) 
  cons(mark(X), Y) >= cons(X, Y) 
  cons(X, mark(Y)) >= cons(X, Y) 
  cons(active(X), Y) >= cons(X, Y) 
  cons(X, active(Y)) >= cons(X, Y) 

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

We consider usable_rules with respect to the following argument filtering:


This leaves the following ordering requirements:

  active#(zeros) > mark#(cons(0, zeros)) 
  mark#(zeros) >= active#(zeros) 
  mark#(cons(X, Y)) >= mark#(X) 

The following interpretation satisfies the requirements:

  0 = 0 
  active = \y0.0 
  active# = \y0.2y0 
  cons = \y0y1.2y0 
  mark = \y0.0 
  mark# = \y0.3y0 
  zeros = 3 

Using this interpretation, the requirements translate to:

  [[active#(zeros)]] = 6 > 0 = [[mark#(cons(0, zeros))]] 
  [[mark#(zeros)]] = 9 > 6 = [[active#(zeros)]] 
  [[mark#(cons(_x0, _x1))]] = 6x0 >= 3x0 = [[mark#(_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_5, R_1, minimal, formative), where P_5 consists of:

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

Thus, the original system is terminating if (P_5, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_5, R_1, 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#(cons(X, Y))) = cons(X, Y) |> X = nu(mark#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_1, minimal, f) by ({}, R_1, 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.