/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  a : [] --> o 
  active : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  mark : [o] --> o 

  active(f(f(a))) => mark(f(g(f(a)))) 
  mark(f(X)) => active(f(X)) 
  mark(a) => active(a) 
  mark(g(X)) => active(g(mark(X))) 
  f(mark(X)) => f(X) 
  f(active(X)) => f(X) 
  g(mark(X)) => g(X) 
  g(active(X)) => g(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#(f(f(a))) =#> mark#(f(g(f(a))))   
    1] active#(f(f(a))) =#> f#(g(f(a)))   
    2] active#(f(f(a))) =#> g#(f(a))   
    3] active#(f(f(a))) =#> f#(a)   
    4] mark#(f(X)) =#> active#(f(X))   
    5] mark#(f(X)) =#> f#(X)   
    6] mark#(a) =#> active#(a)   
    7] mark#(g(X)) =#> active#(g(mark(X)))   
    8] mark#(g(X)) =#> g#(mark(X))   
    9] mark#(g(X)) =#> mark#(X)   
    10] f#(mark(X)) =#> f#(X)   
    11] f#(active(X)) =#> f#(X)   
    12] g#(mark(X)) =#> g#(X)   
    13] g#(active(X)) =#> g#(X)   

  Rules R_0:

    active(f(f(a))) => mark(f(g(f(a)))) 
    mark(f(X)) => active(f(X)) 
    mark(a) => active(a) 
    mark(g(X)) => active(g(mark(X))) 
    f(mark(X)) => f(X) 
    f(active(X)) => f(X) 
    g(mark(X)) => g(X) 
    g(active(X)) => g(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 : 4, 5 
    * 1 :  
    * 2 :  
    * 3 :  
    * 4 : 0, 1, 2, 3 
    * 5 : 10, 11 
    * 6 :  
    * 7 :  
    * 8 : 12, 13 
    * 9 : 4, 5, 6, 7, 8, 9 
    * 10 : 10, 11 
    * 11 : 10, 11 
    * 12 : 12, 13 
    * 13 : 12, 13 

This graph has the following strongly connected components:

  P_1:

    active#(f(f(a))) =#> mark#(f(g(f(a))))   
    mark#(f(X)) =#> active#(f(X))   

  P_2:

    mark#(g(X)) =#> mark#(X)   

  P_3:

    f#(mark(X)) =#> f#(X)   
    f#(active(X)) =#> f#(X)   

  P_4:

    g#(mark(X)) =#> g#(X)   
    g#(active(X)) =#> g#(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(g#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(g#(mark(X))) = mark(X) |> X = nu(g#(X)) 
  nu(g#(active(X))) = active(X) |> X = nu(g#(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(f#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(f#(mark(X))) = mark(X) |> X = nu(f#(X)) 
  nu(f#(active(X))) = active(X) |> X = nu(f#(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(mark#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(mark#(g(X))) = g(X) |> X = nu(mark#(X)) 

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).

The formative rules of (P_1, R_0) are R_1 ::=

  active(f(f(a))) => mark(f(g(f(a)))) 
  mark(f(X)) => active(f(X)) 
  mark(a) => active(a) 
  mark(g(X)) => active(g(mark(X))) 
  f(mark(X)) => f(X) 
  f(active(X)) => f(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 with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_1) are:

  f(mark(X)) => f(X) 
  f(active(X)) => f(X) 

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

  active#(f(f(a))) >? mark#(f(g(f(a)))) 
  mark#(f(X)) >? active#(f(X)) 
  f(mark(X)) >= f(X) 
  f(active(X)) >= f(X) 

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

The following interpretation satisfies the requirements:

  a = 3 
  active = \y0.2y0 
  active# = \y0.3y0 
  f = \y0.y0 
  g = \y0.0 
  mark = \y0.y0 
  mark# = \y0.1 + 3y0 

Using this interpretation, the requirements translate to:

  [[active#(f(f(a)))]] = 9 > 1 = [[mark#(f(g(f(a))))]] 
  [[mark#(f(_x0))]] = 1 + 3x0 > 3x0 = [[active#(f(_x0))]] 
  [[f(mark(_x0))]] = x0 >= x0 = [[f(_x0)]] 
  [[f(active(_x0))]] = 2x0 >= x0 = [[f(_x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1).  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.