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


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


YES
We consider the system theBenchmark.

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

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

  active(c) => mark(f(g(c))) 
  active(f(g(X))) => mark(g(X)) 
  mark(c) => active(c) 
  mark(f(X)) => active(f(X)) 
  mark(g(X)) => active(g(X)) 
  f(mark(X)) => f(X) 
  f(active(X)) => f(X) 
  g(mark(X)) => g(X) 
  g(active(X)) => g(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(c) >? mark(f(g(c))) 
  active(f(g(X))) >? mark(g(X)) 
  mark(c) >? active(c) 
  mark(f(X)) >? active(f(X)) 
  mark(g(X)) >? active(g(X)) 
  f(mark(X)) >? f(X) 
  f(active(X)) >? f(X) 
  g(mark(X)) >? g(X) 
  g(active(X)) >? g(X) 

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

The following interpretation satisfies the requirements:

  active = \y0.1 + y0 
  c = 0 
  f = \y0.2y0 
  g = \y0.2y0 
  mark = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[active(c)]] = 1 >= 1 = [[mark(f(g(c)))]] 
  [[active(f(g(_x0)))]] = 1 + 4x0 >= 1 + 4x0 = [[mark(g(_x0))]] 
  [[mark(c)]] = 1 >= 1 = [[active(c)]] 
  [[mark(f(_x0))]] = 1 + 4x0 >= 1 + 2x0 = [[active(f(_x0))]] 
  [[mark(g(_x0))]] = 1 + 4x0 >= 1 + 2x0 = [[active(g(_x0))]] 
  [[f(mark(_x0))]] = 2 + 4x0 > 2x0 = [[f(_x0)]] 
  [[f(active(_x0))]] = 2 + 2x0 > 2x0 = [[f(_x0)]] 
  [[g(mark(_x0))]] = 2 + 4x0 > 2x0 = [[g(_x0)]] 
  [[g(active(_x0))]] = 2 + 2x0 > 2x0 = [[g(_x0)]] 

We can thus remove the following rules:

  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#(c) =#> mark#(f(g(c)))   
    1] active#(f(g(X))) =#> mark#(g(X))   
    2] mark#(c) =#> active#(c)   
    3] mark#(f(X)) =#> active#(f(X))   
    4] mark#(g(X)) =#> active#(g(X))   

  Rules R_0:

    active(c) => mark(f(g(c))) 
    active(f(g(X))) => mark(g(X)) 
    mark(c) => active(c) 
    mark(f(X)) => active(f(X)) 
    mark(g(X)) => active(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 : 3 
    * 1 : 4 
    * 2 : 0 
    * 3 : 1 
    * 4 :  

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.