/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.

  f : [o] --> o 
  g : [o] --> o 

  f(f(X)) => f(g(f(g(f(X))))) 
  f(g(f(X))) => f(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] f#(f(X)) =#> f#(g(f(g(f(X)))))   
    1] f#(f(X)) =#> f#(g(f(X)))   
    2] f#(f(X)) =#> f#(X)   
    3] f#(g(f(X))) =#> f#(g(X))   

  Rules R_0:

    f(f(X)) => f(g(f(g(f(X))))) 
    f(g(f(X))) => f(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 : 3 
    * 2 : 0, 1, 2, 3 
    * 3 : 3 

This graph has the following strongly connected components:

  P_1:

    f#(f(X)) =#> f#(X)   

  P_2:

    f#(g(f(X))) =#> f#(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) 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_2, R_0) has no usable rules.

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

  f#(g(f(X))) >? f#(g(X)) 

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

The following interpretation satisfies the requirements:

  f = \y0.1 + 3y0 
  f# = \y0.3y0 
  g = \y0.y0 

Using this interpretation, the requirements translate to:

  [[f#(g(f(_x0)))]] = 3 + 9x0 > 3x0 = [[f#(g(_x0))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  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 apply the subterm criterion with the following projection function:

  nu(f#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(f#(f(X))) = f(X) |> X = nu(f#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, 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.