/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.

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

  f(g(X)) => g(f(f(X))) 
  f(h(X)) => h(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#(g(X)) =#> f#(f(X))   
    1] f#(g(X)) =#> f#(X)   

  Rules R_0:

    f(g(X)) => g(f(f(X))) 
    f(h(X)) => h(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).

The formative rules of (P_0, R_0) are R_1 ::=

  f(g(X)) => g(f(f(X))) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_1, 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:

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

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

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, 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.