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

  c : [o * o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  s : [o] --> o 

  f(c(s(X), Y)) => f(c(X, s(Y))) 
  g(c(X, s(Y))) => g(c(s(X), Y)) 
  g(s(f(X))) => g(f(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]):

  f(c(s(X), Y)) >? f(c(X, s(Y))) 
  g(c(X, s(Y))) >? g(c(s(X), Y)) 
  g(s(f(X))) >? g(f(X)) 

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

The following interpretation satisfies the requirements:

  c = \y0y1.y0 + y1 
  f = \y0.2 + 2y0 
  g = \y0.y0 
  s = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[f(c(s(_x0), _x1))]] = 6 + 2x0 + 2x1 >= 6 + 2x0 + 2x1 = [[f(c(_x0, s(_x1)))]] 
  [[g(c(_x0, s(_x1)))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[g(c(s(_x0), _x1))]] 
  [[g(s(f(_x0)))]] = 4 + 2x0 > 2 + 2x0 = [[g(f(_x0))]] 

We can thus remove the following rules:

  g(s(f(X))) => g(f(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#(c(s(X), Y)) =#> f#(c(X, s(Y)))   
    1] g#(c(X, s(Y))) =#> g#(c(s(X), Y))   

  Rules R_0:

    f(c(s(X), Y)) => f(c(X, s(Y))) 
    g(c(X, s(Y))) => g(c(s(X), Y)) 

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 : 0 
    * 1 : 1 

This graph has the following strongly connected components:

  P_1:

    f#(c(s(X), Y)) =#> f#(c(X, s(Y)))   

  P_2:

    g#(c(X, s(Y))) =#> g#(c(s(X), Y))   

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).

This combination (P_2, R_0) has no formative rules!  We will name the empty set of rules:R_1.

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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:

  g#(c(X, s(Y))) >? g#(c(s(X), Y)) 

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

The following interpretation satisfies the requirements:

  c = \y0y1.3y1 
  g# = \y0.2y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[g#(c(_x0, s(_x1)))]] = 18 + 6x1 > 6x1 = [[g#(c(s(_x0), _x1))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_1) by ({}, R_1).  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).

This combination (P_1, R_0) has no formative rules!  This is exactly R_1.

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 [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(c(s(X), Y)) >? f#(c(X, s(Y))) 

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

The following interpretation satisfies the requirements:

  c = \y0y1.3y0 
  f# = \y0.2y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[f#(c(s(_x0), _x1))]] = 18 + 6x0 > 6x0 = [[f#(c(_x0, s(_x1)))]] 

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.