0.00/0.09	YES
0.00/0.09	We consider the system theBenchmark.
0.00/0.09	
0.00/0.09	  Alphabet:
0.00/0.09	
0.00/0.09	    0 : [] --> N 
0.00/0.09	    rec : [] --> (N -> (N -> B) -> N -> B) -> B -> N -> B 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    rec f (g 0) => g 
0.00/0.09	
0.00/0.09	Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term.
0.00/0.09	
0.00/0.09	We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0).  This leads to the following AFSM:
0.00/0.09	
0.00/0.09	  Alphabet:
0.00/0.09	
0.00/0.09	    0 : [] --> N 
0.00/0.09	    rec : [N -> (N -> B) -> N -> B * B] --> N -> B 
0.00/0.09	    ~AP1 : [N -> B * N] --> B 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    rec(F, ~AP1(G, 0)) => G 
0.00/0.09	    ~AP1(F, X) => F X 
0.00/0.09	
0.00/0.09	We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).
0.00/0.09	
0.00/0.09	In order to do so, we start by eta-expanding the system, which gives:
0.00/0.09	
0.00/0.09	  rec(F, ~AP1(G, 0), X) => G X 
0.00/0.09	  ~AP1(F, X) => F X 
0.00/0.09	
0.00/0.09	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
0.00/0.09	
0.00/0.09	  Dependency Pairs P_0:
0.00/0.09	
0.00/0.09	
0.00/0.09	  Rules R_0:
0.00/0.09	
0.00/0.09	    rec(F, ~AP1(G, 0), X) => G X 
0.00/0.09	    ~AP1(F, X) => F X 
0.00/0.09	
0.00/0.09	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
0.00/0.09	
0.00/0.09	We consider the dependency pair problem (P_0, R_0, static, formative).
0.00/0.09	
0.00/0.09	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.09	
0.00/0.09	
0.00/0.09	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
0.00/0.09	
0.00/0.09	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.00/0.09	
0.00/0.09	
0.00/0.09	+++ Citations +++
0.00/0.09	
0.00/0.09	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
0.00/0.09	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.09	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
0.00/0.09	[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.
0.00/0.09	EOF