0.00/0.14	YES
0.00/0.14	We consider the system theBenchmark.
0.00/0.14	
0.00/0.14	  Alphabet:
0.00/0.14	
0.00/0.14	    cons : [nat -> nat * funlist] --> funlist 
0.00/0.14	    false : [] --> bool 
0.00/0.14	    head : [funlist] --> nat -> nat 
0.00/0.14	    if : [bool * nat -> string * nat -> string] --> nat -> string 
0.00/0.14	    nil : [] --> funlist 
0.00/0.14	    s : [nat] --> nat 
0.00/0.14	    tail : [funlist] --> funlist 
0.00/0.14	    test : [nat -> nat] --> bool 
0.00/0.14	    true : [] --> bool 
0.00/0.14	
0.00/0.14	  Rules:
0.00/0.14	
0.00/0.14	    if(true, f, g) => f 
0.00/0.14	    if(false, f, g) => g 
0.00/0.14	    test(/\x.s(x)) => true 
0.00/0.14	    head(cons(f, x)) => f 
0.00/0.14	    tail(cons(f, x)) => x 
0.00/0.14	
0.00/0.14	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.14	
0.00/0.14	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]).
0.00/0.14	
0.00/0.14	In order to do so, we start by eta-expanding the system, which gives:
0.00/0.14	
0.00/0.14	  if(true, F, G, X) => F X 
0.00/0.14	  if(false, F, G, X) => G X 
0.00/0.14	  test(/\x.s(x)) => true 
0.00/0.14	  head(cons(F, X), Y) => F Y 
0.00/0.14	  tail(cons(F, X)) => X 
0.00/0.14	
0.00/0.14	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
0.00/0.14	
0.00/0.14	  Dependency Pairs P_0:
0.00/0.14	
0.00/0.14	
0.00/0.14	  Rules R_0:
0.00/0.14	
0.00/0.14	    if(true, F, G, X) => F X 
0.00/0.14	    if(false, F, G, X) => G X 
0.00/0.14	    test(/\x.s(x)) => true 
0.00/0.14	    head(cons(F, X), Y) => F Y 
0.00/0.14	    tail(cons(F, X)) => X 
0.00/0.14	
0.00/0.14	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
0.00/0.14	
0.00/0.14	We consider the dependency pair problem (P_0, R_0, static, formative).
0.00/0.14	
0.00/0.14	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.14	
0.00/0.14	
0.00/0.14	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
0.00/0.14	
0.00/0.14	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.00/0.14	
0.00/0.14	
0.00/0.14	+++ Citations +++
0.00/0.14	
0.00/0.14	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.14	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
0.00/0.14	[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.
0.00/0.14	EOF