0.00/0.01	YES
0.00/0.01	We consider the system theBenchmark.
0.00/0.01	
0.00/0.01	  Alphabet:
0.00/0.01	
0.00/0.01	    c : [] --> ((C -> L) -> L) -> C 
0.00/0.01	    d : [] --> C 
0.00/0.01	    ex : [] --> C -> L 
0.00/0.01	    nil : [] --> L 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    ex d => nil 
0.00/0.01	    ex (c (/\f.g f)) => g ex 
0.00/0.01	
0.00/0.01	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.
0.00/0.01	
0.00/0.01	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:
0.00/0.01	
0.00/0.01	  Alphabet:
0.00/0.01	
0.00/0.01	    c : [(C -> L) -> L] --> C 
0.00/0.01	    d : [] --> C 
0.00/0.01	    ex : [] --> C -> L 
0.00/0.01	    nil : [] --> L 
0.00/0.01	    ~AP1 : [(C -> L) -> L * C -> L] --> L 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    ex d => nil 
0.00/0.01	    ex c(/\f.~AP1(F, f)) => ~AP1(F, ex) 
0.00/0.01	    ~AP1(F, G) => F G 
0.00/0.01	
0.00/0.01	Symbol ~AP1 is an encoding for application that is only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing it.  This gives:
0.00/0.01	
0.00/0.01	  Alphabet:
0.00/0.01	
0.00/0.01	    c : [(C -> L) -> L] --> C 
0.00/0.01	    d : [] --> C 
0.00/0.01	    ex : [] --> C -> L 
0.00/0.01	    nil : [] --> L 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    ex d => nil 
0.00/0.01	    ex c(/\f.X(f)) => X(ex) 
0.00/0.01	
0.00/0.01	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.01	
0.00/0.01	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.01	
0.00/0.01	  ex d >? nil 
0.00/0.01	  ex c(/\f.X(f)) >? X(ex) 
0.00/0.01	
0.00/0.01	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.01	
0.00/0.01	The following interpretation satisfies the requirements:
0.00/0.01	
0.00/0.01	  c = \G0.3 + G0(\y1.0) 
0.00/0.01	  d = 3 
0.00/0.01	  ex = \y0.0 
0.00/0.01	  nil = 0 
0.00/0.01	
0.00/0.01	Using this interpretation, the requirements translate to:
0.00/0.01	
0.00/0.01	  [[ex d]] = 3 > 0 = [[nil]] 
0.00/0.01	  [[ex c(/\f._x0(f))]] = 3 + F0(\y0.0) > F0(\y0.0) = [[_x0(ex)]] 
0.00/0.01	
0.00/0.01	We can thus remove the following rules:
0.00/0.01	
0.00/0.01	  ex d => nil 
0.00/0.01	  ex c(/\f.X(f)) => X(ex) 
0.00/0.01	
0.00/0.01	All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.
0.00/0.01	
0.00/0.01	
0.00/0.01	+++ Citations +++
0.00/0.01	
0.00/0.01	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
0.00/0.01	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.01	EOF