0.00/0.02	YES
0.00/0.02	We consider the system theBenchmark.
0.00/0.02	
0.00/0.02	  Alphabet:
0.00/0.02	
0.00/0.02	    f : [] --> a -> b -> c 
0.00/0.02	    f1 : [a] --> b -> c 
0.00/0.02	    f2 : [a * b] --> c 
0.00/0.02	
0.00/0.02	  Rules:
0.00/0.02	
0.00/0.02	    f x => f1(x) 
0.00/0.02	    f1(x) y => f2(x, y) 
0.00/0.02	
0.00/0.02	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.02	
0.00/0.02	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.02	
0.00/0.02	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.02	
0.00/0.02	  f(X) >? f1(X) 
0.00/0.02	  f1(X) Y >? f2(X, Y) 
0.00/0.02	
0.00/0.02	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.02	
0.00/0.02	The following interpretation satisfies the requirements:
0.00/0.02	
0.00/0.02	  f = \y0y1.3 + 3y0 + 3y1 
0.00/0.02	  f1 = \y0y1.y0 
0.00/0.02	  f2 = \y0y1.y0 + y1 
0.00/0.02	
0.00/0.02	Using this interpretation, the requirements translate to:
0.00/0.02	
0.00/0.02	  [[f(_x0)]] = \y0.3 + 3y0 + 3x0 > \y0.x0 = [[f1(_x0)]] 
0.00/0.02	  [[f1(_x0) _x1]] = x0 + x1 >= x0 + x1 = [[f2(_x0, _x1)]] 
0.00/0.02	
0.00/0.02	We can thus remove the following rules:
0.00/0.02	
0.00/0.02	  f(X) => f1(X) 
0.00/0.02	
0.00/0.02	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.02	
0.00/0.02	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.02	
0.00/0.02	  f1(X, Y) >? f2(X, Y) 
0.00/0.02	
0.00/0.02	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.02	
0.00/0.02	The following interpretation satisfies the requirements:
0.00/0.02	
0.00/0.02	  f1 = \y0y1.3 + 3y0 + 3y1 
0.00/0.02	  f2 = \y0y1.y0 + y1 
0.00/0.02	
0.00/0.02	Using this interpretation, the requirements translate to:
0.00/0.02	
0.00/0.02	  [[f1(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[f2(_x0, _x1)]] 
0.00/0.02	
0.00/0.02	We can thus remove the following rules:
0.00/0.02	
0.00/0.02	  f1(X, Y) => f2(X, Y) 
0.00/0.02	
0.00/0.02	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.
0.00/0.02	
0.00/0.02	
0.00/0.02	+++ Citations +++
0.00/0.02	
0.00/0.02	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.02	EOF