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	    0 : [] --> a 
0.00/0.01	    add : [] --> a -> a -> a 
0.00/0.01	    curry : [a -> a -> a] --> a -> a -> a 
0.00/0.01	    plus : [] --> a -> a -> a 
0.00/0.01	    s : [a] --> a 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    plus 0 x => x 
0.00/0.01	    plus s(x) y => s(plus x y) 
0.00/0.01	    curry(f) x y => f x y 
0.00/0.01	    add => curry(plus) 
0.00/0.01	
0.00/0.01	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.01	
0.00/0.01	Symbol curry 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	    0 : [] --> a 
0.00/0.01	    add : [] --> a -> a -> a 
0.00/0.01	    plus : [] --> a -> a -> a 
0.00/0.01	    s : [a] --> a 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    plus 0 X => X 
0.00/0.01	    plus s(X) Y => s(plus X Y) 
0.00/0.01	    add => plus 
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	  plus 0 X >? X 
0.00/0.01	  plus s(X) Y >? s(plus X Y) 
0.00/0.01	  add >? plus 
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	  0 = 3 
0.00/0.01	  add = \y0y1.3 + 3y0 + 3y1 
0.00/0.01	  plus = \y0y1.3y0 + 3y1 
0.00/0.01	  s = \y0.3 + y0 
0.00/0.01	
0.00/0.01	Using this interpretation, the requirements translate to:
0.00/0.01	
0.00/0.01	  [[plus 0 _x0]] = 12 + 4x0 > x0 = [[_x0]] 
0.00/0.01	  [[plus s(_x0) _x1]] = 12 + 4x0 + 4x1 > 3 + 4x0 + 4x1 = [[s(plus _x0 _x1)]] 
0.00/0.01	  [[add]] = \y0y1.3 + 3y0 + 3y1 > \y0y1.3y0 + 3y1 = [[plus]] 
0.00/0.01	
0.00/0.01	We can thus remove the following rules:
0.00/0.01	
0.00/0.01	  plus 0 X => X 
0.00/0.01	  plus s(X) Y => s(plus X Y) 
0.00/0.01	  add => plus 
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	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.01	EOF