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	    0 : [] --> a 
0.00/0.02	    id : [] --> a -> a 
0.00/0.02	    plus : [a] --> a -> a 
0.00/0.02	    s : [a] --> a 
0.00/0.02	
0.00/0.02	  Rules:
0.00/0.02	
0.00/0.02	    id x => x 
0.00/0.02	    plus(0) => id 
0.00/0.02	    plus(s(x)) y => s(plus(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	  id X >? X 
0.00/0.02	  plus(0) >? id 
0.00/0.02	  plus(s(X)) Y >? s(plus(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	  0 = 3 
0.00/0.02	  id = \y0.3y0 
0.00/0.02	  plus = \y0y1.3 + 3y0 + 3y1 
0.00/0.02	  s = \y0.3 + y0 
0.00/0.02	
0.00/0.02	Using this interpretation, the requirements translate to:
0.00/0.02	
0.00/0.02	  [[id _x0]] = 4x0 >= x0 = [[_x0]] 
0.00/0.02	  [[plus(0)]] = \y0.12 + 3y0 > \y0.3y0 = [[id]] 
0.00/0.02	  [[plus(s(_x0)) _x1]] = 12 + 3x0 + 4x1 > 6 + 3x0 + 4x1 = [[s(plus(_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	  plus(0) => id 
0.00/0.02	  plus(s(X)) Y => s(plus(X) Y) 
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	  id(X) >? X 
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	  id = \y0.1 + y0 
0.00/0.02	
0.00/0.02	Using this interpretation, the requirements translate to:
0.00/0.02	
0.00/0.02	  [[id(_x0)]] = 1 + x0 > x0 = [[_x0]] 
0.00/0.02	
0.00/0.02	We can thus remove the following rules:
0.00/0.02	
0.00/0.02	  id(X) => X 
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