0.00/0.05	YES
0.00/0.05	We consider the system theBenchmark.
0.00/0.05	
0.00/0.05	  Alphabet:
0.00/0.05	
0.00/0.05	    f : [N -> N * N] --> N 
0.00/0.05	    g : [] --> N -> N 
0.00/0.05	    h : [N] --> N 
0.00/0.05	
0.00/0.05	  Rules:
0.00/0.05	
0.00/0.05	    f(g, x) => h(x) 
0.00/0.05	    f(i, x) => i x 
0.00/0.05	    h(x) => f(/\y.y, x) 
0.00/0.05	
0.00/0.05	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.05	
0.00/0.05	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.05	
0.00/0.05	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.05	
0.00/0.05	  f(g, X) >? h(X) 
0.00/0.05	  f(F, X) >? F X 
0.00/0.05	  h(X) >? f(/\x.x, X) 
0.00/0.05	
0.00/0.05	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.05	
0.00/0.05	The following interpretation satisfies the requirements:
0.00/0.05	
0.00/0.05	  f = \G0y1.y1 + 2G0(y1) 
0.00/0.05	  g = \y0.3 + 3y0 
0.00/0.05	  h = \y0.3y0 
0.00/0.05	
0.00/0.05	Using this interpretation, the requirements translate to:
0.00/0.05	
0.00/0.05	  [[f(g, _x0)]] = 6 + 7x0 > 3x0 = [[h(_x0)]] 
0.00/0.05	  [[f(_F0, _x1)]] = x1 + 2F0(x1) >= x1 + F0(x1) = [[_F0 _x1]] 
0.00/0.05	  [[h(_x0)]] = 3x0 >= 3x0 = [[f(/\x.x, _x0)]] 
0.00/0.05	
0.00/0.05	We can thus remove the following rules:
0.00/0.05	
0.00/0.05	  f(g, X) => h(X) 
0.00/0.05	
0.00/0.05	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.05	
0.00/0.05	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.05	
0.00/0.05	  f(F, X) >? F X 
0.00/0.05	  h(X) >? f(/\x.x, X) 
0.00/0.05	
0.00/0.05	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.05	
0.00/0.05	The following interpretation satisfies the requirements:
0.00/0.05	
0.00/0.05	  f = \G0y1.y1 + G0(y1) 
0.00/0.05	  h = \y0.3 + 3y0 
0.00/0.05	
0.00/0.05	Using this interpretation, the requirements translate to:
0.00/0.05	
0.00/0.05	  [[f(_F0, _x1)]] = x1 + F0(x1) >= x1 + F0(x1) = [[_F0 _x1]] 
0.00/0.05	  [[h(_x0)]] = 3 + 3x0 > 2x0 = [[f(/\x.x, _x0)]] 
0.00/0.05	
0.00/0.05	We can thus remove the following rules:
0.00/0.05	
0.00/0.05	  h(X) => f(/\x.x, X) 
0.00/0.05	
0.00/0.05	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.05	
0.00/0.05	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.05	
0.00/0.05	  f(F, X) >? F X 
0.00/0.05	
0.00/0.05	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.05	
0.00/0.05	The following interpretation satisfies the requirements:
0.00/0.05	
0.00/0.05	  f = \G0y1.1 + y1 + G0(y1) 
0.00/0.05	
0.00/0.05	Using this interpretation, the requirements translate to:
0.00/0.05	
0.00/0.05	  [[f(_F0, _x1)]] = 1 + x1 + F0(x1) > x1 + F0(x1) = [[_F0 _x1]] 
0.00/0.05	
0.00/0.05	We can thus remove the following rules:
0.00/0.05	
0.00/0.05	  f(F, X) => F X 
0.00/0.05	
0.00/0.05	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.
0.00/0.05	
0.00/0.05	
0.00/0.05	+++ Citations +++
0.00/0.05	
0.00/0.05	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.05	EOF